Question

我有一个汽车制造清单

 makes = [acura, honda, ford]

我正在尝试遍历一个字符串数组，并查明单个字符串是否包含其中一个产生AND，如果确实如此，则将该特定产品放入数组中

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make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。因此，如果字符串是“New toyota celica”，则返回应为“toyota”。

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:(参见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

<强>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Regex，正如已经建议的那样：

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

 # => ["hyundai acura ford"]

经过一番讨论，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与正则表达式而不是split一起使用，因为标点符号会成为后者的问题（例如，'Acura，`将不匹配）。< / p>

使用splat运算符的当前参数

3 个答案: