问题
Let's say I have the following data frame:
a <- runif(10)
dd <- as.data.frame(t(a))
names(dd) <- c("ID", "a", "a2", "b", "b2", "f", "XXX", "1", "4", "8")
In dplyr
, there is a nice way to select a number of columns. For example, to select the columns between column a and column f, I can use
dd %>% dplyr::select(a:f)
In my problem, the columns of the last part of the data frame may vary, yet they always have as name a number between 1 and 99. However, I can not seem to be able to do the same trick as above:
> dd %>% select(1:99)
Error: Position must be between 0 and n
> dd %>% select("1":"99")
Error: Position must be between 0 and n
Which is because using select()
tries to select columns by position in this way.
I would like to be able to obtain a data frame with all columns between a and f, and those with labels that are numbers between 1 and 99. Is that possible to do in one go with select()
?
回答1:
Column names starting with a number, such as "1" and "8" in your data, are not syntactically valid names (see ?make.names
). Then see the 'Names and Identifiers' section in ?Quoutes
: "other [syntactically invalid] names can be used provided they are quoted. The preferred quote is the backtick".
Thus, wrap the invalid column names in backticks (`
):
dd %>% dplyr::select(a:f, `1`:`8`)
# a a2 b b2 f 1 4 8
# 1 0.2510023 0.4109819 0.6787226 0.4974859 0.01828614 0.7449878 0.1648462 0.5875638
Another option is to use the SE-version of select
, select_
:
dd %>% dplyr::select_(.dots = c("a", "a2", ..., "1", "4", "8"))
回答2:
We can select columns a:f, and add index of numeric columns by converting colnames to numeric:
dd %>%
select(a:f, which(!is.na(as.numeric(colnames(dd)))))
来源:https://stackoverflow.com/questions/38093584/select-multiple-columns-with-dplyrselect-with-numbers-as-names