I have a vector of different column names and I want to be able to loop over each of them to extract that column from a data.frame. For example, consider the data set mtcars and some variable names stored in a character vector cols. When I try to select a variable from mtcars using a dynamic subset of cols, nether of these work
cols <- c("mpg", "cyl", "am")
col <- cols[1]
col
# [1] "mpg"
mtcars$col
# NULL
mtcars$cols[1]
# NULL
how can I get these to return the same values as
mtcars$mpg
Furthermore how can I loop over all the columns in cols to get the values in some sort of loop.
for(x in seq_along(cols)) {
value <- mtcars[ order(mtcars$cols[x]), ]
}
You can't do that kind of subsetting with $. In the source code (R/src/main/subset.c) it states:
/*The $ subset operator.
We need to be sure to only evaluate the first argument.
The second will be a symbol that needs to be matched, not evaluated.
*/
Second argument? What?! You have to realise that $, like everything else in R, (including for instance ( , + , ^ etc) is a function, that takes arguments and is evaluated. df$V1 could be rewritten as
`$`(df , V1)
or indeed
`$`(df , "V1")
But...
`$`(df , paste0("V1") )
...for instance will never work, nor will anything else that must first be evaluated in the second argument. You may only pass a string which is never evaluated.
Instead use [ (or [[ if you want to extract only a single column as a vector).
For example,
var <- "mpg"
#Doesn't work
mtcars$var
#These both work, but note that what they return is different
# the first is a vector, the second is a data.frame
mtcars[[var]]
mtcars[var]
You can perform the ordering without loops, using do.call to construct the call to order. Here is a reproducible example below:
# set seed for reproducibility
set.seed(123)
df <- data.frame( col1 = sample(5,10,repl=T) , col2 = sample(5,10,repl=T) , col3 = sample(5,10,repl=T) )
# We want to sort by 'col3' then by 'col1'
sort_list <- c("col3","col1")
# Use 'do.call' to call order. Seccond argument in do.call is a list of arguments
# to pass to the first argument, in this case 'order'.
# Since a data.frame is really a list, we just subset the data.frame
# according to the columns we want to sort in, in that order
df[ do.call( order , df[ , match( sort_list , names(df) ) ] ) , ]
col1 col2 col3
10 3 5 1
9 3 2 2
7 3 2 3
8 5 1 3
6 1 5 4
3 3 4 4
2 4 3 4
5 5 1 4
1 2 5 5
4 5 3 5