Question

我想告诉你数组索引列的位置不一样。

import numpy as np
array1 = np.array(list(np.zeros(10))+list(np.ones(10)))
array2 = np.array(list(np.random.randint(2, size=10))+list(np.random.randint(2, size=10)))
matches = array1 == array2
section_sums = np.bincount(np.arange(matches.size)//10,matches)
att = int(section_sums[0])
att2 = int(section_sums[1])

print section_sums
print 'first  10 : '+ str(att)
print 'second 10 : '+ str(att2)

示例：

Array1:
[ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  1.  1.  1.  1.  1.  1.  1.  1. 1. 1.]

Array2:
[ 0.  1.  0 . 1.  1.  1.  0.  1.  0.  1.  1.  1.  1.  0.  1.  0.  1.  1. 1. 0.]

我想要输出：

in section 1 index is not the same: 2,4,5,6,8,10
in section 2 index is not the same: 4,6,10

Answer 1

如果将阵列拆分为两个部分，则可以比较它们。

In [18]: a = np.array(np.split(a, [10]))

In [19]: b = np.array(np.split(b, [10]))

In [23]: ind, items = np.where(a != b)

In [25]: items[ind==0] + 1
Out[25]: array([ 2,  4,  5,  6,  8, 10])

In [26]: items[ind==1] + 1
Out[26]: array([ 4,  6, 10])

Answer 2

这是一种方法 -

idx = np.flatnonzero(~matches)
cut_idx = np.unique(idx//10,return_index=True)[1]
out = np.split(np.mod(idx,10)+1,cut_idx)[1:]

给定输入数组的样本运行 -

In [182]: matches = array1 == array2
     ...: idx = np.flatnonzero(~matches)
     ...: cut_idx = np.unique(idx//10,return_index=True)[1]
     ...: out = np.split(np.mod(idx,10)+1,cut_idx)[1:]
     ...: 

In [183]: out
Out[183]: [array([ 2,  4,  5,  6,  8, 10]), array([ 4,  6, 10])]

比较python中的第二节numpy数组，并显示不同的索引列

2 个答案: