初始化R中的矩阵列表

Question

I was wondering if there's a quick way to initialize a list of matrices in R. For example I'm looking for a (one-liner) to reproduce the same results as the following:

output_array = list()
for(i in 1:10){
output_array[i] = diag(2)
}

Thanks!

Answer 1

还可以尝试以下* apply包装器：

replicate(10, diag(2), simplify=F)

Answer 2

This one liner should work

rep(list(diag(2)), 10)

If you want the contents of the matrices to vary, then something like

lapply(1:10, function(x) matrix(1:x^2, x, x)

will be more appropriate. The contents of the anonymous function will obviously be something a bit more useful than my example, but the principle is the same

Answer 3

Both replicate and rep have been recommended. FYI: The difference is the evaluation of the expression being passed. 'rep' evaluates it's arguments as input, whereas 'replicate' evaluates them inside the 'loop'.

You can see this with random numbers. With replicate the numbers are different because the expression 'diag(rnorm(2))' is evaluated multiple times, whereas with rep, it's only evaluated once and the value is repeated.

rep(list(diag(rnorm(2))),2)

[[1]]

  [,1] [,2] 

[1,] 1.0844 0.0000

[2,] 0.0000 -2.3457

[[2]]

  [,1] [,2] 

[1,] 1.0844 0.0000

[2,] 0.0000 -2.3457

replicate(2,diag(rnorm(2)))

, , 1

  [,1] [,2] 

[1,] 0.42912 0.00000

[2,] 0.00000 0.50606

, , 2

  [,1] [,2] 

[1,] -0.57474 0.00000

[2,] 0.00000 -0.54663

This may or may not matter for you, but there are performance implications.

system.time(replicate(1000, diag(100),simplify=F))

user system elapsed

0.640 0.032 0.674

system.time(rep(list(diag(100)),1000))

user system elapsed

0.072 0.036 0.111

Answer 4

Try

lapply(1:10, diag, 2)

In your for loop you had to write [[ in output_array[[i]] = diag(2)

Answer 5

You can use an array :

 h <- array(1:2, c(2,2,10))

h[,,2]        ####
     [,1] [,2]
[1,]    1    1
[2,]    2    2