R 相当于 python 中的 string.replace()

Question

我需要替换字符向量的某些值:

x <- data.frame(Strings = c("one", "two","three","four","five","four","five","four","five","two","thre","two","three","two","three"), stringsAsFactors = FALSE)
> x
   Strings
1      one
2      two
3    three
4     four
5     five
6     four
7     five
8     four
9     five
10     two
11   three
12     two
13   three
14     two
15   three

在 python 中，我会这样做:

x["Strings"].replace(["one", "two", "thre","three"], ["One","Two","Three","Three"], inplace=True)

但是在 r 函数中 replace()不以同样的简单方式工作。 Stackoverflow 中的字符串替换有很多解决方案，但没有一个这么简单。这在 r 中可能吗？

Answer 1

如果您只想将每个单词的第一个字母大写，我们可以使用 sub :

x$new <- sub('^([a-z])', '\\U\\1', x$Strings, perl = TRUE)

输出:

   Strings   new
1      one   One
2      two   Two
3    three Three
4     four  Four
5     five  Five
6     four  Four
7     five  Five
8     four  Four
9     five  Five
10     two   Two
11    thre  Thre
12     two   Two
13   three Three
14     two   Two
15   three Three

如果已经有新旧词替换列表，我们可以使用str_replace_all ，它与发布的python示例OP具有(某种)相似的风格:

library(stringr)

pattern <- c("one", "two", "thre", "three")
replacements <- c("One", "Two", "Three", "Three")

named_vec <- setNames(replacements, paste0("\\b", pattern, "\\b"))

x$new <- str_replace_all(x$Strings, named_vec)

或与 match或 hashmap :

library(dplyr)

x$new <- coalesce(replacements[match(x$Strings, pattern)], x$new)


library(hashmap)

hash_lookup = hashmap(pattern, replacements)
x$new <- coalesce(hash_lookup[[x$Strings]], x$new)

输出:

   Strings   new
1      one   One
2      two   Two
3    three Three
4     four  four
5     five  five
6     four  four
7     five  five
8     four  four
9     five  five
10     two   Two
11    thre Three
12     two   Two
13   three Three
14     two   Two
15   three Three