"unpacking" a factor list from a data.frame

Question

I'm new to R / having the option to easily re-organize data, and have hunted around for a solution but can't find exactly what I'd like to do. Reshape2's melt/cast doesn't quite seem to work and I haven't mastered plyr well enough to factor it in here.

Basically I have a data.frame with a structure outlined below, with a category column in which each element is a variable-length list of categories (more compact because the # columns is much larger, and I actually have multiple category_lists that I'd like to keep separate):

>mydf
       ID      category_list    xval    yval
1     ID1   cat1, cat2, cat3   xnum1   ynum1
2     ID2         cat2, cat3   xnum2   ynum2
3     ID3               cat1   xnum3   ynum3

I want to do manipulations with the categories as factors (and the values associated, i.e. columns 3/4), so I think I need something like this in the end, where IDs and x/y/other column values are duplicated according to the length of the category list:

       ID           category    xval    yval
1     ID1               cat1   xnum1   ynum1
2     ID1               cat2   xnum1   ynum1
3     ID1               cat3   xnum1   ynum1
4     ID2               cat2   xnum2   ynum2
5     ID2               cat3   xnum2   ynum2
6     ID3               cat3   xnum2   ynum2

If there's another solution to factor/facet on the category_list, that would be a simpler solution but I haven't come across methods that support this, e.g. the following throws an error

>ggplot(mydf, aes(x=x, y=y)) + geom_point() + facet_grid(~cat_list)

Error in layout_base(data, cols, drop = drop) : At least one layer must contain all variables used for facetting

Thanks!

Can you post the output of `dput(mydf)`. – mnel Jan 09 '13 at 00:47 — mnel, Jan 09 '13 at 00:47

mnel · Accepted Answer · 2013-01-09T01:27:11.697

The answer will depend on the format of category_list. If in fact it is a list for each row

Something like

mydf <- data.frame(ID = paste0('ID',1:3), 
 category_list = I(list(c('cat1','cat2','cat3'),  c('cat2','cat3'), c('cat1'))), 
 xval = 1:3, yval = 1:3)

or

library(data.table)
mydf <- as.data.frame(data.table(ID = paste0('ID',1:3), 
 category_list = list(c('cat1','cat2','cat3'),  c('cat2','cat3'), c('cat1')), 
 xval = 1:3, yval = 1:3) )

Then you can use plyr and merge to create your long form data

 newdf <- merge(mydf, ddply(mydf, .(ID), summarize, cat_list = unlist(category_list)), by = 'ID')


   ID    category_list xval yval cat_list
1 ID1 cat1, cat2, cat3    1    1     cat1
2 ID1 cat1, cat2, cat3    1    1     cat2
3 ID1 cat1, cat2, cat3    1    1     cat3
4 ID2       cat2, cat3    2    2     cat2
5 ID2       cat2, cat3    2    2     cat3
6 ID3             cat1    3    3     cat1

or a non-plyr approach that doesn't require merge

 do.call(rbind,lapply(split(mydf, mydf$ID), transform, cat_list = unlist(category_list)))

Accepted for being most concise. I like the use of `I()` and did not know about `merge()`. Thanks! — williaster, Jan 09 '13 at 08:53
for some reason, in my case (unpacking json), the do.call solution worked best. — Andreas, Sep 01 '15 at 10:58

Josh O'Brien · Answer 2 · 2013-01-09T01:08:42.017

A plodding but seemingly robust solution:

## Some example data
df <- as.data.frame(cbind(ID = paste0("ID", 1:2), 
                          category_list = list(4:1, 2:3), 
                          xvar = 8:9, 
                          yvar = 10:9))

## Calculate number of times each row of df will be repeated 
nn <- sapply(df$category_list, length)  
ii <- rep(seq_along(nn), times=nn)       

## Reshape data.frame
transform(df[ii,], 
          category = unlist(df$category_list),
          category_list = NULL, 
          row.names = NULL)
#    ID xvar yvar category
# 1 ID1    8   10        4
# 2 ID1    8   10        3
# 3 ID1    8   10        2
# 4 ID1    8   10        1
# 5 ID2    9    9        2
# 6 ID2    9    9        3

The use of transform, in particular the `df[ii,]` trick to expand it, is really nice. Definitely a useful alternative. This helped me understand `sapply` and `seq_along` better, too. Thanks. — williaster, Jan 09 '13 at 08:57

score 2 · Answer 3 · answered Jan 09 '13 at 00:52

A possibility:

x <- read.table(textConnection('
    ID      category_list    xval    yval
     ID1   "cat1, cat2, cat3"   xnum1   ynum1
     ID2         "cat2, cat3"   xnum2   ynum2
     ID3               "cat1"   xnum3   ynum3'),
          header=TRUE,stringsAsFactors=FALSE)

library(plyr)
ddply(x,"ID",transform,category=strsplit(category_list,",")[[1]])

##    ID    category_list  xval  yval category
## 1 ID1 cat1, cat2, cat3 xnum1 ynum1     cat1
## 2 ID1 cat1, cat2, cat3 xnum1 ynum1     cat2
## 3 ID1 cat1, cat2, cat3 xnum1 ynum1     cat3
## 4 ID2       cat2, cat3 xnum2 ynum2     cat2
## 5 ID2       cat2, cat3 xnum2 ynum2     cat3

score 0 · Answer 4 · answered Jan 09 '13 at 01:10

This will be a non-plyr approach:

cbind( x[ rep(1:nrow(x), 
              times=sapply(x$category_list, 
                            function(xx) sapply( strsplit(xx, ","), length) ) ),
          -2],    # to get rid of the old category column
       new_cats = unlist( strsplit(x$category_list, ",") ) )
 # this used Bolker's example. If these are factor will need to add `as.character`

     ID  xval  yval new_cats
1   ID1 xnum1 ynum1     cat1
1.1 ID1 xnum1 ynum1     cat2
1.2 ID1 xnum1 ynum1     cat3
2   ID2 xnum2 ynum2     cat2
2.1 ID2 xnum2 ynum2     cat3
3   ID3 xnum3 ynum3     cat1

score 0 · Answer 5 · answered Jan 09 '13 at 01:35

Another base R possibility using by:

do.call(rbind,
by(mydf,
   mydf$ID,
   function(x) {
     data.frame(
                ID=x$ID,
                category_list = unlist(strsplit(x$category_list,",")),
                xval=x$xval,
                yval=x$yval
               ) 
   }
  )
)

Result:

       ID category_list  xval  yval
ID1.1 ID1          cat1 xnum1 ynum1
ID1.2 ID1          cat2 xnum1 ynum1
ID1.3 ID1          cat3 xnum1 ynum1
ID2.1 ID2          cat2 xnum2 ynum2
ID2.2 ID2          cat3 xnum2 ynum2
ID3   ID3          cat1 xnum3 ynum3

A5C1D2H2I1M1N2O1R2T1 · Answer 6 · 2015-01-14T02:39:01.507

Note: Original answer deleted as my answer was based on a different data structure than what the OP seems to actually have.

Scenario 1: Column is a `list`

Using @mnel's sample data:

mydf <- data.frame(ID = paste0('ID',1:3), 
 category_list = I(list(c('cat1','cat2','cat3'),  c('cat2','cat3'), c('cat1'))), 
 xval = 1:3, yval = 1:3)

Using listCol_l from my "splitstackshape" package

library(splitstackshape)
listCol_l(mydf, "category_list")
#     ID xval yval category_list_ul
# 1: ID1    1    1             cat1
# 2: ID1    1    1             cat2
# 3: ID1    1    1             cat3
# 4: ID2    2    2             cat2
# 5: ID2    2    2             cat3
# 6: ID3    3    3             cat1

Using unnest from the "tidyr" package

library(tidyr)
unnest(mydf, "category_list")
#    ID category_list xval yval
# 1 ID1          cat1    1    1
# 2 ID1          cat2    1    1
# 3 ID1          cat3    1    1
# 4 ID2          cat2    2    2
# 5 ID2          cat3    2    2
# 6 ID3          cat1    3    3

Scenario 2: Column is a concatenated string

Using @BenBolker's sample data:

x <- read.table(textConnection('
    ID      category_list    xval    yval
     ID1   "cat1, cat2, cat3"   xnum1   ynum1
     ID2         "cat2, cat3"   xnum2   ynum2
     ID3               "cat1"   xnum3   ynum3'),
                header=TRUE,stringsAsFactors=FALSE)

Using cSplit from my "splitstackshape" package

library(splitstackshape)
cSplit(x, "category_list", ",", "long")
#     ID category_list  xval  yval
# 1: ID1          cat1 xnum1 ynum1
# 2: ID1          cat2 xnum1 ynum1
# 3: ID1          cat3 xnum1 ynum1
# 4: ID2          cat2 xnum2 ynum2
# 5: ID2          cat3 xnum2 ynum2
# 6: ID3          cat1 xnum3 ynum3

"unpacking" a factor list from a data.frame

6 Answers6

Scenario 1: Column is a `list`

Scenario 2: Column is a concatenated string

Linked

Related

"unpacking" a factor list from a data.frame

6 Answers6

Scenario 1: Column is a list

Scenario 2: Column is a concatenated string

Linked

Related

Scenario 1: Column is a `list`