string - 如何从字符串中删除单词列表

Question

我想做的事情（在 Clojure 中）：

例如，我有一个需要删除的单词向量：

(def forbidden-words [":)" "the" "." "," " " ...many more...])

...和一个字符串向量：

(def strings ["the movie list" "this.is.a.string" "haha :)" ...many more...])

因此，应该从每个字符串中删除每个禁用词，在这种情况下，结果将是：[“movie list”“thisisastring”“haha”]。

这该怎么做 ？

Answer 1

(def forbidden-words [":)" "the" "." ","])
(def strings ["the movie list" "this.is.a.string" "haha :)"])
(let [pattern (->> forbidden-words (map #(java.util.regex.Pattern/quote %)) 
                (interpose \|)  (apply str))]
  (map #(.replaceAll % pattern "") strings))

Answer 2

(use 'clojure.contrib.str-utils)
(import 'java.util.regex.Pattern)
(def forbidden-words [":)" "the" "." "," " "])
(def strings ["the movie list" "this.is.a.string" "haha :)"])
(def regexes (map #(Pattern/compile % Pattern/LITERAL) forbidden-words))
(for [s strings] (reduce #(re-gsub %2 "" %1) s regexes))

Answer 3

使用函数组合和->宏可以很简单：

(for [s strings] 
  (-> s ((apply comp 
           (for [s forbidden-words] #(.replace %1 s ""))))))

如果您想更“惯用”，可以使用replace-strfrom clojure.contrib.string，而不是#(.replace %1 s "").

这里不需要使用正则表达式。