关于r：创建新的一组变量等于dplyr中一个因子的水平

Create new set of variables equal to the level of a factor in dplyr

我有一个包含 100 列的 data.frame，它们遵循约定 word 和 word_answer

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

我这样设置上述两个因子变量的水平：

1 2	levels(df$apple) <- c("66%","57%","48%","39%","30%","22%","12%" ) levels(df$dog) <- c("82%","71%","60%","49%","38%","27%","16%" )

我正在尝试计算一个距离分数，它是一个 word 因子的数值水平与其对应 word_answer.

的数值水平之间的距离

因此，例如，在"apple"答案的情况下，apple 的第一行是"57%"，这是该因子中的第二个因子水平

1 2	> which(levels(df$apple) =="57%") [1] 2

对应的apple_answer列的因子水平为6

1 2	> which(levels(df$apple) =="22%") [1] 6

所以这种情况下的距离分数是 2-6 = -4

如何计算数据集中每个变量的这些距离分数？

你也可以使用apply函数，像这样：

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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

您可以将数据分为两组，单词及其对应的答案。使用 match 获取它们的位置并从每个值中减去并生成新列。

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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