h = { a: 1 }
h2 = { b: 2 }
h3 = { c: 3 }
Hash#merge适用于2个哈希:h.merge(h2)
如何合并3个哈希?
h.merge(h2).merge(h3)
有效但有更好的方法吗?
答案 0 :(得分:44)
你可以这样做:
h, h2, h3 = { a: 1 }, { b: 2 }, { c: 3 }
a = [h, h2, h3]
p Hash[*a.map(&:to_a).flatten] #= > {:a=>1, :b=>2, :c=>3}
编辑:如果您有许多哈希值,这可能是正确的方法:
a.inject{|tot, new| tot.merge(new)}
# or just
a.inject(&:merge)
答案 1 :(得分:18)
由于 Ruby 2.0 ,可以更加慷慨地完成:
h.merge **h1, **h2
如果密钥重叠 - 后者当然优先:
h = {}
h1 = { a: 1, b: 2 }
h2 = { a: 0, c: 3 }
h.merge **h1, **h2
# => {:a=>0, :b=>2, :c=>3}
h.merge **h2, **h1
# => {:a=>1, :c=>3, :b=>2}
答案 2 :(得分:10)
你可以做到
[*h,*h2,*h3].to_h
# => {:a=>1, :b=>2, :c=>3}
无论密钥是Symbol
s还是
答案 3 :(得分:4)
class Hash
def multi_merge(*args)
args.unshift(self)
args.inject { |accum, ele| accum.merge(ele) }
end
end
应该这样做。正如我所示,你可以很容易地将其打造成Hash。
答案 4 :(得分:3)
newHash = [h, h2, h3].each_with_object({}) { |oh, nh| nh.merge!(oh)}
# => {:a=>1, :b=>2, :c=>3}
答案 5 :(得分:3)
以下是我们在app中使用的2个monkeypatched :: Hash实例方法。以Minitest规格为后盾。出于性能原因,他们在内部使用merge!
而不是merge
。
class ::Hash
# Merges multiple Hashes together. Similar to JS Object.assign.
# Returns merged hash without modifying the receiver.
#
# @param *other_hashes [Hash]
#
# @return [Hash]
def merge_multiple(*other_hashes)
other_hashes.each_with_object(self.dup) do |other_hash, new_hash|
new_hash.merge!(other_hash)
end
end
# Merges multiple Hashes together. Similar to JS Object.assign.
# Modifies the receiving hash.
# Returns self.
#
# @param *other_hashes [Hash]
#
# @return [Hash]
def merge_multiple!(*other_hashes)
other_hashes.each(&method(:merge!))
self
end
end
测试:
describe "#merge_multiple and #merge_multiple!" do
let(:hash1) {{
:a => "a",
:b => "b"
}}
let(:hash2) {{
:b => "y",
:c => "c"
}}
let(:hash3) {{
:d => "d"
}}
let(:merged) {{
:a => "a",
:b => "y",
:c => "c",
:d => "d"
}}
describe "#merge_multiple" do
subject { hash1.merge_multiple(hash2, hash3) }
it "should merge three hashes properly" do
assert_equal(merged, subject)
end
it "shouldn't modify the receiver" do
refute_changes(->{ hash1 }) do
subject
end
end
end
describe "#merge_multiple!" do
subject { hash1.merge_multiple!(hash2, hash3) }
it "should merge three hashes properly" do
assert_equal(merged, subject)
end
it "shouldn't modify the receiver" do
assert_changes(->{ hash1 }, :to => merged) do
subject
end
end
end
end
答案 6 :(得分:3)
Ruby 2.6允许merge
接受多个参数:
h = { a: 1 }
h2 = { b: 2 }
h3 = { c: 3 }
h.merge(h2, h3) # => { a: 1, b: 2, c: 3 }
这也适用于Hash.merge!
和Hash.update
。 here的文档。
简单得多:)
答案 7 :(得分:2)
reduce
(与inject
相同)hash_arr = [{foo: "bar"}, {foo2: "bar2"}, {foo2: "bar2b", foo3: "bar3"}]
hash_arr.reduce { |acc, h| (acc || {}).merge h }
# => {:foo2=>"bar2", :foo3=>"bar3", :foo=>"bar"}
对于那些从Ruby或函数式编程开始的人,我希望这个简短的解释可能有助于理解这里发生的事情。
在Array对象(reduce
)上调用时,hash_arr
方法将迭代遍历数组的每个元素,并将块的返回值存储在累加器(acc
)中。实际上,我的块的h
参数将采用数组中每个哈希的值,acc
参数将采用块通过每次迭代返回的值。
我们使用(acc || {})
来处理acc
为零的初始条件。请注意,merge
方法优先考虑原始哈希中的键/值。这就是为什么"bar2b"
的值不会出现在我的最终哈希值中的原因。
希望有所帮助!
答案 8 :(得分:2)
要建立@Oleg Afanasyev的答案,你也可以做到这一点:
h = { a: 1 }
h2 = { b: 2 }
h3 = { c: 3 }
z = { **h, **h2, **h3 } # => {:a=>1, :b=>2, :c=>3}
干杯!
答案 9 :(得分:1)
只是为了好玩,你也可以这样做:
a = { a: 1 }, { b: 2 }, { c: 3 }
{}.tap { |h| a.each &h.method( :update ) }
#=> {:a=>1, :b=>2, :c=>3}