如何求和几个数组的相应元素？

我有几个不同长度的数组，每个数组都有一个2项数组。例如：如何求和几个数组的相应元素？

[["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]]

我想通过循环，并添加每个阵列的对应元件，进入一个新的数组。输出数组将是我们加在一起的最长阵列的长度。

所以两个数组上述相加将输出以下内容：

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

我不认为我可以使用reduce()或inject()，因为我不想崩溃数组，也不是一个数组简单的元素数组。

我真的不知道我该如何解决这个问题。

来源

2015-03-30 TerryCB

你可以在一行Hash.merge。在合并过程中使用块来合计值。

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

输出：

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]]

综上所述两个以上的阵列，增加一个行：

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

输出：

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

来源

2015-03-30 23:27:29

很好的回答，但我建议（因为OP使用术语“几个”），你将它概括为'sum_arrays（arr）'，其中'arr = [a，b，...]'。你可能不得不将'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）的链接并解释您使用表单该方法使用块来确定正在合并的两个哈希中存在的键的值。 – 2015-03-30 23:51:17

谢谢。多于两个数组的更多行更新了答案。 – 2015-03-31 00:29:37

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

我建议@奥列格的做法，但总有另一种方式：

个

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]]

步骤：

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转换nil到0.0。

要精心b上面的计算：

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计算如下：

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

来源

2015-03-31 00:38:39

如何求和几个数组的相应元素？

回答

相关问题