Parsing a string into a nested data.table












3














I have data in a table in which one cell in every row is a multiline string, which is formatted a a bit like a document with references at the end of it. For example, one of those strings looks like:



item A...1
item B...2
item C...3
item D...2
1=foo
2=bar
3=baz


My eventual goal is to extract foo/bar/baz into columns and count the matching items. So for the above, I'd end up with a row including:



foo | bar | baz
----+-----+----
1 | 2 | 1


I tried to start by extracting the "reference" mappings, as a nested data.table looking like this:



code | reason
-----+-------
1 | foo
2 | bar
3 | baz


Here's how I tried to do it, using data.table and stringr.



encounter_alerts[, whys := lapply(
str_extract_all(text, regex('^[0-9].*$', multiline = TRUE)),
FUN = function (s) { fread(text = s, sep = '=', header = FALSE, col.names = c('code', 'reason')) }
)]


I am very confused by the error message I get when I try to do this:



Error in fread(text = s, sep = "=", header = FALSE, col.names = c("code",  :
file not found: 1=foo


I am explicitly using text rather than file so I'm not sure how it's trying to interpret the line of text as a filename!



When I test this with a single row, it seems to work fine:



> fread(text = str_extract_all(encounter_alerts[989]$text, regex('^[0-9].*$', multiline = TRUE))[[1]], sep = '=', header = FALSE, col.names = c('code', 'reason'))
code reason
1: 1 foo
2: 2 bar


What am I doing wrong? Is there a better way to do this?



Thanks!










share|improve this question





























    3














    I have data in a table in which one cell in every row is a multiline string, which is formatted a a bit like a document with references at the end of it. For example, one of those strings looks like:



    item A...1
    item B...2
    item C...3
    item D...2
    1=foo
    2=bar
    3=baz


    My eventual goal is to extract foo/bar/baz into columns and count the matching items. So for the above, I'd end up with a row including:



    foo | bar | baz
    ----+-----+----
    1 | 2 | 1


    I tried to start by extracting the "reference" mappings, as a nested data.table looking like this:



    code | reason
    -----+-------
    1 | foo
    2 | bar
    3 | baz


    Here's how I tried to do it, using data.table and stringr.



    encounter_alerts[, whys := lapply(
    str_extract_all(text, regex('^[0-9].*$', multiline = TRUE)),
    FUN = function (s) { fread(text = s, sep = '=', header = FALSE, col.names = c('code', 'reason')) }
    )]


    I am very confused by the error message I get when I try to do this:



    Error in fread(text = s, sep = "=", header = FALSE, col.names = c("code",  :
    file not found: 1=foo


    I am explicitly using text rather than file so I'm not sure how it's trying to interpret the line of text as a filename!



    When I test this with a single row, it seems to work fine:



    > fread(text = str_extract_all(encounter_alerts[989]$text, regex('^[0-9].*$', multiline = TRUE))[[1]], sep = '=', header = FALSE, col.names = c('code', 'reason'))
    code reason
    1: 1 foo
    2: 2 bar


    What am I doing wrong? Is there a better way to do this?



    Thanks!










    share|improve this question



























      3












      3








      3







      I have data in a table in which one cell in every row is a multiline string, which is formatted a a bit like a document with references at the end of it. For example, one of those strings looks like:



      item A...1
      item B...2
      item C...3
      item D...2
      1=foo
      2=bar
      3=baz


      My eventual goal is to extract foo/bar/baz into columns and count the matching items. So for the above, I'd end up with a row including:



      foo | bar | baz
      ----+-----+----
      1 | 2 | 1


      I tried to start by extracting the "reference" mappings, as a nested data.table looking like this:



      code | reason
      -----+-------
      1 | foo
      2 | bar
      3 | baz


      Here's how I tried to do it, using data.table and stringr.



      encounter_alerts[, whys := lapply(
      str_extract_all(text, regex('^[0-9].*$', multiline = TRUE)),
      FUN = function (s) { fread(text = s, sep = '=', header = FALSE, col.names = c('code', 'reason')) }
      )]


      I am very confused by the error message I get when I try to do this:



      Error in fread(text = s, sep = "=", header = FALSE, col.names = c("code",  :
      file not found: 1=foo


      I am explicitly using text rather than file so I'm not sure how it's trying to interpret the line of text as a filename!



      When I test this with a single row, it seems to work fine:



      > fread(text = str_extract_all(encounter_alerts[989]$text, regex('^[0-9].*$', multiline = TRUE))[[1]], sep = '=', header = FALSE, col.names = c('code', 'reason'))
      code reason
      1: 1 foo
      2: 2 bar


      What am I doing wrong? Is there a better way to do this?



