Reverse summary to expand comma separated strings in dataframe [duplicate]

This question already has an answer here:

Split comma-separated strings in a column into separate rows

4 answers

I have the following dataframe

group = c("cat", "dog", "horse")

value = c("1", "2", "3")

list = c("siamese,burmese,balinese","corgi,sheltie,collie","arabian,friesian,andalusian" )

df = data.frame(group, value, list)



df

  group value                        list

1   cat     1    siamese,burmese,balinese

2   dog     2        corgi,sheltie,collie

3 horse     3 arabian,friesian,andalusian

and am trying to achieve this:

  group value       list

1   cat     1    siamese

2   cat     1    burmese

3   cat     1   balinese

4   dog     2      corgi

5   dog     2    sheltie

6   dog     2     collie

7 horse     3    arabian

8 horse     3   friesian

9 horse     3 andalusian

I know how to summarize a dataframe, but I now realize that I don't know how to "unsummarize" one with comma separated strings.

asked Nov 25 '18 at 23:51

Helen K.

333

marked as duplicate by Henrik r
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Nov 26 '18 at 0:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1

Welcome to SO! Great first quetion! 👍🏼! Try: tidyr::separate_rows(df, list, sep=",")

– hrbrmstr
Nov 25 '18 at 23:57

I always forget about separate_rows and start reaching for unnest and strsplit. Thanks for the reminder.

– neilfws
Nov 25 '18 at 23:59

@neilfws I regularly have to reference the help on both gather and separate as — for some reason — I can't keep the parameters in active memory and I tend to start from the pkg help index page so I see separate_rows alot :-) Perhaps disunite (or something like that) might be a better name or alias for it.

– hrbrmstr
Nov 26 '18 at 0:03

add a comment |

This question already has an answer here:

Split comma-separated strings in a column into separate rows

4 answers

I have the following dataframe

group = c("cat", "dog", "horse")

value = c("1", "2", "3")

list = c("siamese,burmese,balinese","corgi,sheltie,collie","arabian,friesian,andalusian" )

df = data.frame(group, value, list)



df

  group value                        list

1   cat     1    siamese,burmese,balinese

2   dog     2        corgi,sheltie,collie

3 horse     3 arabian,friesian,andalusian

and am trying to achieve this:

  group value       list

1   cat     1    siamese

2   cat     1    burmese

3   cat     1   balinese

4   dog     2      corgi

5   dog     2    sheltie

6   dog     2     collie

7 horse     3    arabian

8 horse     3   friesian

9 horse     3 andalusian

I know how to summarize a dataframe, but I now realize that I don't know how to "unsummarize" one with comma separated strings.

asked Nov 25 '18 at 23:51

Helen K.

333

marked as duplicate by Henrik r
Users with the r badge can single-handedly close r questions as duplicates and reopen them as needed.

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Nov 26 '18 at 0:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1

Welcome to SO! Great first quetion! 👍🏼! Try: tidyr::separate_rows(df, list, sep=",")

– hrbrmstr
Nov 25 '18 at 23:57

I always forget about separate_rows and start reaching for unnest and strsplit. Thanks for the reminder.

– neilfws
Nov 25 '18 at 23:59

@neilfws I regularly have to reference the help on both gather and separate as — for some reason — I can't keep the parameters in active memory and I tend to start from the pkg help index page so I see separate_rows alot :-) Perhaps disunite (or something like that) might be a better name or alias for it.

