合并两个数组时，哈希

我有红宝石两个数组保留了重复键的键值对：合并两个数组时，哈希

array_one = ["farmer_joe", "farmer_judy", "farmer_crazy_eyes", "farmer_joe"] 

array_two = ["pigs", "chickens", "elephants", "cows"]

如果我使用压缩功能我失去了重复的值，密钥对农民乔。

hash_one = Hash[array_one.zip array_two] 

=> {"farmer_joe"=>"cows", "farmer_judy"=>"chickens", "farmer_crazy_eyes"=>"elephants"}

理想我想一个可以让我以一记漂亮的红宝石味的ONELINE方法克服这一点。也许像这样合并重复键并将它们的值添加到数组。

hash_one = Hash[array_one.super_special_zip array_two] 

=> {"farmer_joe"=>["pigs","cows"], "farmer_judy"=>["chickens"], "farmer_crazy_eyes"=>["elephants"]}

是否有这样的super_special_zip方法？还是有一个很好的理由，为什么这是一个傻瓜差事呢？

来源

2015-05-24 Huw

'array_one.zip array_two'不输重复，变成一个哈希不结果。 – steenslag

有三种标准的做法。

a1 = ["farmer_joe", "farmer_judy", "farmer_crazy_eyes", "farmer_joe"] 
a2 = ["pigs", "chickens", "elephants", "cows"] 
pairs = a1.zip(a2) # or [a1,a2].transpose 
    #=> [["farmer_joe", "pigs"], ["farmer_judy", "chickens"], 
    # ["farmer_crazy_eyes", "elephants"], ["farmer_joe", "cows"]]

1.使用Hash.new创建散列与一个空数组

pairs.each_with_object(Hash.new { |h,k| h[k]=[] }) { |(f,l),h| h[f] << l } 
    # => {"farmer_joe"=>["pigs", "cows"], "farmer_judy"=>["chickens"], 
    #  "farmer_crazy_eyes"=>["elephants"]}

这方面的一个变体（其倾向于稍快）的缺省值是：

pairs.each_with_object({}) { |(f,l),h| (h[f] ||= []) << l }

2.使用Hash#update（又名合并！）的形式，它需要一个块来确定ke的值目前在这两个哈希YS被合并

pairs.each_with_object({}) { |(f,l),h| h.update(f=>[l]) { |_,o,n| o+n } } 
    #=> {"farmer_joe"=>["pigs", "cows"], "farmer_judy"=>["chickens"], 
    # "farmer_crazy_eyes"=>["elephants"]}

3.使用Enumerable#group_by

h = pairs.group_by(&:first) 
    #=> {"farmer_joe"=>[["farmer_joe", "pigs"], ["farmer_joe", "cows"]], 
    # "farmer_judy"=>[["farmer_judy", "chickens"]], 
    # "farmer_crazy_eyes"=>[["farmer_crazy_eyes", "elephants"]]} 
h.keys.each { |k| h[k] = h[k].map(&:last) } 
h 
    #=> {"farmer_joe"=>["pigs", "cows"], "farmer_judy"=>["chickens"], 
    # "farmer_crazy_eyes"=>["elephants"]}

还有许多其他的最后两行，其中之一是：

h.merge(h) { |*_,v| v.map(&:last) }

来源

2015-05-24 12:54:34

的一种方式做到这一点

array_one = ["farmer_joe", "farmer_judy", "farmer_crazy_eyes", "farmer_joe"] 
array_two = ["pigs", "chickens", "elephants", "cows"] 

hash_one = {} 

array_one.each_with_index do |farmer,i| 
    if hash_one.has_key?(farmer) 
     hash_one[farmer] << array_two[i] 
    else 
     hash_one[farmer] = [array_two[i]] 
    end 
end 

hash_one # => {"farmer_joe"=>["pigs", "cows"], "farmer_judy"=>["chickens"], "farmer_crazy_eyes"=>["elephants"]}

来源

2015-05-24 13:05:36

类似于上面，我会说这有点更清楚，但我喜欢卡里的单线。如果我无法获得以'super_special_zip'方法构建的魔法，上面的方法可能会奏效。非常感谢您的贡献！ – Huw

这是更可读和简单的+1 – Ashik

而另一种方式来做到这将是（无轨）

a1 = ["farmer_joe", "farmer_judy", "farmer_crazy_eyes", "farmer_joe"] 
a2= ["pigs", "chickens", "elephants", "cows"] 

a1.zip(a2).group_by(&:first).map{|key, value| [key, value.map(&:last)]}.to_h 

# => {"farmer_joe"=>["pigs", "cows"], "farmer_judy"=>["chickens"], "farmer_crazy_eyes"=>["elephants"]}

来源

2016-04-14 03:10:41 t3pleni9

合并两个数组时，哈希

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