'mydata$column <- NULL' not working in function - r

Question

This is the function:
remove_column <- function(column_vector) {
for (column in column_vector) {
if (grepl('.y$', column)) {
mydata$column <- NULL
}
}
}
What I think it'd doing: I'm passing a vector of my column names to the function, it's looping through list of names and asking whether the last characters of each column name are ".y". If that is the case, the function eliminates the column.
I've tried putting prints here and there to see my vector and to see whether the conditional evaluates to TRUE or FALSE, and everything seems to be working fine, but for some reason, it doesn't get rid of the column.
The following function returns my column vector:
duplicate_names <- function(col_names) {
duplicates <- c()
for (name in col_names) {
# split by period i.e. colname.x would be [colname, x]
if (lengths(strsplit(name, '\\.')) > 1) {
duplicates <- c(duplicates, name)
}
}
return(duplicates)
}
I usually call it like this:
duplicate_names(names(mydata))
This is what the vector of columns looks like:
c('v1.x', 'v2.y')
When I print the function it returns the following:
[1] "v1.x" "v2.y"
As requested by a user, the dput(droplevels(horsedata[1:5, 1:5])) (data that I am using for this):
dput(droplevels(horsedata[1:5, 1:5]))
structure(list(ÿþhorse_name = structure(c(3L, 1L, 2L, 4L, 5L), .Label = c("IM PRETTY FAMES",
"JESS ROYAL BUCKS", "KISS ME IM SUGAR", "LOLAMO", "RUN MADISON RUN"
), class = "factor"), owner_name = structure(c(3L, 2L, 1L, 5L,
4L), .Label = c("Christine Tavares", "Heste Sport, Inc.", "Picov Cattle Co.",
"Procter, Wayne and Carol", "Ruth F. Barbour"), class = "factor"),
program = structure(1:5, .Label = c("1", "2", "3", "4", "5"
), class = "factor"), pp = 1:5, todays_cls = c(61L, 61L,
61L, 61L, 61L)), .Names = c("ÿþhorse_name", "owner_name",
"program", "pp", "todays_cls"), row.names = c(NA, 5L), class = "data.frame")

Answer 1

We don't need a loop to subset the columns.
mydata[!grepl('\\.y$', column_list)]
If there are other columns not in the column_list and we want to keep them (assuming that the 'column_list' is ordered)
mydata[setdiff(1:ncol(mydata), grep('\\.y$', column_list))]
We can modify your function by
changing .y$ to \\.y$ as . means any character and not just the dot
Instead of $, we use [ to subset the dataset
Return the dataset after the assignment
remove_column <- function(dat, column_vec) {
for (column in column_vec) {
if (grepl('\\.y$', column, perl=TRUE)) {
dat[column] <- NULL
}
}
dat
}
remove_column(mydata, column_list)
# v1.x v2.x v3
#1 6 1 9
#2 4 11 7
#3 14 15 5
#4 10 2 4
#5 13 4 0
#6 19 14 1
#7 5 1 8
#8 16 12 7
#9 16 13 5
#10 5 0 7
data
mydata <- structure(list(v1.x = c(6L, 4L, 14L, 10L, 13L, 19L, 5L, 16L,
16L, 5L), v1.y = c(12L, 7L, 14L, 14L, 6L, 18L, 4L, 0L, 10L, 2L
), v2.x = c(1L, 11L, 15L, 2L, 4L, 14L, 1L, 12L, 13L, 0L), v2.y = c(6L,
5L, 7L, 3L, 19L, 4L, 15L, 13L, 14L, 20L), v3 = c(9L, 7L, 5L,
4L, 0L, 1L, 8L, 7L, 5L, 7L)), .Names = c("v1.x", "v1.y", "v2.x",
"v2.y", "v3"), row.names = c(NA, -10L), class = "data.frame")
column_list <- c('v1.x', 'v1.y', 'v2.x', 'v2.y')