Question

说我有很多句子：

sentences = ["Tom is a good person", "Jack spent some time", "Kat did something wrong"]

我有一个黑名单名称：

blacklist = ["Jack", "Kat"]

现在，我需要将sentences过滤到一个数组中，以删除所有包含黑名单名称的句子，因此：

["Tom is a good person"]

我将如何在Ruby中做到这一点？

谢谢！

Answer 1

sentences = ["Tom is a good person", "Jack spent some time", "Kat did something wrong",
             "Kathy knows her stuff"]

blacklist = ["Jack", "Kat"]

r = /\b#{Regexp.union(blacklist)}\b/
  #=> /\b(?-mix:Jack|Kat)\b/
sentences.reject { |s| s.match?(r) }
  #=> ["Tom is a good person", "Kathy knows her stuff"]

在正则表达式中需要分词（\b），以便"Kat"与"Kathy"的前三个字母不匹配。可以这样写：

r = /\b#{blacklist.join('|')}\b/
  #=> /\bJack|Kat\b/

Answer 2

您只需要拒绝记录

sentences.reject!{|sentence| sentence.match(blacklist.join('|'))}

您将获得所需的输出-

["Tom is a good person"]

拒绝文档！ -https://ruby-doc.org/core-2.2.0/Array.html#method-i-reject-21

reject!将更新同一数组，如果您不希望这样做，则可以使用reject并将其存储在新数组中。

返回=~ ：）

sentences.reject!{|sentence| !((Regexp.new(blacklist.join('|')) =~ sentence).nil?) }

如果`=〜'找到匹配项，则拒绝该句子（使用的代码-！nil？）

这基本上会做同样的事情。尽管红宝石说，比赛更快

Answer 3

sentences = ["Tom is a good person", "Jack spent some time", "Kat did something wrong"]

blacklist = ["Jack", "Kat"]

程序

p sentences.filter{|string|!(blacklist.map{|x|string.match?(x)}.any?)}

如果您使用的是

p sentences.select{|string|!(blacklist.map{|x|string.include?(x)}.any?)}

结果

["Tom is a good person"]

Ruby：过滤器数组删除列入黑名单的名称

3 个答案: