循环使用条件 r 的列表

Question

下面给出了一个数据框DF和一个列表mappingList。

DF <- data.frame(
           "colors number 3 former" = c("r","r","?","l","?","r","?","?","r","?"),
           "music number 3 latter" = c("r","l","r","l","r","r","l","l","r","l"),
           "genres number 3 latter" = c("l","r","?","l","?","r","?","l","l","r"),
           "genres number 12 former" = c("r","r","?","l","l","r","l","?","r","?"),
           "music number 12 latter" = c("r","l","?","l","?","r","l","l","r","?"),
           "fabric number 12 latter" = c("l","r","?","l","r","r","r","l","l","r"),
           "colors number 12 latter" = c("r","r","?","r","?","r","?","r","r","?"),
           check.names = FALSE
           )

mappingList <- list("number 3",
                    "genres",
                    "music",
                    "number 12",
                    "music",
                    "fabric",
                    "colors")

在DF中，当一列以former结尾并包含值“?”时，需要从以latter结尾的列编码。通过编码，我的意思是，former 列中的? 值需要填充其对应的latter 列中的任何值。 former 列可以有多个 latter 列。从mappingList 中找到former 列的对应latter 列。例如对于colors number 3 former，mappingList 中有 2 个列指示符：genres 和 music，因为它们在 number 3 下，colors number 3 former 属于并包含子字符串 number 3。在 for 循环中 colors number 3 former 应该首先从 genres number 3 latter 编码，对于具有值 ? 的行。如果former 列中仍然存在?，则应使用第二个选项进行映射，即“音乐编号 3 later(the next element under genres in number 3). The loop should stop if there are no more ?left in theformercolumn, if not it should move down in themappingList` 用于该数字。原始数据帧大得多，所以手动映射不是首选。预期的输出是：

expectedDF <- data.frame(
           "colors number 3 former" = c("r","r","r","l","r","r","l","l","r","r"),
           "music number 3 latter" = c("r","l","r","l","r","r","l","l","r","l"),
           "genres number 3 latter" = c("l","r","?","l","?","r","?","l","l","r"),
           "genres number 12 former" = c("r","r","?","l","l","r","l","l","r","r"),
           "music number 12 latter" =  c("r","l","?","l","?","r","l","l","r","?"),
           "fabric number 12 latter" = c("l","r","?","l","r","r","r","l","l","r"),
           "colors number 12 latter" = c("r","r","?","r","?","r","?","r","r","?"),
           check.names = FALSE
           )

我用嵌套循环尝试了这种方法，但是一旦循环到达下一个数字，我就找不到停止循环的方法：

# Take columns with that end with "former"
# Populate former columns in columnsToBeEncoded
columnsToBeEncoded <- list()
for(col in names(DF)){
  if(grepl("former", col)){
    columnsToBeEncoded <- append(columnsToBeEncoded, col)
  }
}

#columnsToBeEncoded

# Encode  "former" columns where row is "?" from "latter" columns by the order in mappingList
for(col in columnsToBeEncoded){
  # extract column number from former column
  colNumber <- paste(strsplit(col, " ")[[1]][2:3], collapse = " ")
  # Find indices where former column has "?"
  j <- which(DF[, col] == "?")
  for(element in mappingList){
    # I think the if statement below is not working
    # Inside the if statement I see elements with "number" in it are involved too
    if(!grepl(colNumber, element)){
      elementNameinColumnForm <- paste(c(element, colNumber, "latter"), collapse = " ")
      print(elementNameinColumnForm)
      DF[j,col] <- DF[j,elementNameinColumnForm]

    }
  }
}

Answer 1

试试下面的代码是否有效。它适用于您提供的示例，因此希望它也适用于更多示例：

columnsToBeEncoded = names(DF)[grepl("former", names(DF))] # alternative, vectorised form

