看看np.bincount:

http://docs.scipy.org/doc/numpy/reference/generated/numpy.bincount.html

import numpy as np
x = np.array([1,1,1,2,2,2,5,25,1,1])
y = np.bincount(x)
ii = np.nonzero(y)[0]

然后:

zip(ii,y[ii]) 
# [(1, 5), (2, 3), (5, 1), (25, 1)]

or:

np.vstack((ii,y[ii])).T
# array([[ 1,  5],
         [ 2,  3],
         [ 5,  1],
         [25,  1]])

或者你想结合计数和唯一值。

你可以这样写freq_count:

def freq_count(data):
    mp = dict();
    for i in data:
        if i in mp:
            mp[i] = mp[i]+1
        else:
            mp[i] = 1
    return mp

numpy。Bincount可能是最好的选择。如果你的数组除了包含小的密集整数之外还包含任何东西，那么像这样包装它可能是有用的:

def count_unique(keys):
    uniq_keys = np.unique(keys)
    bins = uniq_keys.searchsorted(keys)
    return uniq_keys, np.bincount(bins)

例如:

>>> x = array([1,1,1,2,2,2,5,25,1,1])
>>> count_unique(x)
(array([ 1,  2,  5, 25]), array([5, 3, 1, 1]))

Most of simple problems get complicated because simple functionality like order() in R that gives a statistical result in both and descending order is missing in various python libraries. But if we devise our thinking that all such statistical ordering and parameters in python are easily found in pandas, we can can result sooner than looking in 100 different places. Also, development of R and pandas go hand-in-hand because they were created for same purpose. To solve this problem I use following code that gets me by anywhere:

unique, counts = np.unique(x, return_counts=True)
d = {'unique':unique, 'counts':count}  # pass the list to a dictionary
df = pd.DataFrame(d) #dictionary object can be easily passed to make a dataframe
df.sort_values(by = 'count', ascending=False, inplace = True)
df = df.reset_index(drop=True) #optional only if you want to use it further

老问题，但我想提供我自己的解决方案，这是最快的，使用普通列表而不是np。数组作为输入(或首先转移到列表)，基于我的台架测试。

如果你也遇到这种情况，请检查一下。

def count(a):
    results = {}
    for x in a:
        if x not in results:
            results[x] = 1
        else:
            results[x] += 1
    return results

例如,

>>>timeit count([1,1,1,2,2,2,5,25,1,1]) would return:

100000个循环，最好的3:2.26µs每循环

>>>timeit count(np.array([1,1,1,2,2,2,5,25,1,1]))

100000个回路，最好的3:8.8µs每回路

>>>timeit count(np.array([1,1,1,2,2,2,5,25,1,1]).tolist())

100000个回路，最佳3:5.85µs每回路

而公认的答案会更慢，而scipy.stats.itemfreq解决方案更糟糕。

更深入的测试并没有证实所制定的期望。

from zmq import Stopwatch
aZmqSTOPWATCH = Stopwatch()

aDataSETasARRAY = ( 100 * abs( np.random.randn( 150000 ) ) ).astype( np.int )
aDataSETasLIST  = aDataSETasARRAY.tolist()

import numba
@numba.jit
def numba_bincount( anObject ):
    np.bincount(    anObject )
    return

aZmqSTOPWATCH.start();np.bincount(    aDataSETasARRAY );aZmqSTOPWATCH.stop()
14328L

aZmqSTOPWATCH.start();numba_bincount( aDataSETasARRAY );aZmqSTOPWATCH.stop()
592L

aZmqSTOPWATCH.start();count(          aDataSETasLIST  );aZmqSTOPWATCH.stop()
148609L

参考下面关于影响小型数据集大量重复测试结果的缓存和其他ram内副作用的评论。

使用pandas模块:

>>> import pandas as pd
>>> import numpy as np
>>> x = np.array([1,1,1,2,2,2,5,25,1,1])
>>> pd.value_counts(x)
1     5
2     3
25    1
5     1
dtype: int64