如何使用map函数依次用R中另一个向量中的字符串替换向量中的子串？

Question

我有一个包含三个字符串的向量，它们都包含子串“X”：

v<- c("kpX_43", "kpX_10", "kpX_11")

“X”是一个占位符，由“1”，“2”，“3”或“a1”填充。现在，我尝试做的是将“X”替换为“1”，“2”，“3”和“a1”。此外，我必须添加一个“a”和一个“b”，导致以下向量：

> v_new
[1] "kp1_43a" "kp1_10a" "kp1_11a" "kp2_43a" "kp2_10a" "kp2_11a" "kp3_43a" "kp3_10a" "kp3_11a" "kp1_43b" "kpa1_10a" "kpa1_11a" "kpa1_43b" "kp1_10b" "kp1_11b" "kp2_43b" "kp2_10b" "kp2_11b" "kp3_43b" "kp3_10b" "kp3_11b" "kpa1_10b" "kpa1_11b" "kpa1_43b"

我应该使用dyplr和purrr这样做。

好的，到目前为止我尝试过的内容如下：

strings<- as.matrix(c(seq(1:3), "a1"))

v_new <- v %>% 
  map(., ~str_c(., letters[seq(from = 1, to = 2)])) %>% 
  map(function(x) {
    str_replace_all(., "X" , strings)}) %>%
  unlist()

...导致：

> v_new
 [1] "c(\"kp1_43a\", \"kp1_43b\")"   "c(\"kp2_10a\", \"kp2_10b\")"   "c(\"kp3_11a\", \"kp3_11b\")"   "c(\"kpa1_43a\", \"kpa1_43b\")"
 [5] "c(\"kp1_43a\", \"kp1_43b\")"   "c(\"kp2_10a\", \"kp2_10b\")"   "c(\"kp3_11a\", \"kp3_11b\")"   "c(\"kpa1_43a\", \"kpa1_43b\")"
 [9] "c(\"kp1_43a\", \"kp1_43b\")"   "c(\"kp2_10a\", \"kp2_10b\")"   "c(\"kp3_11a\", \"kp3_11b\")"   "c(\"kpa1_43a\", \"kpa1_43b\")"

...有几个警告，所有阅读：

Warning messages:
1: In stri_replace_all_regex(string, pattern, fix_replacement(replacement),  :argument is not an atomic vector; coercing

因此，虽然结果已经接近我想要的输出，但我的代码/我的方法显然有问题。谁能帮我吗？

Answer 1

您需要在最后一张地图中添加额外的map来迭代a和b并用字符串替换X

library(purrr)
v %>% map(., ~str_c(., letters[seq(from = 1, to = 2)])) %>%
      map(. %>% map(.,~str_replace_all(.,'X',strings))) %>% unlist()
      #OR map(~map(.,~str_replace_all(.,'X',strings)))

Answer 2

library(purrr)

v <- c("kpX_43", "kpX_10", "kpX_11")
strings <- c(1:3, "a1")
suffixes <- c("a", "b")

cross(tibble::lst(v, strings, suffixes)) %>% 
  map_chr(~ paste0(sub("X", .$strings, .$v), .$suffixes))
#>  [1] "kp1_43a"  "kp1_10a"  "kp1_11a"  "kp2_43a"  "kp2_10a"  "kp2_11a" 
#>  [7] "kp3_43a"  "kp3_10a"  "kp3_11a"  "kpa1_43a" "kpa1_10a" "kpa1_11a"
#> [13] "kp1_43b"  "kp1_10b"  "kp1_11b"  "kp2_43b"  "kp2_10b"  "kp2_11b" 
#> [19] "kp3_43b"  "kp3_10b"  "kp3_11b"  "kpa1_43b" "kpa1_10b" "kpa1_11b"

由reprex package创建于2019-04-15（v0.2.1）

Answer 3

如果你不是非常依赖dplyr和purrr你可以使用gsub在基地R中做到这一点

as.vector(sapply(strings, function(x) 
  paste0(gsub("X", x, v), rep(letters[1:2], 3))))
# [1] "kp1_43a"  "kp1_10b"  "kp1_11a"  "kp1_43b"  "kp1_10a" 
# [6] "kp1_11b"  "kp2_43a"  "kp2_10b"  "kp2_11a"  "kp2_43b" 
# [11] "kp2_10a"  "kp2_11b"  "kp3_43a"  "kp3_10b"  "kp3_11a" 
# [16] "kp3_43b"  "kp3_10a"  "kp3_11b"  "kpa1_43a" "kpa1_10b"
# [21] "kpa1_11a" "kpa1_43b" "kpa1_10a" "kpa1_11b"

Answer 4

使用expand.grid，然后粘贴：

apply(expand.grid("kp",
                  c("1", "2", "3", "a1"),
                  "_",
                  c(43, 10, 11),
                  c("a", "b")), 1, paste, collapse = "")

#  [1] "kp1_43a"  "kp2_43a"  "kp3_43a"  "kpa1_43a" "kp1_10a"  "kp2_10a" 
#  [7] "kp3_10a"  "kpa1_10a" "kp1_11a"  "kp2_11a"  "kp3_11a"  "kpa1_11a"
# [13] "kp1_43b"  "kp2_43b"  "kp3_43b"  "kpa1_43b" "kp1_10b"  "kp2_10b" 
# [19] "kp3_10b"  "kpa1_10b" "kp1_11b"  "kp2_11b"  "kp3_11b"  "kpa1_11b"

---

编辑：正如Marcus在评论中所建议的那样：

do.call(paste0, expand.grid("kp",
                            c("1", "2", "3", "a1"),
                            "_",
                            c(43, 10, 11),
                            c("a", "b")))