Question

如何使用以下数组中的键初始化哈希？

keys = [ 'a' , 'b' , 'c' ]

所需的哈希h应为：

puts h 
# { 'a' => nil , 'b' => nil , 'c' => nil }

Answer 1

我们使用Enumerable#each_with_object和Hash::[]。

 keys = [ 'a' , 'b' , 'c' ]
 Hash[keys.each_with_object(nil).to_a]
 # => {"a"=>nil, "b"=>nil, "c"=>nil}

或使用Array#product

keys = [ 'a' , 'b' , 'c' ]
Hash[keys.product([nil])]
# => {"a"=>nil, "b"=>nil, "c"=>nil}

Answer 2

使用Array#zip的另一种选择：

Hash[keys.zip([])]
# => {"a"=>nil, "b"=>nil, "c"=>nil}

更新：正如Arup Rakshit所建议的那样，建议的解决方案之间的性能比较：

require 'fruity'

ary = *(1..10_000)

compare do
  each_with_object {  ary.each_with_object(nil).to_a  }
  product          {  ary.product([nil])  }
  zip              {  ary.zip([])  }
  map              {  ary.map { |k| [k, nil] }  }
end

结果：

Running each test once. Test will take about 1 second.
zip is faster than product by 19.999999999999996% ± 1.0%
product is faster than each_with_object by 30.000000000000004% ± 1.0%
each_with_object is similar to map

Answer 3

使用新的（Ruby 2.1）to_h：

keys.each_with_object(nil).to_h

Answer 4

=> keys = [ 'a' , 'b' , 'c' ]
=> Hash[keys.map { |x, z| [x, z] }]
# {"a"=>nil, "b"=>nil, "c"=>nil}

=> Hash[keys.map { |x| [x, nil] }]
# {"a"=>nil, "b"=>nil, "c"=>nil}

=> Hash[keys.map { |x, _| [x, _] }]
# {"a"=>nil, "b"=>nil, "c"=>nil}

Answer 5

我认为你应该尝试：

x = {}
keys.map { |k, _| x[k.to_s] = _ }

Answer 6

如果您需要使用特定的内容初始化哈希，则还可以将Array#zip与Array#cycle结合使用：

Hash[keys.zip([''].cycle)]
# => {"a"=>"", "b"=>"", "c"=>""}

如何使用数组中的键初始化哈希？

6 个答案: