dplyr: nonstandard column names (white space, punctuation, starts with numbers)

df <- structure(list(`a a` = 1:3, `a b` = 2:4), .Names = c("a a", "a b"
), row.names = c(NA, -3L), class = "data.frame")

and the data looks like

  a a a b
1   1   2
2   2   3
3   3   4

Following call to select

select(df, 'a a')

gives

Error in abs(ind[ind < 0]) : 
  non-numeric argument to mathematical function

How can I select "a a" and/or rename it to something without space using select? I know the following approaches:

1. names(df)[1] <- "a"
2. select(df, a=1)
3. select(df, ends_with("a"))

but if I am working on a large data set, how can I get an exact match without knowing the index numer or similar column names?

Solution 1:

You may select the variable by using backticks `.

select(df, `a a`)
#   a a
# 1   1
# 2   2
# 3   3

However, if your main objective is to rename the column, you may use rename in plyr package, in which you can use both "" and ``.

rename(df, replace = c("a a" = "a"))
rename(df, replace = c(`a a` = "a"))

Or in base R:

names(df)[names(df) == "a a"] <- "a"

For a more thorough description on the use of various quotes, see ?Quotes. The 'Names and Identifiers' section is especially relevant here:

other [syntactically invalid] names can be used provided they are quoted. The preferred quote is the backtick".

See also ?make.names about valid names.

See also this post about renaming in dplyr

Solution 2:

Some alternatives to backticks, good as of dplyr 0.5.0, the current version as of this writing.

If you're trying to programmatically select an argument as a column and you don't want to rename or do something like paste/sprintf the column name into backticks, you can use as.name in conjunction with the non-standard evaluation version of select, which is select_:

dplyr::select_(df, as.name("a a"))

Many of the dplyr functions have non-standard versions. In the case of select specifically, you can also use the standard version in conjunction with the select helper one_of. See ?dplyr::select_helpers for documentation:

dplyr::select(df, dplyr::one_of("a a"))