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.

另一个简单的解决方案可能是使用numpy.count_nonzero():

import numpy as np
y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
y_nonzero_num = np.count_nonzero(y==1)
y_zero_num = np.count_nonzero(y==0)
y_nonzero_num
4
y_zero_num
8

不要让这个名字误导了你，如果你像例子中那样使用布尔值，它会达到目的的。

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

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)) !

使用numpy.unique:

import numpy
a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
unique, counts = numpy.unique(a, return_counts=True)

>>> dict(zip(unique, counts))
{0: 7, 1: 4, 2: 1, 3: 2, 4: 1}

使用collections.Counter的非numpy方法;

import collections, numpy
a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
counter = collections.Counter(a)

>>> counter
Counter({0: 7, 1: 4, 3: 2, 2: 1, 4: 1})

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