Question

我有defaultdict，然后转换为字典。现在我正在寻找一种方法来计算按值分组的项目。我的意思是：

dict = {1: [103], 2: [9], 3: [3, 4], 4: [103, 106], 10: [3, 4], 11: [1, 9], 28: [1]}

for key in dict.items():
    print key

(1, [103])
(2, [9])
(3, [3, 4])
(4, [103, 106])
(10, [3, 4])
(11, [1, 9])
(28, [1])

item : total
{4 : 2, 106 : 1} ...

我该怎么做？

Answer 1

您可以使用collections.Counter来计算值列表中每个元素的出现次数，例如

>>> from collections import Counter
>>> Counter(value for values in d.itervalues() for value in values)
Counter({1: 2, 3: 2, 4: 2, 103: 2, 9: 2, 106: 1})

Counter只是字典的子类。所以，你可以像任何其他字典一样使用它

可以使用itertools.chain.from_iterable完成值的展平，就像这样

>>> from collections import Counter
>>> from itertools import chain
>>> Counter(chain.from_iterable(d.itervalues()))
Counter({1: 2, 3: 2, 4: 2, 103: 2, 9: 2, 106: 1})

Answer 2

这是另一种方法，除了基础知识之外什么都不用。

d = {1: [103], 2: [9], 3: [3, 4], 4: [103, 106], 10: [3, 4], 11: [1, 9], 28: [1]}


answer = {} # store the counts here
for key in d:
    for value in d[key]:
        if value in answer:
            # if the key is already in the answer
            answer[value] += 1
        else:
            # if the key has never been counted before
            # then add it to the answer and set its value to 1
            answer.update({value:1}) 


for key in answer:
    print key, answer[key]

打印：

Answer 3

m_dict = dict
n = {}
for k, v in m_dict.iteritems():
    for j in v:
        n[j] = (n[j] + 1 if n.get(j) != None else 1)
print n

结果：

{1: 2, 3: 2, 4: 2, 9: 2, 103: 2, 106: 1}

或者：

m_dict = dict
n = sum([v for k,v in m_dict.iteritems()], [])
c = {j : n.count(j) for j in n}
print c

结果：

{1: 2, 3: 2, 4: 2, 9: 2, 103: 2, 106: 1}

在defaultdict中计算项目

3 个答案: