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I have used replicate() to produce a list of n_iter elements. A minimal working example with hacked code follows. I seek to produce the same output more efficiently, perhaps with Reduce() and do.call('rbind', list).

I begin with a list of n_iter elements. Each element contains two elements, each of which contains two additional elements (which contain vectors). That implies four vector "leaves" for each iteration.

As output, I would like to produce a four-element list, with each element containing a matrix of n_iter rows corresponding to all iterations of one of the four "leaf" vectors (i.e., each row is a "leaf" vector, and a single matrix represents all iterations of that vector).

# Create data
data <- as.data.frame(matrix(rnorm(6000, 0, 1), nrow = 1000, ncol = 6))
names(data) <- c("outcome_1", "outcome_2", "predictor_1a",
                 "predictor_1b", "predictor_2a", "predictor_2b")

# Specify number of iterations.
n_iter <- 100

# Run bootstrap with 10,000 replicates.
replicates <- replicate(n_iter, {
        samp <- data[sample(nrow(data), replace = TRUE), ]
        res <- lapply(list(c('predictor_1a', 'predictor_1b'), c('predictor_2a', 'predictor_2b')), function (t) {
                    lapply(c('outcome_1', 'outcome_2'), function (y) {
                        form <- as.formula(paste(y, paste(t, collapse = " + "), sep = " ~ "))
                        res <- lm(form, data = data)$coef
                    })
                })
            }, simplify = FALSE)

# Store results for each of the four comparisons in a list object.
repl_version <- list(list(), list(), list(), list())
for (i in 1:n_iter) {
    repl_version[[1]][[i]] <- replicates[[i]][[1]][[1]]
    repl_version[[2]][[i]] <- replicates[[i]][[1]][[2]]
    repl_version[[3]][[i]] <- replicates[[i]][[2]][[1]]
    repl_version[[4]][[i]] <- replicates[[i]][[2]][[2]]
}

# Produce SE estimates for each comparison.
repl_version <- lapply(repl_version, function (version) {
    version_matrix <- do.call('rbind', version)
})
ironchefsakai
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  • So what is the issue in your code? You want to convert the *repl_version* `list` or `lapply` loop? – Parfait Apr 05 '19 at 16:10
  • Basically, yes. An `lapply()` with `Reduce()` or something—combining the for loop and the `lapply()` that produces estimates (obviously, not here, since I removed the sampling for the purposes of a MWE)! – ironchefsakai Apr 05 '19 at 16:52
  • What is the problem currently with code? – Parfait Apr 05 '19 at 19:59

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