r语言 - 将原始数据列分割为n个分隔符

Question

您好，我正在尝试分割列
下面是我的df

value = c("AB/cc/dd/id,1,3,33","CC/DD/EE/F,F/GG,22,33,4","AB/cc,22,2,34","KK/SS/G,G,3,22,41")
df = data.frame(value)

我正试图分割列并获得字符串，直到第3 &;逗号(，)&;From last
我的输出df应该如下所示

value1 = c("AB/cc/dd/id","CC/DD/EE/F,F/GG","AB/cc","KK/SS/G,G")
df_out = data.frame(value1)

我使用字符串包来完成它

library(stringr)
df[c('col1', 'col2')] <- str_split_fixed(df$value, ',', 2)

Thanks in advance

Answer 1

这里有另一种方法可以获取从最后一个到第3个逗号的字符串，而不需要regex:

df$value |> 
str_split(",") |> 
map(function(x)  x[1: (length(x)-3)] |> 
str_c(collapse = ","))  |> 
map_df(as.data.frame) |> 
setNames("value1")
#           value1
#1     AB/cc/dd/id
#2 CC/DD/EE/F,F/GG
#3           AB/cc
#4       KK/SS/G,G

Answer 2

为了防止最后3个逗号之间不仅有数字，而且可能有任何其他字母数字(包括/)，您可以使用:

a <- "AB/cc/dd/id,1,/gg/,33"
stringr::str_extract(a, ".*(?=(\,[/A-z0-9]+){3})")
#> [1] "AB/cc/dd/id"

或其他base R解:

gsub("(\,.*){3}$", "", a)

Answer 3

使用gsub:

gsub("[^[:alpha:],/]", "", value) |> gsub(",+$", "", .)
[1] "AB/cc/dd/id"     "CC/DD/EE/F,F/GG" "AB/cc"           "KK/SS/G,G"

正则表达式解释:

"[^[:alpha:],/]"

• []:定义字符列表
• ^:否定该列表，gsub将查找匹配列表中没有的任何内容
• [:alpha:],/:列表内容、字母、逗号和&;/&;;

",+$"

• ,:匹配逗号
• +:可能出现一次或多次
• $:只在字符串
末尾

Answer 4

您可以在base R内尝试gsub

> gsub("(,[^,]+){3}$", "", value)
[1] "AB/cc/dd/id"     "CC/DD/EE/F,F/GG" "AB/cc"           "KK/SS/G,G"