      Thanks!










      share|improve this question















      I have data in a table in which one cell in every row is a multiline string, which is formatted a a bit like a document with references at the end of it. For example, one of those strings looks like:



      item A...1
      item B...2
      item C...3
      item D...2
      1=foo
      2=bar
      3=baz


      My eventual goal is to extract foo/bar/baz into columns and count the matching items. So for the above, I'd end up with a row including:



      foo | bar | baz
      ----+-----+----
      1 | 2 | 1


      I tried to start by extracting the "reference" mappings, as a nested data.table looking like this:



      code | reason
      -----+-------
      1 | foo
      2 | bar
      3 | baz


      Here's how I tried to do it, using data.table and stringr.



      encounter_alerts[, whys := lapply(
      str_extract_all(text, regex('^[0-9].*$', multiline = TRUE)),
      FUN = function (s) { fread(text = s, sep = '=', header = FALSE, col.names = c('code', 'reason')) }
      )]


      I am very confused by the error message I get when I try to do this:



      Error in fread(text = s, sep = "=", header = FALSE, col.names = c("code",  :
      file not found: 1=foo


      I am explicitly using text rather than file so I'm not sure how it's trying to interpret the line of text as a filename!



      When I test this with a single row, it seems to work fine:



      > fread(text = str_extract_all(encounter_alerts[989]$text, regex('^[0-9].*$', multiline = TRUE))[[1]], sep = '=', header = FALSE, col.names = c('code', 'reason'))
      code reason
      1: 1 foo
      2: 2 bar


      What am I doing wrong? Is there a better way to do this?



      Thanks!







      r data.table






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 14 at 1:53

























      asked Nov 13 at 3:55









      Nicholas Riley

      37k586114




      37k586114
























          4 Answers
          4






          active

          oldest

          votes


















          2














          Note: Edited after reading comments



          From your comment, I tried to reproduce what I understand your data might look like.



          library(tidyverse)

          df <- tibble(
          strings = c("item A...1
          item B...2
          item C...3
          item D...2
          1=foo
          2=bar
          3=baz",
          "item A...2
          item B...2
          item C...3
          item D...1
          1=toto
          2=foo
          3=lala",
          "item A...3
          item B...3
          item C...3
          item D...1
          1=tutu
          3=ttt")
          )


          Code:



          get_ref <- function(string) {
          string %>%
          str_split("n") %>%
          unlist() %>%
          str_subset("=") %>%
          str_split_fixed("=", 2) %>%
          as_tibble() %>%
          rename(code = V1, reason = V2)
          }

          list1 <- map(df$strings, get_ref)

          get_value <- function(string) {
          string %>%
          str_split("n") %>%
          unlist() %>%
          str_subset("\.\.\.") %>%
          str_replace_all(".*\.\.\.", "") %>%
          as_tibble() %>%
          rename(code = value)
          }

          list2 <- map(df$strings, get_value)

          get_result <- function(df1, df2) {
          left_join(df1, df2) %>%
          count(reason) %>%
          spread(reason, n)
          }

          result <- map2_df(list1, list2, get_result)

          result[is.na(result)] <- 0

          result


          Result



          # A tibble: 3 x 7
          bar baz foo lala toto ttt tutu
          <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
          1 2 1 1 0 0 0 0
          2 0 0 2 1 1 0 0
          3 0 0 0 0 0 3 1





          share|improve this answer























          • I'm sorry my question was confusing! You are correct in 'note 2'; there is such a multi-line string in every row of my data. I think this approach will work for me but need to correct for some assumptions you (understandably!) made that aren't the case in my real data…
            – Nicholas Riley
            Nov 13 at 14:32










          • So, is your data in a data frame since you talk about "row"? Sorry I am still a little confused about what your data looks like
            – prosoitos
            Nov 13 at 16:05










          • Would you mind giving a little more information about its structure so that I could help you adapt my code to match your data structure?
            – prosoitos
            Nov 13 at 16:06












          • If you have a data frame with one variable consisting of a vector of strings similar to the one you pasted in your question, you could wrap my code in a function and pass it to pmap_df() to apply it to every row and output a data frame with one result per row. I'll be happy to write this up if you give me enough info on your data frame.
            – prosoitos
            Nov 13 at 16:09












          • I edited my answer to match what I now understand your data might look like
            – prosoitos
            Nov 13 at 17:01



















          0














          using stringr and dplyr you can do it easily



          library(stringr)
          library(dplyr)
          v <- as.data.frame(c( "item A...1",
          "item B...2",
          "item C...3",
          "item D...2"))
          colnames(v)<- "items"

          matching <- c( "1",
          "2",
          "3")
          Mapping <- read.table(text="code reason
          1 foo
          2 bar
          3 baz
          ", header = T)

