Question

我正在尝试编写一些代码来转换此嵌套数组的元素：

array = [
  [['number'], ['name'], ['postion'], ['points']],
  [['91'], ['dave'], ['center'], ['42']],
  [['82'], ['sanjay'], ['behind'], ['14']],
  [['133'], ['ayman'], ['front'], ['23']]
]

到哈希：

user1 = {number: 91, name: dave, postion: center, points: 42}
user2 = {number: 82, name: Sanjay, ...}

如果有人能够轻松理解代码，那就太感激了。

Answer 1

header, *data = array.map(&:flatten)
user1, user2, user3 = data.map { |row| header.zip(row).to_h }

Answer 2

要使用任意数量的字段和用户将数据传输到示例输出，您可能会执行以下操作：

array =  [
[['number'], ['name'], ['postion'], ['points']],
[  ['91'],  ['dave'], ['center'],  ['42']],
[  ['82'],   ['sanjay'],  ['behind'],    ['14']],
[  ['133'], ['ayman'], ['front'],    ['23']]]

temp, data={},{}
array.transpose.map { |(h, *rest)| temp[h]=rest.flatten }

(1..temp.max_by { |k,v| v.length }[1].length)
             .each_with_index { |n, i| data["user%d" % [n]]=temp.map { |k,l| [k[0], l[i]] } }
data.map { |k, v| data[k]=v.to_h }

然后你得到：

> data 
{"user1"=>{"number"=>"91", "name"=>"dave", "postion"=>"center", "points"=>"42"}, "user2"=>{"number"=>"82", "name"=>"sanjay", "postion"=>"behind", "points"=>"14"}, "user3"=>{"number"=>"133", "name"=>"ayman", "postion"=>"front", "points"=>"23"}}

Answer 3

使用数组users（而不是变量user1，user2等）更有意义，因此代码可以与任意数量的用户一起使用。然后，用户i的值为user[i]（从用户0开始计算）。

keys, *values = array.map(&:flatten)
users = [keys.map(&:to_sym)].product(values).map { |pair| pair.transpose.to_h }
  #=> [{:number=>"91", :name=>"dave", :postion=>"center", :points=>"42"},
  #    {:number=>"82", :name=>"sanjay", :postion=>"behind", :points=>"14"},
  #    {:number=>"133", :name=>"ayman", :postion=>"front", :points=>"23"}]

步骤如下。

keys, *values = array.map(&:flatten)
  #=> [["number", "name", "postion", "points"],
  #    ["91", "dave", "center", "42"],
  #    ["82", "sanjay", "behind", "14"],
  #    ["133", "ayman", "front", "23"]]
keys
  #=> ["number", "name", "postion", "points"]
values
  #=> [["91", "dave", "center", "42"],
  #    ["82", "sanjay", "behind", "14"],
  #    ["133", "ayman", "front", "23"]]
a = keys.map(&:to_sym)
  #=> [:number, :name, :postion, :points]
b = [a]
  #=> [[:number, :name, :postion, :points]]
c = b.product(values)
  #=> [[[:number, :name, :postion, :points], ["91", "dave", "center", "42"]],
       [[:number, :name, :postion, :points], ["82", "sanjay", "behind", "14"]],
     [[:number, :name, :postion, :points], ["133", "ayman", "front", "23"]]]
c.map { |pair| pair.transpose.to_h }
  # <return value above>

执行最后一步时，c的每个元素都传递给块，块变量pair被赋值给该值，并执行块计算。当c的第一个元素传递给块时，将执行以下步骤。

pair = c.first
  #=> [[:number, :name, :postion, :points], ["91", "dave", "center", "42"]]
d = pair.transpose
  #=> [[:number, "91"], [:name, "dave"], [:postion, "center"], [:points, "42"]]
d.to_h
  #=> {:number=>"91", :name=>"dave", :postion=>"center", :points=>"42"}

其余的计算方法类似。

Answer 4

您可以获取下一个哈希的每个键，将第一个数组放在主数组中，如：

array.first.flatten
# => ["number", "name", "postion", "points"]

那么，你可以得到数组的其余部分，从索引1开始到最后一个元素，＆＃34;展平＆＃34;它得到一个数组，每4个值，如：

p array[1..-1].flatten.each_slice(4).to_a
[["91", "dave", "center", "42"], ["82", "sanjay", "behind", "14"], ["133", "ayman", "front", "23"]]

那么你就可以为每个数组创建一个哈希值，每个数组中包含4个值，并使用keys数组中的每个值，如哈希键：

p array[1..-1].flatten.each_slice(4).map.with_index{|e,i|
  Hash[e.map.with_index{|f,i| [keys[i], f]}]
}
# [
#   {"number"=>"91",  "name"=>"dave",   "postion"=>"center", "points"=>"42"}, 
#   {"number"=>"82",  "name"=>"sanjay", "postion"=>"behind", "points"=>"14"}, 
#   {"number"=>"133", "name"=>"ayman",  "postion"=>"front",  "points"=>"23"}
# ]

拥有它可以迭代它，或根据需要创建本地或实例变量。

将嵌套数组中的某些元素转换为哈希

4 个答案: