It is possible to do this via
subset(df, select = grepl("1", names(df)))
For automating this as a function, one can use use [
to do the subsetting. Couple that with one of R's regular expression functions and you have all you need.
By way of an example, here is a custom function implementing the ideas I mentioned above.
Subset <- function(df, pattern) {
ind <- grepl(pattern, names(df))
df[, ind]
}
Note this does not error checking etc and just relies upon grepl
to return a logical vector indicating which columns match pattern
, which is then passed to [
to subset by columns. Applied to your df
this gives:
> Subset(df, pattern = "1")
a1 b1
1 1 2
2 2 3
3 3 4
4 4 5
5 5 6
6 6 7
7 7 8
8 8 9
9 9 10
10 10 11
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