# Create a nested list, where each key is a "number XX" and its elements are the variables needed
number_smth = which(grepl("number", mappingList))
mappingListNested = lapply(seq_along(number_smth), function(i){
  if (i+1 <= length(number_smth)){
    return(mappingList[(number_smth[i]+1):(number_smth[i+1]-1)])
  } else {
    return(mappingList[(number_smth[i]+1):length(mappingList)])
  }
})

names(mappingListNested) = paste('number', str_extract(mappingList[number_smth], "[[:digit:]]+"))

# Encode  "former" columns where row is "?" from "latter" columns by the order in mappingList
for(col in columnsToBeEncoded){
  # extract column number from former column
  colNumber <- paste(strsplit(col, " ")[[1]][2:3], collapse = " ")
  # Find indices where former column has "?"
  replacementVariables = mappingListNested[[colNumber]]
  for (var in replacementVariables){
    varNameinColumnForm <- paste(c(var, colNumber, "latter"), collapse = " ")
    DF[, col] = ifelse(
      DF[, col] == "?", # which elements are "?"
      DF[, varNameinColumnForm], # replace those which are "?" with the values of var
      DF[, col] # otherwise leave unchanged
    )
  }
}

一步一步：

1. 我将 mappingList 转换为嵌套列表，以使迭代更容易。嵌套列表将包含与 mappingList 中的“编号 XX”项一样多的元素。每个元素都是一个字符向量，其中包含“数字 XX”和下一个“数字 XX”之间的变量，如果我理解正确的话，这应该是替换方案。

number_smth = which(grepl("number", mappingList))
mappingListNested = lapply(seq_along(number_smth), function(i){
  if (i+1 <= length(number_smth)){
    return(mappingList[(number_smth[i]+1):(number_smth[i+1]-1)])
  } else {
    return(mappingList[(number_smth[i]+1):length(mappingList)])
  }
})

1. 用相应的列名命名列表，以使索引更容易。为此，我以与您类似的方式提取每个“数字”中的数字，只是我在这里使用正则表达式（为此您需要 stringr 包）。如果您愿意，实际上可以使用strsplit() 自行处理

names(mappingListNested) = paste('number', stringr::str_extract(mappingList[number_smth], "[[:digit:]]+"))

1. for 循环方案看起来与您的几乎相同，只是我使用ifelse() 直接替换值。 ifelse() 是一种矢量化方式，用于遍历向量、检查某些条件并将满足该条件的值替换为其他值。语法为ifelse(logical_vector, replacement_TRUE, replacement_FALSE)。
   1. 在我的例子中，logical_vector 是DF[, col] == "?"，它检查DF 的col 列的每个元素是否等于“？”。这将给出TRUEs 和FALSEs 的向量。
   2. 下一个参数用于替换 DF[,col] 的那些元素 TRUE（即带有“？”的元素），在这种情况下，它是 mappingListNested 中的任何变量。
   3. 下一个参数（在本例中为 DF[,col] 列本身）将替换为 FALSE 的元素，换句话说，它将使列保持不变，只要它不是“？”

# Encode  "former" columns where row is "?" from "latter" columns by the order in mappingList
for(col in columnsToBeEncoded){
  # extract column number from former column
  colNumber <- paste(strsplit(col, " ")[[1]][2:3], collapse = " ")
  # Find indices where former column has "?"
  replacementVariables = mappingListNested[[colNumber]]
  for (var in replacementVariables){
    varNameinColumnForm <- paste(c(var, colNumber, "latter"), collapse = " ")
    DF[, col] = ifelse(
      DF[, col] == "?", # which elements are "?"
      DF[, varNameinColumnForm], # replace those which are "?" with the values of var
      DF[, col] # otherwise leave unchanged
    )
  }
}

由于我对 mappingListNested 的元素进行了迭代，因此我确保当变量结束时迭代将停止。此外，由于 DF[,col] 列在每次迭代时都会更改，因此请确保先替换为第一个变量，然后替换为下一个变量，依此类推。

Answer 2

这是另一个解决方案。说明请参考代码中的cmets。

#----

#Your data.

DF <- data.frame(
  "colors number 3 former" = c("r","r","?","l","?","r","?","?","r","?"),
  "music number 3 latter" = c("r","l","r","l","r","r","l","l","r","l"),
  "genres number 3 latter" = c("l","r","?","l","?","r","?","l","l","r"),
  "genres number 12 former" = c("r","r","?","l","l","r","l","?","r","?"),
  "music number 12 latter" = c("r","l","?","l","?","r","l","l","r","?"),
  "fabric number 12 latter" = c("l","r","?","l","r","r","r","l","l","r"),
  "colors number 12 latter" = c("r","r","?","r","?","r","?","r","r","?"),
  check.names = FALSE
)

mappingList <- list("number 3",
                    "genres",
                    "music",
                    "number 12",
                    "music",
                    "fabric",
                    "colors")


expectedDF <- data.frame(
  "colors number 3 former" = c("r","r","r","l","r","r","l","l","r","r"),
  "music number 3 latter" = c("r","l","r","l","r","r","l","l","r","l"),
  "genres number 3 latter" = c("l","r","?","l","?","r","?","l","l","r"),
  "genres number 12 former" = c("r","r","?","l","l","r","l","l","r","r"),
  "music number 12 latter" =  c("r","l","?","l","?","r","l","l","r","?"),
  "fabric number 12 latter" = c("l","r","?","l","r","r","r","l","l","r"),
  "colors number 12 latter" = c("r","r","?","r","?","r","?","r","r","?"),
  check.names = FALSE
)


#--------

#Solution.

library(stringr)
library(magrittr)
library(tidyr)
library(dplyr)

#The mappingList isn't handy.
#Converting this into a data.frame with two columns: 
#"former", which indicates the former column in DF, 
#and "latter", which indicates the corresponding latter 
#column in DF from which the data in the former column 
#needs to be filled in.

mlist <- data.frame(latter = unlist(mappingList), stringsAsFactors = FALSE)

#A loop to identify former and latter values from the 
#data.frame's one available column.
j <- 0
for(i in 1:nrow(mlist)){
  if(str_detect(mlist$latter[i], "number [0-9]+")){
    j <- j + 1
  }
  mlist$type[i] <- j
}
rm(j)


#Munging to create the former and latter columns 
#properly.
mlist %<>% 
  group_by(type) %>% 
  mutate(former = latter[1]) %>% 
  ungroup()

mlist %<>% filter(latter != former)

mlist %<>% 
  group_by(type) %>% 
  mutate(ord = row_number()) %>% 
  ungroup()

mlist %<>% select(c(former, latter, ord))

#For ease of use, bringing the former and latter 
#columns contents as close to the column names in 
#DF as is possible.
mlist %<>% 
  mutate(latter = paste0(latter, " ", former, " latter"), 
         former = paste0(former, " former"))


#Nested loops to fill in the DF rows.
#Basic logic is: take a row in DF.
#Loop through the rows of mlist.
#mlist basically holds the fill-in relationship's 
#column names. So extract the former and latter 
#(fcol and lcol) column names respectively.
#Then check if that particular former column in the 
#ith row of DF is a "?". If it is, fill it in with 
#the value from the cell corresponding to the column 
#name indicated by the jth row of mlist and the ith row 
#of DF.
#This also automatically takes care of the fact that the 
#next latter column's value gets used if a "?" remains.

for(i in 1:nrow(DF)){
  
  for(j in 1:nrow(mlist)){
    
    fcol <- colnames(DF)[str_detect(colnames(DF), mlist$former[j])]
    lcol <- colnames(DF)[str_detect(colnames(DF), mlist$latter[j])]
    
    if(DF[i, fcol] == "?"){
      DF[i, fcol] <- DF[i, lcol]
    }
    
  }
  
}


identical(DF, expectedDF)

# [1] TRUE

#----