在R中的数组中，我们如何在数组的每个元素中进行减法？

I currently have an array which is:

我目前有一个数组是:

vector1 <- c(5,9,3)
vector2 <- c(10,11,12,13,14,15)
new.array <- array(c(vector1,vector2),dim = c(3,3,100))
print(new.array)
, , 1

     [,1] [,2] [,3]
[1,]    5   10   13
[2,]    9   11   14
[3,]    3   12   15

, , 2

     [,1] [,2] [,3]
[1,]    5   10   13
[2,]    9   11   14
[3,]    3   12   15
...

I am wondering how I can subtract just the second from the third row in each matrix, while preserving the array structure. For example I would like to obtain:

我想知道如何从每个矩阵的第三行中减去第二行,同时保留数组结构。例如,我想获得:

, , 1

     [,1] [,2] [,3]
[1,]   -6    1    1

, , 2

     [,1] [,2] [,3]
[1,]   -6    1    1
...

Thanks.

2 个解决方案

#1

This will do it:

这样做:

new.array[3, , , drop = FALSE] - new.array[2, , , drop = FALSE]

The drop = FALSE is to make sure that the subset of array has the same dimmensions as the original array.

drop = FALSE是为了确保数组的子集具有与原始数组相同的dimmensions。

#2

You can do this:

你可以这样做:

If you run a dim(new.array) it will return three dimesions in order of row, column and last dimension which represents your number of matrices .To access any particular matrix in order. You have to do : array[,,1] will fetch first matrix of that particular array and so on. In this case the total number of matrices formed is 100. Run a dim(new.array) and see for yourself. Using lapply on these 100 small matrices. you can subtract the third row from second row. like below.

如果运行dim(new.array),它将按行,列和最后一维的顺序返回三个维数,表示矩阵的数量。按顺序访问任何特定的矩阵。你必须这样做:array [,, 1]将获取该特定数组的第一个矩阵,依此类推。在这种情况下,形成的矩阵总数为100.运行一个暗淡(new.array)并亲眼看看。在这100个小矩阵上使用lapply。你可以从第二行中减去第三行。如下。

> lapply(1:100, function(x)new.array[,,x][2,] - new.array[,,x][3,])

#1