Question

我有一个像这样的列表

L=['d','f','d','c','c','f','d','f']

我想知道L中有多少d，f和c出现并存储结果如下：

R=[['d',3],['f',3],['c',2]]

什么是最好的approch（算法）？

Answer 1

最好的方法（算法），就是不要自己动手！

>>> from collections import Counter
>>> L=['d','f','d','c','c','f','d','f']
>>> Counter(L)
Counter({'d': 3, 'f': 3, 'c': 2})

如果你坚持列表：

>>> Counter(L).items()
[('c', 2), ('d', 3), ('f', 3)]

Answer 2

我觉得字典对此更好：

>>> from collections import Counter
>>> L = ['d','f','d','c','c','f','d','f']
>>> Counter(L)
Counter({'d': 3, 'f': 3, 'c': 2})

但是，如果你坚持列表清单：

>>> L=['d','f','d','c','c','f','d','f']
>>> from collections import Counter
>>> var = Counter(L)
>>> [[key, value] for key, value in var.items()]
[['c', 2], ['d', 3], ['f', 3]]

Answer 3

L=['d','f','d','c','c','f','d','f']
from collections import Counter
print Counter(L)

<强>输出

Counter({'d': 3, 'f': 3, 'c': 2})

您可以使用Counter.most_common方法获取此结果

print Counter(L).most_common()

<强>输出

[('d', 3), ('f', 3), ('c', 2)]

Answer 4

在排序数据上使用itertools.groupby的可能解决方案

<强>实施

from itertools import groupby
[[k,  len(list(v))] for k, v in groupby(sorted(L))]

<强>输出

[['c', 2], ['d', 3], ['f', 3]]

效果比较

    In [9]: L = [choice(ascii_letters) for _ in range(1000)]

    In [10]: %timeit [[k,  len(list(v))] for k, v in groupby(sorted(L))]
    1000 loops, best of 3: 271 us per loop

    In [11]: %timeit Counter(L).items()
    1000 loops, best of 3: 306 us per loop

注意的

应该注意的是，计数器解决方案中散列数据的开销超过了Tim's Sort

中的排序复杂度

python计数列表项目出现并将结果放入列表中

4 个答案: