数组算法中出现频率最高的元素，需要 python 代码转换为 ruby

Question

下面是我写的 python 代码，它将找到数组中最常见的元素，我想将下面的代码转换为 ruby 具有与 ruby 等价物相同的格式到 python 关键字，即对于 python 语法 "not in" 我想知道 ruby 等同于它等..，所有测试用例都应该有效。谢谢！

def most_frequent(given_list):
    max_count = -1
    max_item = None
    count = {}
    for i in given_list:
        if i not in count:
            count[i] = 1
        else:
            count[i] += 1
        if count[i] > max_count:
            max_count = count[i]
            max_item = i
    return max_item

# NOTE: The following input values will be used for testing your solution.
# most_frequent(list1) should return 1.
list1 = [1, 3, 1, 3, 2, 1]
# most_frequent(list2) should return 3.
list2 = [3, 3, 1, 3, 2, 1]
# most_frequent(list3) should return None.
list3 = []
# most_frequent(list4) should return 0.
list4 = [0]
# most_frequent(list5) should return -1.
list5 = [0, -1, 10, 10, -1, 10, -1, -1, -1, 1]

result1 = most_frequent(list1)
result2 = most_frequent(list2)
result3 = most_frequent(list3)
result4 = most_frequent(list4)
result5 = most_frequent(list5)
print result1, result2, result3, result4, result5

Answer 1

已转换 Ruby 代码：

def most_frequent(given_list)
  max_count = -1
  max_item = nil
  count = {}
  for i in given_list
    count[i] ||= 1
    count[i] += 1
    if count[i] > max_count
      max_count = count[i]
      max_item = i
    end
  end
  max_item
end

注意：Ruby 伙计们不会像这样编写代码，ruby 中提供了更简洁的方法来编写此代码。

在 ruby 而不是 None 使用 nil 如果你想在输出中使用 None 那么最后将 max_item 替换为 max_items || 'None'

要打印 result 的值，请使用 puts

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