D*********2
发帖数: 535 | 1 不好意思麻烦下各位R高手～
我现在有两个矩阵，A, B, 还有一个同样的概率矩阵，要从A,B中抽样建一个新的矩阵U
。U中的每一个element, 都是从A ,B中的对应位置依bernoulli(P)抽样，P也是同样的
对应位置。
按理说是很整齐的格式。但R里面这个sample好像只能处理按元素处理？
有没有什么省时的方法，谢谢谢谢。
> (A <- matrix(1:12, 4, 3))
[,1] [,2] [,3]
[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12
> (B <- matrix(rep(c(1:3),4), 4, 3))
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 2 3 1
[3,] 3 1 2
[4,] 1 2 3
> (P <- matrix(runif(12), 4, 3))
[,1] [,2] [, | t**i
发帖数: 688 | 2 (1) Based on P matrix, generate a matrix of Bernouli sample, point-by-point
in correspondence with P matrix. This matrix, S, has either 0 or 1 at every
element.
(2) Obtain a SS matrix such that SS = matrix of 1's everywhere - S
(3) Obtain A*S + B*SS. | D*********2
发帖数: 535 | 3
point
every
Can (1) be done in matrix form? I mean if there is still a loop involved, it
is relatively slow.
【在 t**i 的大作中提到】
: (1) Based on P matrix, generate a matrix of Bernouli sample, point-by-point
: in correspondence with P matrix. This matrix, S, has either 0 or 1 at every
: element.
: (2) Obtain a SS matrix such that SS = matrix of 1's everywhere - S
: (3) Obtain A*S + B*SS.
| o****o
发帖数: 8077 | 4 for (1), you can do
(matrix(runif(12), 4, 3)>P)*1
it
【在 D*********2 的大作中提到】
:
: point
: every
: Can (1) be done in matrix form? I mean if there is still a loop involved, it
: is relatively slow.
| D*********2
发帖数: 535 | 5
Thank you!!!!! Exactly what I need.
【在 o****o 的大作中提到】
: for (1), you can do
: (matrix(runif(12), 4, 3)>P)*1
:
: it
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