          ## Answer
          df1<- v %>%
          mutate(code = str_extract(v$items, str_c(matching, collapse = "|")))
          str(df1)
          str(Mapping)
          df1$code <- as.numeric(df1$code )

          df1 <- left_join(df1,Mapping)


          please have a look






          share|improve this answer





























            0














            There's probably a nicer way to do this, but here's a solution that doesn't require any additional libraries (beyond stringr, which you're already using).



            sample_str <- 'item A...1
            item B...2
            item C...3
            item D...2
            1=foo
            2=bar
            3=baz'

            lines <- stringr::str_split(sample_str, 'n', simplify = T)

            extracted_strs <- lines[stringr::str_detect(lines, '^\d=\w+$')]

            dfs_list <- lapply(extracted_strs, function(x) {
            str_parts <- stringr::str_split(x, '=', simplify = T)
            df_args = list()
            df_args[[str_parts[2]]] = as.integer(str_parts[1])
            df_args[['stringsAsFactors']] = F

            do.call(data.frame, df_args)
            })


            df <- do.call(cbind, dfs)





            share|improve this answer





















            • Thanks. This creates a data.frame with a variable for foo, bar and bar. Would this be easier to use later on than what I created as my example?
              – Nicholas Riley
              Nov 13 at 14:30



















            0














            Thanks so much to @prosoitos for helping with this. Here's the final code I ended up using, highly based on the accepted answer — it's a mix of different packages and so forth which I hope to clean up eventually, but deadlines happen...



            get_code_reason_mapping <- function(alert_text) {
            alert_text %>%
            str_extract_all(regex('^[0-9]=(.*)$', multiline = T)) %>%
            unlist() %>%
            str_split_fixed("=", 2) %>%
            as.data.table() %>%
            setnames(c('code', 'reason'))
            }

            encounter_alerts$code_reason_mapping <- map(encounter_alerts$alert_text, get_code_reason_mapping)

            get_why_codes <- function(alert_text) {
            alert_text %>%
            str_extract_all(regex('[/n][0-9e][0-9>][0-9]$', multiline = TRUE)) %>%
            unlist() %>%
            str_sub(-1) %>%
            as.data.table() %>%
            setnames(c('code'))
            }

            encounter_alerts$why_codes <- map(encounter_alerts$alert_text, get_why_codes)

            get_code_counts <- function(df1, df2) {
            left_join(df1, df2) %>%
            count(reason) %>%
            spread(reason, n)
            }

            code_counts <- map2_df(encounter_alerts$code_reason_mapping, encounter_alerts$why_codes, get_code_counts)

            code_counts[is.na(code_counts)] <- 0

            code_counts





            share|improve this answer





















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              4 Answers
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              active

              oldest

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              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              Note: Edited after reading comments



              From your comment, I tried to reproduce what I understand your data might look like.



              library(tidyverse)

              df <- tibble(
              strings = c("item A...1
              item B...2
              item C...3
              item D...2
              1=foo
              2=bar
              3=baz",
              "item A...2
              item B...2
              item C...3
              item D...1
              1=toto
              2=foo
              3=lala",
              "item A...3
              item B...3
              item C...3
              item D...1
              1=tutu
              3=ttt")
              )


              Code:



              get_ref <- function(string) {
              string %>%
              str_split("n") %>%
              unlist() %>%
              str_subset("=") %>%
              str_split_fixed("=", 2) %>%
              as_tibble() %>%
              rename(code = V1, reason = V2)
              }

              list1 <- map(df$strings, get_ref)

              get_value <- function(string) {
              string %>%
              str_split("n") %>%
              unlist() %>%
              str_subset("\.\.\.") %>%
              str_replace_all(".*\.\.\.", "") %>%
              as_tibble() %>%
              rename(code = value)
              }

              list2 <- map(df$strings, get_value)

              get_result <- function(df1, df2) {
              left_join(df1, df2) %>%
              count(reason) %>%
              spread(reason, n)
              }

              result <- map2_df(list1, list2, get_result)

              result[is.na(result)] <- 0

              result


              Result



              # A tibble: 3 x 7
              bar baz foo lala toto ttt tutu
              <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
              1 2 1 1 0 0 0 0
              2 0 0 2 1 1 0 0
              3 0 0 0 0 0 3 1





              share|improve this answer























              • I'm sorry my question was confusing! You are correct in 'note 2'; there is such a multi-line string in every row of my data. I think this approach will work for me but need to correct for some assumptions you (understandably!) made that aren't the case in my real data…
                – Nicholas Riley
                Nov 13 at 14:32










              • So, is your data in a data frame since you talk about "row"? Sorry I am still a little confused about what your data looks like
                – prosoitos
                Nov 13 at 16:05










              • Would you mind giving a little more information about its structure so that I could help you adapt my code to match your data structure?
                – prosoitos
                Nov 13 at 16:06












              • If you have a data frame with one variable consisting of a vector of strings similar to the one you pasted in your question, you could wrap my code in a function and pass it to pmap_df() to apply it to every row and output a data frame with one result per row. I'll be happy to write this up if you give me enough info on your data frame.
                – prosoitos
                Nov 13 at 16:09












              • I edited my answer to match what I now understand your data might look like
                – prosoitos
                Nov 13 at 17:01
















              2














              Note: Edited after reading comments



              From your comment, I tried to reproduce what I understand your data might look like.



              library(tidyverse)

              df <- tibble(
              strings = c("item A...1
              item B...2
              item C...3
              item D...2
              1=foo
              2=bar
              3=baz",
              "item A...2
              item B...2
              item C...3
              item D...1
              1=toto
              2=foo
              3=lala",
              "item A...3
              item B...3
              item C...3
              item D...1
              1=tutu
              3=ttt")
              )


              Code:



              get_ref <- function(string) {
              string %>%
              str_split("n") %>%
              unlist() %>%
              str_subset("=") %>%
              str_split_fixed("=", 2) %>%
              as_tibble() %>%
              rename(code = V1, reason = V2)
              }

              list1 <- map(df$strings, get_ref)

              get_value <- function(string) {
              string %>%
              str_split("n") %>%
              unlist() %>%
              str_subset("\.\.\.") %>%
              str_replace_all(".*\.\.\.", "") %>%
              as_tibble() %>%
              rename(code = value)
              }

              list2 <- map(df$strings, get_value)

              get_result <- function(df1, df2) {
              left_join(df1, df2) %>%
              count(reason) %>%
              spread(reason, n)
              }

              result <- map2_df(list1, list2, get_result)

              result[is.na(result)] <- 0

              result


              Result



              # A tibble: 3 x 7
              bar baz foo lala toto ttt tutu
              <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
              1 2 1 1 0 0 0 0
              2 0 0 2 1 1 0 0
              3 0 0 0 0 0 3 1





              share|improve this answer























              • I'm sorry my question was confusing! You are correct in 'note 2'; there is such a multi-line string in every row of my data. I think this approach will work for me but need to correct for some assumptions you (understandably!) made that aren't the case in my real data…
                – Nicholas Riley
                Nov 13 at 14:32










              • So, is your data in a data frame since you talk about "row"? Sorry I am still a little confused about what your data looks like
                – prosoitos
                Nov 13 at 16:05










              • Would you mind giving a little more information about its structure so that I could help you adapt my code to match your data structure?
                – prosoitos
                Nov 13 at 16:06












              • If you have a data frame with one variable consisting of a vector of strings similar to the one you pasted in your question, you could wrap my code in a function and pass it to pmap_df() to apply it to every row and output a data frame with one result per row. I'll be happy to write this up if you give me enough info on your data frame.
                – prosoitos
                Nov 13 at 16:09












              • I edited my answer to match what I now understand your data might look like
                – prosoitos
                Nov 13 at 17:01














              2












              2








              2






              Note: Edited after reading comments



              From your comment, I tried to reproduce what I understand your data might look like.



              library(tidyverse)

              df <- tibble(
              strings = c("item A...1
              item B...2
              item C...3
              item D...2
              1=foo
              2=bar
              3=baz",
              "item A...2
              item B...2
              item C...3
              item D...1
              1=toto
              2=foo
              3=lala",
              "item A...3
              item B...3
              item C...3
              item D...1
              1=tutu
              3=ttt")
              )


              Code:



              get_ref <- function(string) {
              string %>%
              str_split("n") %>%
              unlist() %>%
              str_subset("=") %>%
              str_split_fixed("=", 2) %>%
              as_tibble() %>%
              rename(code = V1, reason = V2)
              }

              list1 <- map(df$strings, get_ref)

              get_value <- function(string) {
              string %>%
              str_split("n") %>%
              unlist() %>%
              str_subset("\.\.\.") %>%
              str_replace_all(".*\.\.\.", "") %>%
              as_tibble() %>%
              rename(code = value)
              }

              list2 <- map(df$strings, get_value)

              get_result <- function(df1, df2) {
              left_join(df1, df2) %>%
              count(reason) %>%
              spread(reason, n)
              }

              result <- map2_df(list1, list2, get_result)

              result[is.na(result)] <- 0

              result


              Result



              # A tibble: 3 x 7
              bar baz foo lala toto ttt tutu
              <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
              1 2 1 1 0 0 0 0
              2 0 0 2 1 1 0 0
              3 0 0 0 0 0 3 1





              share|improve this answer














              Note: Edited after reading comments



              From your comment, I tried to reproduce what I understand your data might look like.



              library(tidyverse)

              df <- tibble(
              strings = c("item A...1
              item B...2
              item C...3
              item D...2
              1=foo
              2=bar
              3=baz",
              "item A...2
              item B...2
              item C...3
              item D...1
              1=toto
              2=foo
              3=lala",
              "item A...3
              item B...3
              item C...3
              item D...1
              1=tutu
              3=ttt")
              )


              Code:



              get_ref <- function(string) {
              string %>%
              str_split("n") %>%
              unlist() %>%
              str_subset("=") %>%
              str_split_fixed("=", 2) %>%
              as_tibble() %>%
              rename(code = V1, reason = V2)
              }

              list1 <- map(df$strings, get_ref)

              get_value <- function(string) {
              string %>%
              str_split("n") %>%
              unlist() %>%
              str_subset("\.\.\.") %>%
              str_replace_all(".*\.\.\.", "") %>%
              as_tibble() %>%
              rename(code = value)
              }

              list2 <- map(df$strings, get_value)

              get_result <- function(df1, df2) {
              left_join(df1, df2) %>%
              count(reason) %>%
              spread(reason, n)
              }

              result <- map2_df(list1, list2, get_result)

              result[is.na(result)] <- 0

              result


              Result



              # A tibble: 3 x 7
              bar baz foo lala toto ttt tutu
              <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
              1 2 1 1 0 0 0 0
              2 0 0 2 1 1 0 0
              3 0 0 0 0 0 3 1






              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Nov 13 at 17:00

























              answered Nov 13 at 7:16









              prosoitos

              910219




              910219












              • I'm sorry my question was confusing! You are correct in 'note 2'; there is such a multi-line string in every row of my data. I think this approach will work for me but need to correct for some assumptions you (understandably!) made that aren't the case in my real data…
                – Nicholas Riley
                Nov 13 at 14:32










              • So, is your data in a data frame since you talk about "row"? Sorry I am still a little confused about what your data looks like
                – prosoitos
                Nov 13 at 16:05










              • Would you mind giving a little more information about its structure so that I could help you adapt my code to match your data structure?
                – prosoitos
                Nov 13 at 16:06












              • If you have a data frame with one variable consisting of a vector of strings similar to the one you pasted in your question, you could wrap my code in a function and pass it to pmap_df() to apply it to every row and output a data frame with one result per row. I'll be happy to write this up if you give me enough info on your data frame.
                – prosoitos
                Nov 13 at 16:09












              • I edited my answer to match what I now understand your data might look like
                – prosoitos
                Nov 13 at 17:01


















              • I'm sorry my question was confusing! You are correct in 'note 2'; there is such a multi-line string in every row of my data. I think this approach will work for me but need to correct for some assumptions you (understandably!) made that aren't the case in my real data…
                – Nicholas Riley
                Nov 13 at 14:32










              • So, is your data in a data frame since you talk about "row"? Sorry I am still a little confused about what your data looks like
                – prosoitos
                Nov 13 at 16:05










              • Would you mind giving a little more information about its structure so that I could help you adapt my code to match your data structure?
                – prosoitos
                Nov 13 at 16:06












              • If you have a data frame with one variable consisting of a vector of strings similar to the one you pasted in your question, you could wrap my code in a function and pass it to pmap_df() to apply it to every row and output a data frame with one result per row. I'll be happy to write this up if you give me enough info on your data frame.
                – prosoitos
                Nov 13 at 16:09












              • I edited my answer to match what I now understand your data might look like
                – prosoitos
                Nov 13 at 17:01
















              I'm sorry my question was confusing! You are correct in 'note 2'; there is such a multi-line string in every row of my data. I think this approach will work for me but need to correct for some assumptions you (understandably!) made that aren't the case in my real data…
              – Nicholas Riley
              Nov 13 at 14:32




              I'm sorry my question was confusing! You are correct in 'note 2'; there is such a multi-line string in every row of my data. I think this approach will work for me but need to correct for some assumptions you (understandably!) made that aren't the case in my real data…
              – Nicholas Riley
              Nov 13 at 14:32












              So, is your data in a data frame since you talk about "row"? Sorry I am still a little confused about what your data looks like
              – prosoitos
              Nov 13 at 16:05




              So, is your data in a data frame since you talk about "row"? Sorry I am still a little confused about what your data looks like
              – prosoitos
              Nov 13 at 16:05












              Would you mind giving a little more information about its structure so that I could help you adapt my code to match your data structure?
              – prosoitos
              Nov 13 at 16:06






              Would you mind giving a little more information about its structure so that I could help you adapt my code to match your data structure?
              – prosoitos
              Nov 13 at 16:06














              If you have a data frame with one variable consisting of a vector of strings similar to the one you pasted in your question, you could wrap my code in a function and pass it to pmap_df() to apply it to every row and output a data frame with one result per row. I'll be happy to write this up if you give me enough info on your data frame.
              – prosoitos
              Nov 13 at 16:09






              If you have a data frame with one variable consisting of a vector of strings similar to the one you pasted in your question, you could wrap my code in a function and pass it to pmap_df() to apply it to every row and output a data frame with one result per row. I'll be happy to write this up if you give me enough info on your data frame.
              – prosoitos
              Nov 13 at 16:09














              I edited my answer to match what I now understand your data might look like
              – prosoitos
              Nov 13 at 17:01




              I edited my answer to match what I now understand your data might look like
              – prosoitos
              Nov 13 at 17:01













              0














              using stringr and dplyr you can do it easily



              library(stringr)
              library(dplyr)
              v <- as.data.frame(c( "item A...1",
              "item B...2",
              "item C...3",
              "item D...2"))
              colnames(v)<- "items"

              matching <- c( "1",
              "2",
              "3")
              Mapping <- read.table(text="code reason
              1 foo
              2 bar
              3 baz
              ", header = T)

              ## Answer
              df1<- v %>%
              mutate(code = str_extract(v$items, str_c(matching, collapse = "|")))
              str(df1)
              str(Mapping)
              df1$code <- as.numeric(df1$code )

              df1 <- left_join(df1,Mapping)


              please have a look






              share|improve this answer


























                0














                using stringr and dplyr you can do it easily



                library(stringr)
                library(dplyr)
                v <- as.data.frame(c( "item A...1",
                "item B...2",
                "item C...3",
                "item D...2"))
                colnames(v)<- "items"

                matching <- c( "1",
                "2",
                "3")
                Mapping <- read.table(text="code reason
                1 foo
                2 bar
                3 baz
                ", header = T)

                ## Answer
                df1<- v %>%
                mutate(code = str_extract(v$items, str_c(matching, collapse = "|")))
                str(df1)
                str(Mapping)
                df1$code <- as.numeric(df1$code )

                df1 <- left_join(df1,Mapping)


                please have a look






                share|improve this answer
























                  0












                  0








                  0






                  using stringr and dplyr you can do it easily



                  library(stringr)
                  library(dplyr)
                  v <- as.data.frame(c( "item A...1",
                  "item B...2",
                  "item C...3",
                  "item D...2"))
                  colnames(v)<- "items"

                  matching <- c( "1",
                  "2",
                  "3")
                  Mapping <- read.table(text="code reason
                  1 foo
                  2 bar
                  3 baz
                  ", header = T)

                  ## Answer
                  df1<- v %>%
                  mutate(code = str_extract(v$items, str_c(matching, collapse = "|")))
                  str(df1)
                  str(Mapping)
                  df1$code <- as.numeric(df1$code )

                  df1 <- left_join(df1,Mapping)


                  please have a look






                  share|improve this answer












                  using stringr and dplyr you can do it easily



                  library(stringr)
                  library(dplyr)
                  v <- as.data.frame(c( "item A...1",
                  "item B...2",
                  "item C...3",
                  "item D...2"))
                  colnames(v)<- "items"

                  matching <- c( "1",
                  "2",
                  "3")
                  Mapping <- read.table(text="code reason
                  1 foo
                  2 bar
                  3 baz
                  ", header = T)

                  ## Answer
                  df1<- v %>%
                  mutate(code = str_extract(v$items, str_c(matching, collapse = "|")))
                  str(df1)
                  str(Mapping)
                  df1$code <- as.numeric(df1$code )

                  df1 <- left_join(df1,Mapping)


                  please have a look







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 13 at 4:05









                  Hunaidkhan

                  755113




                  755113























                      0














                      There's probably a nicer way to do this, but here's a solution that doesn't require any additional libraries (beyond stringr, which you're already using).



                      sample_str <- 'item A...1
                      item B...2
                      item C...3
                      item D...2
                      1=foo
                      2=bar
                      3=baz'

                      lines <- stringr::str_split(sample_str, 'n', simplify = T)

                      extracted_strs <- lines[stringr::str_detect(lines, '^\d=\w+$')]

                      dfs_list <- lapply(extracted_strs, function(x) {
                      str_parts <- stringr::str_split(x, '=', simplify = T)
                      df_args = list()
                      df_args[[str_parts[2]]] = as.integer(str_parts[1])
                      df_args[['stringsAsFactors']] = F

                      do.call(data.frame, df_args)
                      })


                      df <- do.call(cbind, dfs)





                      share|improve this answer





















                      • Thanks. This creates a data.frame with a variable for foo, bar and bar. Would this be easier to use later on than what I created as my example?
                        – Nicholas Riley
                        Nov 13 at 14:30
















