利用a系列提供的方法:

>>> import pandas as pd
>>> y = [0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]
>>> pd.Series(y).value_counts()
0    8
1    4
dtype: int64

它还涉及一个步骤，但更灵活的解决方案也适用于2d数组和更复杂的过滤器，即创建一个布尔掩码，然后在掩码上使用.sum()。

>>>>y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
>>>>mask = y == 0
>>>>mask.sum()
8

就我个人而言，我认为: (y == 0).sum() and (y == 1).sum()

E.g.

import numpy as np
y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
num_zeros = (y == 0).sum()
num_ones = (y == 1).sum()

要计算出现的次数，可以使用np。独特的(数组,return_counts = True):

In [75]: boo = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
 
# use bool value `True` or equivalently `1`
In [77]: uniq, cnts = np.unique(boo, return_counts=1)
In [81]: uniq
Out[81]: array([0, 1])   #unique elements in input array are: 0, 1

In [82]: cnts
Out[82]: array([8, 4])   # 0 occurs 8 times, 1 occurs 4 times

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

一个普遍而简单的答案是:

numpy.sum(MyArray==x)   # sum of a binary list of the occurence of x (=0 or 1) in MyArray

这将导致这完整的代码作为例子

import numpy
MyArray=numpy.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])  # array we want to search in
x=0   # the value I want to count (can be iterator, in a list, etc.)
numpy.sum(MyArray==0)   # sum of a binary list of the occurence of x in MyArray

现在，如果MyArray是多维的，你想要计算值在直线(= pattern以后)上分布的次数。

MyArray=numpy.array([[6, 1],[4, 5],[0, 7],[5, 1],[2, 5],[1, 2],[3, 2],[0, 2],[2, 5],[5, 1],[3, 0]])
x=numpy.array([5,1])   # the value I want to count (can be iterator, in a list, etc.)
temp = numpy.ascontiguousarray(MyArray).view(numpy.dtype((numpy.void, MyArray.dtype.itemsize * MyArray.shape[1])))  # convert the 2d-array into an array of analyzable patterns
xt=numpy.ascontiguousarray(x).view(numpy.dtype((numpy.void, x.dtype.itemsize * x.shape[0])))  # convert what you search into one analyzable pattern
numpy.sum(temp==xt)  # count of the searched pattern in the list of patterns