Using apply function over the row margin with expectation of stacked results

问题

What is going on with the apply function below? Why doesn't it produce the desired output? Is there a simple way to produce the desired output using the apply function?

I feel like I'm missing something really fundamental in how the apply function is working. I think this basically comes down to the following lines in the apply function.

Line #67: else length(ans <- unlist(ans, recursive = FALSE))

Line #83: array(ans, c(len.a%/%d2, d.ans))

But I'm still not really getting what's going on. I know there are other, non-apply, ways to produce the desired output. But at this point I'm just trying to understand what's going on with apply.

ff <- matrix(1:6,ncol=2)
# [,1] [,2]
# [1,]    1    4
# [2,]    2    5
# [3,]    3    6

bb <- matrix(7:10,ncol=2)
# [,1] [,2]
# [1,]    7    9
# [2,]    8   10

# DESIRE:
#  7 36
# 14 45
# 21 54
#  8 40
# 16 50
# 24 60

# This works (but isn't the general solution I'm looking for)
rr1 <- t(t(ff) * bb[1,])
rr2 <- t(t(ff) * bb[2,])
rbind(rr1,rr2)
# [,1] [,2]
# [1,]    7   36
# [2,]   14   45
# [3,]   21   54
# [4,]    8   40
# [5,]   16   50
# [6,]   24   60

# This produces the right numbers but swaps the off-diagonal blocks. Why?
apply(bb, MARGIN=1, function(x,y) t(t(y) * x), y=ff)
# [,1] [,2]
# [1,]    7    8
# [2,]   14   16
# [3,]   21   24
# [4,]   36   40
# [5,]   45   50
# [6,]   54   60

# Other formulations like these don't make a difference
apply(bb, MARGIN=2, function(x,y) t(t(y) * as.vector(t(x))), y=ff)
apply(bb, MARGIN=2, function(x) t(t(ff) * x))

来源：https://stackoverflow.com/questions/54538730/using-apply-function-over-the-row-margin-with-expectation-of-stacked-results