dplyr: repeating rows based on splitted factor

up vote
0
down vote

favorite

I was wondering if it is possible to use dplyr to repeat rows based on the result of a function.

If I have a data frame that looks like this:

df <- data.frame(count = 1:4, type = LETTERS[1:4], subtype = letters[1:4], 

             stringsAsFactors = F)

df[2,"subtype"] <- "a,b"



  count type subtype

1     1    A       a

2     2    B     a,b

3     3    C       c

4     4    D       d

Is it possible to repeat the same row for each subtype letter after splitting them according the the comma? I would like to get something like this:

  count type subtype

1     1    A       a

2     2    B       a

3     2    B       b

4     3    C       c

5     4    D       d

I tried using rowwise + do but I'm still struggling with no results!

Thanks in advance,

Giovanni

asked Nov 8 at 11:49

Giovanni

6117

Try separate_rows function from tidyr: rdrr.io/cran/tidyr/man/separate_rows.html
– AntoniosK
Nov 8 at 12:00

add a comment |

up vote
0
down vote

favorite

I was wondering if it is possible to use dplyr to repeat rows based on the result of a function.

If I have a data frame that looks like this:

df <- data.frame(count = 1:4, type = LETTERS[1:4], subtype = letters[1:4], 

             stringsAsFactors = F)

df[2,"subtype"] <- "a,b"



  count type subtype

1     1    A       a

2     2    B     a,b

3     3    C       c

4     4    D       d

Is it possible to repeat the same row for each subtype letter after splitting them according the the comma? I would like to get something like this:

  count type subtype

1     1    A       a

2     2    B       a

3     2    B       b

4     3    C       c

5     4    D       d

I tried using rowwise + do but I'm still struggling with no results!

Thanks in advance,

Giovanni

asked Nov 8 at 11:49

Giovanni

6117

Try separate_rows function from tidyr: rdrr.io/cran/tidyr/man/separate_rows.html
– AntoniosK
Nov 8 at 12:00

add a comment |

up vote
0
down vote

favorite

I was wondering if it is possible to use dplyr to repeat rows based on the result of a function.

If I have a data frame that looks like this:

df <- data.frame(count = 1:4, type = LETTERS[1:4], subtype = letters[1:4], 

             stringsAsFactors = F)

df[2,"subtype"] <- "a,b"



  count type subtype

1     1    A       a

2     2    B     a,b

3     3    C       c

4     4    D       d

Is it possible to repeat the same row for each subtype letter after splitting them according the the comma? I would like to get something like this:

  count type subtype

1     1    A       a

2     2    B       a

3     2    B       b

4     3    C       c

5     4    D       d

I tried using rowwise + do but I'm still struggling with no results!

Thanks in advance,

Giovanni

asked Nov 8 at 11:49

Giovanni

6117

I was wondering if it is possible to use dplyr to repeat rows based on the result of a function.

If I have a data frame that looks like this:

df <- data.frame(count = 1:4, type = LETTERS[1:4], subtype = letters[1:4], 

             stringsAsFactors = F)

df[2,"subtype"] <- "a,b"



  count type subtype

1     1    A       a

2     2    B     a,b

3     3    C       c

4     4    D       d

Is it possible to repeat the same row for each subtype letter after splitting them according the the comma? I would like to get something like this:

  count type subtype

1     1    A       a

2     2    B       a

3     2    B       b

4     3    C       c

5     4    D       d

I tried using rowwise + do but I'm still struggling with no results!

Thanks in advance,

Giovanni

r dplyr duplicates rows rowwise

asked Nov 8 at 11:49

Giovanni

6117

asked Nov 8 at 11:49

Giovanni

6117

asked Nov 8 at 11:49

Giovanni

6117

asked Nov 8 at 11:49

Giovanni

6117

asked Nov 8 at 11:49

Giovanni

6117

Try separate_rows function from tidyr: rdrr.io/cran/tidyr/man/separate_rows.html
– AntoniosK
Nov 8 at 12:00

add a comment |

Try separate_rows function from tidyr: rdrr.io/cran/tidyr/man/separate_rows.html
– AntoniosK
Nov 8 at 12:00

Try separate_rows function from tidyr: rdrr.io/cran/tidyr/man/separate_rows.html
– AntoniosK
Nov 8 at 12:00

add a comment |

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

df <- data.frame(count = 1:4, type = LETTERS[1:4], subtype = letters[1:4], 

                 stringsAsFactors = F)

df[2,"subtype"] <- "a,b"



library(tidyverse)



df %>% separate_rows(subtype)



#   count type subtype

# 1     1    A       a

# 2     2    B       a

# 3     2    B       b

# 4     3    C       c

# 5     4    D       d

answered Nov 8 at 12:03

AntoniosK

12k1822

1

Brilliant! It worked, thanks a lot!
– Giovanni
Nov 8 at 12:57

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53207154%2fdplyr-repeating-rows-based-on-splitted-factor%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

df <- data.frame(count = 1:4, type = LETTERS[1:4], subtype = letters[1:4], 

                 stringsAsFactors = F)

df[2,"subtype"] <- "a,b"



library(tidyverse)



df %>% separate_rows(subtype)



#   count type subtype

# 1     1    A       a

# 2     2    B       a

# 3     2    B       b

# 4     3    C       c

# 5     4    D       d

answered Nov 8 at 12:03

AntoniosK

12k1822

1

Brilliant! It worked, thanks a lot!
– Giovanni
Nov 8 at 12:57

add a comment |

up vote
1
down vote

accepted

df <- data.frame(count = 1:4, type = LETTERS[1:4], subtype = letters[1:4], 

                 stringsAsFactors = F)

df[2,"subtype"] <- "a,b"



library(tidyverse)



df %>% separate_rows(subtype)



#   count type subtype

# 1     1    A       a

# 2     2    B       a

# 3     2    B       b

# 4     3    C       c

# 5     4    D       d

answered Nov 8 at 12:03

AntoniosK

12k1822

1

Brilliant! It worked, thanks a lot!
– Giovanni
Nov 8 at 12:57

add a comment |

up vote
1
down vote

accepted

df <- data.frame(count = 1:4, type = LETTERS[1:4], subtype = letters[1:4], 

                 stringsAsFactors = F)

df[2,"subtype"] <- "a,b"



library(tidyverse)



df %>% separate_rows(subtype)



#   count type subtype

# 1     1    A       a

# 2     2    B       a

# 3     2    B       b

# 4     3    C       c

# 5     4    D       d

answered Nov 8 at 12:03

AntoniosK

12k1822

df <- data.frame(count = 1:4, type = LETTERS[1:4], subtype = letters[1:4], 

                 stringsAsFactors = F)

df[2,"subtype"] <- "a,b"



library(tidyverse)



df %>% separate_rows(subtype)



#   count type subtype

# 1     1    A       a

# 2     2    B       a

# 3     2    B       b

# 4     3    C       c

# 5     4    D       d

answered Nov 8 at 12:03

AntoniosK

12k1822

answered Nov 8 at 12:03

AntoniosK

12k1822

answered Nov 8 at 12:03

AntoniosK

12k1822

answered Nov 8 at 12:03

AntoniosK

12k1822

1

Brilliant! It worked, thanks a lot!
– Giovanni
Nov 8 at 12:57

add a comment |

1

Brilliant! It worked, thanks a lot!
– Giovanni
Nov 8 at 12:57

Brilliant! It worked, thanks a lot!
– Giovanni
Nov 8 at 12:57

add a comment |

draft saved

draft discarded

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Agfdhyk