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}
有什么办法吗?
答案 0 :(得分:189)
h = Hash[a.zip b] # => {:baz=>1, :bof=>2, :bar=>"world", :foo=>"hello"}
...该死的,我喜欢Ruby。
答案 1 :(得分:31)
只是想指出这样做的方法稍微清晰一点:
h = a.zip(b).to_h # => {:foo=>"hello", :bar=>"world", :baz=>1, :bof=>2}
必须同意"我喜欢Ruby"部分虽然!
答案 2 :(得分:14)
这个怎么样?
[a, b].transpose.to_h
如果你使用Ruby 1.9:
Hash[ [a, b].transpose ]
我觉得a.zip(b)
看起来a
是主人,b
是奴隶,但是这种风格是平的。
答案 3 :(得分:0)
只是出于好奇:
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%