我如何计算以下数组中的0和1的数量?
y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
y.count(0)为:
numpy。Ndarray对象没有属性计数
我如何计算以下数组中的0和1的数量?
y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
y.count(0)为:
numpy。Ndarray对象没有属性计数
当前回答
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]}
其他回答
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。
我会使用np.where:
how_many_0 = len(np.where(a==0.)[0])
how_many_1 = len(np.where(a==1.)[0])
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
如果你确切地知道你要找的数字,你可以使用下面的方法;
lst = np.array([1,1,2,3,3,6,6,6,3,2,1])
(lst == 2).sum()
返回2在数组中出现的次数。