CODEWITHSUNDEEP
r
zebrainatree
zebrainatree

Reputation: 361

is.element on column of lists in data frame

I have a data frame with a column that contains some elements that are lists. I would like to find out which rows of the data frame contain a keyword in that column.

The data frame, df, looks a bit like this

idstr                tag
1                 wl
2           other.to
3         other.from
4 c("wl","other.to")
5                 wl
6           other.wl
7 c("ll","other.to")

The goal is to assign all of the rows with 'wl' in their tag to a new data frame. In this example, I would want a new data frame that looks like:

idstr  tag
1      wl
4      c("wl","other.to")
5      wl

I tried something like this

df_wl <- df[which(is.element('wl',df$tag)),]

but this only returns the first element of the data frame (whether or not it contains 'wl'). I think the trouble lies in iterating through the rows and implementing the "is.element" function. Here are two implementations of the function and it's results:

is.element('wl',df$tag[[4]]) >  TRUE
is.element('wl',df$tag[4]) >  FALSE

How do you suggest I iterate through the dataframe to assign df_wl with it's proper values?

PS: Here's the dput:

structure(list(idstr = 1:7, tag = structure(c(6L, 5L, 4L, 2L, 6L, 3L, 1L), .Label =     c("c(\"ll\",\"other.to\")", "c(\"wl\",\"other.to\")", "other.wl", "other.from", "other.to", "wl"), class = "factor")), .Names = c("idstr", "tag"), row.names = c(NA, -7L), class = "data.frame")

Upvotes: 1

Views: 1390

Answers (1)

Rich Scriven
Rich Scriven

Reputation: 99331

Based on your dput data. this may work. The regular expression (^wl$)|(\"wl\") matches wl from beginning to end, or any occurrence of "wl" (wrapped in double quotes)

df[grepl("(^wl$)|(\"wl\")", df$tag),]
#   idstr                tag
# 1     1                 wl
# 4     4 c("wl","other.to")
# 5     5                 wl

Upvotes: 2

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