Question

假设我有以下数据框：

df <- data.frame(x = c(1,1,1,3,3,3,4,4,4),
                 y = c(1,3,4,1,3,4,1,3,4),
                 z = c(6,3,7,8,2,3,0,1,2))

我想根据另一个数据框中包含的映射重命名列x和y中的值：

df.aux <- data.frame(id = c(0,1,2,3,4),
                     desc = c('zero', 'one', 'two', 'three', 'four))

获得：

      x     y z
1   one   one 6
2   one three 3
3   one  four 7
4 three   one 8
5 three three 2
6 three  four 3
7  four   one 0
8  four three 1
9  four  four 2

请注意，df.aux中的映射可能包含比df更多的级别。我可以通过将列名称df $ x更改为df $ id来进行映射，合并两个df＆＃39;然后重复df $ y。但它非常笨拙，所以必须有更好的方法。

Answer 1

您可以使用匹配功能进行映射：

df.new <- data.frame(x=df.aux$desc[match(df$x, df.aux$id)], y=df.aux$desc[match(df$y, df.aux$id)], z=df$z)

Answer 2

factor函数允许直接分配映射和标签。

> df <- data.frame(x = c(1,1,1,3,3,3,4,4,4),
+                  y = c(1,3,4,1,3,4,1,3,4),
+                  z = c(6,3,7,8,2,3,0,1,2))

> df.aux <- data.frame(id = c(0,1,2,3,4), desc = c('zero', 'one', 'two', 'three', 'four'))

> df$x <- factor(df$x, levels = df.aux$id, labels = df.aux$desc)
> df$y <- factor(df$y, levels = df.aux$id, labels = df.aux$desc)
> df
      x     y z
1   one   one 6
2   one three 3
3   one  four 7
4 three   one 8
5 three three 2
6 three  four 3
7  four   one 0
8  four three 1
9  four  four 2

如果df.aux太大而无法快速完成，您可以先使用

对其进行子集化

df.aux[df.aux$id %in% union(df$x, df$y),]

根据另一个数据框重命名多列中的记录

2 个答案: