映射列值

Question

我想使用一些映射函数转换给定列的值。例如：

df <- data.frame(A = 1:5, B = sample(1:20, 10))
df
   A  B
1  1 17
2  2  5
3  3  3
4  4 11
5  5 19
6  1 16
7  2  4
8  3  7
9  4  6
10 5  9

我的目标是将A列的所有元素映射如下：

1 -> "tt"
2 -> "ff"
3 -> "ss"
4 -> "fs"
5 -> "sf"

我写了以下内容：

mappingList <- c("tt", "ff", "ss", "fs", "sf")
df$A <- unlist(lapply(df$A, function(x){replace(x, x>0, mappingList[x])}))
df
  A  B
1  tt 17
2  ff  5
3  ss  3
4  fs 11
5  sf 19
6  tt 16
7  ff  4
8  ss  7
9  fs  6
10 sf  9

上面的代码工作正常。

现在让我们假设另一个数据框，其中A列不是由整数1,2,3,4,5组成，而是由任何其他的＆＃39;泛型＆＃39;项目，说：

df <- data.frame(A = paste("str",1:5,sep=""), B = sample(1:20, 10))

或

df <- data.frame(A = seq(5, 25, by=5), B = sample(1:20, 10))

问题：您如何编写映射？

Answer 1

你看过factor吗？

df$A_2 <- factor(df$A, levels = 1:5, labels = c("tt", "ff", "ss", "fs", "sf"))
df
#    A  B A_2
# 1  1 17  tt
# 2  2  5  ff
# 3  3  3  ss
# 4  4 11  fs
# 5  5 19  sf
# 6  1 16  tt
# 7  2  4  ff
# 8  3  7  ss
# 9  4  6  fs
# 10 5  9  sf

基本上，您的levels参数应该具有匹配的原始值，并且您的labels参数应该具有替换值。

---

您还可以创建一个带有命名向量的查找表。

示例：

df <- data.frame(A = paste("str",1:5,sep=""), B = sample(1:20, 10))

NamedVec <- setNames(paste("str",1:5,sep=""), c("tt", "ff", "ss", "fs", "sf"))
NamedVec
#     tt     ff     ss     fs     sf
# "str1" "str2" "str3" "str4" "str5" 
NamedVec[df$A]
#     tt     ff     ss     fs     sf     tt     ff     ss     fs     sf 
# "str1" "str2" "str3" "str4" "str5" "str1" "str2" "str3" "str4" "str5" 
names(NamedVec[df$A])
#  [1] "tt" "ff" "ss" "fs" "sf" "tt" "ff" "ss" "fs" "sf"

Answer 2

尝试：

mappingList[df$A]
#[1] "tt" "ff" "ss" "fs" "sf" "tt" "ff" "ss" "fs" "sf"

对于其他两个数据集：

df1 <-  data.frame(A = paste("str",1:5,sep=""), B = sample(1:20, 10))
df2 <- data.frame(A = seq(5, 25, by=5), B = sample(1:20, 10))

mappingList[as.numeric(df1$A)]
#[1] "tt" "ff" "ss" "fs" "sf" "tt" "ff" "ss" "fs" "sf"

mappingList[as.numeric(factor(df2$A))]
#[1] "tt" "ff" "ss" "fs" "sf" "tt" "ff" "ss" "fs" "sf"