Question

我试图通过迭代将数组转换为哈希。我有以下数组：

weekdays = [["Monday",2],["Tuesday",4],["Thursday",5]]

我循环遍历数组以创建一个像这样的哈希：

hash_weekdays = Hash.new
weekdays.each do |item|
  hash_weekdays["weekday"] = item[0]
  hash_weekdays["number"] = item[1]
end

然而，这仅显示我最后一个工作日。关于从哪里去的任何想法？谢谢！

Answer 1

散列包含键值对。每个键仅引用一个值，每个键在散列中是uniq。您的密钥weekday和number的值将在每次迭代时被覆盖。这就是你只得到最后一次迭代结果的原因。

您可以将其转换为哈希数组作为替代。

weekdays.map{|day, number| {weekday:day, number: number}}
# => [{:weekday=>"Monday", :number=>2}, {:weekday=>"Tuesday", :number=>4}, {:weekday=>"Thursday", :number=>5}]

如果您的工作日或数字在主阵列中是唯一的，您可以使用其中一种作为键，另一种作为哈希值。

weekdays.to_h
# => {"Monday"=>2, "Tuesday"=>4, "Thursday"=>5}

或

weekdays.to_h.invert
# => {2=>"Monday", 4=>"Tuesday", 5=>"Thursday"}

Answer 2

你可以这样做：

weekdays.to_h #=> {"Monday"=>2, "Tuesday"=>4, "Thursday"=>5}

或修复您的代码：

weekdays = [["Monday",2],["Tuesday",4],["Thursday",5]]

hash_weekdays = Hash.new
weekdays.each do |item|
  hash_weekdays[item[0]] = item[1]
end

p hash_weekdays #=> {"Monday"=>2, "Tuesday"=>4, "Thursday"=>5}

Answer 3

您需要将代码更改为以下内容：

hash_weekdays = Hash.new
weekdays.each do |item|
  hash_weekdays[item[0]] = item[1]
end
hash_weekdays
 #=> {"Monday"=>2, "Tuesday"=>4, "Thursday"=>5}

当weekdays（["Monday",2]）的第一个元素传递给块时，块变量被赋值为：

item = ["Monday",2]

由于您需要引用item的每个元素，因此使用两个块变量是很常见的，这两个块变量的值是使用并行赋值分配的（又名多个赋值< / em>的）：

day, nbr = ["Monday",2] #=> ["Monday", 2] day #=> "Monday" nbr #=> 2

这允许你写

hash_weekdays = {} # the more common way of writing hash_weekdays = Hash.new weekdays.each { |day, nbr| hash_weekdays[day] = nbr } # used {...} rather than do..end hash_weekdays

这可以说是更清楚。

请注意，您首先将hash_weekdays初始化为空哈希，然后如果您希望获取哈希值的新值（作为方法的最后一行，则需要最后一行hash_weekdays，例如）。您可以使用方法Enumerable#each_with_object：
将其减少为单行（非常类似于Ruby）
weekdays.each_with_object({}) { |item, hash_weekdays| hash_weekdays[item[0]] = item[1] } #=> {"Monday"=>2, "Tuesday"=>4, "Thursday"=>5}

请注意，这使用并行分配。传递给块[["Monday", 2], {}]的第一个元素按如下方式分配：

item, hash_weekdays = weekdays.each_with_object({}).next #=> [["Monday", 2], {}] item #=> ["Monday", 2] hash_weekdays #=> {}

Ruby方式是以稍微复杂的方式使用并行赋值：

weekdays.each_with_object({}) { |(day, nbr), hash_weekdays| hash_weekdays[day] = nbr } #=> {"Monday"=>2, "Tuesday"=>4, "Thursday"=>5}

正如其他人所说，最直接的答案是使用方法Hash::[]或（在v2.0中引入）Array#to_h：

Hash[weekdays] #=> {"Monday"=>2, "Tuesday"=>4, "Thursday"=>5} Hash[*weekdays.flatten] #=> Hash["Monday", 2, "Tuesday", 4, "Thursday", 5] #=> {"Monday"=>2, "Tuesday"=>4, "Thursday"=>5} weekdays.to_h #=> {"Monday"=>2, "Tuesday"=>4, "Thursday"=>5}

通过循环将Ruby数组转换为哈希

3 个答案: