Create multiple new variables that are calculated from existing variables
I'm trying to figure out how to create multiple new variables that are calculated using variables currently in my dataset
Here's some example data
library(tidyverse)
df <- data.frame(
a1 = rnorm(100),
a2 = rnorm(100),
b1 = rnorm(100),
b2 = rnorm(100),
c1 = rnorm(100),
c2 = rnorm(100)
)
Essentially, I want to create a new variable for each a, b, c pair that divides a1 by a2, b1 by b2, etc. For example:
df <- df %>%
mutate(a3 = a1/a2)
The variables in my dataset don't follow these naming conventions, so I feel like I need to assign names to vectors:
numerators <- c('a1', 'b1', 'c1')
denominators <- c('a2', 'b2', 'c2')
And then creating a new variable would follow the basic convention
mutate(newvars = numerators/denominators)
But I'm stuck as to how to actually do this. Any help would be much appreciated -- thanks!
r dplyr iteration purrr
asked Nov 14 '18 at 0:57
Ekholme
254
add a comment |
I'm trying to figure out how to create multiple new variables that are calculated using variables currently in my dataset
Here's some example data
library(tidyverse)
df <- data.frame(
a1 = rnorm(100),
a2 = rnorm(100),
b1 = rnorm(100),
b2 = rnorm(100),
c1 = rnorm(100),
c2 = rnorm(100)
)
Essentially, I want to create a new variable for each a, b, c pair that divides a1 by a2, b1 by b2, etc. For example:
df <- df %>%
mutate(a3 = a1/a2)
The variables in my dataset don't follow these naming conventions, so I feel like I need to assign names to vectors:
numerators <- c('a1', 'b1', 'c1')
denominators <- c('a2', 'b2', 'c2')
And then creating a new variable would follow the basic convention
mutate(newvars = numerators/denominators)
But I'm stuck as to how to actually do this. Any help would be much appreciated -- thanks!
r dplyr iteration purrr
asked Nov 14 '18 at 0:57
Ekholme
254
add a comment |
I'm trying to figure out how to create multiple new variables that are calculated using variables currently in my dataset
Here's some example data
library(tidyverse)
df <- data.frame(
a1 = rnorm(100),
a2 = rnorm(100),
b1 = rnorm(100),
b2 = rnorm(100),
c1 = rnorm(100),
c2 = rnorm(100)
)
Essentially, I want to create a new variable for each a, b, c pair that divides a1 by a2, b1 by b2, etc. For example:
df <- df %>%
mutate(a3 = a1/a2)
The variables in my dataset don't follow these naming conventions, so I feel like I need to assign names to vectors:
numerators <- c('a1', 'b1', 'c1')
denominators <- c('a2', 'b2', 'c2')
And then creating a new variable would follow the basic convention
mutate(newvars = numerators/denominators)
But I'm stuck as to how to actually do this. Any help would be much appreciated -- thanks!
r dplyr iteration purrr
asked Nov 14 '18 at 0:57
Ekholme
254
I'm trying to figure out how to create multiple new variables that are calculated using variables currently in my dataset
Here's some example data
library(tidyverse)
df <- data.frame(
a1 = rnorm(100),
a2 = rnorm(100),
b1 = rnorm(100),
b2 = rnorm(100),
c1 = rnorm(100),
c2 = rnorm(100)
)
Essentially, I want to create a new variable for each a, b, c pair that divides a1 by a2, b1 by b2, etc. For example:
df <- df %>%
mutate(a3 = a1/a2)
The variables in my dataset don't follow these naming conventions, so I feel like I need to assign names to vectors:
numerators <- c('a1', 'b1', 'c1')
denominators <- c('a2', 'b2', 'c2')
And then creating a new variable would follow the basic convention
mutate(newvars = numerators/denominators)
But I'm stuck as to how to actually do this. Any help would be much appreciated -- thanks!
r dplyr iteration purrr
r dplyr iteration purrr
asked Nov 14 '18 at 0:57
Ekholme
254
asked Nov 14 '18 at 0:57
Ekholme
254
asked Nov 14 '18 at 0:57
Ekholme
254
asked Nov 14 '18 at 0:57
Ekholme
254
asked Nov 14 '18 at 0:57
Ekholme
254
254
add a comment |
add a comment |
1 Answer
1
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Here is an option to split the dataset into a list of data.frame based on the column name pattern, then reduce it by dividing elementwise on each pair of columns in each of the dataset and bind with the original dataset
library(tidyverse)
df %>%
split.default(sub("\d+", "", names(.))) %>%
map_df(reduce, `/`) %>%
rename_all(~paste0(., 3)) %>%
bind_cols(df, .)
answered Nov 14 '18 at 4:07
akrun
397k13187260
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here is an option to split the dataset into a list of data.frame based on the column name pattern, then reduce it by dividing elementwise on each pair of columns in each of the dataset and bind with the original dataset
library(tidyverse)
df %>%
split.default(sub("\d+", "", names(.))) %>%
map_df(reduce, `/`) %>%
rename_all(~paste0(., 3)) %>%
bind_cols(df, .)
answered Nov 14 '18 at 4:07
akrun
397k13187260
add a comment |
Here is an option to split the dataset into a list of data.frame based on the column name pattern, then reduce it by dividing elementwise on each pair of columns in each of the dataset and bind with the original dataset
library(tidyverse)
df %>%
split.default(sub("\d+", "", names(.))) %>%
map_df(reduce, `/`) %>%
rename_all(~paste0(., 3)) %>%
bind_cols(df, .)
answered Nov 14 '18 at 4:07
akrun
397k13187260
add a comment |
Here is an option to split the dataset into a list of data.frame based on the column name pattern, then reduce it by dividing elementwise on each pair of columns in each of the dataset and bind with the original dataset
library(tidyverse)
df %>%
split.default(sub("\d+", "", names(.))) %>%
map_df(reduce, `/`) %>%
rename_all(~paste0(., 3)) %>%
bind_cols(df, .)
answered Nov 14 '18 at 4:07
akrun
397k13187260
Here is an option to split the dataset into a list of data.frame based on the column name pattern, then reduce it by dividing elementwise on each pair of columns in each of the dataset and bind with the original dataset
library(tidyverse)
df %>%
split.default(sub("\d+", "", names(.))) %>%
map_df(reduce, `/`) %>%
rename_all(~paste0(., 3)) %>%
bind_cols(df, .)
answered Nov 14 '18 at 4:07
akrun
397k13187260
answered Nov 14 '18 at 4:07
akrun
397k13187260
answered Nov 14 '18 at 4:07
akrun
397k13187260
answered Nov 14 '18 at 4:07
akrun
397k13187260
397k13187260
add a comment |
add a comment |
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