Split delimited strings in a column and insert as new rows

106,139

Solution 1

Here is another way of doing it..

df <- read.table(textConnection("1|a,b,c\n2|a,c\n3|b,d\n4|e,f"), header = F, sep = "|", stringsAsFactors = F)

df
##   V1    V2
## 1  1 a,b,c
## 2  2   a,c
## 3  3   b,d
## 4  4   e,f

s <- strsplit(df$V2, split = ",")
data.frame(V1 = rep(df$V1, sapply(s, length)), V2 = unlist(s))
##   V1 V2
## 1  1  a
## 2  1  b
## 3  1  c
## 4  2  a
## 5  2  c
## 6  3  b
## 7  3  d
## 8  4  e
## 9  4  f

Solution 2

As of Dec 2014, this can be done using the unnest function from Hadley Wickham's tidyr package (see release notes http://blog.rstudio.org/2014/12/08/tidyr-0-2-0/)

> library(tidyr)
> library(dplyr)
> mydf

  V1    V2
2  1 a,b,c
3  2   a,c
4  3   b,d
5  4   e,f
6  .     .


> mydf %>% 
    mutate(V2 = strsplit(as.character(V2), ",")) %>% 
    unnest(V2)

   V1 V2
1   1  a
2   1  b
3   1  c
4   2  a
5   2  c
6   3  b
7   3  d
8   4  e
9   4  f
10  .  .

Update 2017: note the separate_rows function as described by @Tif below.

It works so much better, and it allows to "unnest" multiple columns in a single statement:

> head(mydf)
geneid              chrom    start  end strand  length  gene_count
ENSG00000223972.5   chr1;chr1;chr1;chr1;chr1;chr1;chr1;chr1;chr1    11869;12010;12179;12613;12613;12975;13221;13221;13453   12227;12057;12227;12721;12697;13052;13374;14409;13670   +;+;+;+;+;+;+;+;+   1735    11
ENSG00000227232.5   chr1;chr1;chr1;chr1;chr1;chr1;chr1;chr1;chr1;chr1;chr1  14404;15005;15796;16607;16858;17233;17606;17915;18268;24738;29534   14501;15038;15947;16765;17055;17368;17742;18061;18366;24891;29570   -;-;-;-;-;-;-;-;-;-;-   1351    380
ENSG00000278267.1   chr1    17369   17436   -   68  14
ENSG00000243485.4   chr1;chr1;chr1;chr1;chr1    29554;30267;30564;30976;30976   30039;30667;30667;31097;31109   +;+;+;+;+   1021    22
ENSG00000237613.2   chr1;chr1;chr1  34554;35277;35721   35174;35481;36081   -;-;-   1187    24
ENSG00000268020.3   chr1    52473   53312   +   840 14


> mydf %>% separate_rows(strand, chrom, gene_start, gene_end)
geneid  length  gene_count  strand  chrom   start   end
ENSG00000223972.5   1735    11  +   chr1    11869   12227
ENSG00000223972.5   1735    11  +   chr1    12010   12057
ENSG00000223972.5   1735    11  +   chr1    12179   12227
ENSG00000223972.5   1735    11  +   chr1    12613   12721
ENSG00000223972.5   1735    11  +   chr1    12613   12697
ENSG00000223972.5   1735    11  +   chr1    12975   13052
ENSG00000223972.5   1735    11  +   chr1    13221   13374
ENSG00000223972.5   1735    11  +   chr1    13221   14409
ENSG00000223972.5   1735    11  +   chr1    13453   13670
ENSG00000227232.5   1351    380 -   chr1    14404   14501
ENSG00000227232.5   1351    380 -   chr1    15005   15038
ENSG00000227232.5   1351    380 -   chr1    15796   15947
ENSG00000227232.5   1351    380 -   chr1    16607   16765
ENSG00000227232.5   1351    380 -   chr1    16858   17055
ENSG00000227232.5   1351    380 -   chr1    17233   17368
ENSG00000227232.5   1351    380 -   chr1    17606   17742
ENSG00000227232.5   1351    380 -   chr1    17915   18061
ENSG00000227232.5   1351    380 -   chr1    18268   18366
ENSG00000227232.5   1351    380 -   chr1    24738   24891
ENSG00000227232.5   1351    380 -   chr1    29534   29570
ENSG00000278267.1   68  5   -   chr1    17369   17436
ENSG00000243485.4   1021    8   +   chr1    29554   30039
ENSG00000243485.4   1021    8   +   chr1    30267   30667
ENSG00000243485.4   1021    8   +   chr1    30564   30667
ENSG00000243485.4   1021    8   +   chr1    30976   31097
ENSG00000243485.4   1021    8   +   chr1    30976   31109
ENSG00000237613.2   1187    24  -   chr1    34554   35174
ENSG00000237613.2   1187    24  -   chr1    35277   35481
ENSG00000237613.2   1187    24  -   chr1    35721   36081
ENSG00000268020.3   840 0   +   chr1    52473   53312

Solution 3

Now you can use tidyr 0.5.0's separate_rows is in place of strsplit + unnest.

For example:

library(tidyr)
(df <- read.table(textConnection("1|a,b,c\n2|a,c\n3|b,d\n4|e,f"), header = F, sep = "|", stringsAsFactors = F))
  V1    V2
1  1 a,b,c
2  2   a,c
3  3   b,d
4  4   e,f
separate_rows(df, V2)

Gives:

  V1 V2
1  1  a
2  1  b
3  1  c
4  2  a
5  2  c
6  3  b
7  3  d
8  4  e
9  4  f

See reference: https://blog.rstudio.org/2016/06/13/tidyr-0-5-0/

Solution 4

Here's a data.table solution:

d.df <- read.table(header=T, text="V1 | V2
1 | a,b,c
2 | a,c
3 | b,d
4 | e,f", stringsAsFactors=F, sep="|", strip.white = TRUE)
require(data.table)
d.dt <- data.table(d.df, key="V1")
out <- d.dt[, list(V2 = unlist(strsplit(V2, ","))), by=V1]

#    V1 V2
# 1:  1  a
# 2:  1  b
# 3:  1  c
# 4:  2  a
# 5:  2  c
# 6:  3  b
# 7:  3  d
# 8:  4  e
# 9:  4  f

> sapply(out$V2, nchar) # (or simply nchar(out$V2))
# a b c a c b d e f 
# 1 1 1 1 1 1 1 1 1 

Solution 5

You can consider cSplit with direction = "long" from my "splitstackshape" package.

Usage would be:

cSplit(mydf, "V2", ",", "long")
##    V1 V2
## 1:  1  a
## 2:  1  b
## 3:  1  c
## 4:  2  a
## 5:  2  c
## 6:  3  b
## 7:  3  d
## 8:  4  e
## 9:  4  f

Old answer....

Here is one approach using base R. It assumes we're starting with a data.frame named "mydf". It uses read.csv to read in the second column as a separate data.frame, which we combine with the first column from your source data. Finally, you use reshape to convert the data into a long form.

temp <- data.frame(Ind = mydf$V1, 
                   read.csv(text = as.character(mydf$V2), header = FALSE))
temp1 <- reshape(temp, direction = "long", idvar = "Ind", 
                 timevar = "time", varying = 2:ncol(temp), sep = "")
temp1[!temp1$V == "", c("Ind", "V")]
#     Ind  V
# 1.1   1  a
# 2.1   2  a
# 3.1   3  b
# 4.1   4  e
# 1.2   1  b
# 2.2   2  c
# 3.2   3  d
# 4.2   4  f
# 1.3   1  c

Another fairly direct alternative is:

stack(
  setNames(
    sapply(strsplit(mydf$V2, ","), 
           function(x) gsub("^\\s|\\s$", "", x)), mydf$V1))
  values ind
1      a   1
2      b   1
3      c   1
4      a   2
5      c   2
6      b   3
7      d   3
8      e   4
9      f   4
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106,139
Boxuan
Author by

Boxuan

Updated on July 10, 2022

Comments

  • Boxuan
    Boxuan almost 2 years

    I have a data frame as follow:

    +-----+-------+
    |  V1 |  V2   |
    +-----+-------+
    |  1  | a,b,c |
    |  2  | a,c   |
    |  3  | b,d   |
    |  4  | e,f   |
    |  .  | .     |
    +-----+-------+
    

    Each of the alphabet is a character separated by comma. I would like to split V2 on each comma and insert the split strings as new rows. For instance, the desired output will be:

    +----+----+
    | V1 | V2 |
    +----+----+
    |  1 |  a |
    |  1 |  b |
    |  1 |  c |
    |  2 |  a |
    |  2 |  c |
    |  3 |  b |
    |  3 |  d |
    |  4 |  e |
    |  4 |  f |
    +----+----+
    

    I am trying to use strsplit() to spit V2 first, then cast the list into a data frame. It didn't work. Any help will be appreciated.

  • A5C1D2H2I1M1N2O1R2T1
    A5C1D2H2I1M1N2O1R2T1 about 11 years
    It appears that you would have to throw a gsub or something in there to strip whitespace with data.table too, but it's somewhat strange that it isn't visible in the output. print(as.data.frame(d.dt), quote=TRUE) shows that the whitespace is still there. +1 though.
  • Arun
    Arun about 11 years
    Thanks. added strip.white = TRUE.
  • indra_patil
    indra_patil about 8 years
    Hey what if I have other column in this data frame and I do want those columns also in final splitted data frame?
  • cloudscomputes
    cloudscomputes over 6 years
    This one is not simple but it is thoughtful
  • Jonathan Rauscher
    Jonathan Rauscher over 6 years
    @cloudscomputes this actually is a pretty simple answer. Thank you.
  • ersan
    ersan about 3 years
    How can i reverse the function? When output is the input, and input will be my desired output.
  • PesKchan
    PesKchan almost 3 years
    wow from biostar to here ..found you