将三维数组中的n维相加

everyone! I have a very straight problem. I have a three dimensional array called w, like this:

每个人!我有一个很直接的问题。我有一个叫做w的三维数组，像这样

> w
, , 1

     [,1] [,2] [,3]
[1,]  0.5  0.5  0.5
[2,]  0.5  0.5  0.5

, , 2

     [,1] [,2] [,3]
[1,]  0.1  0.1  1.0
[2,]  0.5  0.5  0.5

Now, it is just a representation, so this is not the actual data.

它只是一个表示，所以这不是实际的数据。

The thing is that I have add together the elements from the third dimensions, like w[1, 1, 1] + w[1, 1, 2] + w[1, 1, 3], but I don't know how many members the third dimension will have. I cannot do it in a for loop because it is within a nested for loop already (two for loops).

我把第三维的元素加起来，比如w[1,1,1] + w[1, 1, 2] + w[1, 1, 3]，但我不知道第三维会有多少个元素。我不能在for循环中这样做，因为它已经在一个嵌套的for循环中(两个for循环)。

So, I basically have to add together w[, , 1] + w[, , 2] + w[, , 3]....

所以,我基本上加一起w[1],+ w(,2)+ w[3],....

I tried something like

我试着像

for (k in 1:dims(w)[3]) # it is one of the for loops
lapply(w[, , k], '+')

but it only prints the w[, , 1] and that is it.

但它只打印w[，， 1]，就是这样。

In c++, I think you would simply write y += w[, , n].

在c++中，我认为你只需写出y += w[， n]。

I would really appreciate some thoughts on how I should approach this or maybe a solution :).

我真的很想知道我该如何解决这个问题，或者也许是一个解决方案:)。

*edit: a very embarrassing typo.

编辑:一个非常尴尬的错误。

1 个解决方案

#1

Looks like this does what you want:

看起来这是你想要的:

# sample data
w<-array(sample(1:4),dim=c(3,3,3))

# sum over dimensions 1 and 2
apply(w, MARGIN=c(1, 2), sum)

Hope this helps!

希望这可以帮助!

#1