Question

我有一个python字典对象，看起来有点像这样：

shutdown()

现在对于每个'sign'键，我想计算每个值出现的次数。

HazelcastInstance

然而，这会打印字典中出现'符号'的次数，而不是获取[{"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"}]的值并计算特定值出现的次数。

例如，我想看到的输出是：

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

等等。我应该改变什么来获得所需的输出？

Answer 1

使用collections.Counter及其most_common method：

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)

Answer 2

您可以使用collections.Counter模块，使用简单的生成器表达式，例如

>>> from collections import Counter
>>> Counter(k['sign'] for k in data if k.get('sign'))
Counter({'Sagittarius': 2, 'Capricorn': 2, 'Aquarius': 2, 'Leo': 2, 'Scorpio': 1, 'Gemini': 1})

这将为您提供一个字典，其中signs为关键字，其出现次数为值。

您可以使用普通字典执行相同的操作，例如

>>> result = {}
>>> for k in data:
...     if 'sign' in k:
...         result[k['sign']] = result.get(k['sign'], 0) + 1
>>> result
{'Sagittarius': 2, 'Capricorn': 2, 'Aquarius': 2, 'Leo': 2, 'Scorpio': 1, 'Gemini': 1}

dictionary.get方法接受第二个参数，如果在字典中找不到该键，则该参数将是返回的默认值。因此，如果当前符号不在result中，则会改为0。

Answer 3

def counter(my_list):
    my_list = sorted(my_list)
    first_val, *all_val = my_list
    p_index = my_list.index(first_val)
    my_counter = {}
    for item in all_val:
         c_index = my_list.index(item)
         diff = abs(c_index-p_index)
         p_index = c_index
         my_counter[first_val] = diff 
         first_val = item
    c_index = my_list.index(item)
    diff = len(my_list) - c_index
    my_counter[first_val] = diff 
    return my_counter

>>> counter([list(i.values())[1] for i in my_list])
{'Aquarius': 2,
 'Capricorn': 2,
 'Gemini': 1,
 'Leo': 2,
 'Sagittarius': 2,
 'Scorpio': 1}

计算python字典中每个键的出现次数

3 个答案: