一个例子可以更好地说明我的目的：

z <- list(z1 = list(a = 1, b = 2, c = 3), z2 = list(b = 4, a = 1, c = 0))

想要一个相当于随后的R对象的输出：

o <- list(a = list(z1 = 1, z2 = 1), b = list(z1 = 2, z2 = 4), c = list(z1 = 3, z2 = 0))

创建了自己的解决方案

revert_list_str_1 <- function(ls) {
  res <- lapply(names(ls[[1]]), function(n, env) {
    name <- paste(n, 'elements', sep = '_')
    assign(name, vector('list', 0))
    inner <- sapply(ls, function(x) {
      assign(name, c(get(name), x[which(names(x) == n)]))
    })
    names(inner) <- names(ls)

    inner
  })
  names(res) <- names(ls[[1]])

  res
}

执行str(revert_list_str_1(z))获得后续的输出

List of 3
 $ a:List of 2
  ..$ z1: num 1
  ..$ z2: num 1
 $ b:List of 2
  ..$ z1: num 2
  ..$ z2: num 4
 $ c:List of 2
  ..$ z1: num 3
  ..$ z2: num 0

# yes 1.1
# yes 1.2
# yes 2.1, not 2.2, not 2.3
revert_list_str_1 <- function(ls) { # @leodido
    # see above
}

# not 1.1
# not 1.2
# not 2.1, not 2.2, not 2.3
revert_list_str_2 <- function(ls) { # @mnel
  # convert each component of list to a data.frame
  # so rbind.data.frame so named elements are matched
  x <- data.frame((do.call(rbind, lapply(ls, data.frame))))
  # convert each column into an appropriately named list
  o <- lapply(as.list(x), function(i, nam) as.list(`names<-`(i, nam)), nam = rownames(x))

  o
}

# yes 1.1
# yes 1.2
# yes 2.1, not 2.2, yes 2.3
revert_list_str_3 <- function(ls) { # @mnel
  # unique names
  nn <- Reduce(unique, lapply(ls, names))
  # convert from matrix to list `[` used to ensure correct ordering
  as.list(data.frame(do.call(rbind,lapply(ls, `[`, nn))))
}

# yes 1.1
# yes 1.2
# yes 2.1, not 2.2, yes 2.3
revert_list_str_4 <- function(ls) { # @Josh O'Brien
  # get sub-elements in same order
  x <- lapply(ls, `[`, names(ls[[1]]))
  # stack and reslice
  apply(do.call(rbind, x), 2, as.list) 
}

# not 1.1
# not 1.2
# not 2.1, not 2.2, not 2.3
revert_list_str_5 <- function(ls) { # @mnel
  apply(data.frame((do.call(rbind, lapply(ls, data.frame)))), 2, as.list)
}

# not 1.1
# not 1.2
# not 2.1, yes 2.2, yes 2.3
revert_list_str_6 <- function(ls) { # @baptiste + @Josh O'Brien
  b <- recast(z, L2 ~ L1)
  apply(b, 1, as.list)
}

# yes 1.1
# yes 1.2
# not 2.1, yes 2.2, yes 2.3
revert_list_str_7 <-  function(ll) { # @Josh O'Brien
  nms <- unique(unlist(lapply(ll, function(X) names(X))))
  ll <- lapply(ll, function(X) setNames(X[nms], nms))
  ll <- apply(do.call(rbind, ll), 2, as.list)
  lapply(ll, function(X) X[!sapply(X, is.null)])
}

输入：

list(z1 = list(a = 1, b = 2, c = 3), z2 = list(a = 0, b = 3, d = 22, f = 9)).

结果：

Unit: microseconds
    expr       min         lq     median         uq       max
1 func_1   250.069   467.5645   503.6420   527.5615  2028.780
2 func_3   204.386   393.7340   414.5485   429.6010  3517.438
3 func_4    89.922   173.7030   189.0545   194.8590  1669.178
4 func_6 11295.463 20985.7525 21433.8680 21934.5105 72476.316
5 func_7   348.585   387.0265   656.7270   691.2060  2393.988

基准2

输入：

list(z1 = list(a = 1, b = 2, c = 'ciao'), z2 = list(a = 0, b = 3, c = 5)).

revert_list_str_6

结果：

Unit: microseconds
    expr     min       lq   median       uq      max
1 func_1 249.558 483.2120 502.0915 550.7215 2096.978
2 func_3 210.899 387.6835 400.7055 447.3785 1980.912
3 func_4  92.420 170.9970 182.0335 192.8645 1857.582
4 func_7 257.772 469.9280 477.8795 487.3705 2035.101

revert_list_str_4。

基准3

输入：

list(z1 = list(a = 1, b = m, c = 'ciao'), z2 = list(a = 0, b = 3, c = m)).

m是一个3x3的整数矩阵，并revert_list_str_6已被排除在外。

结果：

Unit: microseconds
expr     min       lq   median       uq      max
1 func_1 261.173 484.6345 503.4085 551.6600 2300.750
2 func_3 209.322 393.7235 406.6895 449.7870 2118.252
3 func_4  91.556 174.2685 184.5595 196.2155 1602.983
4 func_7 252.883 474.1735 482.0985 491.9485 2058.306