subset with pattern

Question

Say I have a data frame df

df <- data.frame( a1 = 1:10, b1 = 2:11, c2 = 3:12 )

I wish to subset the columns, but with a pattern

Answer 1

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

Answer 2

Base R now has a convenience function endsWith():

df[, endsWith(names(df), "1")]

In data.table you can do:

library(data.table)
setDT(df)
df[, .SD, .SDcols = patterns("1")]
# Or more precisely
df[, .SD, .SDcols = patterns("1$")]

In dplyr:

library(dplyr)
select(df, contains("1"))
# Or more precisely
select(df, ends_with("1"))

Answer 3

Same same but different:

df2 <- df[, grep("1", names(df))]
   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