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
使用numpy怎么样?count_non0,类似的
>>> import numpy as np
>>> y = np.array([1, 2, 2, 2, 2, 0, 2, 3, 3, 3, 0, 0, 2, 2, 0])
>>> np.count_nonzero(y == 1)
1
>>> np.count_nonzero(y == 2)
7
>>> np.count_nonzero(y == 3)
3
其他回答
using numpy.count
$ a = [0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]
$ np.count(a, 1)
这个函数返回变量在数组中出现的次数:
def count(array,variable):
number = 0
for i in range(array.shape[0]):
for j in range(array.shape[1]):
if array[i,j] == variable:
number += 1
return number
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()
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]}