– hrbrmstr
Nov 26 '18 at 0:03

add a comment |

This question already has an answer here:

Split comma-separated strings in a column into separate rows

4 answers

I have the following dataframe

group = c("cat", "dog", "horse")

value = c("1", "2", "3")

list = c("siamese,burmese,balinese","corgi,sheltie,collie","arabian,friesian,andalusian" )

df = data.frame(group, value, list)



df

  group value                        list

1   cat     1    siamese,burmese,balinese

2   dog     2        corgi,sheltie,collie

3 horse     3 arabian,friesian,andalusian

and am trying to achieve this:

  group value       list

1   cat     1    siamese

2   cat     1    burmese

3   cat     1   balinese

4   dog     2      corgi

5   dog     2    sheltie

6   dog     2     collie

7 horse     3    arabian

8 horse     3   friesian

9 horse     3 andalusian

I know how to summarize a dataframe, but I now realize that I don't know how to "unsummarize" one with comma separated strings.

asked Nov 25 '18 at 23:51

Helen K.

333

This question already has an answer here:

Split comma-separated strings in a column into separate rows

4 answers

I have the following dataframe

group = c("cat", "dog", "horse")

value = c("1", "2", "3")

list = c("siamese,burmese,balinese","corgi,sheltie,collie","arabian,friesian,andalusian" )

df = data.frame(group, value, list)



df

  group value                        list

1   cat     1    siamese,burmese,balinese

2   dog     2        corgi,sheltie,collie

3 horse     3 arabian,friesian,andalusian

and am trying to achieve this:

  group value       list

1   cat     1    siamese

2   cat     1    burmese

3   cat     1   balinese

4   dog     2      corgi

5   dog     2    sheltie

6   dog     2     collie

7 horse     3    arabian

8 horse     3   friesian

9 horse     3 andalusian

I know how to summarize a dataframe, but I now realize that I don't know how to "unsummarize" one with comma separated strings.

This question already has an answer here:

Split comma-separated strings in a column into separate rows

4 answers

asked Nov 25 '18 at 23:51

Helen K.

333

asked Nov 25 '18 at 23:51

Helen K.

333

asked Nov 25 '18 at 23:51

Helen K.

333

asked Nov 25 '18 at 23:51

Helen K.

333

asked Nov 25 '18 at 23:51

Helen K.

333

marked as duplicate by Henrik r
Users with the r badge can single-handedly close r questions as duplicates and reopen them as needed.

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Nov 26 '18 at 0:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Henrik r
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Nov 26 '18 at 0:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1

Welcome to SO! Great first quetion! 👍🏼! Try: tidyr::separate_rows(df, list, sep=",")

– hrbrmstr
Nov 25 '18 at 23:57

I always forget about separate_rows and start reaching for unnest and strsplit. Thanks for the reminder.

– neilfws
Nov 25 '18 at 23:59

@neilfws I regularly have to reference the help on both gather and separate as — for some reason — I can't keep the parameters in active memory and I tend to start from the pkg help index page so I see separate_rows alot :-) Perhaps disunite (or something like that) might be a better name or alias for it.

– hrbrmstr
Nov 26 '18 at 0:03

add a comment |

1

Welcome to SO! Great first quetion! 👍🏼! Try: tidyr::separate_rows(df, list, sep=",")

– hrbrmstr
Nov 25 '18 at 23:57

I always forget about separate_rows and start reaching for unnest and strsplit. Thanks for the reminder.

– neilfws
Nov 25 '18 at 23:59

@neilfws I regularly have to reference the help on both gather and separate as — for some reason — I can't keep the parameters in active memory and I tend to start from the pkg help index page so I see separate_rows alot :-) Perhaps disunite (or something like that) might be a better name or alias for it.

– hrbrmstr
Nov 26 '18 at 0:03

Welcome to SO! Great first quetion! 👍🏼! Try: tidyr::separate_rows(df, list, sep=",")

– hrbrmstr
Nov 25 '18 at 23:57

I always forget about separate_rows and start reaching for unnest and strsplit. Thanks for the reminder.

– neilfws
Nov 25 '18 at 23:59

@neilfws I regularly have to reference the help on both gather and separate as — for some reason — I can't keep the parameters in active memory and I tend to start from the pkg help index page so I see separate_rows alot :-) Perhaps disunite (or something like that) might be a better name or alias for it.

