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使用apply函数迭代两个列表

mapply lapply list r

Paul van Oppen · 技术社区 · 7 年前

我有一个问题,我有一个数据帧列表,其中数据帧的每列在第一行有一个名称,在列中的某些位置有x-s。如果有x,则第一行中的名称被视为选中。在实际问题中,我阅读了一个包含许多工作表的xlsx文件,其中每个工作表都包含一个大矩阵:每列的第一行有一个名称,在一个稀疏的矩阵中有许多x-s。每个工作表都成为数据帧列表中的一个数据帧。行名称包含一个标识符,该标识符与查找相关,但与此处描述的我的问题无关。

data1 <- data.frame(Col1 = c("Mark", "x", "", "x", "", ""),
                    Col2 = c("Paul", "", "", "", "x", ""),
                    Col3 = c("Jane", "", "", "", "", ""),
                    Col4 = c("Mary", "x", "x", "x", "", ""),
                    Col5 = c("Peter", "x", "x", "x", "", ""),
                    stringsAsFactors = FALSE)

data2 <- data.frame(Col1 = c("Mark", "x", "x", "", "", ""),
                    Col2 = c("Paul", "", "", "", "", ""),
                    Col3 = c("Jane", "", "", "", "", ""),
                    Col4 = c("Mary", "x", "", "x", "", ""),
                    Col5 = c("Peter", "x", "x", "", "", ""),
                             stringsAsFactors = FALSE)

data <- list(data1 = data1, data2 = data2)

列表中的每个数据帧具有以下结构(为方便起见,显示为矩阵),其中列表中每个数据帧的名称相同。只有x-s不同:

> as.matrix(data1)
     Col1   Col2   Col3   Col4   Col5   
[1,] "Mark" "Paul" "Jane" "Mary" "Peter"
[2,] "x"    ""     ""     "x"    "x"    
[3,] ""     ""     ""     "x"    "x"    
[4,] "x"    ""     ""     "x"    "x"    
[5,] ""     "x"    ""     ""     ""     
[6,] ""     ""     ""     ""     ""

我想在列表中的每个数据框中添加一列(“批准人”),如果列中有一个“x”,则该列是第1行中名称的串联,如下所示:

     Col1   Col2   Col3   Col4   Col5    Approvers          
[1,] "Mark" "Paul" "Jane" "Mary" "Peter" ""                 
[2,] "x"    ""     ""     "x"    "x"     "Mark; Mary; Peter"
[3,] ""     ""     ""     "x"    "x"     "Mary; Peter"      
[4,] "x"    ""     ""     "x"    "x"     "Mark; Mary; Peter"
[5,] ""     "x"    ""     ""     ""      "Paul"             
[6,] ""     ""     ""     ""     ""      ""

目前,我分两步解决这个问题:

在嵌套的for循环中,我查找第一行中的所有名称并将它们连接起来。

position <- lapply(data, function(x) apply(x, 1, function(y) which(y %in% "x")))
position <- lapply(position, function(x) lapply(x, function(y) {if (length(y) == 0L) return(0) else return(y)})) # remove int(0) and replace with 0
position <- lapply(position, function(x) lapply(x, function(x) paste(x, collapse = ","))) # flatten second level list into string


for (i in 1:length(data)) {
  for (j in 1:nrow(data[[i]])) {
    if (as.numeric(unlist(strsplit(position[[i]][[j]], ",")))[[1]] == 0) {
      data[[i]][j, "Approvers"] <- ""
    } else {
      data[[i]][j, "Approvers"] <- paste(data[[i]][1, as.numeric(unlist(strsplit(position[[i]][[j]], ",")))], collapse = "; ")
    }
  }
}

对我来说,这很笨拙,我想使用lappy和mapply同时遍历两个列表来实现这一点,但我不知道如何做到这一点。此外,创建position对象并将x-s的列索引折叠成字符串,然后在循环中将其分离,这过于复杂。

2 回复 | 直到 7 年前

akrun 7 年前

我们可以使用 lapply 在 list 然后使用 apply 在行和上循环 paste x :

res <- lapply(data, function(x) {
       x$Approvers <- apply(x, 1, FUN = function(y) paste(x[1,][y =="x"], collapse=";"))
       x})
res
#$data1
#  Col1 Col2 Col3 Col4  Col5       Approvers
#1 Mark Paul Jane Mary Peter                
#2    x              x     x Mark;Mary;Peter
#3                   x     x      Mary;Peter
#4    x              x     x Mark;Mary;Peter
#5         x                            Paul
#6                                          

#$data2
#  Col1 Col2 Col3 Col4  Col5       Approvers
#1 Mark Paul Jane Mary Peter                
#2    x              x     x Mark;Mary;Peter
#3    x                    x      Mark;Peter
#4                   x                  Mary
#5                                          
#6

注意:看起来 names 数据集的值应为“Mark”、“Paul”等,而不是“Col1”、“Col2”,。。

conor 7 年前

作为替代方案,可能值得整理这些数据,以便更容易操作和推理。此外,您想要的输出可能并不总是理想的,因为它返回了完整的行 NAs . 这里的代码重构了数据帧,使列名成为人名。然后对数据进行整形,使其有两列 name row_ix )其中,该名称列显示“x”。然后我放下 NAs 然后将名称粘贴在一起,返回更整洁的数据帧。

我很欣赏这一点,但从长远来看,以更整洁的方式存储数据可能会为您节省问题。

library(dplyr)
library(purrr)
library(tidyr)
library(magrittr)

data %>% 
  map(function(x) #map function to all dataframes in list
  x %>% set_colnames(.[1, ]) %>% # set column names equal to first row values
  dmap(~ifelse(. == "x", seq_along(.), NA)) %>% # check for "x" in all rows of all columns
  gather(name, row_ix) %>% # reshape from wide to long, call new columns name and row_ix
  drop_na() %>% # drop NAs in the dataframe
  group_by(row_ix) %>% # group by row index
  summarise(approvers = paste0(name, collapse = ";")) # concatenate names from each group
  )

$data1
# A tibble: 4 Ã 2
  row_ix       approvers
   <int>           <chr>
1      2 Mark;Mary;Peter
2      3      Mary;Peter
3      4 Mark;Mary;Peter
4      5            Paul

$data2
# A tibble: 3 Ã 2
  row_ix       approvers
   <int>           <chr>
1      2 Mark;Mary;Peter
2      3      Mark;Peter
3      4            Mary