Question

我的脚本将一个元素（ticket [4]）分配给一个哈希。

哈希数组看起来像这样

all = [{"hello"=>{"2014-01-02"=>0, "2014-01-03"=>0}}, {"bye"=>{"2014-01-02"=>0, "2014-01-03"=>0}}]

数组故障单看起来像这样

tickets = [["hello","2014-01-02","1","Clôturé","2"], ["hello","2014-01-03","1","Clôturé","1"]]

脚本运行后，应该以

结果

all = [{"hello"=>{"2014-01-02"=>2, "2014-01-03"=>1}}, {"bye"=>{"2014-01-02"=>0, "2014-01-03"=>0}}]

相反，我有

all = [{"hello"=>{"2014-01-02"=>2, "2014-01-03"=>1}}, {"bye"=>{"2014-01-02"=>2, "2014-01-03"=>1}}]

为所有键分配值2。我希望我的脚本只分配给一个特定的键＆＃34;你好＆＃34;在这种情况下而不是＆＃34;你好＆＃34;和＆＃34; bye＆＃34;

 tickets.each do |t|
    d_d = Date.strptime(t[1],"%d-%m-%Y")
    n_c = t[0].to_s
    all.each do |e|
        e.each do |nom,pair|
            pair.each do |d,tick|
                 d1 = Date.strptime(d,"%d-%m-%Y")
                 if n_c == nom.to_s && d1 == d_d
                 p pair[d] = t[4]
                end
            end
        end
    end
end

Answer 1

你可以这样做：

all = [{"hello"=>{"2014-01-02"=>0, "2014-01-03"=>0}}, {"bye"=>{"2014-01-02"=>0, "2014-01-03"=>0}}]
tickets = [["hello","2014-01-02","1","Clôturé","2"], ["hello","2014-01-03","1","Clôturé","1"]]

tickets.each do |ticket|
  # This is the same as:
  #  name = ticket[0]
  #  date = ticket[1]
  name, date = ticket[0..1]

  # Here we're finding the correct ticket in "all" array.
  # The "correct ticket" is the hash whose key is equal the "name"
  all_ticket = all.select { |h| h.keys.first == name }.first

  # If the "correct ticket" still doesn't have a count, we attribute 0 to it
  all_ticket[name][date] ||= 0

  # And here we're incrementing its count
  all_ticket[name][date] += ticket.last.to_i
end

# Prints the result
p all

Answer 2

<强>代码

def aggregate(all, tickets)
  tickets_tot = tickets.each_with_object({}) { |(word, date, *_, nbr),h|
    h[[word, date]] = nbr.to_i }     
  all.map do |h|
    word, g = h.to_a.first
    f = g.each_with_object({}) { |(date,nbr),f| f[date] = nbr.to_i +
      tickets_tot.fetch([word, date], 0) }
    { word=>f }
  end
end

示例

对于问题中给出的all和tickets，

aggregate(all, tickets) #=> [{"hello"=>{"2014-01-02"=>2, "2014-01-03"=>1}}, # {"bye"=>{"2014-01-02"=>0, "2014-01-03"=>0}}]

下面

tickets_tot #=> {["hello", "2014-01-02"]=>2, ["hello", "2014-01-03"]=>1}

<强>解释

让我们逐步完成计算。

＆GT;计算tickets_tot

出于效率原因，我们首先计算tickets_tot，因为这只需要通过tickets。另一种方法是为tickets的每个值枚举all。

首先定义枚举器

enum = tickets.each_with_object({}) #=> #<Enumerator: [["hello", "2014-01-02", "1", "Clôturé", "2"], # ["hello", "2014-01-03", "1", "Clôturé", "1"]]:each_with_object({})>

生成enum的第一个元素，传递给块，并为块变量赋值。

(word, date, *_, nbr),h = enum.next #=> [["hello", "2014-01-02", "1", "Clôturé", "2"], {}] word #=> "hello" date #=> "2014-01-02" _ #=> ["1", "Clôturé"] nbr #=> "2" h #=> {}

*_将变量_分配给date和nbr之间所有值的数组。（请注意，IRB将变量_用于其自身目的，因此它不会正确报告其值。）下划线用于表示此块变量未在块计算中使用。

哈希h最初为空，但在执行计算时会发生变化。

接下来执行块计算。

h[[word, date]] = nbr.to_i h[["hello", "2014-01-02"]] = 2

enum生成的下一个和最后一个值现在传递给块。

(word, date, *_, nbr),h = enum.next #=> [["hello", "2014-01-03", "1", "Clôturé", "1"], # {["hello", "2014-01-02"]=>2}] word #=> "hello" date #=> "2014-01-03" _ #=> ["1", "Clôturé"] nbr #=> "1" h #=> {["hello", "2014-01-02"]=>2}

请注意h已更新。然后我们执行块计算。

h[[word, date]] = nbr.to_i #=> h[["hello", "2014-01-03"]] = 1

最后，

tickets_tot = h #=> {["hello", "2014-01-02"]=>2, ["hello", "2014-01-03"]=>1}

＆GT;将all映射到所需的哈希数组

我们现在转到计算的第二部分。 all的第一个值传递给map块，块变量被赋值其值并执行块。

h = all[0] #=> {"hello"=>{"2014-01-02"=>0, "2014-01-03"=>0}} a = h.to_a #=> [["hello", {"2014-01-02"=>0, "2014-01-03"=>0}]] word, g = a.first #=> ["hello", {"2014-01-02"=>0, "2014-01-03"=>0}] word #=> "hello" g #=> {"2014-01-02"=>0, "2014-01-03"=>0} f = g.each_with_object({}) { |(date,nbr),f| f[date] = nbr.to_i + tickets_tot.fetch([word, date], 0) } #=> {"2014-01-02"=>2, "2014-01-03"=>1}

请参阅Hash#fetch。

{ word=>f } #=> {"hello"=>{"2014-01-02"=>2, "2014-01-03"=>1}}

这更新了第一个哈希值。最后，

h = all[1] #=> {"bye"=>{"2014-01-02"=>0, "2014-01-03"=>0}} a = h.to_a #=> [["bye", {"2014-01-02"=>0, "2014-01-03"=>0}]] word, g = a.first #=> ["bye", {"2014-01-02"=>0, "2014-01-03"=>0}] word #=> "bye" g #=> {"2014-01-02"=>0, "2014-01-03"=>0} f = g.each_with_object({}) { |(date,nbr),f| f[date] = nbr.to_i + tickets_tot.fetch([word, date], 0) } #=> {"2014-01-02"=>0, "2014-01-03"=>0}

此时fetch的默认值为0。

{ word=>f } # => {"bye"=>{"2014-01-02"=>0, "2014-01-03"=>0}}

此映射与初始哈希值保持不变。因此map返回给定的两个哈希数组。

Answer 3

好的！该脚本运行良好，但ruby将日期视为同一个对象。我不得不添加＆＃34; .clone＆＃34;在创建包含所有日期的哈希时。

nom.each do |n|
    hash = {}
    hash = {n => dates.clone}
    all << hash
end

分配给所有哈希的值

3 个答案: