Question

示例：

array1 = ["budget2017.doc", "accounting2017.doc", "mydogisdumb.doc"]
array2 = ["budget.doc", "accounting.doc", "imstupid.doc"]

我想比较两个数组的相似性，并从array1返回相关元素。

array1.select { |x| x.include?(array2) }

我需要将结果作为["budget2017.doc", "accounting2017.doc"]

的新数组

但很明显上面的工作没有成功，因为"budget.doc"与"budget2017.doc"不匹配。如果我可以匹配每个元素的前几个字符并从array1返回相关元素，我就可以完成我需要的工作。

Answer 1

array1 = %w[budget2017.doc accounting2017.doc mydogisdumb.doc]
array2 = %w[budget.doc accounting.doc imstupid.doc] 

array1.select do |elem|
  array2.any? do |ee|
    s, e = ee.split('.')
    elem.start_with?(s) && elem.end_with?(e)
  end
end
#⇒ ["budget2017.doc", "accounting2017.doc"]

或者，效率更高：

selectors = array2.map { |e| e.split('.') }
array1.select do |elem|
  selectors.any? do |(s, e)|
    elem.start_with?(s) && elem.end_with?(e)
  end
end
#⇒ ["budget2017.doc", "accounting2017.doc"]

Answer 2

根据评论，查找array1的所有元素，它们与array2中的元素具有相同的前7个字符：

array1.select do |element|
  array2.any? { |match_candidate| match_candidate.start_with? element[0...7] }
end

Answer 3

arr1 = ["budget2017.doc", "acc2017.doc", "acc.doc", "budget2016.doc", "foo.doc"] 
arr2 = ["budget.doc", "acc.doc", "foo.docx,", "goo.doc"] 

a2 = arr2.map { |s| s.split('.') }
  #=> [["budget", "doc"], ["acc", "doc"], ["foo", "docx,"], ["goo", "doc"]] 
arr1.select { |s1| a2.any? { |pfx, sfx| s1 =~ /\A#{pfx}.*\.#{sfx}\z/ } }
  #=> ["budget2017.doc", "acc2017.doc", "acc.doc", "budget2016.doc"]

Ruby - 如果数组中的元素与另一个数组中的元素部分匹配

3 个答案: