如何从两个大小相等的数组中构建Ruby哈希？

Question

我有两个数组

a = [:foo, :bar, :baz, :bof]

和

b = ["hello", "world", 1, 2]

我想要

{:foo => "hello", :bar => "world", :baz => 1, :bof => 2}

任何方式吗？

Answer 1

h = Hash[a.zip b] # => {:baz=>1, :bof=>2, :bar=>"world", :foo=>"hello"}

...该死，我爱露比。

Answer 2

只是想指出，有一种更简洁的方法：

h = a.zip(b).to_h # => {:foo=>"hello", :bar=>"world", :baz=>1, :bof=>2}

虽然必须同意“我爱Ruby”这一部分！

Answer 3

这个怎么样？

[a, b].transpose.to_h

如果使用Ruby 1.9：

Hash[ [a, b].transpose ]

[我觉得a.zip(b)看起来像a是主控，b是从属，但在这种样式下它们是扁平的。

Answer 4

出于好奇的缘故：

require 'fruity'

a = [:foo, :bar, :baz, :bof]
b = ["hello", "world", 1, 2]

compare do
  jtbandes { h = Hash[a.zip b] }
  lethjakman { h = a.zip(b).to_h }
  junichi_ito1 { [a, b].transpose.to_h }
  junichi_ito2 { Hash[ [a, b].transpose ] } 
end

# >> Running each test 8192 times. Test will take about 1 second.
# >> lethjakman is similar to junichi_ito1
# >> junichi_ito1 is similar to jtbandes
# >> jtbandes is similar to junichi_ito2

compare do 
  junichi_ito1 { [a, b].transpose.to_h }
  junichi_ito2 { Hash[ [a, b].transpose ] } 
end

# >> Running each test 8192 times. Test will take about 1 second.
# >> junichi_ito1 is faster than junichi_ito2 by 19.999999999999996% ± 10.0%