我有一个列表,其中每个元素都是 5*5 矩阵。例如
[[1]]
V1 V2 V3 V4 V5
[1,] 0.000000 46.973700 21.453500 338.547000 10.401600
[2,] 43.020500 0.000000 130.652000 840.526000 56.363700
[3,] 12.605600 173.238000 0.000000 642.075000 19.628100
[4,] 217.946000 626.368000 481.329000 0.000000 642.341000
[5,] 217.946000 626.368000 481.329000 0.000000 642.341000
[[2]]
V1 V2 V3 V4 V5
[1,] 0.000000 47.973700 21.453500 338.547000 10.401600
[2,] 143.020500 0.000000 130.652000 840.526000 56.363700
[3,] 312.605600 17.238000 0.000000 642.075000 19.628100
[4,] 17.946000 126.368000 481.329000 0.000000 642.341000
[5,] 217.946000 626.368000 481.329000 0.000000 642.341000
...
如何使用类似 apply 的函数对矩阵 [1] 到 [n] 求和,并返回一个 5*5 矩阵作为结果(每个元素是每个矩阵中对应元素的总和)列表)?
最佳答案
使用减少
。
## dummy data
.list <- list(matrix(1:25, ncol = 5), matrix(1:25, ncol = 5))
Reduce('+', .list)
## [,1] [,2] [,3] [,4] [,5]
## [1,] 2 12 22 32 42
## [2,] 4 14 24 34 44
## [3,] 6 16 26 36 46
## [4,] 8 18 28 38 48
## [5,] 10 20 30 40 50
关于r - 对矩阵列表求和,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11641701/