Question

我想通过将“X”或“Y”列中的文本相应地匹配为“x”或“y”来替换由“/”分隔的元素，而忽略“ - ”。

df <- data.frame("X"=c("A","AB","CD","E","T"),
"Y"=c("AT","A","CDCD","F","A"),
"R1"=c("A/A","AB/AB","CD/CD","E/E","-/-"),
"R2"=c("A/AT","AB/AB","CD/CDCD","F/F","T/T"),
"R3"=c("AT/AT","A/AB","CDCD/CDCD","E/F","A/T"),
"R4"=c("AT/A","A/A","-/-","F/E","T/A"),
"R5"=c("-/-","A/AB","CDCD/CD","F/F","A/A"),
"R6"=c("A/A","-/-","CD/CD","E/E","-/-"))

预期结果是：

X  Y    R1  R2  R3  R4  R5  R6
A  AT   x/x x/y y/y y/x -/- x/x
AB A    x/x x/x y/x y/y y/x -/-
CD CDCD x/x x/y y/y -/- y/x x/x
E  F    x/x y/y x/y y/x y/y x/x
T  A    -/- x/x y/x x/y y/y -/-

我不知道如何以有效的方式做到这一点。我感谢任何帮助。

Answer 1

非常棘手的问题！这是一个解决方案：

df[,3:ncol(df)] <- t(apply(df,1,function(R) sapply(strsplit(R[3:ncol(df)],'/'),function(S) paste0(gsub(paste0('^',R[1],'$'),'x',gsub(paste0('^',R[2],'$'),'y',S)),collapse='/'))));
df;
##    X    Y  R1  R2  R3  R4  R5  R6
## 1  A   AT x/x x/y y/y y/x -/- x/x
## 2 AB    A x/x x/x y/x y/y y/x -/-
## 3 CD CDCD x/x x/y y/y -/- y/x x/x
## 4  E    F x/x y/y x/y y/x y/y x/x
## 5  T    A -/- x/x y/x x/y y/y -/-

Answer 2

这是 dplyr 和 stringr 的一种方式。

library(dplyr)
library(stringr)

df[] <- lapply(df, as.character)
df %>% 
    rowwise() %>%
    do({
        vars <- c(.$X, .$Y)
        # ordering gives precedence to longer vars 
        replacements <- setNames(c('x', 'y'), vars)[order(nchar(vars), decreasing=TRUE)]
        setNames(data.frame(.[1:2], as.list(str_replace_all(unlist(tail(., -2)), replacements))), names(.))
    })

#    X    Y  R1  R2  R3  R4  R5  R6
# 1  A   AT x/x x/y y/y y/x -/- x/x
# 2 AB    A x/x x/x y/x y/y y/x -/-
# 3 CD CDCD x/x x/y y/y -/- y/x x/x
# 4  E    F x/x y/y x/y y/x y/y x/x
# 5  T    A -/- x/x y/x x/y y/y -/-

如何用r中的条件替换文本元素

2 个答案: