从列表子元素列表中提取矩阵，保持矩阵的列表/子列表索引

Question

New to r and was hoping to find an elegant way of solving what seems like a simple problem. The context of the problem is as follows: I am running regressions for a set of companies at rolling periods of time. I am storing the summary of each regression in a list of lists. So, for example:

results[[i]][[t]] = summary(lm(y~x)) , where y and x are the associated vectors for company i at time t . I would like to extract matrices like sigma from results such that:

sigma[i,t] = results [[i]] [[t]]$sigma

Clearly I can do this with nested loops, but it seems that there must be a simple way of extracting this matrix in one step with something like lapply, sapply, etc. I have seen similar problems answered throughout the web and this blog, but have not been able to correctly adapt them to this problem. Another twist is some of the entries in results are 'Null', which happens when there is insufficient data for a specific company at a specific time to run the regression.

Any help or direction would be much appreciated.

Answer 1

You can use lapply and do.call :

First create some sample data:

results <- list()
for (i in 1:5){
  results[[i]] <- list()
  for (t in 1:3){
    x <- sample(10)
    y <- sample(10)
    results[[i]][[t]] <- summary(lm(x~y))
  }
}

Then create the new matrix with the sigmas:

sigma <- do.call(rbind, lapply(results, function(x)lapply(x, function(y)y$sigma)))
colnames(sigma) <- paste("t", 1:ncol(sigma), sep="")
rownames(sigma) <- paste("c", 1:nrow(sigma), sep="")

the matrix looks as follows:

> sigma
   t1       t2       t3      
c1 2.302831 3.201325 3.154122
c2 3.066436 3.179956 3.146427
c3 2.752409 3.189946 2.819306
c4 3.211249 3.210777 2.983795
c5 3.179956 3.179956 2.340034

Answer 2

或者另一种方式：

sigma <- apply(simplify2array(results),1:2,function(v)v[[1]]$sigma)

Answer 3

And another couple of ways, why not...

## seed used to generate data
set.seed(1)
sigs <- unlist(results)
sigma <- sigs[ names(sigs) %in% "sigma"]
sigma <- matrix(sigma , length( results ) )
#        [,1]     [,2]     [,3]
#[1,] 3.206527 2.797726 3.100342
#[2,] 3.208417 3.138230 3.138230
#[3,] 2.819306 3.138230 3.201325
#[4,] 3.179956 3.209833 3.194218
#[5,] 2.983795 2.652614 3.174233

Thanks to @user1981275 for providing some reproducible data.

Time is in columns.

A variation on lapply is to use sapply as it's return is already in the form you require:

t(sapply( results , function(x) sapply( x , function(y) y$sigma ) ) )
#        [,1]     [,2]     [,3]
#[1,] 3.206527 2.797726 3.100342
#[2,] 3.208417 3.138230 3.138230
#[3,] 2.819306 3.138230 3.201325
#[4,] 3.179956 3.209833 3.194218
#[5,] 2.983795 2.652614 3.174233