使用 splat 运算符的当前参数

Question

我有一个汽车品牌列表

 makes = [acura, honda, ford]

我正在尝试遍历一个字符串数组，并找出单个字符串是否包含其中一个 make AND 如果是，则将该特定 make 放入数组中

所以我有

strings.each do |string|
  if string.include?(*makes)
  else
  end
end

如何使用 splat 进程的当前参数来确定哪个使与字符串匹配？有没有办法做到这一点？

编辑：正如我在下面的评论中发布的那样，我正在寻找要返回的特定品牌，而不是真/假答案。所以如果字符串是"新丰田赛利卡"，则返回应该是"丰田"。

Answer 1

使用 Enumerable#any? ：

makes = ['acura', 'honda', 'ford']
strings = ['hyundai acura ford', 'sports car']
strings.each do |string|
  p makes.any? { |make| string.include? make }
end

使用正则表达式的替代方法：（请参阅Regexp::union）

strings = ['hyundai acura ford', 'sports car']
makes = ['acura', 'honda', 'ford']
pattern = Regexp.union(makes)
strings.each do |string|
  p string.match(pattern) != nil
end

---

更新

strings.each do |string|
  p makes.find { |make| string.include? make }
end

或

strings.each do |string|
  p makes.select { |make| string.include? make }
end

Answer 2

如果你的makes不是很长，那么最短的事情之一就是使用正则表达式，如前所述：

makes = ['acura', 'honda', 'ford']
strings = ['hyundai acura ford', 'sports car']
strings.grep(/#{makes.join('|')}/)
 # => ["hyundai acura ford"]

经过轻微的讨论，我们认为这是最好的选择之一：

strings.grep(Regexp.union(makes))

Answer 3

另一种方式，通过交叉数组：

makes = ["acura", "honda", "ford"]
strings = [
"I own a Toyota and a Ford",
"My friend Becky loves her Acura",
"I plan to buy a BMW",
"I now have an Acura, but have had both a honda and a Ford"
]
strings.each do |s|
  a = s.scan(/(w+)/).flatten.map(&:downcase) & makes
  puts "#{s}n" + (a.empty? ? "  No matches" : "  Matches: #{a.join}")
end
I own a Toyota and a Ford
  Matches: ford
My friend Becky loves her Acura
  Matches: acura
I plan to buy a BMW
  No matches
I now have an Acura, but have had both a honda and a Ford
  Matches: acura honda ford

请注意，有必要将scan与正则表达式一起使用，而不是split，因为标点符号将是后者的问题（例如，"Acura"将不匹配）。