在我的数据框中以其他列为条件填充一列，并使用第三列中的值

Question

警告：这个问题看起来太简单了，我作为初学者可能没能在 SO 上更复杂的主题中找到正确的解决方案（查看 , , here 和更多地方）

我想根据另一列在我的数据框中填充一列，并将更多列用作输入。举个例子就更清楚了：

  Version1 Version2 Version3 Version4 Presented_version Color
1     blue      red    green   yellow                 1    NA
2      red     blue   yellow    green                 4    NA
3   yellow    green      red     blue                 3    NA

我想用 Version1/Version2/Version3/Version 的值填充“Color”列 4. Presented_version 告诉我需要这四个值中的哪一个。例如，在第 1 行中，Presented_version 为 1，因此需要的值在 "Version1"（"blue"）中。第 1 行的颜色应为蓝色。

有人可以告诉我一种无需使用大量 "if" 语句遍历数据帧即可执行此操作的方法吗？

structure(list(Version1 = structure(1:3, .Label = c("blue", "red", 
"yellow"), class = "factor"), Version2 = structure(c(3L, 1L, 
2L), .Label = c("blue", "green", "red"), class = "factor"), Version3 = structure(c(1L, 
3L, 2L), .Label = c("green", "red", "yellow"), class = "factor"), 
    Version4 = structure(3:1, .Label = c("blue", "green", "yellow"
    ), class = "factor"), Presented_version = c(1L, 4L, 3L), 
    Color = c(NA, NA, NA)), class = "data.frame", row.names = c(NA, 
-3L))

======================= 已编辑！

我简化了示例来解释我的问题，但上面的示例在几个方面与我的实际数据集不同，因此解决方案做出了我的数据实际上不符合的假设。这是 data.frame 的更准确表示。特别是 Presented_version 和 Version1...Version 4 列的内容之间没有固定匹配（根据额外的列而有所不同，我现在称之为 Painter），并且 Version1 到 Version4 不一定在我数据集中的第 1 到 4 列。

  FillerColumn Painter Version1 Version2 Version3 Version4 Version_presented Color FillerColumn.1
1           77       A     blue      red    green   yellow                 1    NA             77
2           77       B      red     blue   yellow    green                 4    NA             77
3           77       C   yellow    green      red     blue                 3    NA             77
4           77       D      red     blue   yellow    green                 1    NA             77

structure(list(FillerColumn = c(77L, 77L, 77L, 77L), Painter = structure(1:4, .Label = c("A", 
"B", "C", "D"), class = "factor"), Version1 = structure(c(1L, 
2L, 3L, 2L), .Label = c("blue", "red", "yellow"), class = "factor"), 
    Version2 = structure(c(3L, 1L, 2L, 1L), .Label = c("blue", 
    "green", "red"), class = "factor"), Version3 = structure(c(1L, 
    3L, 2L, 3L), .Label = c("green", "red", "yellow"), class = "factor"), 
    Version4 = structure(c(3L, 2L, 1L, 2L), .Label = c("blue", 
    "green", "yellow"), class = "factor"), Version_presented = c(1L, 
    4L, 3L, 1L), Color = c(NA, NA, NA, NA), FillerColumn.1 = c(77L, 
    77L, 77L, 77L)), class = "data.frame", row.names = c(NA, 
-4L))

Answer 1

使用mapply

的一种方式

cols <- grep("^Version", names(df))
df$Color <- unlist(mapply(function(x, y) df[x, cols][y], 
                   1:nrow(df),df$Presented_version))

df
#  Version1 Version2 Version3 Version4 Presented_version Color
#1     blue      red    green   yellow                 1  blue
#2      red     blue   yellow    green                 4 green
#3   yellow    green      red     blue                 3   red

和 apply

apply(df, 1, function(x) x[cols][as.numeric(x["Presented_version"])])
#[1] "blue"  "green" "red"

Answer 2

我们可以使用带有 row/column 索引的矢量化选项来提取值而不是任何循环

df1$color <- df1[1:4][cbind(1:nrow(df1), df1$Presented_version)]
df1$color
#[1] "blue"  "green" "red"

基准

dfN <- df1[rep(seq_len(nrow(df1)), 1e6),]


system.time({
   dfN[1:4][cbind(1:nrow(dfN), dfN$Presented_version)]

 })
# user  system elapsed 
#   1.216   0.110   1.321


system.time({
 cols <- grep("^Version", names(dfN))
 unlist(mapply(function(x, y) dfN[x, cols][y], 
                    1:nrow(dfN),dfN$Presented_version))

 })
#  user  system elapsed 
# 319.907   1.644 322.418

现在，让我们看看 apply

的另一个选项

system.time({
  apply(dfN, 1, function(x) x[cols][as.numeric(x["Presented_version"])])
 }) 
#  user  system elapsed 
# 14.240   0.365  14.550

Answer 3

我喜欢弄乱数据集。尝试 data.table melt 方法

df <- setDT(df)

df1 <- melt.data.table(df,
                       id.vars = c('Presented_version'),
                       measure.vars = patterns('Version'),
                       value.name = 'Color',
                       variable.name = 'Version')[
  , version1 := str_extract(Version, '\d+')][
    Presented_version == version1][
      version1 := NULL]

导致

   Presented_version  Version Color 
1:                 1 Version1  blue        
2:                 3 Version3   red        
3:                 4 Version4 green

而且，如果您希望信息采用相同的原始结构

merge(df, 
      df1[, .(Presented_version, Color)],
      by = 'Presented_version')

   Presented_version Version1 Version2 Version3 Version4 Color
1:                 1     blue      red    green   yellow  blue
2:                 3   yellow    green      red     blue   red
3:                 4      red     blue   yellow    green green

在我的数据框中以其他列为条件填充一列，并使用第三列中的值

fill a column in my data frame conditional on other column, with values from third column

conditional

r

multiple-columns

dataframe

基准