问题
Apologies, I just realised that this has already been answered here.
This should be pretty basic but I do not really understand why it is happening. Can someone help? This is the simple code with the example 'data':
applyDirichletPrior <- function (row_vector) {
row_vector_added <- row_vector + min (row_vector)
row_vector_result <- row_vector_added / sum(row_vector_added)
}
data <- matrix(c(1,2,3,4,5,6,7,8), nrow=2, ncol=4)
applied <- apply(data, 1, function(x) applyDirichletPrior(x))
The output is given as:
> applied
[,1] [,2]
[1,] 0.1 0.1428571
[2,] 0.2 0.2142857
[3,] 0.3 0.2857143
[4,] 0.4 0.3571429
Whereas I expect an output in the same format as the input data, like:
> applied
[,1] [,2] [,3] [,4]
[1,] 0.1000000 0.2000000 0.3000000 0.4000000
[2,] 0.1428571 0.2142857 0.2857143 0.3571429
Why and at what stage during apply
is the transpose happening?
回答1:
Combining the results of the individual apply steps is somewhat arbitrary. On what basis was your expectation different? The behaviour you see is how the documentation describes it:
If each call to ‘FUN’ returns a vector of length ‘n’, then ‘apply’ returns an array of dimension ‘c(n, dim(X)[MARGIN])’ if ‘n > 1’.
Note that you can easily perform the transpose afterwards:
> t(applied)
[,1] [,2] [,3] [,4]
[1,] 0.1000000 0.2000000 0.3000000 0.4000000
[2,] 0.1428571 0.2142857 0.2857143 0.3571429
来源:https://stackoverflow.com/questions/18312389/results-transposed-with-r-apply