如何在 R 中的 list() 上执行嵌套循环？

Question

Suppose I have a list like this in R:

> L
[[1]]
[1] 0.6876619 0.7847888 0.6377801 0.2078056 0.8981001

[[2]]
[1] 0.9358160 0.8905056 0.7715877 0.8648426 0.4915060

[[3]]
[1] 0.88095630 0.08010288 0.15140700 0.35400865 0.60317717

[[4]]
[1] 0.07436267 0.85873209 0.49881141 0.92363954 0.87208334

And I want to find the correlation coefficient between each pair of vectors, eg, cor(L[[i]],L[[j]]) . Is there any solution to perform it with the apply family function? Please take it as a specific case of a general question: What if we need to do a triple nested loop over a List() in R?

Answer 1

You can nest lapply calls:

lapply(L, function(x) lapply(L, function(y) cor(x,y))))

If you want the results presented more nicely, put them in a matrix:

    L <- list(rnorm(10), rnorm(10), rnorm(10))
    matrix(unlist(lapply(L, 
                         function(x) lapply(L, 
                                            function(y) cor(x,y)))),
           length(L))
#>            [,1]       [,2]       [,3]
#> [1,]  1.0000000 -0.3880931 -0.4164212
#> [2,] -0.3880931  1.0000000  0.4158335
#> [3,] -0.4164212  0.4158335  1.0000000

^{Created on 2021-05-31 by the reprex package (v2.0.0)}

Answer 2

You can cbind the list and call cor with the resulting matirx .

cor(do.call(cbind, L))
#           [,1]        [,2]        [,3]        [,4]
#[1,]  1.0000000 -0.46988357  0.14151672  0.14151672
#[2,] -0.4698836  1.00000000 -0.09177819 -0.09177819
#[3,]  0.1415167 -0.09177819  1.00000000  1.00000000
#[4,]  0.1415167 -0.09177819  1.00000000  1.00000000

In case there is one more level in the list use unlist .

L2 <- lapply(L, list) #Create list with one more level.
cor(do.call(cbind, unlist(L2, FALSE)))

In case it is unknown or mixed, a recursive call of a function could be used:

L3 <- list(L[[1]], L[[2]], L2[[3]], L2[[4]])
f <- function(x) {
  if(is.list(x)) sapply(x, f)
  else x
}
cor(f(L3))

Data:

L <- list(c(0.6876619,0.7847888,0.6377801,0.2078056,0.8981001)
        , c(0.9358160,0.8905056,0.7715877,0.8648426,0.4915060)
        , c(0.88095630,0.08010288,0.15140700,0.35400865,0.60317717)
        , c(0.88095630,0.08010288,0.15140700,0.35400865,0.60317717))

Answer 3

You could use mapply. Generate all the combinations of interest (pairs, triples, ...) and then apply

L=replicate(5,rnorm(5),simplify=F)

tmp=expand.grid(1:length(L),1:length(L))

tmp$cor=mapply(
  function(y,x){cor(L[[y]],L[[x]])},
  tmp$Var1,
  tmp$Var2
)

   Var1 Var2        cor
1     1    1  1.0000000
2     2    1  0.1226881
3     3    1 -0.2871613
4     4    1  0.4746545
5     5    1  0.9779644
6     1    2  0.1226881
7     2    2  1.0000000
...