Question

我被朋友给了编码挑战，我一直在努力执行它。我应该采用数组，并操纵数据并根据他的规范构建自定义哈希。

开始数据

    [
        ['first_name', 'last_name', 'phone_number', 'level_of_caringness'],
        ['Phteven', 'Cartwright', 911, 'None'],
        ['Carol', 'Smellsbad', 666, 'Hateful'],
        ['Bambam', 'Idontwanttomaketheseanymore', 134, 'QWERQWER']
    ]

结束目标

{911: { first_name: "Phteven", last_name: 'Cartwright', phone_number: 911, level_of_caringness: 'None'},
 666: {first_name: "Carol", last_name: 'Smellsbad', phone_number: 666, level_of_caringness: 'Hateful'}
 134: {first_name: "Bambam", last_name: 'Idontwanttomaketheseanymore', phone_number: 134, level_of_caringness: 'QWERQWER'}

}

def array_to_hash3(array)
 keys = array.shift
 array.each_with_object({}) {|v, h| h[[array][2]] = {keys[0] => v[0], keys[1] => v[1], keys[2] => v[2], keys[3] => v[3]}}

端

这让我：

{nil=>{"first_name"=>"Bambam", "last_name"=>"Idontwanttomaketheseanymore", "phone_number"=>134, "level_of_caringness"=>"QWERQWER"}}

所以我在球场，但不是那里。

感谢您的投入！

修改

我能够更加接近这个

def array_to_hash3(array)
 key = array.shift
 array.each_with_object({}) {|v, h| h[v[2]] = {key[0] => v[0], key[1] => v[1], key[2] => v[2], key[3] => v[3]}}
end

结果

{911=>{"first_name"=>"Phteven", "last_name"=>"Cartwright", "phone_number"=>911, "level_of_caringness"=>"None"}, 
 666=>{"first_name"=>"Carol", "last_name"=>"Smellsbad", "phone_number"=>666, "level_of_caringness"=>"Hateful"}, 
 134=>{"first_name"=>"Bambam", "last_name"=>"Idontwanttomaketheseanymore", "phone_number"=>134, "level_of_caringness"=>"QWERQWER"}}

所以我留下的一件事是原始请求想要使用电话号码作为符号而不是fixnum / integer键。有什么想法吗？

修改

终于搞定了

def array_to_hash4(array)
  key = array.shift
  array.each_with_object({}) do |record, h|
    h[record[2]] = Hash[key.zip(record)]
  end
end

{
 911=>{"first_name"=>"Phteven", 
       "last_name"=>"Cartwright", 
       "phone_number"=>911, 
       "level_of_caringness"=>"None"}, 
 666=>{"first_name"=>"Carol", 
       "last_name"=>"Smellsbad", 
       "phone_number"=>666, 
       "level_of_caringness"=>"Hateful"}, 
 134=>{"first_name"=>"Bambam",                  
       "last_name"=>"Idontwanttomaketheseanymore", 
       "phone_number"=>134, 
       "level_of_caringness"=>"QWERQWER"}
}

Answer 1

arr = [
  ['first_name', 'last_name', 'phone_number', 'level_of_caringness'],
  ['Phteven', 'Cartwright', 911, 'None'],
  ['Carol', 'Smellsbad', 666, 'Hateful'],
  ['Bambam', 'Idontwanttomaketheseanymore', 134, 'QWERQWER']
]

[arr.first].product(arr[1..-1]).
     map { |keys, values| keys.zip(values).to_h }.
     each_with_object({}) { |g,h| h.update(g["phone_number"]=>g) } 
  #=> [{911=>{"first_name"=>"Phteven", "last_name"=>"Cartwright",
  #           "phone_number"=>911, "level_of_caringness"=>"None"}},
  #    {666=>{"first_name"=>"Carol", "last_name"=>"Smellsbad",
  #          "phone_number"=>666, "level_of_caringness"=>"Hateful"}}, 
  #    {134=>{"first_name"=>"Bambam", "last_name"=>"Idontwanttomaketheseanymore",
  #           "phone_number"=>134, "level_of_caringness"=>"QWERQWER"}}]

Array#to_h附带Ruby v.2.0，因此如果您使用的是早期版本，请在块中使用Hash[keys.zip(values)]。

步骤：

a = [arr.first].product(arr[1..-1])
  #=> [[["first_name", "last_name", "phone_number", "level_of_caringness"],
  #     ["Phteven", "Cartwright", 911, "None"]],
  #    [["first_name", "last_name", "phone_number", "level_of_caringness"],
  #     ["Carol", "Smellsbad", 666, "Hateful"]],
  #    [["first_name", "last_name", "phone_number", "level_of_caringness"],
  #     ["Bambam", "Idontwanttomaketheseanymore", 134, "QWERQWER"]]]

b = a.map { |keys, values| keys.zip(values).to_h }
  #=> [{"first_name"=>"Phteven", "last_name"=>"Cartwright",
  #     "phone_number"=>911, "level_of_caringness"=>"None"},
  #    {"first_name"=>"Carol", "last_name"=>"Smellsbad",
  #     "phone_number"=>666, "level_of_caringness"=>"Hateful"},
  #    {"first_name"=>"Bambam", "last_name"=>"Idontwanttomaketheseanymore",
  #     "phone_number"=>134, "level_of_caringness"=>"QWERQWER"}]

传递给a的{{1}}的第一个值的块计算如下：

map

最后一步只是将哈希数组转换为前面显示的哈希值：

keys, values = [["first_name", "last_name", "phone_number", "level_of_caringness"],
                ["Phteven", "Cartwright", 911, "None"]]
  #=> [["first_name", "last_name", "phone_number", "level_of_caringness"],
  #    ["Phteven", "Cartwright", 911, "None"]] 
keys
  #=> ["first_name", "last_name", "phone_number", "level_of_caringness"] 
values
  #=> ["Phteven", "Cartwright", 911, "None"] 
c = keys.zip(values)
  #=> [["first_name", "Phteven"], ["last_name", "Cartwright"],
  #    ["phone_number", 911], ["level_of_caringness", "None"]] 
c.to_h
  #=> {"first_name"=>"Phteven", "last_name"=>"Cartwright",
  #    "phone_number"=>911, "level_of_caringness"=>"None"}

Answer 2

假设您拥有变量master_list中的数据：

master_list[1..-1].each_with_object({}) do |arr, m|
  m[arr[2]] = Hash[master_list[0].map(&:to_sym).zip(arr)]
end

=> {911=>
     {:first_name=>"Phteven",
      :last_name=>"Cartwright",
      :phone_number=>911,
      :level_of_caringness=>"None"},
    666=>
     {:first_name=>"Carol",
      :last_name=>"Smellsbad",
      :phone_number=>666,
      :level_of_caringness=>"Hateful"},
    134=>
     {:first_name=>"Bambam",
      :last_name=>"Idontwanttomaketheseanymore",
      :phone_number=>134,
      :level_of_caringness=>"QWERQWER"}}

Answer 3

require 'pp'

data = [
  ['first_name', 'last_name', 'phone_number', 'level_of_caringness'],
  ['Phteven', 'Cartwright', 911, 'None'],
  ['Carol', 'Smellsbad', 666, 'Hateful'],
  ['Bambam', 'Idontwanttomaketheseanymore', 134, 'QWERQWER']
]

result = {}
cols = data[0]

data[1..-1].each do |row|
  inner_hash = {}

  cols.each_with_index do |col, i|
    inner_hash[col] = row[i] 
  end

  result[row[2].to_s.to_sym] = inner_hash
end

pp result

--output:--
{:"911"=>
  {"first_name"=>"Phteven",
   "last_name"=>"Cartwright",
   "phone_number"=>911,
   "level_of_caringness"=>"None"},
 :"666"=>
  {"first_name"=>"Carol",
   "last_name"=>"Smellsbad",
   "phone_number"=>666,
   "level_of_caringness"=>"Hateful"},
 :"134"=>
  {"first_name"=>"Bambam",
   "last_name"=>"Idontwanttomaketheseanymore",
   "phone_number"=>134,
   "level_of_caringness"=>"QWERQWER"}}

使用外部迭代器：

require 'pp'

data = [
  ['first_name', 'last_name', 'phone_number', 'level_of_caringness'],
  ['Phteven', 'Cartwright', 911, 'None'],
  ['Carol', 'Smellsbad', 666, 'Hateful'],
  ['Bambam', 'Idontwanttomaketheseanymore', 134, 'QWERQWER']
]

result = {}
keys = data[0].each

data[1..-1].each do |row|
  values = row.each
  inner_hash = {}

  loop do
    inner_hash[keys.next] = values.next
  end

  result[row[2].to_s.to_sym] = inner_hash
  keys.rewind
end

pp result

--output:--
{:"911"=>
  {"first_name"=>"Phteven",
   "last_name"=>"Cartwright",
   "phone_number"=>911,
   "level_of_caringness"=>"None"},
 :"666"=>
  {"first_name"=>"Carol",
   "last_name"=>"Smellsbad",
   "phone_number"=>666,
   "level_of_caringness"=>"Hateful"},
 :"134"=>
  {"first_name"=>"Bambam",
   "last_name"=>"Idontwanttomaketheseanymore",
   "phone_number"=>134,
   "level_of_caringness"=>"QWERQWER"}}

Answer 4

这是一个易于阅读和理解的解决方案。

def array_to_hash(list)
  result = {}
  keys = list.shift.map(&:to_sym)
  list.each{ |ar| result[ar[2]] = Hash[keys.zip(ar)]}
  result
end

使用Ruby

4 个答案: