R将用户定义的函数应用于数据框的所有行

Question

我正在努力遍历数据帧中某一列的行，然后使用当前行来定义将在函数中使用的参数。这是示例数据框：

df <- 
structure(list(child = c("A268", "A268497", "A268497BOX", "A268497BOX2", 
"A268497BOX218", "A277", "A277A79", "A277A79091", "A277A790911", 
"A277A79091144", "A492", "A492586", "A492586BOX", "A492586BOX1", 
"A492586BOX144", "A492A69", "A492A69027", "A492A690271", "A492A69027144", 
"A492A6902715K", "A492A6902719Y", "A492A690271BH", "A492A690271BI", 
"A492A690271CQ", "A492A690271CS", "A492A690271CT", "A492A690271CU", 
"A492A690271CV", "A492A690271CW", "A492A690271CX", "A492A690271CY", 
"A492A690271DA", "A492A69028", "A492A690281", "A492A69028144", 
"A492A69402", "A492A694021", "A492A69402144", "A492A70", "A492A70033", 
"A492A700331", "A492A70033144", "A492A700332", "A492A70033244", 
"A492A70034", "A492A700341", "A492A70034144", "A492A70035", "A492A700351", 
"A492A70035144"), clvl = c(2, 3, 4, 5, 6, 2, 3, 4, 5, 6, 2, 3, 
4, 5, 6, 3, 4, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 4, 
5, 6, 4, 5, 6, 3, 4, 5, 6, 5, 6, 4, 5, 6, 4, 5, 6), parent = c("A", 
"A268", "A268497", "A268497BOX", "A268497BOX2", "A", "A277", 
"A277A79", "A277A79091", "A277A790911", "A", "A492", "A492586", 
"A492586BOX", "A492586BOX1", "A492", "A492A69", "A492A69027", 
"A492A690271", "A492A690271", "A492A690271", "A492A690271", "A492A690271", 
"A492A690271", "A492A690271", "A492A690271", "A492A690271", "A492A690271", 
"A492A690271", "A492A690271", "A492A690271", "A492A690271", "A492A69", 
"A492A69028", "A492A690281", "A492A69", "A492A69402", "A492A694021", 
"A492", "A492A70", "A492A70033", "A492A700331", "A492A70033", 
"A492A700332", "A492A70", "A492A70034", "A492A700341", "A492A70", 
"A492A70035", "A492A700351"), plvl = c(1, 2, 3, 4, 5, 1, 2, 3, 
4, 5, 1, 2, 3, 4, 5, 2, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 
5, 5, 5, 3, 4, 5, 3, 4, 5, 2, 3, 4, 5, 4, 5, 3, 4, 5, 3, 4, 5
)), row.names = c(NA, 50L), class = "data.frame")

我的目标是生成此：

我尝试通过循环并在循环内使用不同版本的apply函数来执行此操作，但我无法正确执行。在这里，我需要定义每次迭代时x和y将是当前行中的child和pathString。有没有一种干净便捷的方法？

df[] <- apply(df,1,function(x,y) sub(x,y,x))

Answer 1

假设child（或pathString）中的字符数将继续增加，如共享数据所示，一种方式是使用purrr::accumulate，该方法允许从先前的输出中获取输入并按组应用它。

library(dplyr)

df %>%
  group_by(gr = cumsum(c(TRUE, diff(nchar(child)) < 0))) %>%
  mutate(ans = purrr::accumulate(pathString, ~sub(".*(/.*)",paste0(.x, "\\1"),.y))) 

#   child         pathString        gr ans               
#   <chr>         <chr>          <int> <chr>             
# 1 A268          A/268              1 A/268             
# 2 A268497       A268/497           1 A/268/497         
# 3 A268497BOX    A268497/BOX        1 A/268/497/BOX     
# 4 A268497BOX2   A268497BOX/2       1 A/268/497/BOX/2   
# 5 A268497BOX218 A268497BOX2/18     1 A/268/497/BOX/2/18
# 6 A277          A/277              2 A/277             
# 7 A277A79       A277/A79           2 A/277/A79         
# 8 A277A79091    A277A79/091        2 A/277/A79/091     
# 9 A277A790911   A277A79091/1       2 A/277/A79/091/1   
#10 A277A79091144 A277A790911/44     2 A/277/A79/091/1/44

保留最终输出中组的gr列以阐明如何创建组。

---

我们也可以使用Reduce在基R中实现相同的逻辑

apply_fun <- function(x, y) sub(".*(/.*)", paste0(x, "\\1"), y)

df$ans <- with(df, ave(pathString, cumsum(c(TRUE, diff(nchar(child)) < 0)), 
FUN = function(x) Reduce(apply_fun, x, accumulate = TRUE)))

Answer 2

我设法使用下面的代码块完成了它，但是循环需要75-80秒，我想可能会有更快的方法：

for(row in 1:nrow(df5)) {

  x=df5[row,2] #child
  y=df5[row,3] #pathString
  g=df5[row,c('gr')]

  df5$pathString[df5$gr==g] <- sub(x,y,df5$pathString[df5$gr==g])
  df5$child[df5$gr==g] <- sub(x,y,df5$child[df5$gr==g])

}

注意，gr是根据clvl=2填充的：

library(zoo)
df4$gr <- ifelse(df4$clvl==2,df4$child,NA)
df4$gr <- na.locf(df4$gr)

这就是df4的制作方式：

df4 <- sqldf("select  *, parent || replace(child,parent,'/') AS pathString FROM df ORDER BY child")