如果您想一次计算所有值，可以使用numpy数组和bincount非常快速地完成，如下所示

import numpy as np
a = np.array([1, 2, 3, 4, 1, 4, 1])
np.bincount(a)

这给出了

>>> array([0, 3, 1, 1, 2])

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}

使用%timeit查看哪个操作更有效。np.array计数操作应该更快。

 from collections import Counter
 mylist = [1,7,7,7,3,9,9,9,7,9,10,0] 
 types_counts=Counter(mylist)
 print(types_counts)

我已经将所有建议的解决方案（以及一些新的解决方案）与perfplot（我的一个小项目）进行了比较。

清点一项

对于足够大的阵列，事实证明

numpy.sum(numpy.array(a) == 1)

比其他解决方案稍快。

清点所有项目

如前所述，

numpy.bincount(a)

是你想要的。

再现绘图的代码：

from collections import Counter
from collections import defaultdict
import numpy
import operator
import pandas
import perfplot


def counter(a):
    return Counter(a)


def count(a):
    return dict((i, a.count(i)) for i in set(a))


def bincount(a):
    return numpy.bincount(a)


def pandas_value_counts(a):
    return pandas.Series(a).value_counts()


def occur_dict(a):
    d = {}
    for i in a:
        if i in d:
            d[i] = d[i]+1
        else:
            d[i] = 1
    return d


def count_unsorted_list_items(items):
    counts = defaultdict(int)
    for item in items:
        counts[item] += 1
    return dict(counts)


def operator_countof(a):
    return dict((i, operator.countOf(a, i)) for i in set(a))


perfplot.show(
    setup=lambda n: list(numpy.random.randint(0, 100, n)),
    n_range=[2**k for k in range(20)],
    kernels=[
        counter, count, bincount, pandas_value_counts, occur_dict,
        count_unsorted_list_items, operator_countof
        ],
    equality_check=None,
    logx=True,
    logy=True,
    )

from collections import Counter
from collections import defaultdict
import numpy
import operator
import pandas
import perfplot


def counter(a):
    return Counter(a)


def count(a):
    return dict((i, a.count(i)) for i in set(a))


def bincount(a):
    return numpy.bincount(a)


def pandas_value_counts(a):
    return pandas.Series(a).value_counts()


def occur_dict(a):
    d = {}
    for i in a:
        if i in d:
            d[i] = d[i] + 1
        else:
            d[i] = 1
    return d


def count_unsorted_list_items(items):
    counts = defaultdict(int)
    for item in items:
        counts[item] += 1
    return dict(counts)


def operator_countof(a):
    return dict((i, operator.countOf(a, i)) for i in set(a))


b = perfplot.bench(
    setup=lambda n: list(numpy.random.randint(0, 100, n)),
    n_range=[2 ** k for k in range(20)],
    kernels=[
        counter,
        count,
        bincount,
        pandas_value_counts,
        occur_dict,
        count_unsorted_list_items,
        operator_countof,
    ],
    equality_check=None,
)
b.save("out.png")
b.show()

如果您想一次计算所有值，可以使用numpy数组和bincount非常快速地完成，如下所示

import numpy as np
a = np.array([1, 2, 3, 4, 1, 4, 1])
np.bincount(a)

这给出了

>>> array([0, 3, 1, 1, 2])

虽然这是一个很古老的问题，但由于我没有找到一个单行，所以我做了一个。

# original numbers in list
l = [1, 2, 2, 3, 3, 3, 4]

# empty dictionary to hold pair of number and its count
d = {}

# loop through all elements and store count
[ d.update( {i:d.get(i, 0)+1} ) for i in l ]

print(d)
# {1: 1, 2: 2, 3: 3, 4: 1}