测试一个数组是否包含第二个数组中的任何元素的最有效方法是什么?
下面的两个示例试图回答以下问题:foods
是否包含来自cheeses
的任何元素
cheeses = %w(chedder stilton brie mozzarella feta haloumi reblochon)
foods = %w(pizza feta foods bread biscuits yoghurt bacon)
puts cheeses.collect{|c| foods.include?(c)}.include?(true)
puts (cheeses - foods).size < cheeses.size
发布于 2010-10-15 19:55:24
(cheeses & foods).empty?
正如马克-安德烈·拉富特在评论中所说,&
在线性时间内工作,而any?
+ include?
将是二次时间。对于更大的数据集,线性时间会更快。对于较小的数据集,any?
+ include?
可能更快,如Lee Jarvis的答案所示--可能是因为&
分配了一个新的数组,而另一个解决方案没有分配新的数组,并作为一个简单的嵌套循环返回布尔值。
发布于 2010-10-15 19:53:52
Enumerable#any?怎么样?
>> cheeses = %w(chedder stilton brie mozzarella feta haloumi)
=> ["chedder", "stilton", "brie", "mozzarella", "feta", "haloumi"]
>> foods = %w(pizza feta foods bread biscuits yoghurt bacon)
=> ["pizza", "feta", "foods", "bread", "biscuits", "yoghurt", "bacon"]
>> foods.any? {|food| cheeses.include?(food) }
=> true
基准测试脚本:
require "benchmark"
N = 1_000_000
puts "ruby version: #{RUBY_VERSION}"
CHEESES = %w(chedder stilton brie mozzarella feta haloumi).freeze
FOODS = %w(pizza feta foods bread biscuits yoghurt bacon).freeze
Benchmark.bm(15) do |b|
b.report("&, empty?") { N.times { (FOODS & CHEESES).empty? } }
b.report("any?, include?") { N.times { FOODS.any? {|food| CHEESES.include?(food) } } }
end
结果:
ruby version: 2.1.9
user system total real
&, empty? 1.170000 0.000000 1.170000 ( 1.172507)
any?, include? 0.660000 0.000000 0.660000 ( 0.666015)
发布于 2010-10-15 19:58:12
您可以检查交叉点是否为空。
cheeses = %w(chedder stilton brie mozzarella feta haloumi)
foods = %w(pizza feta foods bread biscuits yoghurt bacon)
foods & cheeses
=> ["feta"]
(foods & cheeses).empty?
=> false
https://stackoverflow.com/questions/3941945
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