映射列的值

Question

我想變換使用一些映射函數的給定列的值。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 

我的目標是列A的所有元素映射如下：

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

我寫了下面的：

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 

代碼爲上述工作的罰款。

現在讓我們假設另一個數據幀，其中A列不言整數1,2,3,4,5，而是任何其他「通用」的項目，說：

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)) 

問題：你會如何編寫映射？

Answer 1

你看factor？

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，你的levels參數應具有的原始值相匹配，並且您labels的說法應該有替換值。

---

您還可以創建一個帶有命名向量的查找表。

實施例：

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

嘗試：

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

對於兩個其他數據集：

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"