如何使用分隔符连接多列，但根据 R 中的条件忽略某些列？答案

【问题标题】：How to concatenate multiple columns with separators but ignore some of columns based on condition in R?如何使用分隔符连接多列，但根据 R 中的条件忽略某些列？
【发布时间】：2020-02-21 14:09:13
【问题描述】：

您好，我想将包含字符串或空格或 NA 的列与“;”连接起来。让我们举个例子：


Actor1<- c("Driver","NA","","")
Actor2<- c("President","Zombie","","")
Actor3<- c("CEO","Devil","","")
Actor4<-c("Priest","","Killer","Mayor")

df_ex <-data.frame(Actor1, Actor2, Actor3, Actor4)

我试过这个：

df_ex %>%
  mutate(combined= paste0(Actor1,";",Actor2,";",Actor3,";",Actor4))

但显然结果是错误的，例如：

df_ex[3,]

合并列的结果是这样的： ;;;杀手

我希望结果是：杀手。

注意：还有 NA 和空白 "" 以及哪些 id 可以忽略。

提前致谢，干杯

【问题讨论】：

标签： r concatenation multiple-columns

【解决方案1】：

我离成为regex 专家还很遥远，但我会在这里提出tidyverse 方法：

Actor1 <- c("Driver","NA","","")
Actor2 <- c("President","Zombie","","")
Actor3 <- c("CEO","Devil","","")
Actor4 <-c("Priest","","Killer","Mayor")

library(tidyverse)

data.frame(Actor1, Actor2, Actor3, Actor4) %>%
  mutate_all(~str_replace(., pattern = "NA", replacement = "")) %>% 
  unite(col = "combined", sep = ";", remove = F) %>% 
  mutate(combined = str_replace_all(combined, pattern = "^[:punct:]|[:punct:]$|[:punct:]{2,}", replacement = "")) %>% 
  select(-combined, everything(.), combined)

#>   Actor1    Actor2 Actor3 Actor4                    combined
#> 1 Driver President    CEO Priest Driver;President;CEO;Priest
#> 2           Zombie  Devil                       Zombie;Devil
#> 3                         Killer                      Killer
#> 4                          Mayor                       Mayor

如果你只想要其中的一些列，你可以在unite 中传递它们：

data.frame(Actor1, Actor2, Actor3, Actor4) %>%
  mutate_all(~str_replace(., pattern = "NA", replacement = "")) %>% 
  unite(Actor2, Actor4, col = "combined", sep = ";", remove = F) %>% 
  mutate(combined = str_replace_all(combined, pattern = "^[:punct:]|[:punct:]$|[:punct:]{2,}", replacement = "")) %>% 
  select(-combined, everything(.), combined)

#>   Actor1    Actor2 Actor3 Actor4         combined
#> 1 Driver President    CEO Priest President;Priest
#> 2           Zombie  Devil                  Zombie
#> 3                         Killer           Killer
#> 4                          Mayor            Mayor

【讨论】：

【解决方案2】：

Actor1<- c("Driver","NA","","")
Actor2<- c("President","Zombie","","")
Actor3<- c("CEO","Devil","","")
Actor4<-c("Priest","","Killer","Mayor")

matrix_ex <-cbind(Actor1, Actor2, Actor3, Actor4)
#apply(df_ex,1,paste,collapse=";")
x<-apply(matrix_ex,1,function(x){paste(x[!(is.na(x)|x==""|x=="NA")],collapse=";")})
x

[1] "Driver;President;CEO;Priest" "Zombie;Devil"                "Killer"                      "Mayor"                                    
> cat(paste(x,collapse="\n"))
#Driver;President;CEO;Priest
#Zombie;Devil
#Killer
#Mayor

回答cmets：


df_ex <-data.frame(Actor1=Actor1, Actor2=Actor2, Actor3=Actor3, Actor4=Actor4,rnorm(4))

df_ex$concat<-apply(df_ex[c("Actor1","Actor3")],1,function(x){paste(x[!(is.na(x)|x==""|x=="NA")],collapse=";")})
df_ex$concat

df_ex$concat2<-apply(df_ex[c(1,3)],1,function(x){paste(x[!(is.na(x)|x==""|x=="NA")],collapse=";")})
df_ex$concat2

【讨论】：

这正是我想要避免的：由于 na 或空白而导致前面那些不必要的分隔符
还有一个问题：假设我有更多列并且只想连接选定的列，如何将其应用于您的代码？
假设我只想连接 Actor1 和 Actor 3
```selectedcolumns=c(1,3);x

【解决方案3】：

你可以试试下面的代码，使用do.call + paste

df_ex$combine <- gsub("\\bNA;?\\b|;{2,}|;$","",do.call(paste,c(df_ex,sep = ";")))

这样

> df_ex
  Actor1    Actor2 Actor3 Actor4                     combine
1 Driver President    CEO Priest Driver;President;CEO;Priest
2     NA    Zombie  Devil                       Zombie;Devil
3                         Killer                      Killer
4                          Mayor                       Mayor

【讨论】：

Thomas，在第 2 行末尾还有一个 ;。