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python - Python 字典中的求和值?

转载 作者:太空宇宙 更新时间:2023-11-04 07:09:52 25 4
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当我添加相同的键时,如何对 python 字典中的值求和?

d = {'key1':10,'key2':14,'key3':47}
d['key1'] = 20

在上面之后 d['key1'] 的值应该是 30。

这可能吗?

最佳答案

您可以使用collections.Counter:

>>> from collections import Counter
>>> d =Counter()
>>> d.update({'key1':10,'key2':14,'key3':47})
>>> d['key1'] += 20
>>> d['key4'] += 50 # Also works for keys that are not present
>>> d
Counter({'key4': 50, 'key3': 47, 'key1': 30, 'key2': 14})

计数器有一些优点:

>>> d1 = Counter({'key4': 50, 'key3': 4})
#You can add two counters
>>> d.update(d1)
>>> d
Counter({'key4': 100, 'key3': 51, 'key1': 30, 'key2': 14})

您可以使用 most_common() 获取已排序项目的列表(基于值):

>>> d.most_common()
[('key4', 100), ('key3', 51), ('key1', 30), ('key2', 14)]

时序比较:

>>> keys = [ random.randint(0,1000) for _ in xrange(10**4)]
>>> def dd():
d = defaultdict(int)
for k in keys:
d[k] += 10
...
>>> def count():
d = Counter()
for k in keys:
d[k] += 10
...
>>> def simple_dict():
... d = {}
... for k in keys:
... d[k] = d.get(k,0) + 10
...
>>> %timeit dd()
100 loops, best of 3: 3.47 ms per loop
>>> %timeit count()
100 loops, best of 3: 10.1 ms per loop
>>> %timeit simple_dict()
100 loops, best of 3: 5.01 ms per loop

关于python - Python 字典中的求和值?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17043207/

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