蟒蛇计数列表项出现次数和列表

Question

把结果我有这样

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

名单，我想看看有多少d，f和c出现次数都为L并将结果类似：

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

什么是最好的接近（算法）？

Answer 1

最好的办法（算法），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}) 

如果名单上坚持：

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

Answer 2

我觉得一本字典是为了这个美好的：

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

不过，如果你是坚定的关于列表的列表：

>>> 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) 

输出

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

您可以使用Counter.most_common方法得到的结果就像一个排序DA使用itertools.groupby一种可能的解决这个

print Counter(L).most_common() 

输出

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

Answer 4

TA

实施

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

输出

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

性能比较

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

应当指出的是，在散列数据在柜台解决方案的开销，过冲排序复杂性Tim's Sort