tidyr use separate_rows over multiple columns

Question

I have a data.frame where some cells contain strings of comma separate values:

d <- data.frame(a=c(1:3), 
       b=c("name1, name2, name3", "name4", "name5, name6"),
       c=c("name7","name8, name9", "name10" ))

I want to separate those strings where each name is split into its own cell. This is easy with

tidyr::separate_rows(d, b, sep=",") 

if it is done for one column a time. But I can't do this for both columns "b" and "c" at the same time, since it requires that the number of names in each string is the same. Instead of writing

tidyr::separate_rows(d, b, sep=",") 
tidyr::separate_rows(d, c, sep=",") 

Is there a way to do this in a one-liner, for e.g. with apply? Something like

apply(d, 2, separate_rows(...)) 

Not sure how to pass the arguments to the separate_rows() function.

Answer 1

You can use a pipe. Note that sep = ", " is automatically detected.

d %>% separate_rows(b) %>% separate_rows(c)
#   a     b      c
# 1 1 name1  name7
# 2 1 name2  name7
# 3 1 name3  name7
# 4 2 name4  name8
# 5 2 name4  name9
# 6 3 name5 name10
# 7 3 name6 name10

Note: Using tidyr version 0.6.0, where the %>% operator is included in the package.

---

Update: Using @doscendodiscimus comment, we could use a for() loop and reassign d in each iteration. This way we can have as many columns as we like. We will use a character vector of column names, so we'll need to switch to the standard evaluation version, separate_rows_.

cols <- c("b", "c")
for(col in cols) {
    d <- separate_rows_(d, col)
}

which gives the updated d

  a     b      c
1 1 name1  name7
2 1 name2  name7
3 1 name3  name7
4 2 name4  name8
5 2 name4  name9
6 3 name5 name10
7 3 name6 name10

Answer 2

Here's an alternative approach using splitstackshape::cSplit and zoo::na.locf.

library(splitstackshape)
library(zoo)

df <- cSplit(d, 1:ncol(d), "long", sep = ",")
na.locf(df[rowSums(is.na(df)) != ncol(df),])
#    a     b      c
#1:  1 name1  name7
#2:  1 name2  name7
#3:  1 name3  name7
#4:  2 name4  name8
#5:  2 name4  name9
#6:  3 name5 name10
#7:  3 name6 name10