                      0














                      There's probably a nicer way to do this, but here's a solution that doesn't require any additional libraries (beyond stringr, which you're already using).



                      sample_str <- 'item A...1
                      item B...2
                      item C...3
                      item D...2
                      1=foo
                      2=bar
                      3=baz'

                      lines <- stringr::str_split(sample_str, 'n', simplify = T)

                      extracted_strs <- lines[stringr::str_detect(lines, '^\d=\w+$')]

                      dfs_list <- lapply(extracted_strs, function(x) {
                      str_parts <- stringr::str_split(x, '=', simplify = T)
                      df_args = list()
                      df_args[[str_parts[2]]] = as.integer(str_parts[1])
                      df_args[['stringsAsFactors']] = F

                      do.call(data.frame, df_args)
                      })


                      df <- do.call(cbind, dfs)





                      share|improve this answer





















                      • Thanks. This creates a data.frame with a variable for foo, bar and bar. Would this be easier to use later on than what I created as my example?
                        – Nicholas Riley
                        Nov 13 at 14:30














                      0












                      0








                      0






                      There's probably a nicer way to do this, but here's a solution that doesn't require any additional libraries (beyond stringr, which you're already using).



                      sample_str <- 'item A...1
                      item B...2
                      item C...3
                      item D...2
                      1=foo
                      2=bar
                      3=baz'

                      lines <- stringr::str_split(sample_str, 'n', simplify = T)

                      extracted_strs <- lines[stringr::str_detect(lines, '^\d=\w+$')]

                      dfs_list <- lapply(extracted_strs, function(x) {
                      str_parts <- stringr::str_split(x, '=', simplify = T)
                      df_args = list()
                      df_args[[str_parts[2]]] = as.integer(str_parts[1])
                      df_args[['stringsAsFactors']] = F

                      do.call(data.frame, df_args)
                      })


                      df <- do.call(cbind, dfs)





                      share|improve this answer












                      There's probably a nicer way to do this, but here's a solution that doesn't require any additional libraries (beyond stringr, which you're already using).



                      sample_str <- 'item A...1
                      item B...2
                      item C...3
                      item D...2
                      1=foo
                      2=bar
                      3=baz'

                      lines <- stringr::str_split(sample_str, 'n', simplify = T)

                      extracted_strs <- lines[stringr::str_detect(lines, '^\d=\w+$')]

                      dfs_list <- lapply(extracted_strs, function(x) {
                      str_parts <- stringr::str_split(x, '=', simplify = T)
                      df_args = list()
                      df_args[[str_parts[2]]] = as.integer(str_parts[1])
                      df_args[['stringsAsFactors']] = F

                      do.call(data.frame, df_args)
                      })


                      df <- do.call(cbind, dfs)






                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Nov 13 at 5:33









                      Eric Burden

                      512




                      512












                      • Thanks. This creates a data.frame with a variable for foo, bar and bar. Would this be easier to use later on than what I created as my example?
                        – Nicholas Riley
                        Nov 13 at 14:30


















                      • Thanks. This creates a data.frame with a variable for foo, bar and bar. Would this be easier to use later on than what I created as my example?
                        – Nicholas Riley
                        Nov 13 at 14:30
















                      Thanks. This creates a data.frame with a variable for foo, bar and bar. Would this be easier to use later on than what I created as my example?
                      – Nicholas Riley
                      Nov 13 at 14:30




                      Thanks. This creates a data.frame with a variable for foo, bar and bar. Would this be easier to use later on than what I created as my example?
                      – Nicholas Riley
                      Nov 13 at 14:30











                      0














                      Thanks so much to @prosoitos for helping with this. Here's the final code I ended up using, highly based on the accepted answer — it's a mix of different packages and so forth which I hope to clean up eventually, but deadlines happen...



                      get_code_reason_mapping <- function(alert_text) {
                      alert_text %>%
                      str_extract_all(regex('^[0-9]=(.*)$', multiline = T)) %>%
                      unlist() %>%
                      str_split_fixed("=", 2) %>%
                      as.data.table() %>%
                      setnames(c('code', 'reason'))
                      }

                      encounter_alerts$code_reason_mapping <- map(encounter_alerts$alert_text, get_code_reason_mapping)

                      get_why_codes <- function(alert_text) {
                      alert_text %>%
                      str_extract_all(regex('[/n][0-9e][0-9>][0-9]$', multiline = TRUE)) %>%
                      unlist() %>%
                      str_sub(-1) %>%
                      as.data.table() %>%
                      setnames(c('code'))
                      }

                      encounter_alerts$why_codes <- map(encounter_alerts$alert_text, get_why_codes)

                      get_code_counts <- function(df1, df2) {
                      left_join(df1, df2) %>%
                      count(reason) %>%
                      spread(reason, n)
                      }

                      code_counts <- map2_df(encounter_alerts$code_reason_mapping, encounter_alerts$why_codes, get_code_counts)

                      code_counts[is.na(code_counts)] <- 0

                      code_counts





                      share|improve this answer


























                        0














                        Thanks so much to @prosoitos for helping with this. Here's the final code I ended up using, highly based on the accepted answer — it's a mix of different packages and so forth which I hope to clean up eventually, but deadlines happen...



                        get_code_reason_mapping <- function(alert_text) {
                        alert_text %>%
                        str_extract_all(regex('^[0-9]=(.*)$', multiline = T)) %>%
                        unlist() %>%
                        str_split_fixed("=", 2) %>%
                        as.data.table() %>%
                        setnames(c('code', 'reason'))
                        }

                        encounter_alerts$code_reason_mapping <- map(encounter_alerts$alert_text, get_code_reason_mapping)

                        get_why_codes <- function(alert_text) {
                        alert_text %>%
                        str_extract_all(regex('[/n][0-9e][0-9>][0-9]$', multiline = TRUE)) %>%
                        unlist() %>%
                        str_sub(-1) %>%
                        as.data.table() %>%
                        setnames(c('code'))
                        }

                        encounter_alerts$why_codes <- map(encounter_alerts$alert_text, get_why_codes)

                        get_code_counts <- function(df1, df2) {
                        left_join(df1, df2) %>%
                        count(reason) %>%
                        spread(reason, n)
                        }

                        code_counts <- map2_df(encounter_alerts$code_reason_mapping, encounter_alerts$why_codes, get_code_counts)

                        code_counts[is.na(code_counts)] <- 0

                        code_counts





                        share|improve this answer
























                          0












                          0








                          0






                          Thanks so much to @prosoitos for helping with this. Here's the final code I ended up using, highly based on the accepted answer — it's a mix of different packages and so forth which I hope to clean up eventually, but deadlines happen...



                          get_code_reason_mapping <- function(alert_text) {
                          alert_text %>%
                          str_extract_all(regex('^[0-9]=(.*)$', multiline = T)) %>%
                          unlist() %>%
                          str_split_fixed("=", 2) %>%
                          as.data.table() %>%
                          setnames(c('code', 'reason'))
                          }

                          encounter_alerts$code_reason_mapping <- map(encounter_alerts$alert_text, get_code_reason_mapping)

                          get_why_codes <- function(alert_text) {
                          alert_text %>%
                          str_extract_all(regex('[/n][0-9e][0-9>][0-9]$', multiline = TRUE)) %>%
                          unlist() %>%
                          str_sub(-1) %>%
                          as.data.table() %>%
                          setnames(c('code'))
                          }

                          encounter_alerts$why_codes <- map(encounter_alerts$alert_text, get_why_codes)

                          get_code_counts <- function(df1, df2) {
                          left_join(df1, df2) %>%
                          count(reason) %>%
                          spread(reason, n)
                          }

                          code_counts <- map2_df(encounter_alerts$code_reason_mapping, encounter_alerts$why_codes, get_code_counts)

                          code_counts[is.na(code_counts)] <- 0

                          code_counts





                          share|improve this answer












                          Thanks so much to @prosoitos for helping with this. Here's the final code I ended up using, highly based on the accepted answer — it's a mix of different packages and so forth which I hope to clean up eventually, but deadlines happen...



                          get_code_reason_mapping <- function(alert_text) {
                          alert_text %>%
                          str_extract_all(regex('^[0-9]=(.*)$', multiline = T)) %>%
                          unlist() %>%
                          str_split_fixed("=", 2) %>%
                          as.data.table() %>%
                          setnames(c('code', 'reason'))
                          }

                          encounter_alerts$code_reason_mapping <- map(encounter_alerts$alert_text, get_code_reason_mapping)

                          get_why_codes <- function(alert_text) {
                          alert_text %>%
                          str_extract_all(regex('[/n][0-9e][0-9>][0-9]$', multiline = TRUE)) %>%
                          unlist() %>%
                          str_sub(-1) %>%
                          as.data.table() %>%
                          setnames(c('code'))
                          }

                          encounter_alerts$why_codes <- map(encounter_alerts$alert_text, get_why_codes)

                          get_code_counts <- function(df1, df2) {
                          left_join(df1, df2) %>%
                          count(reason) %>%
                          spread(reason, n)
                          }

                          code_counts <- map2_df(encounter_alerts$code_reason_mapping, encounter_alerts$why_codes, get_code_counts)

                          code_counts[is.na(code_counts)] <- 0

                          code_counts






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 15 at 2:10









                          Nicholas Riley

                          37k586114




                          37k586114






























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