I have a df that contain colnames like this:
[1] "lab_id" "weeks" "group"
[4] "level" "id_row" "id"
[7] "number" "tube" "dp"
[10] "time" "label" "age"
[13] "gender" "wtspike_ab1" "wtrbd_ab1"
[16] "wts1_ab1" "wts2_ab1" "wtntd_ab2"
[19] "wtn_ab2" "alphaspike_ab2" "alpharbd_ab2"
[22] "betaspike_ab2" "betarbd_ab2" "gammaspike_ab2"
My goal is filter this df by column names, using the following approach:
test1 <- test[, test <- grepl('wtspike|wtrbd|wts1|wts2|wtntd|wtn', colnames(test))]
This works filtering the colnames that not match the pattern, which is fine. However, is taking off the metadata columns 1:14, so I tried to use something like this to avoid perform the operation in the columns 1:14.
filtered_test <- test[, 1:14(test) & grepl('wtspike|wtrbd|wts1|wts2|wtntd|wtn', colnames(test))]
Which gives me the following error:
Error: attempt to apply non-function
My question is related to what alternative can I use to perform the same operation without take the columns 1:14.
I think you're trying to keep all columns matching your grepl as well as columns 1-14 (whether or not they match). Perhaps this?
mt <- mtcars[1:3,]
mt[, grepl("d|w", colnames(mt))]
# disp drat wt
# Mazda RX4 160 3.90 2.620
# Mazda RX4 Wag 160 3.90 2.875
# Datsun 710 108 3.85 2.320
If we also want to keep the first two columns (akin to your 1:14), then
mt[, replace(grepl("d|w", colnames(mt)), 1:2, TRUE)]
# mpg cyl disp drat wt
# Mazda RX4 21.0 6 160 3.90 2.620
# Mazda RX4 Wag 21.0 6 160 3.90 2.875
# Datsun 710 22.8 4 108 3.85 2.320
Or another approach (same results)):
mt[, seq_along(mt) %in% 1:2 | grepl("d|w", colnames(mt))]
# mpg cyl disp drat wt
# Mazda RX4 21.0 6 160 3.90 2.620
# Mazda RX4 Wag 21.0 6 160 3.90 2.875
# Datsun 710 22.8 4 108 3.85 2.320