如何求和幾個數組的相應元素？

我有幾個不同長度的數組，每個數組都有一個2項數組。例如：如何求和幾個數組的相應元素？

[["12:00", 7.0], ["01:00", 3.3], ["02:00", 11.9], ["03:00", 56.5]]

或

[["12:00", 44.3], ["01:00", 2.25], ["02:00", 2.44], ["03:00", 46.11], ["04:00", 8.9], ["05:00", 18.187]]

我想通過循環，並添加每個陣列的對應元件，進入一個新的數組。輸出數組將是我們加在一起的最長陣列的長度。

所以兩個數組上述相加將輸出以下內容：

[["12:00", 51.3], ["01:00", 5.55], ["02:00", 14.34], ["03:00", 102.61], ["04:00", 8.9], ["05:00", 18.187]]

我不認爲我可以使用reduce()或inject()，因爲我不想崩潰數組，也不是一個數組簡單的元素數組。

我真的不知道我該如何解決這個問題。

來源

2015-03-30 TerryCB

你可以在一行Hash.merge。在合併過程中使用塊來合計值。

def sum_arrays(a, b) 
    Hash[a].merge(Hash[b]){|k, i, j| i + j}.to_a 
end

輸出：

a = [["12:00", 7.0], ["01:00", 3.3], ["02:00", 11.9], ["03:00", 56.5]] 
b = [["12:00", 44.3], ["01:00", 2.25], ["02:00", 2.44], ["03:00", 46.11], ["04:00", 8.9], ["05:00", 18.187]]  
sum_arrays(a,b) 

=> [["12:00", 51.3], ["01:00", 5.55], ["02:00", 14.34], ["03:00", 102.61], ["04:00", 8.9], ["05:00", 18.187]]

綜上所述兩個以上的陣列，增加一個行：

def sum_many_arrays(*a) 
    a.reduce{|s, i| sum_arrays(s, i)} 
end

輸出：

sum_many_arrays([[:a,1],[:b,2]],[[:a,2],[:b,2],[:c,1]],[[:a,5],[:b,2]]) 
=> [[:a, 8], [:b, 6], [:c, 1]]

來源

2015-03-30 23:27:29

很好的回答，但我建議（因爲OP使用術語「幾個」），你將它概括爲'sum_arrays（arr）'，其中'arr = [a，b，...]'。你可能不得不將'merge'改成'merge！'（又名'update'）。此外，提供指向方法[Hash＃merge]（http://ruby-doc.org/core-2.2.0/Hash.html#method-i-merge）的鏈接並解釋您使用表單該方法使用塊來確定正在合併的兩個哈希中存在的鍵的值。 – 2015-03-30 23:51:17

謝謝。多於兩個數組的更多行更新了答案。 – 2015-03-31 00:29:37

很酷的使用哈希＃合併和#to_a。 – Puhlze 2015-03-31 02:24:55

我建議@奧列格的做法，但總有另一種方式：

個

arr = [[["12:00", 7.0], ["01:00", 3.3], ["02:00", 11.9], ["03:00", 56.5]], 
     [["12:00", 44.3], ["01:00", 2.25], ["02:00", 2.44], ["03:00", 46.11], 
     ["04:00", 8.9]], 
     [["01:00", 22.25], ["02:00", 1.84], ["12:00", 13.3]]] 

keys = arr.reduce([]) { |keys,a| keys | a.map(&:first) } 
arr.map { |a| a.to_h.values_at(*keys) } 
    .transpose 
    .map { |e| [keys.shift, e.reduce(0) { |tot,x| tot + x.to_f }] } 
    #=> [["12:00", 64.6], ["01:00", 27.8], ["02:00", 16.18], 
    # ["03:00", 102.61], ["04:00", 8.9]]

步驟：

keys = arr.reduce([]) { |keys,a| keys | a.map(&:first) } 
    #=> ["12:00", "01:00", "02:00", "03:00", "04:00"] 
b = arr.map { |a| a.to_h.values_at(*keys) } 
    #=> [[ 7.0, 3.3, 11.9, 56.5, nil], 
    # [44.3, 2.25, 2.44, 46.11, 8.9], 
    # [13.3, 22.25, 1.84, nil, nil]] 
c = b.transpose 
    #=> [[7.0, 44.3, 13.3], 
    # [3.3, 2.25, 22.25], 
    # [11.9, 2.44, 1.84], 
    # [56.5, 46.11, nil], 
    # [nil, 8.9, nil]] 
c.map { |e| [keys.shift, e.reduce(0) { |tot,x| tot + x.to_f }] } 
    #=> [["12:00", 64.6], ["01:00", 27.8], ["02:00", 16.18], 
    # ["03:00", 102.61], ["04:00", 8.9]]

注意NilClass#to_f轉換nil到0.0。

要精心b上面的計算：

d = arr.map 
    #=> #<Enumerator: [[["12:00", 7.0], ["01:00", 3.3], ["02:00", 11.9], 
    #     ["03:00", 56.5]], 
    #     [["12:00", 44.3], ["01:00", 2.25], ["02:00", 2.44], 
    #     ["03:00", 46.11], ["04:00", 8.9]], 
    #     [["01:00", 22.25], ["02:00", 1.84], ["12:00", 13.3]]]:map> 

a = d.next 
    #=> [["12:00", 7.0], ["01:00", 3.3], ["02:00", 11.9], ["03:00", 56.5]] 
e = a.to_h 
    #=> {"12:00"=>7.0, "01:00"=>3.3, "02:00"=>11.9, "03:00"=>56.5} 
f = e.values_at(*keys) 
    #=> e.values_at(*["12:00", "01:00", "02:00", "03:00", "04:00"]) 
    #=> [7.0, 3.3, 11.9, 56.5, nil] 

a = d.next 
    #=> [["12:00", 44.3], ["01:00", 2.25], ["02:00", 2.44], 
    # ["03:00", 46.11], ["04:00", 8.9]] 
e = a.to_h 
    #=> {"12:00"=> 44.3, "01:00"=>2.25, "02:00"=>2.44, 
    # "03:00"=>46.11, "04:00"=> 8.9} 
f = e.values_at(*keys) 
    #=> [44.3, 2.25, 2.44, 46.11, 8.9] 

a = d.next 
    #=> [["01:00", 22.25], ["02:00", 1.84], ["12:00", 13.3]] 
e = a.to_h 
    #=> {"01:00"=>22.25, "02:00"=>1.84, "12:00"=>13.3} 
f = e.values_at(*keys) 
    #=> [13.3, 22.25, 1.84, nil, nil]

c.map計算如下：

keys = ["12:00", "01:00", "02:00", "03:00", "04:00"] 

d = c.map 
    #=> #<Enumerator: [[ 7.0, 44.3, 13.3], [3.3, 2.25, 22.25], 
    #     [11.9, 2.44, 1.84], [56.5, 46.11, nil], 
    #     [nil, 8.9, nil]]:map> 

e = d.next 
    #=> [7.0, 44.3, 13.3] 
f = keys.shift 
    #=> "12:00" 
keys 
    #=> ["01:00", "02:00", "03:00", "04:00"] 
e.reduce(0) { |tot,x| tot + x.to_f } 
    #=> 64.6 

e = d.next 
    #=> [3.3, 2.25, 22.25] 
f = keys.shift 
    #=> "01:00" 
keys 
    #=> ["02:00", "03:00", "04:00"] 
e.reduce(0) { |tot,x| tot + x.to_f } 
    #=> 27.8

e = d.next 
    #=> [11.9, 2.44, 1.84] 
f = keys.shift 
    #=> "02:00" 
keys 
    #=> ["03:00", "04:00"] 
e.reduce(0) { |tot,x| tot + x.to_f } 
    #=> 16.18 

e = d.next 
    #=> [56.5, 46.11, nil] 
f = keys.shift 
    #=> "03:00" 
keys 
    #=> ["04:00"] 
e.reduce(0) { |tot,x| tot + x.to_f } 
    #=> 102.61 

e = d.next 
    #=> [nil, 8.9, nil] 
f = keys.shift 
    #=> "04:00" 
keys 
    #=> [] 
e.reduce(0) { |tot,x| tot + x.to_f } 
    #=> 8.9

來源

2015-03-31 00:38:39

如何求和幾個數組的相應元素？

回答

相關問題