创建一组新的变量，这些变量等于dplyr中某个因子的级别

我有一个data.frame，它有100列，遵循惯例

word

和

word\u-answer

df <- data.frame(apple = "57%", apple_answer = "22%", dog = "82%", dog_answer = "16%")

相应的

apple\u-answer

列的因子级别为6

> which(levels(df$apple) == "22%")
[1] 6

因此，在这种情况下，距离得分为2-6=-4

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如何计算数据集中每个变量的距离分数？

您可以将数据分为两组，word和相应的答案。使用

match

获取它们的位置并从每个值中减去，然后生成新列

answer_cols <- grep('_answer', names(df))
new_cols <- paste0(names(df)[-answer_cols], '_dist')

df[new_cols] <- Map(function(x, y) match(x, levels(x)) - match(y, levels(x)),
                                     df[-answer_cols], df[answer_cols])

df
#  apple apple_answer dog dog_answer apple_dist dog_dist
#1   57%          22% 82%        16%         -4       -6

answer\u cols您也可以使用应用功能，如下所示：

df$apple_dist = apply(df[,1:2], 1, function(x) {
    which(levels(df$apple) == x[1]) - which(levels(df$apple) == x[2])
})

df$dog_dist = apply(df[,3:4], 1, function(x) {
    which(levels(df$dog) == x[1]) - which(levels(df$dog) == x[2])
})

> df
  apple apple_answer dog dog_answer apple_dist dog_dist
1   57%          22% 82%        16%         -4       -6

啊，谢谢你！非常令人印象深刻。我花了太多时间试图弄清楚如何在dplyr中实现这一点，但这是一种更简单的方法。谢谢
answer_cols <- grep('_answer', names(df))
new_cols <- paste0(names(df)[-answer_cols], '_dist')

df[new_cols] <- Map(function(x, y) match(x, levels(x)) - match(y, levels(x)),
                                     df[-answer_cols], df[answer_cols])

df
#  apple apple_answer dog dog_answer apple_dist dog_dist
#1   57%          22% 82%        16%         -4       -6

df$apple_dist = apply(df[,1:2], 1, function(x) {
    which(levels(df$apple) == x[1]) - which(levels(df$apple) == x[2])
})

df$dog_dist = apply(df[,3:4], 1, function(x) {
    which(levels(df$dog) == x[1]) - which(levels(df$dog) == x[2])
})

> df
  apple apple_answer dog dog_answer apple_dist dog_dist
1   57%          22% 82%        16%         -4       -6