Subset one column from data frame keeping the subset as a data frame

Question

I have a table containing filters, such as

> X = data.frame(filter1=c(1,1,1),filter2=c(1,1,0),filter3=c(1,0,0))
> X
  filter1 filter2 filter3
1       1       1       1
2       1       1       0
3       1       0       0

I like to select the filters based on conditionals:

> selected_filters = c(1:2)
> X[,paste0("filter",c(selected_filters))]
  filter1 filter2
1       1       1
2       1       1
3       1       0

However, if I have only one column selected, the data frame converts into a list

> selected_filters = c(2)
> X[,paste0("filter",c(selected_filters))]
[1] 1 1 0

How to select also only one column keeping the result as a data frame? Desired result would look like

Part of the problem was actually to try to find the correct documentation, because it is not obvious that for the square bracket syntax `X[...]` one cannot look for help using syntax `?[]`. As @acrun points out in his answer, one should seek for help with syntax `?Extract`. — Heikki, Apr 16 '19 at 06:42
We can get help as with any other function: `?"["` will take us to *Extract* manual. Similarly `?"+"` would take us to *Arithmetic*, etc. — zx8754, Apr 16 '19 at 06:44
Note that you can use `selected_filters` instead of `c(selected_filters)` in the solutions proposed. — LocoGris, Apr 16 '19 at 06:48

score 2 · Accepted Answer · answered Apr 16 '19 at 06:33

2

We need drop = FALSE

X[, paste0("filter",c(selected_filters)), drop = FALSE]
#   filter2
#1       1
#2       1
#3       0

If we look at ?Extract, the Usage shows

x[i, j, ... , drop = TRUE]

and in the description, it says

drop - For matrices and arrays. If TRUE the result is coerced to the lowest possible dimension (see the examples). This only works for extracting elements, not for the replacement. See drop for further details.

Note that subset behavior is different because it is by default drop = FALSE

subset(X, select = paste0("filter",c(selected_filters)))

answered Apr 16 '19 at 06:33

akrun

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I am not sure If my answer is the same as yours...i.e. `drop = TRUE` is the same as selecting without the comma...? – Sotos Apr 16 '19 at 06:40
1

@Sotos, No, it is fine :=) – akrun Apr 16 '19 at 06:42

score 2 · Answer 2 · answered Apr 16 '19 at 06:39

You will get a data frame If you avoid the comma.

X[paste0("filter", c(selected_filters))]
#  filter2
#1       1
#2       1
#3       0

str(X[paste0("filter", c(selected_filters))])
'data.frame':   3 obs. of  1 variable:
 $ filter2: num  1 1 0

str(X[,paste0("filter", c(selected_filters))])
 num [1:3] 1 1 0

Subset one column from data frame keeping the subset as a data frame

2 Answers2