r - str_extract_all : return all patterns found in string concatenated as vector

Question

我想提取除模式之外的所有内容并将其返回到字符串中。

我试图将 str_extract_all 与 sapply 和 cat 结合在一起

x = c("a_1","a_20","a_40","a_30","a_28")
data <- tibble(age = x)


# extracting just the first pattern is easy
data %>% 
  mutate(age_new = str_extract(age,"[^a_]"))
# combining str_extract_all and sapply doesnt work
data %>% 
  mutate(age_new = sapply(str_extract_all(x,"[^a_]"),function(x) cat(x,sep="")))


class(str_extract_all(x,"[^a_]"))
sapply(str_extract_all(x,"[^a_]"),function(x) cat(x,sep=""))

返回 NULL 而不是串联模式

Answer 1

而不是 cat ，我们可以使用 paste .另外，还有 tidyverse , 可以利用 map和 str_c (代替 paste - 来自 stringr )

library(tidyverse)
data %>% 
  mutate(age_new = map_chr(str_extract_all(x, "[^a_]+"), ~ str_c(.x, collapse="")))

使用`OP的代码

data %>%
    mutate(age_new = sapply(str_extract_all(x,"[^a_]"),
               function(x) paste(x,collapse="")))

如果目的是获取数字

library(readr)
data %>%
     mutate(age_new = parse_number(x))