要计算具有共同类型的不同元素的数量，请执行以下操作：

li = ['A0','c5','A8','A2','A5','c2','A3','A9']

print sum(1 for el in li if el[0]=='A' and el[1] in '01234')

给予

3，而不是6

为什么不使用Pandas？

import pandas as pd

my_list = ['a', 'b', 'c', 'd', 'a', 'd', 'a']

# converting the list to a Series and counting the values
my_count = pd.Series(my_list).value_counts()
my_count

输出：

a    3
d    2
b    1
c    1
dtype: int64

如果您正在查找特定元素的计数，例如a，请尝试：

my_count['a']

输出：

3

# Python >= 2.6 (defaultdict) && < 2.7 (Counter, OrderedDict)
from collections import defaultdict
def count_unsorted_list_items(items):
    """
    :param items: iterable of hashable items to count
    :type items: iterable

    :returns: dict of counts like Py2.7 Counter
    :rtype: dict
    """
    counts = defaultdict(int)
    for item in items:
        counts[item] += 1
    return dict(counts)


# Python >= 2.2 (generators)
def count_sorted_list_items(items):
    """
    :param items: sorted iterable of items to count
    :type items: sorted iterable

    :returns: generator of (item, count) tuples
    :rtype: generator
    """
    if not items:
        return
    elif len(items) == 1:
        yield (items[0], 1)
        return
    prev_item = items[0]
    count = 1
    for item in items[1:]:
        if prev_item == item:
            count += 1
        else:
            yield (prev_item, count)
            count = 1
            prev_item = item
    yield (item, count)
    return


import unittest
class TestListCounters(unittest.TestCase):
    def test_count_unsorted_list_items(self):
        D = (
            ([], []),
            ([2], [(2,1)]),
            ([2,2], [(2,2)]),
            ([2,2,2,2,3,3,5,5], [(2,4), (3,2), (5,2)]),
            )
        for inp, exp_outp in D:
            counts = count_unsorted_list_items(inp) 
            print inp, exp_outp, counts
            self.assertEqual(counts, dict( exp_outp ))

        inp, exp_outp = UNSORTED_WIN = ([2,2,4,2], [(2,3), (4,1)])
        self.assertEqual(dict( exp_outp ), count_unsorted_list_items(inp) )


    def test_count_sorted_list_items(self):
        D = (
            ([], []),
            ([2], [(2,1)]),
            ([2,2], [(2,2)]),
            ([2,2,2,2,3,3,5,5], [(2,4), (3,2), (5,2)]),
            )
        for inp, exp_outp in D:
            counts = list( count_sorted_list_items(inp) )
            print inp, exp_outp, counts
            self.assertEqual(counts, exp_outp)

        inp, exp_outp = UNSORTED_FAIL = ([2,2,4,2], [(2,3), (4,1)])
        self.assertEqual(exp_outp, list( count_sorted_list_items(inp) ))
        # ... [(2,2), (4,1), (2,1)]

给定列表X

 import numpy as np
 X = [1, -1, 1, -1, 1]

显示此列表元素的i:frequency（i）的字典为：

{i:X.count(i) for i in np.unique(X)}

输出：

{-1: 2, 1: 3}

mot = ["compte", "france", "zied"]
lst = ["compte", "france", "france", "france", "france"]
dict((x, lst.count(x)) for x in set(mot))

这给了

{'compte': 1, 'france': 4, 'zied': 0}

如果您使用的是Python 2.7或3.x，并且希望每个元素的出现次数，请使用Counter：

>>> from collections import Counter
>>> z = ['blue', 'red', 'blue', 'yellow', 'blue', 'red']
>>> Counter(z)
Counter({'blue': 3, 'red': 2, 'yellow': 1})