how to drop columns by passing variable name with dplyr?

Question

I have a df as follows:

a <- data_frame(keep=c("hello", "world"),drop = c("nice", "work"))
a
Source: local data frame [2 x 2]
   keep  drop
  (chr) (chr)
1 hello  nice
2 world  work

I can use a %>% select(-drop) to drop the column without problem. however, if I want to pass a variable to present drop column, then it returns error.

name <- "drop"
a  %>% select(-(name))

Error in -(name) : invalid argument to unary operator

Psidom · Accepted Answer · 2017-01-24T04:05:44.083

46

You can use one_of to find the column positions and then use - to drop it, select(-one_of(name)), if you check ?select, the usage is documented in the Drop variable section in the Examples:

name <- "drop"
a %>% select(-one_of(name))

# A tibble: 2 × 1
#   keep
#  <chr>
#1 hello
#2 world

Or with select_, you need to paste - with the column names to drop them and pass the pasted column names to the .dots parameter if there are more than one column to be dropped:

name <- "drop"
a %>% select_(.dots = paste("-", name))

# A tibble: 2 × 1
#   keep
#  <chr>
#1 hello
#2 world

edited Jan 24 '17 at 04:05

answered Jan 24 '17 at 03:29

Psidom

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a simple question - generally when column names are in strings then we use `select_()` version right? when i tried this, didn't work – joel.wilson Jan 24 '17 at 03:37
select_ is no longer advice. Can you update the solution? – xiaodai Nov 22 '19 at 00:32

score 11 · Answer 2 · edited Jan 09 '19 at 09:25

You can simple use

a <- data_frame(keep=c("hello", "world"),drop = c("nice", "work"))
select(a, -starts_with('drop'))
#   Source: local data frame [2 x 1]
#
#   keep
#   (chr)
# 1 hello
# 2 world

you have to search for some previously written solutions too. Please read the document here Select/rename variables by name.DPLYR

I hope that does the job for you :) @Psidom thanx for updating your answer.. but I will request upvoters for vote for me too as I recently became an active user and still am to get basic privileges on stackoverflow.

score 4 · Answer 3 · edited Aug 27 '23 at 18:02

4

We can use select_ with setdiff

a %>%
    select_(setdiff(names(.), name))
# A tibble: 2 × 1
#   keep
#   <chr>
#1 hello
#2 world

edited Aug 27 '23 at 18:02

robertspierre

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answered Jan 24 '17 at 05:05

akrun

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Notice that `select_` is not available anymore with modern versions of dplyr (as of August 2023). See [James Hirschorn's answer](https://stackoverflow.com/a/70359311/1719931) for how to do it with modern versions of dplyr. – robertspierre Aug 27 '23 at 18:10

moodymudskipper · Answer 4 · 2018-08-21T07:29:19.513

4

A few more possibilities:

name <- "drop"
a %>% `[<-`(name, value=NULL)
a %>% magrittr::inset(name,value=NULL)
a %>% purrr::modify_at(name,~NULL)

edited Aug 21 '18 at 07:29

answered Oct 30 '17 at 03:00

moodymudskipper

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score 1 · Answer 5 · answered May 25 '19 at 16:05

I could only get these solutions to work by first ungrouping the data using ungroup:

df <- df  %>%  ungroup  %>%  select(-hello)

Notice no quotation marks on the column name you want to drop (hello). Also, to remove multiple columns, just place a , after hello and add the second column.

score 1 · Answer 6 · edited Aug 27 '23 at 18:11

From the ?select_ help:

dplyr used to offer twin versions of each verb suffixed with an underscore [...] However, dplyr now uses tidy evaluation semantics. [...] Thus, the underscored versions are now superfluous.

The example given in vignette("programming"), similar to @Psidom's answer, is:

name <- "drop"
a %>% select(!all_of(name))

Alternatively, this one could create a function to drop columns, so that drop does not need quoting:

drop_columns <- function(data, cols) {
  data %>% select(!{{cols}})
}

drop_columns(a, drop)

how to drop columns by passing variable name with dplyr?

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