Question

我试图隔离数组中的某些字符串，例如，

["banana man", "apple", "banana woman"]

可由字符串的开头（即子字符串）识别。我想保留'banana man'和'banana woman'，但请删除'apple'。任何帮助将不胜感激。

Answer 1

这是grep

的一个很好的用例

ary = ["banana man", "apple", "banana woman"]
ary.grep(/^banana/) # => ["banana man", "banana woman"]

Answer 2

尝试使用start_with?和select来选择那些以您想要的字符串开头的内容：

["banana man", "apple", "banana woman"].select { |i| i.start_with?("bana") }
=> ["banana man", "banana woman"]

Answer 3

基准时间：

require 'fruity'

ARY = ["banana man", "apple", "banana woman"]

ARY.grep(/^banana/) # => ["banana man", "banana woman"]
ARY.select { |i| i.start_with?("bana") } # => ["banana man", "banana woman"]

compare do
  grep_only { ARY.grep(/^banana/) }
  select_start_with { ARY.select { |i| i.start_with?("bana") } }
end
# >> Running each test 4096 times. Test will take about 1 second.
# >> select_start_with is faster than grep_only by 3x ± 1.0

扩展ARY：

ARY = ["banana man", "apple", "banana woman"] * 1000

compare do
  grep_only { ARY.grep(/^banana/) }
  select_start_with { ARY.select { |i| i.start_with?("bana") } }
end
# >> Running each test 8 times. Test will take about 1 second.
# >> select_start_with is faster than grep_only by 3x ± 0.1

Answer 4

在Tin Mans的要求下，由于您的实际规格非常模糊，我会将此替代解决方案添加为可能但容易出错的选项。

ARY = ["banana man", "apple", "banana woman"] 
ARY.join('  ').scan(/banana\s\w+/)
#=> ["banana man", "banana woman"]

这将执行grep，但速度只有select和start_with?组合的一半，并且仅基于帖子，而不是您潜在的潜在意图。

如何过滤字符串数组

4 个答案: