比较两个哈希数组并根据结果对一个数组进行更改[Ruby]答案

【问题标题】：comparing two arrays of hashes and making changes to one based off of result[Ruby]比较两个哈希数组并根据结果对一个数组进行更改[Ruby]
【发布时间】：2014-03-01 10:04:19
【问题描述】：

我有两个数组：

a = [{:name=>"John Doe", :number=>"555-234-5678", :count=>30}, {:name=>"Jane Doe", :number=>"555-456-1234", :count=>12}, {:name=>"Kobie Bryant", :number=>"555-621-9876", :count=>8}, {:name=>"JD", :number=>"555-234-5678", :count=>3}, {:name=>"KB", :number=>"555-621-9876", :count=>6}]

b = [{:name=>"JD", :number=>"555-234-5678", :count=>3}, {:name=>"KB", :number=>"555-621-9876", :count=>6}]

当 a[:name] 的首字母匹配 b[:name] 并且 a[:number] 匹配 b[:number] 时，我想将两个计数的总和添加到 a[:count]，然后删除任何a 数组中 b 中元素的实例。

所以 a 的结果是：

a = [{:name=>"John Doe", :number=>"555-234-5678", :count=>33}, {:name=>"Jane Doe", :number=>"555-456-1234", :count=>12}, {:name=>"Kobie Bryant", :number=>"555-621-9876", :count=>14}]

到目前为止，我有：

h = []
b.each do |double|
    a.each do |conn|
        if (double[:name][0,1] == conn[:name][0,1].split[0,1]) && (double[:name][1,1] == conn[:name][1,1].split[0,1])
            h[double] += conn[:count] + b[:count]
            a.delete(conn)
        end
    end
end

非常感谢任何帮助。谢谢大家！

【问题讨论】：

再次检查您的a 和b 数组。输出没有说明确切的逻辑。怎么33这里{:name=>"John Doe", :number=>"555-234-5678", :count=>33}？
John Doe 的数量 = 30 + JD 的数量 = 3。由于它们的数量相同，所以 John Doe 的数量就是总和。

标签： ruby arrays hash

【解决方案1】：

我建议你把这个过程分解一下。

def initials_from(name)
  name.split(" ").map { |name| name[0] }.join("")
end

def is_dup?(first, second)
  first == second
end

def matches?(first, second)
  initials_from(first[:name]) == second[:name] &&
    first[:number] == second[:number]
end

combined_result = a.each_with_object([]) { |first, result|
  # Skip duplicates
  duplicate = b.detect { |b_member| is_dup?(first, b_member) }
  next unless duplicate.nil?

  matching = b.detect { |b_member| matches?(first, b_member) }

  if matching
    combined = first.dup
    combined[:count] = first[:count] + matching[:count]
    result.push(combined)
  else
    result.push(first)
  end
}

combined_result
# [
#   {:name=>"John Doe",     :number=>"555-234-5678", :count=>33},
#   {:name=>"Jane Doe",     :number=>"555-456-1234", :count=>12},
#   {:name=>"Kobie Bryant", :number=>"555-621-9876", :count=>14}
# ]

然后，如果您真的想就地修改 a 而不是获取新数组：

a.replace(combined_result)

【讨论】：

【解决方案2】：

我会这样做：

a = [{:name=>"John Doe", :number=>"555-234-5678", :count=>30}, {:name=>"Jane Doe", :number=>"555-456-1234", :count=>12}, {:name=>"Kobie Bryant", :number=>"555-621-9876", :count=>8}, {:name=>"JD", :number=>"555-234-5678", :count=>3}, {:name=>"KB", :number=>"555-621-9876", :count=>6}]

b = [{:name=>"JD", :number=>"555-234-5678", :count=>3}, {:name=>"KB", :number=>"555-621-9876", :count=>6}]

ary = a.group_by do |inner_hash|
  name = inner_hash[:name]
  [/([A-Z])(?:\w+\s+)?([A-Z])/.match(name).captures.join , inner_hash[:number]]
end

array_of_hash = ary.flat_map do |k,v|
  bol = b.select do |hash|
    name = hash[:name]
    [/([A-Z])(?:\w+\s+)?([A-Z])/.match(name).captures.join , hash[:number]] == k
  end.empty?
  unless bol
    count = v.inject(0) { |sum,h1| sum = sum + h1[:count] }
    max_val_string = v.max_by { |h| h[:name].size }[:name]
  end
  next {:name => max_val_string, :number => k.last , :count => count} unless bol
  v
end
array_of_hash
# => [{:name=>"John Doe", :number=>"555-234-5678", :count=>33},
#     {:name=>"Jane Doe", :number=>"555-456-1234", :count=>12},
#     {:name=>"Kobie Bryant", :number=>"555-621-9876", :count=>14}]

【讨论】：

【解决方案3】：

假设b 中的[:name, :number] 是唯一的（如果不是，则需要稍作修改）：

b_index = Hash[b.map { |x| [x.values_at(:name, :number), x[:count]] }]
b_index.default = 0

r = 
  a.reject { |x| 
    b_index.has_key?(x.values_at(:name, :number)) 
  } \
  .map { |x| 
    name, number = x.values_at(:name, :number)
    initials = name.split(/\s+/).map { |y| y[0] }.join 
    x[:count] += b_index[[initials, number]]
    x
  }

【讨论】：