计算python字典中每个键的出现次数

本文介绍了计算python字典中每个键的出现次数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个看起来像这样的python字典对象:

I have a python dictionary object that looks somewhat like this:

[{"house": 4, "sign": "Aquarius"}, {"house": 2, "sign": "Sagittarius"}, {"house": 8, "sign": "Gemini"}, {"house": 3, "sign": "Capricorn"}, {"house": 2, "sign": "Sagittarius"}, {"house": 3, "sign": "Capricorn"}, {"house": 10, "sign": "Leo"}, {"house": 4, "sign": "Aquarius"}, {"house": 10, "sign": "Leo"}, {"house": 1, "sign": "Scorpio"}]

现在，对于每个符号"键，我想计算每个值出现多少次.

Now for each 'sign' key, I'd like to count how many times each value occurs.

def predominant_sign(data): signs = [k['sign'] for k in data if k.get('sign')] print len(signs)

但是，这会打印出"sign"出现在字典中的次数，而不是获取sign的值并计算出现特定值的次数.

This however, prints number of times 'sign' appears in the dictionary, instead of getting the value of the sign and counting the number of times a particular value appears.

例如，我想看到的输出是:

For example, the output I'd like to see is:

Aquarius: 2 Sagittarius: 2 Gemini: 1 ...

以此类推.我应该改变什么以获得期望的输出?

And so on. What should I change to get the desired output?

推荐答案

使用 collections.Counter 及其 most_common方法:

from collections import Counter def predominant_sign(data): signs = Counter(k['sign'] for k in data if k.get('sign')) for sign, count in signs.most_common(): print(sign, count)