– hrbrmstr
Nov 26 '18 at 0:03

add a comment |

2 Answers
2

active

oldest

votes

data.frame(

  group = c("cat", "dog", "horse"),

  value = c("1", "2", "3"),

  list = c("siamese,burmese,balinese","corgi,sheltie,collie","arabian,friesian,andalusian"),

  stringsAsFactors = FALSE

) -> xdf

tidyverse:

tidyr::separate_rows(xdf, list, sep=",")

##   group value       list

## 1   cat     1    siamese

## 2   cat     1    burmese

## 3   cat     1   balinese

## 4   dog     2      corgi

## 5   dog     2    sheltie

## 6   dog     2     collie

## 7 horse     3    arabian

## 8 horse     3   friesian

## 9 horse     3 andalusian

Base R:

do.call(

  rbind.data.frame,

  lapply(1:nrow(xdf), function(idx) {



    data.frame(

      group = xdf[idx, "group"],

      value = xdf[idx, "value"],

      list = strsplit(xdf[idx, "list"], ",")[[1]],

      stringsAsFactors = FALSE

    )



  })

)

##   group value       list

## 1   cat     1    siamese

## 2   cat     1    burmese

## 3   cat     1   balinese

## 4   dog     2      corgi

## 5   dog     2    sheltie

## 6   dog     2     collie

## 7 horse     3    arabian

## 8 horse     3   friesian

## 9 horse     3 andalusian

The shootout:

microbenchmark::microbenchmark(



  unnest = transform(xdf, list = strsplit(list, ",")) %>%

    tidyr::unnest(list),



  separate_rows = tidyr::separate_rows(xdf, list, sep=","),



  base = do.call(

    rbind.data.frame,

    lapply(1:nrow(xdf), function(idx) {



      data.frame(

        group = xdf[idx, "group"],

        value = xdf[idx, "value"],

        list = strsplit(xdf[idx, "list"], ",")[[1]],

        stringsAsFactors = FALSE

      )



    })

  )

)

## Unit: microseconds

##           expr      min        lq     mean   median        uq       max neval

##         unnest 3689.890 4280.7045 6326.231 4881.160  6428.508 16670.715   100

##  separate_rows 5093.618 5602.2510 8479.712 6289.193 10352.847 24447.528   100

##           base  872.343  975.1615 1589.915 1099.391  1660.324  6663.132   100

I'm constantly surprised at the horrible performance of tidyr ops.

edited Nov 26 '18 at 0:19

answered Nov 26 '18 at 0:01

hrbrmstr

61.7k693153

add a comment |

IIUC we have unnest in R

library(dplyr)

library(tidyr)



df = data.frame(group, value, list,stringsAsFactors = F)

df %>%

   transform(list = strsplit(list, ",")) %>%

   unnest(list)

  group value       list

1   cat     1    siamese

2   cat     1    burmese

3   cat     1   balinese

4   dog     2      corgi

5   dog     2    sheltie

6   dog     2     collie

7 horse     3    arabian

8 horse     3   friesian

9 horse     3 andalusian

answered Nov 26 '18 at 0:10

Wen-Ben

119k83469

add a comment |

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

data.frame(

  group = c("cat", "dog", "horse"),

  value = c("1", "2", "3"),

  list = c("siamese,burmese,balinese","corgi,sheltie,collie","arabian,friesian,andalusian"),

  stringsAsFactors = FALSE

) -> xdf

tidyverse:

tidyr::separate_rows(xdf, list, sep=",")

##   group value       list

## 1   cat     1    siamese

## 2   cat     1    burmese

## 3   cat     1   balinese

## 4   dog     2      corgi

## 5   dog     2    sheltie

## 6   dog     2     collie

## 7 horse     3    arabian

## 8 horse     3   friesian

## 9 horse     3 andalusian

Base R:

do.call(

  rbind.data.frame,

  lapply(1:nrow(xdf), function(idx) {



    data.frame(

      group = xdf[idx, "group"],

      value = xdf[idx, "value"],

      list = strsplit(xdf[idx, "list"], ",")[[1]],

      stringsAsFactors = FALSE

    )



  })

)

##   group value       list

## 1   cat     1    siamese

## 2   cat     1    burmese

## 3   cat     1   balinese

## 4   dog     2      corgi

## 5   dog     2    sheltie

## 6   dog     2     collie

## 7 horse     3    arabian

## 8 horse     3   friesian

## 9 horse     3 andalusian

The shootout:

microbenchmark::microbenchmark(



  unnest = transform(xdf, list = strsplit(list, ",")) %>%

    tidyr::unnest(list),



  separate_rows = tidyr::separate_rows(xdf, list, sep=","),



  base = do.call(

    rbind.data.frame,

    lapply(1:nrow(xdf), function(idx) {



      data.frame(

        group = xdf[idx, "group"],

        value = xdf[idx, "value"],

        list = strsplit(xdf[idx, "list"], ",")[[1]],

        stringsAsFactors = FALSE

      )



    })

  )

)

## Unit: microseconds

##           expr      min        lq     mean   median        uq       max neval

##         unnest 3689.890 4280.7045 6326.231 4881.160  6428.508 16670.715   100

##  separate_rows 5093.618 5602.2510 8479.712 6289.193 10352.847 24447.528   100

##           base  872.343  975.1615 1589.915 1099.391  1660.324  6663.132   100

I'm constantly surprised at the horrible performance of tidyr ops.

edited Nov 26 '18 at 0:19

answered Nov 26 '18 at 0:01

hrbrmstr

61.7k693153

add a comment |

data.frame(

  group = c("cat", "dog", "horse"),

  value = c("1", "2", "3"),

  list = c("siamese,burmese,balinese","corgi,sheltie,collie","arabian,friesian,andalusian"),

  stringsAsFactors = FALSE

) -> xdf

tidyverse:

tidyr::separate_rows(xdf, list, sep=",")

##   group value       list

## 1   cat     1    siamese

## 2   cat     1    burmese

## 3   cat     1   balinese

## 4   dog     2      corgi

## 5   dog     2    sheltie

## 6   dog     2     collie

## 7 horse     3    arabian

## 8 horse     3   friesian

## 9 horse     3 andalusian

Base R:

do.call(

  rbind.data.frame,

  lapply(1:nrow(xdf), function(idx) {



    data.frame(

      group = xdf[idx, "group"],

      value = xdf[idx, "value"],

      list = strsplit(xdf[idx, "list"], ",")[[1]],

      stringsAsFactors = FALSE

    )



  })

)

##   group value       list

## 1   cat     1    siamese

## 2   cat     1    burmese

## 3   cat     1   balinese

## 4   dog     2      corgi

## 5   dog     2    sheltie

## 6   dog     2     collie

## 7 horse     3    arabian

## 8 horse     3   friesian

## 9 horse     3 andalusian

The shootout:

microbenchmark::microbenchmark(



  unnest = transform(xdf, list = strsplit(list, ",")) %>%

    tidyr::unnest(list),



  separate_rows = tidyr::separate_rows(xdf, list, sep=","),



  base = do.call(

    rbind.data.frame,

    lapply(1:nrow(xdf), function(idx) {



      data.frame(

        group = xdf[idx, "group"],

        value = xdf[idx, "value"],

        list = strsplit(xdf[idx, "list"], ",")[[1]],

        stringsAsFactors = FALSE

      )



    })

  )

)

## Unit: microseconds

##           expr      min        lq     mean   median        uq       max neval

##         unnest 3689.890 4280.7045 6326.231 4881.160  6428.508 16670.715   100

##  separate_rows 5093.618 5602.2510 8479.712 6289.193 10352.847 24447.528   100

##           base  872.343  975.1615 1589.915 1099.391  1660.324  6663.132   100

I'm constantly surprised at the horrible performance of tidyr ops.

edited Nov 26 '18 at 0:19

answered Nov 26 '18 at 0:01

hrbrmstr

61.7k693153

add a comment |

data.frame(

  group = c("cat", "dog", "horse"),

  value = c("1", "2", "3"),

  list = c("siamese,burmese,balinese","corgi,sheltie,collie","arabian,friesian,andalusian"),

  stringsAsFactors = FALSE

) -> xdf

tidyverse:

tidyr::separate_rows(xdf, list, sep=",")

##   group value       list

## 1   cat     1    siamese

## 2   cat     1    burmese

## 3   cat     1   balinese

## 4   dog     2      corgi

## 5   dog     2    sheltie

## 6   dog     2     collie

## 7 horse     3    arabian

## 8 horse     3   friesian

## 9 horse     3 andalusian

Base R:

do.call(

  rbind.data.frame,

  lapply(1:nrow(xdf), function(idx) {



    data.frame(

      group = xdf[idx, "group"],

      value = xdf[idx, "value"],

      list = strsplit(xdf[idx, "list"], ",")[[1]],

      stringsAsFactors = FALSE

    )



  })

)

##   group value       list

## 1   cat     1    siamese

## 2   cat     1    burmese

## 3   cat     1   balinese

## 4   dog     2      corgi

## 5   dog     2    sheltie

## 6   dog     2     collie

## 7 horse     3    arabian

## 8 horse     3   friesian

## 9 horse     3 andalusian

The shootout:

microbenchmark::microbenchmark(



  unnest = transform(xdf, list = strsplit(list, ",")) %>%

    tidyr::unnest(list),



  separate_rows = tidyr::separate_rows(xdf, list, sep=","),



  base = do.call(

    rbind.data.frame,

    lapply(1:nrow(xdf), function(idx) {



      data.frame(

        group = xdf[idx, "group"],

        value = xdf[idx, "value"],

        list = strsplit(xdf[idx, "list"], ",")[[1]],

        stringsAsFactors = FALSE

      )



    })

  )

)

## Unit: microseconds

##           expr      min        lq     mean   median        uq       max neval

##         unnest 3689.890 4280.7045 6326.231 4881.160  6428.508 16670.715   100

##  separate_rows 5093.618 5602.2510 8479.712 6289.193 10352.847 24447.528   100

##           base  872.343  975.1615 1589.915 1099.391  1660.324  6663.132   100

I'm constantly surprised at the horrible performance of tidyr ops.

edited Nov 26 '18 at 0:19

answered Nov 26 '18 at 0:01

hrbrmstr

61.7k693153

data.frame(

  group = c("cat", "dog", "horse"),

  value = c("1", "2", "3"),

  list = c("siamese,burmese,balinese","corgi,sheltie,collie","arabian,friesian,andalusian"),

  stringsAsFactors = FALSE

) -> xdf

tidyverse:

tidyr::separate_rows(xdf, list, sep=",")

##   group value       list

## 1   cat     1    siamese

## 2   cat     1    burmese

## 3   cat     1   balinese

## 4   dog     2      corgi

## 5   dog     2    sheltie

## 6   dog     2     collie

## 7 horse     3    arabian

## 8 horse     3   friesian

## 9 horse     3 andalusian

Base R:

do.call(

  rbind.data.frame,

  lapply(1:nrow(xdf), function(idx) {



    data.frame(

      group = xdf[idx, "group"],

      value = xdf[idx, "value"],

      list = strsplit(xdf[idx, "list"], ",")[[1]],

      stringsAsFactors = FALSE

    )



  })

)

##   group value       list

## 1   cat     1    siamese

## 2   cat     1    burmese

## 3   cat     1   balinese

## 4   dog     2      corgi

## 5   dog     2    sheltie

## 6   dog     2     collie

## 7 horse     3    arabian

## 8 horse     3   friesian

## 9 horse     3 andalusian

The shootout:

microbenchmark::microbenchmark(



  unnest = transform(xdf, list = strsplit(list, ",")) %>%

    tidyr::unnest(list),



  separate_rows = tidyr::separate_rows(xdf, list, sep=","),



  base = do.call(

    rbind.data.frame,

    lapply(1:nrow(xdf), function(idx) {



      data.frame(

        group = xdf[idx, "group"],

        value = xdf[idx, "value"],

        list = strsplit(xdf[idx, "list"], ",")[[1]],

        stringsAsFactors = FALSE

      )



    })

  )

)

## Unit: microseconds

##           expr      min        lq     mean   median        uq       max neval

##         unnest 3689.890 4280.7045 6326.231 4881.160  6428.508 16670.715   100

##  separate_rows 5093.618 5602.2510 8479.712 6289.193 10352.847 24447.528   100

##           base  872.343  975.1615 1589.915 1099.391  1660.324  6663.132   100

I'm constantly surprised at the horrible performance of tidyr ops.

edited Nov 26 '18 at 0:19

answered Nov 26 '18 at 0:01

hrbrmstr

61.7k693153

edited Nov 26 '18 at 0:19

answered Nov 26 '18 at 0:01

hrbrmstr

61.7k693153

answered Nov 26 '18 at 0:01

hrbrmstr

61.7k693153

answered Nov 26 '18 at 0:01

hrbrmstr

61.7k693153

add a comment |

IIUC we have unnest in R

library(dplyr)

library(tidyr)



df = data.frame(group, value, list,stringsAsFactors = F)

df %>%

   transform(list = strsplit(list, ",")) %>%

   unnest(list)

  group value       list

1   cat     1    siamese

2   cat     1    burmese

3   cat     1   balinese

4   dog     2      corgi

5   dog     2    sheltie

6   dog     2     collie

7 horse     3    arabian

8 horse     3   friesian

9 horse     3 andalusian

answered Nov 26 '18 at 0:10

Wen-Ben

119k83469

add a comment |

IIUC we have unnest in R

library(dplyr)

library(tidyr)



df = data.frame(group, value, list,stringsAsFactors = F)

df %>%

   transform(list = strsplit(list, ",")) %>%

   unnest(list)

  group value       list

1   cat     1    siamese

2   cat     1    burmese

3   cat     1   balinese

4   dog     2      corgi

5   dog     2    sheltie

6   dog     2     collie

7 horse     3    arabian

8 horse     3   friesian

9 horse     3 andalusian

answered Nov 26 '18 at 0:10

Wen-Ben

119k83469

add a comment |

IIUC we have unnest in R

library(dplyr)

library(tidyr)



df = data.frame(group, value, list,stringsAsFactors = F)

df %>%

   transform(list = strsplit(list, ",")) %>%

   unnest(list)

  group value       list

1   cat     1    siamese

2   cat     1    burmese

3   cat     1   balinese

4   dog     2      corgi

5   dog     2    sheltie

6   dog     2     collie

7 horse     3    arabian

8 horse     3   friesian

9 horse     3 andalusian

answered Nov 26 '18 at 0:10

Wen-Ben

119k83469

IIUC we have unnest in R

library(dplyr)

library(tidyr)



df = data.frame(group, value, list,stringsAsFactors = F)

df %>%

   transform(list = strsplit(list, ",")) %>%

   unnest(list)

  group value       list

1   cat     1    siamese

2   cat     1    burmese

3   cat     1   balinese

4   dog     2      corgi

5   dog     2    sheltie

6   dog     2     collie

7 horse     3    arabian

8 horse     3   friesian

9 horse     3 andalusian

answered Nov 26 '18 at 0:10

Wen-Ben

119k83469

answered Nov 26 '18 at 0:10

Wen-Ben

119k83469

answered Nov 26 '18 at 0:10

Wen-Ben

119k83469

answered Nov 26 '18 at 0:10

Wen-Ben

119k83469

add a comment |

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