解析嵌套列表的元素的更有效方法

Question

我正在開發a function that parses a nested list。不幸的是，由於原始數據的性質，我無法想象如何繞過這樣做。函數中的最後三位代碼嚇了我一下，但他們確實完成了工作。在這裏，他們是：

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，你是正確的看到的是，這是對lapply連續5個電話。是的，你猜對了，read.as.csv還呼籲lapply

---

，使一個小重複的例子，考慮嵌套列表x和下一個「雙」 lapply塊。結果正是我想要的，但我很好奇

是否有更好，更有效的方法將函數應用於嵌套列表的內部列表？

內部列表元素是不同字符串長度的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家庭鮮爲人知的功能是rapply - 用於「遞歸lapply」。好像你正在試圖做的：

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" 

對於這個特殊的例子，它是你的做法背後陰影，但是當你擴展的例子了，它被證明是更有效的。