使用 recode() 和通过 paste() 生成的变量名

Question

我正在尝试将包含作为字符的答案标签的变量重新编码为数值。为此，我正在使用来自 dplyr 的recode()。

为了自动执行此操作，我想使用 paste() 生成变量名称，但显然 recode() 无法从 paste 获取输出。

我已经尝试过noquote() 和as.name()，但是对于这两个 R 都告诉我重新编码不能使用“noquote”/“name”类的对象。

例子：

item1 <- c("Don't agree at all", "Totally agree")
item2 <- c("Indifferent", "Totally agree")

for (i in 1:2) {
recode(paste("item", i, sep=""), "Totally agree"=1, "Indifferent"=2, "Don't agree at all"=3)
}

我会期待

> item1
  [1] 3 1

我该如何解决这个问题？

更新

我找到了一种解决方法，首先将相关列提取到另一个数据框中，然后将recode() 函数与sapply() 一起应用。现在我可以重新合并数据框了。

Answer 1

您只需将命名向量 v 与来自 mget 的变量列表一起放入 Map 并对其进行子集化。

v <- c("Totally agree"=1, "Indifferent"=2, "Don't agree at all"=3)

Map(function(x, y) unname(y[x]), mget(ls(pattern="^item")), list(v))
# $item1
# [1] 3 1
# 
# $item2
# [1] 2 1

或者，假设您有一个像这样的数据框，

head(dat1)
#   id         item1              item2         x
# 1  1 Totally agree      Totally agree 0.0356312
# 2  2 Totally agree      Totally agree 1.3149588
# 3  3 Totally agree        Indifferent 0.9781675
# 4  4 Totally agree        Indifferent 0.8817912
# 5  5   Indifferent        Indifferent 0.4822047
# 6  6   Indifferent Don't agree at all 0.9657529

那么你可以用类似的方式来做这件事。我们甚至可以简化代码，因为我们不再需要 Map 来返回 unnamed 对象了。

v1 <- c("Totally agree"=1, "Indifferent"=2, "Don't agree at all"=3)

item_nm <- c("item1", "item2")
dat1[item_nm] <- Map(`[`, list(v1), dat2[item_nm])
dat1
#    id item1 item2          x
# 1   1     1     1  0.0356312
# 2   2     1     1  1.3149588
# 3   3     1     2  0.9781675
# 4   4     1     2  0.8817912
# 5   5     2     2  0.4822047
# 6   6     2     3  0.9657529
# 7   7     2     3 -0.8145709
# 8   8     1     1  0.2839578
# 9   9     3     1 -0.1616986
# 10 10     3     3  1.9355718

第二个参数在每次 Map 迭代时被回收（即 list(v1, v1) 也可以使用）。

更一般地说，对于您要以数字方式重新编码的每一列，list 在Map 的第二个参数中多一个向量。

head(dat2)
#   id         item1  item2         x
# 1  1 Totally agree Always 0.0356312
# 2  2 Totally agree Always 1.3149588
# 3  3 Totally agree   Both 0.9781675
# 4  4 Totally agree   Both 0.8817912
# 5  5   Indifferent   Both 0.4822047
# 6  6   Indifferent  Never 0.9657529

v2 <- c("Always"=1, "Both"=2, "Never"=3)

dat2[item_nm] <- Map(`[`, list(v1, v2), dat2[item_nm])
dat2
#    id item1 item2          x
# 1   1     1     1  0.0356312
# 2   2     1     1  1.3149588
# 3   3     1     2  0.9781675
# 4   4     1     2  0.8817912
# 5   5     2     2  0.4822047
# 6   6     2     3  0.9657529
# 7   7     2     3 -0.8145709
# 8   8     1     1  0.2839578
# 9   9     3     1 -0.1616986
# 10 10     3     3  1.9355718

---

数据：

dat1 <- structure(list(id = 1:10, item1 = c("Totally agree", "Totally agree", 
"Totally agree", "Totally agree", "Indifferent", "Indifferent", 
"Indifferent", "Totally agree", "Don't agree at all", "Don't agree at all"
), item2 = c("Totally agree", "Totally agree", "Indifferent", 
"Indifferent", "Indifferent", "Don't agree at all", "Don't agree at all", 
"Totally agree", "Totally agree", "Don't agree at all"), x = c(0.0356311982051355, 
1.31495884897891, 0.978167526364279, 0.881791226863203, 0.482204688262918, 
0.965752878105794, -0.814570938270238, 0.283957806364306, -0.161698647607024, 
1.93557176599585)), class = "data.frame", row.names = c(NA, -10L
))

dat2 <- structure(list(id = 1:10, item1 = c("Totally agree", "Totally agree", 
"Totally agree", "Totally agree", "Indifferent", "Indifferent", 
"Indifferent", "Totally agree", "Don't agree at all", "Don't agree at all"
), item2 = c("Always", "Always", "Both", "Both", "Both", "Never", 
"Never", "Always", "Always", "Never"), x = c(0.0356311982051355, 
1.31495884897891, 0.978167526364279, 0.881791226863203, 0.482204688262918, 
0.965752878105794, -0.814570938270238, 0.283957806364306, -0.161698647607024, 
1.93557176599585)), class = "data.frame", row.names = c(NA, -10L
))

Answer 2

关于“但显然 recode() 无法从粘贴中获取输出。”：这与 recode 无关，但（几乎）any R 函数以这种方式工作。 paste 返回一个字符串，recode 期望一个向量作为它的第一个参数...（值得注意的例外，其中包括：library 我们可以传递一个字符串 或 库名称作为对象）。

如果您坚持使用“for 循环”方法，您可以做的是使用 assign 和 eval(sym("a string")) 之类的组合：

item1 <- c("Don't agree at all", "Totally agree")
item2 <- c("Indifferent", "Totally agree")


library(dplyr)
for (i in 1:2) {
  assign(paste("item", i, sep="") , recode(eval(sym(paste("item", i, sep=""))), "Totally agree"=1, "Indifferent"=2, "Don't agree at all"=3))
}

结果：

item1

[1] 3 1

item2

[1] 2 1

编辑：

一种可能更直接且更“dplyr”-y 的方法可能是：

tdd <- data.frame(item1, item2) %>% 
  mutate_at(vars(starts_with("item")), ~recode(., "Totally agree"=1, "Indifferent"=2, "Don't agree at all"=3))

tdd 现在是：

tdd

  item1 item2
1     3     2
2     1     1