老实说，我发现最容易转换为熊猫系列或DataFrame:

import pandas as pd
import numpy as np

df = pd.DataFrame({'data':np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])})
print df['data'].value_counts()

或者是Robert Muil的一句俏皮话:

pd.Series([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]).value_counts()

对于您的情况，还可以查看numpy.bincount

In [56]: a = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])

In [57]: np.bincount(a)
Out[57]: array([8, 4])  #count of zeros is at index 0, i.e. 8
                        #count of ones is at index 1, i.e. 4

筛选并使用len

使用len是另一种选择。

A = np.array([1,0,1,0,1,0,1])

假设我们想要0的出现次数。

A[A==0]  # Return the array where item is 0, array([0, 0, 0])

现在，用len把它包起来。

len(A[A==0])  # 3
len(A[A==1])  # 4
len(A[A==7])  # 0, because there isn't such item.

对于一般条目:

x = np.array([11, 2, 3, 5, 3, 2, 16, 10, 10, 3, 11, 4, 5, 16, 3, 11, 4])
n = {i:len([j for j in np.where(x==i)[0]]) for i in set(x)}
ix = {i:[j for j in np.where(x==i)[0]] for i in set(x)}

将输出一个计数:

{2: 2, 3: 4, 4: 2, 5: 2, 10: 2, 11: 3, 16: 2}

和指标:

{2: [1, 5],
3: [2, 4, 9, 14],
4: [11, 16],
5: [3, 12],
10: [7, 8],
11: [0, 10, 15],
16: [6, 13]}

你有一个只有1和0的特殊数组。一个技巧就是使用

np.mean(x)

也就是数组中1的百分比。另外,使用

np.sum(x)
np.sum(1-x)

会给出数组中1和0的绝对值。

没有人建议使用numpy。Bincount (input, minlength)与minlength = np.size(input)，但这似乎是一个很好的解决方案，而且绝对是最快的:

In [1]: choices = np.random.randint(0, 100, 10000)

In [2]: %timeit [ np.sum(choices == k) for k in range(min(choices), max(choices)+1) ]
100 loops, best of 3: 2.67 ms per loop

In [3]: %timeit np.unique(choices, return_counts=True)
1000 loops, best of 3: 388 µs per loop

In [4]: %timeit np.bincount(choices, minlength=np.size(choices))
100000 loops, best of 3: 16.3 µs per loop

numpy之间的加速太疯狂了。unique(x, return_counts=True)和numpy。Bincount (x, minlength=np.max(x)) !