我有以下输入:
array = [{:year=>2015, :platform_id=>2},
{:year=>nil, :platform_id=>2},
{:year=>nil, :platform_id=>4},
{:year=>2015, :platform_id=>4}]
我需要预期的结果:
[{platform_id=>2, year=>[2015, nil]},
{platform_id=>4, year=>[nil, 2015]}]
我的代码是:
array.inject(:merge)
但这让我得到了这个结果,这不是我想要的结果:
{:year=>2015, :platform_id=>4}
收到回复后更新如下:
在看到答案后我会进行性能测试,结果如下:
arr = [
{:year => 2015, :platform_id => 2},
{:year => nil, :platform_id => 2},
{:year => nil, :platform_id => 4},
{:year => 2015, :platform_id => 4}
]
#approach 1
x1 = Time.now.to_f
exp = arr.each_with_object({}) do |h, exp|
exp[h[:platform_id]] ||= {:platform_id => h[:platform_id], :year => []}
exp[h[:platform_id]][:year] << h[:year]
end.values
x2 = Time.now.to_f
p x2-x1
#approach 2
x3 = Time.now.to_f
new_data = arr.group_by { |d| d[:platform_id] }
new_arr = []
new_data.each do |k,v|
t2 = v.map{|x| x[:year]}
temp = {"platform_id": k, "years": t2}
new_arr.push(temp)
end
x4 = Time.now.to_f
p x4-x3
#approach 3
x5 = Time.now.to_f
f = arr.each_with_object({}) { |g,h|
h.update(g[:platform_id]=>[g[:year]]) { |_,o,n| o+n } }
#=> {2=>[2015, nil], 4=>[nil, 2015]}
f.map { |k,v| { :platform_id=>k, :year=>v } }
x6 = Time.now.to_f
p x6-x5
#output is:
9.059906005859375e-06
6.4373016357421875e-06
9.775161743164062e-06
答案 0 :(得分:2)
arr = [
{ :year=>2015, :platform_id=>2 },
{ :year=>nil, :platform_id=>2 },
{ :year=>nil, :platform_id=>4 },
{ :year=>2015, :platform_id=>4 }
]
arr.each_with_object({}) { |g,h|
h.update(g[:platform_id]=>[g[:year]]) { |_,o,n| o+n } }.
map { |k,v| { :platform_id=>k, :year=>v } }
#=> [{:platform_id=>2, :year=>[2015, nil]},
# {:platform_id=>4, :year=>[nil, 2015]}]
这两个步骤如下。
f = arr.each_with_object({}) { |g,h|
h.update(g[:platform_id]=>[g[:year]]) { |_,o,n| o+n } }
#=> {2=>[2015, nil], 4=>[nil, 2015]}
f.map { |k,v| { :platform_id=>k, :year=>v } }
#=> [{:platform_id=>2, :year=>[2015, nil]},
# {:platform_id=>4, :year=>[nil, 2015]}]
第一步使用Hash#update(aka merge!
)的形式,它使用一个块(此处为{ |_,o,n| o+n }
)来计算两个哈希值中合并的键值。