Question

如果我有一个数组

private void readItems() {
        File filesDir = getFilesDir();
        File locationFile = new File(filesDir, "locations.txt");
        try {
            items = new ArrayList<UserLocation>(FileUtils.readLines(locationFile));
        } catch (IOException e) {
            items = new ArrayList<locationFile>();
        }
    }

如何沿着以下行返回一些东西，它通过数组并根据数组中的一个属性计算总数？

apple_array = [
  #<name: "Bob", apples_eaten: 3>,
  #<name: "Robert", apples_eaten: 7>,
  #<name: "Bob", apples_eaten: 5>,
  #<name: "Rebecca", apples_eaten: 2>,
  #<name: "Robert", apples_eaten: 3>
]

现在我知道如何通过调用name: Bob, apples_eaten: 8 name: Robert, apples_eaten: 10 name: Rebecca, apples_eaten: 2来返回apples_eaten的总金额，有没有办法将该行映射到每个姓名？

Answer 1

将它们全部放入映射名称=＆gt;的新哈希中individual_total：

totals = Hash.new(0)
apple_array.each do |a|
  totals[a.name] += a.apples_eaten
end

Answer 2

apple_array.inject(Hash.new 0) { |sum, e| sum[e.name] += e.apples_eaten;sum }

Answer 3

具有多个属性：

# list the fruit
attributes = %i(apples_eaten pears_eaten)
# define the struct
Meal = Struct.new(:name, *attributes)
# create sample data
apple_array = [Meal.new("bob", 1, 2), Meal.new("marge", 5, 4), Meal.new("gomer", 7, 1), Meal.new("bob", 5, 0), Meal.new("gomer", 3, 9)]

totals = apple_array.inject(
  # start from a hash that defaults to empty stomach
  Hash.new { |h, k| h[k] = Meal.new(k, 0, 0) }
) { |r, e|
  # then for each value in apple_array, add each attribute to the total
  attributes.each do |a|
    r[e.name][a] += e[a]
  end
  # mustn't forget to return the accumulator
  r
# hash was just for easy lookup, but we want an array, so...
}.values

require 'pp'
pp totals
# => [#<struct Meal name="bob", apples_eaten=6, pears_eaten=2>,
#     #<struct Meal name="marge", apples_eaten=5, pears_eaten=4>,
#     #<struct Meal name="gomer", apples_eaten=10, pears_eaten=10>]

Answer 4

这是我的解决方案。它按:name对数组进行排序，然后将数组划分为子数组，其中每个项具有相同的:name，然后将子数组减少为单个项，其总和为:apples_eaten。显而易见，如何总结其他属性......

apple_array = apple_array
  .sort_by { |a| a[:name] }
  .slice_when { |a, b| a[:name] != b[:name] }
  .map do |l|
    l.reduce do |a, b|
      a[:apples_eaten] += b[:apples_eaten]
      a
    end
  end

Ruby：返回数组中每个项目的总计数

4 个答案: