填充缺少的条目在Ruby中

比较数组时，我有计数日期的三个数组：填充缺少的条目在Ruby中

first = [["July 01", "2"]["July 03", "2"]] 
second = [["June 30", "2"]["July 01", "2"]["July 02", "2"]] 
third = [["July 01", "2"]["July 02", "2"]]

我想（没有成功）三个阵列相比，得到了全范围的日期，并注入失踪者为0结果存入别人......所以，在年底的每个阵列将具有启动于6月30日期条目，直到7月3日，像这样：

first = [["June 30", "0"]["July 01", "2"]["July 02", "0"]["July 03", "2"]] 
second = [["June 30", "2"]["July 01", "2"]["July 02", "2"]["July 03", "0"]] 
third = [["June 30", "0"]["July 01", "2"]["July 02", "2"]["July 03", "0"]]

我尝试了一堆非常复杂的比较（像做演绎，存储为一个新的数组，然后使用该数组添加到缺少的但是当有两个以上的数组进行比较时，它变得非常复杂）和注入来做到这一点，但我认为必须有一个相对简单的方法来使用Ruby或Rails来实现。有任何想法吗？

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2017-07-31 Ben

这里的另一种方式，用Date：

require 'date' 

def compare_dates(*items) 
    all_dates = items.flatten(1).map { |d| Date.parse(d.first) } 
    str_dates = (all_dates.min..all_dates.max).map { |d| d.strftime("%B %d") } 

    items.map do |arr| 
    str_dates.map do |date| 
     current = arr.select { |e| e[0] == date }.flatten 
     current.empty? ? [date, "0"] : current 
    end 
    end 
end 

compare_dates(first, second, third) 
#=> [[["June 30", "0"], ["July 01", "2"], ["July 02", "0"], ["July 03", "2"]], 
# [["June 30", "2"], ["July 01", "2"], ["July 02", "2"], ["July 03", "0"]], 
# [["June 30", "0"], ["July 01", "2"], ["July 02", "2"], ["July 03", "0"]]]

如果要覆盖每个数组的值，你可以这样做：

first, second, third = compare_dates(first, second, third) 

first 
#=> [["June 30", "0"], ["July 01", "2"], ["July 02", "0"], ["July 03", "2"]] 

second 
#=> [["June 30", "2"], ["July 01", "2"], ["July 02", "2"], ["July 03", "0"]] 

third 
#=> [["June 30", "0"], ["July 01", "2"], ["July 02", "2"], ["July 03", "0"]]

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2017-07-31 19:48:51 Gerry

干净，简洁，美丽。谢谢格里。 – Ben

我有一个问题，为什么这是一个数组数组？如果你可以把它变成散列，问题处理起来很简单。它可以是这个样子：

first = {"july 01" => 2, "july 02" => 1} 
second = {"june 31" => 1, "july 01" => 1} 


keys = first.keys 
keys << second.keys 

keys.each do |key| 
    first[key] = first[key] || 0 
end

我没有测试过这一点，这可能不是最有效的方式，但你可以在此之上进行优化。我希望有所帮助。

你也可以使用类似Convert array of 2-element arrays into a hash, where duplicate keys append additional values的东西将它转换成哈希。

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2017-07-31 18:27:03 aks

谢谢，但这并没有真正的帮助。我已经写了一些可以处理两个数组（或散列，无所谓）的东西。我需要一种方法来比较三个或更多数组并返回所需的结果，而不会太复杂。我想我可能会错过一个核心ruby方法开始。 – Ben

你可以把所有数组的联合，然后从它计算。

all = first | second | third 
#=> [["July 01", "2"], ["July 03", "2"], ["June 30", "2"], ["July 02", "2"]] 

(first | all).map { |k, v| first.include?([k, v]) ? [k, v] : [k, "0"] } 
      .sort_by { |i| [ Time.new(0, i[0][0..2]).month, i[0][-2..-1] ] } 

#=> [["June 30", "0"], ["July 01", "2"], ["July 02", "0"], ["July 03", "2"]]

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2017-07-31 19:22:54

arr = [[["July 01", "2"], ["July 03", "2"]], 
     [["June 30", "2"], ["July 01", "2"], ["July 02", "2"]], 
     [["July 01", "2"], ["July 02", "2"]]] 

require 'date' 

default = arr.flatten(1). 
       map(&:first). 
       uniq. 
       sort_by { |s| Date.strptime(s, '%B %d') }. 
       product(['0']). 
       to_h 
    #=> {"June 30"=>"0", "July 01"=>"0", "July 02"=>"0", "July 03"=>"0"} 

arr.map { |a| default.merge(a.to_h).to_a } 
    #=> [[["June 30", "0"], ["July 01", "2"], ["July 02", "0"], ["July 03", "2"]], 
    # [["June 30", "2"], ["July 01", "2"], ["July 02", "2"], ["July 03", "0"]], 
    # [["June 30", "0"], ["July 01", "2"], ["July 02", "2"], ["July 03", "0"]]]

的步骤如下。

b = arr.flatten(1) 
    #=> [["July 01", "2"], ["July 03", "2"], ["June 30", "2"], ["July 01", "2"], 
    # ["July 02", "2"], ["July 01", "2"], ["July 02", "2"]] 
c = b.map(&:first) 
    #=> ["July 01", "July 03", "June 30", "July 01", "July 02", "July 01", "July 02"] 
d = c.uniq 
    #=> ["July 01", "July 03", "June 30", "July 02"] 
e = d.sort_by { |s| Date.strptime(s, '%B %d') } 
    #=> ["June 30", "July 01", "July 02", "July 03"] 
f = e.product(['0']) 
    #=> [["June 30", "0"], ["July 01", "0"], ["July 02", "0"], ["July 03", "0"]] 
default = f.to_h 
    #=> {"June 30"=>"0", "July 01"=>"0", "July 02"=>"0", "July 03"=>"0"}

为了计算

arr.map { |a| default.merge(a.to_h).to_a }

的arr第一值被传递到块和块变量a被设定为等于该值，并且执行块的计算。

a = arr.first 
    #=> [["July 01", "2"], ["July 03", "2"]] 
    g = a.to_h 
    #=> {"July 01"=>"2", "July 03"=>"2"} 
    h = default.merge(g) 
    #=> {"June 30"=>"0", "July 01"=>"2", "July 02"=>"0", "July 03"=>"2"} 
    h.to_a 
    #=> [["June 30", "0"], ["July 01", "2"], ["July 02", "0"], ["July 03", "2"]]

arr其他值的计算是相似的。

计算d参见Date::strptime。

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2017-08-01 04:52:10

填充缺少的条目在Ruby中

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