使用numpy。唯一的return_counts=True (NumPy 1.9+):

import numpy as np

x = np.array([1,1,1,2,2,2,5,25,1,1])
unique, counts = np.unique(x, return_counts=True)

>>> print(np.asarray((unique, counts)).T)
 [[ 1  5]
  [ 2  3]
  [ 5  1]
  [25  1]]

与scipy.stats.itemfreq相比:

In [4]: x = np.random.random_integers(0,100,1e6)

In [5]: %timeit unique, counts = np.unique(x, return_counts=True)
10 loops, best of 3: 31.5 ms per loop

In [6]: %timeit scipy.stats.itemfreq(x)
10 loops, best of 3: 170 ms per loop

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

多维频率计数，即计数数组。

>>> print(color_array    )
  array([[255, 128, 128],
   [255, 128, 128],
   [255, 128, 128],
   ...,
   [255, 128, 128],
   [255, 128, 128],
   [255, 128, 128]], dtype=uint8)


>>> np.unique(color_array,return_counts=True,axis=0)
  (array([[ 60, 151, 161],
    [ 60, 155, 162],
    [ 60, 159, 163],
    [ 61, 143, 162],
    [ 61, 147, 162],
    [ 61, 162, 163],
    [ 62, 166, 164],
    [ 63, 137, 162],
    [ 63, 169, 164],
   array([     1,      2,      2,      1,      4,      1,      1,      2,
         3,      1,      1,      1,      2,      5,      2,      2,
       898,      1,      1,  

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

from collections import Counter
x = array( [1,1,1,2,2,2,5,25,1,1] )
mode = counter.most_common(1)[0][0]

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。