R如何用新子字符串替换较长字符串中的子字符串

Question

我有一个长向量。每个元素都是一个字符串。 每个字符串都可以拆分为以','分隔的子字符串。

我想检查向量中的每个字符串是否至少包含一个“坏”字符串。如果是这样，则应将包含该“坏”字符串的整个 SUBstring 替换为新字符串。我用循环写了一个很长的函数。但我可以发誓一定有一种更简单的方法——也许用 stringr？ 非常感谢您的建议！

# Create an example data frame:
test <- data.frame(a = c("str1_element_1_aaa, str1_element_2",
                         "str2_element_1",
                         "str3_element_1, str3_element_2_aaa, str3_element_3"),
                   stringsAsFactors = F)
test
str(test)

# Defining my long function that checks if each string in a
# vector contains a substring with a "bad" string in it.
# If it does, that whole substring is replaced with a new string:
library(stringr)
mystring_replace = function(strings_vector, badstring, newstring){
  with_string <- grepl(badstring, strings_vector)  # what elements contain badstring?
  mysplits <- str_split(string = test$a[with_string], pattern = ', ') # split those elements with badstring based on ', '
  for (i in 1:length(mysplits)) {   # loop through the list of splits:
    allstrings <- mysplits[[i]]
    for (ii in 1:length(allstrings)) {  # loop through substrings
      if (grepl(badstring, allstrings[ii])) mysplits[[i]][ii] <- newstring
    }
  }
  for (i in seq_along(mysplits)) {  # merge the split elements back together
    mysplits[[i]] <- paste(mysplits[[i]], collapse = ", ")
  }
  strings_vector[with_string] <- unlist(mysplits)
  return(strings_vector)
}
# Test
mystring_replace(test$a, badstring = '_aaa', newstring = "NEW")

Answer 1

你觉得这样可以吗？

new_str_replace <- function(strings_vector, badstring, newstring){
  split.dat <- strsplit(strings_vector,', ')[[1]]
  split.dat[grepl(badstring, split.dat)] <- newstring
  return(paste(split.dat, collapse = ', '))
}

results <- unname(sapply(test$a, new_str_replace, badstring = '_aaa', newstring = 'NEW'))
results
#[1] "NEW, str1_element_2"                 "str2_element_1"                     
#[3] "str3_element_1, NEW, str3_element_3"

Answer 2

我以分而治之的方式做到了。首先我写了一个只对一个字符串进行操作的函数，然后将它向量化。

# does the operation for a string only. divide-and-conquer
replace_one = function(string, badstring, newstring) {
  # split it at ", "
  strs = str_split(string, ", ")[[1]]
  # an ifelse to find the ones containing badstring and replacing them
  strs = ifelse(grepl(badstring, strs, fixed = TRUE), newstring, strs)
  # join them again
  paste0(strs, collapse = ", ")
}

# vectorizes it
my_replace = Vectorize(replace_one, "string", USE.NAMES = FALSE)

Answer 3

这是一种使用tidyverse、purrr 和stringr 的方法：

library(tidyverse)
library(stringr)

# Small utility function
find_and_replace <- function(string, bad_string, replacement_string) {
  ifelse(str_detect(string, bad_string), replacement_string, string)
}

str_split(test$a, ", ") %>%                 
  map(find_and_replace, "aaa", "NEW") %>%   
  map_chr(paste, collapse = ", ") %>%
  unlist

基本上：将向量拆分为一个列表，将find_and_replace 映射到该列表上，然后折叠结果。我建议在每个管道 %>% 之后单独查看结果。