使用Ruby将“主列表”样式嵌套数组转换为嵌套哈希

Question

我被朋友给了一个编码挑战，我一直在努力执行它。我应该采取一个数组，并操纵数据，并构建一个自定义哈希到他的规格。

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'] 
    ] 

最终目标

{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"}} 

所以我在园球，但也不能令人信服。

感谢您的输入！

编辑

我能得到很多与此

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

​​

所以有一件事我留下的，是更接近原始请求想要将电话号码用作符号而不是fixnum /整数键。那里有任何想法？

编辑

终于得到它

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来用Ruby 2.0版，因此，如果您使用的是早期版本，在该块使用Hash[keys.zip(values)]。

的步骤：

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map的a第一值的块计算如下：

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"} 

最后一步仅仅散列阵列转换为所示的散列更早：

b.each_with_object({}) { |g,h| h.update(g["phone_number"]=>g) } 

Answer 2

假设你hav E在可变master_list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：

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

下面是一个容易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