解析嵌套列表的元素的更有效方法

Question

我正在开发a function that parses a nested list。不幸的是，由于原始数据的性质，我无法想象如何绕过这样做。函数中的最后三位代码吓了我一下，但他们确实完成了工作。在这里，他们是：

mkList <- lapply(rec, function(x){ 
     lapply(regex, function(y) grep(y, x, value = TRUE)) }) 
rem <- lapply(mkList, function(x){ 
     lapply(x, function(y) sub("[a-z]+,", "", y)) }) 
lapply(rem, read.as.csv) 

是的，你是正确的看到的是，这是对lapply连续5个电话。是的，你猜对了，read.as.csv还呼吁lapply

---

，使一个小重复的例子，考虑嵌套列表x和下一个“双” lapply块。结果正是我想要的，但我很好奇

是否有更好，更有效的方法将函数应用于嵌套列表的内部列表？

内部列表元素是不同字符串长度的csv向量。

> (x <- list(list(a = c("a,b,c", "d,e,f"), 
        b = c("1,2,a,b,c,d", "3,4,e,f,g,h")))) 

# [[1]] 
# [[1]]$a 
# [1] "a,b,c" "d,e,f" 
# 
# [[1]]$b 
# [1] "1,2,a,b,c,d" "3,4,e,f,g,h" 

> lapply(x, function(y){ 
     lapply(y, function(z) do.call(rbind, strsplit(z, ","))) 
    }) 

# [[1]] 
# [[1]]$a 
#  [,1] [,2] [,3] 
# [1,] "a" "b" "c" 
# [2,] "d" "e" "f" 
# 
# [[1]]$b 
#  [,1] [,2] [,3] [,4] [,5] [,6] 
# [1,] "1" "2" "a" "b" "c" "d" 
# [2,] "3" "4" "e" "f" "g" "h" 

Answer 1

其中在*apply家庭鲜为人知的功能是rapply - 用于“递归lapply”。好像你正在试图做的：

rapply(x, function(y) do.call(rbind, strsplit(y, ",", TRUE)), how = "replace") 
# [[1]] 
# [[1]]$a 
#  [,1] [,2] [,3] 
# [1,] "a" "b" "c" 
# [2,] "d" "e" "f" 
# 
# [[1]]$b 
#  [,1] [,2] [,3] [,4] [,5] [,6] 
# [1,] "1" "2" "a" "b" "c" "d" 
# [2,] "3" "4" "e" "f" "g" "h" 

对于这个特殊的例子，它是你的做法背后阴影，但是当你扩展的例子了，它被证明是更有效的。