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R:分两种情况,保留除沫器

strsplit split string r

Makoto Miyazaki · 技术社区 · 7 年前

我试着根据不同的标准分句。我想在“traction”和“ramasse”后面分句。我查了grepl的语法规则,但不太明白。

一个称为 export 有专栏 ref ,其str值以“traction”或“ramasse”结尾。

>export$ref
                        ref
[1] "62133130_074_traction"
[2]  "62156438_074_ramasse"
[3]  "62153874_070_ramasse"
[4] "62138861_074_traction"

我想把ref列中的str值分成两个。

                ref           R&T
[1] "62133130_074_"    "traction"
[2] "62156438_074_"     "ramasse"
[3]  "62153874_070_"    "ramasse"
[4] "62138861_074_"    "traction"

我试过的(没有一个是好的)

strsplit(export$ref, c("traction", "ramasse"))
strsplit(export$ref, "\\_(?<=\\btraction)|\\_(?<=\\bramasse)", perl = TRUE)
strsplit(export$ref, "(?=['traction''ramasse'])", perl = TRUE)

任何帮助都将不胜感激!

2 回复 | 直到 7 年前

Maurits Evers 7 年前

下面是另一个使用 stringr::str_split :

library(stringr);
str_split(ref, pattern = "_(?=[A-Za-z]+)", simplify = T)
#    [,1]           [,2]
#[1,] "62133130_074" "traction"
#[2,] "62156438_074" "ramasse"
#[3,] "62153874_070" "ramasse"
#[4,] "62138861_074" "traction"

样本数据

ref <- c(
    "62133130_074_traction",
    "62156438_074_ramasse",
    "62153874_070_ramasse",
    "62138861_074_traction")

talat 7 年前

这里有一个不同的方法:

strsplit(x, "_(?=[^_]+$)", perl = TRUE)

[[1]]
[1] "62133130_074" "traction"    

[[2]]
[1] "62156438_074" "ramasse"     

[[3]]
[1] "62153874_070" "ramasse"     

[[4]]
[1] "62138861_074" "traction"

这意味着将列/向量拆分为一个下划线(“uu”),后跟不包含另一个下划线的任意数量的符号。

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