Hi,
Although you did not provide any reproducible example, it seems you
store the same type of values in your data.frames. If this is true, it
is much more efficient to store your data in an array:
mylist <- list(a = data.frame(week1 = rnorm(24), week2 = rnorm(24)),
b = data.frame(week1 = rnorm(24), week2 = rnorm(24)))
myarray <- unlist(mylist, use.names = FALSE)
dim(myarray) <- c(nrow(mylist$a), ncol(mylist$a), length(mylist))
dimnames(myarray) <- list(hour = rownames(mylist$a),
week = colnames(mylist$a),
other = names(mylist))
# now you can do:
mean(myarray[, "week1", "a"])
# or:
colMeans(myarray)
Cheers,
Denes
On 02/08/2016 02:33 PM, Wolfgang Waser wrote:
Hello,
I have a list of 7 data frames, each data frame having 24 rows (hour of
the day) and 5 columns (weeks) with a total of 5 x 24 values
I would like to combine all 7 columns of week 1 (and 2 ...) in a
separate data frame for hourly calculations, e.g.
apply(new.data.frame,1,mean)
In some way sapply (lapply) works, but I cannot directly select columns
of the original data frames in the list. As a workaround I have to
select a range of values:
sapply(list_of_dataframes,"[",1:24)
Values 1:24 give the first column, 25:48 the second and so on.
Is there an easier / more direct way to select for specific columns
instead of selecting a range of values, avoiding loops?
Cheers,
Wolfgang
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