Question

对于此矩阵K =

 [[-1.  1.  0.]
 [ 0.  0.  0.]
 [ 0. -1.  1.]
 [ 0.  0.  0.]
 [ 0. -1.  1.]
 [ 0.  0.  0.]]

任务是在数组中存储非零唯一行的索引（此处答案为{0,2}），以便

K([0,2],:)

可用于线性代数运算。我的尝试是：

myList = []    
for i in range(len(K)): #generate pairs
    for j in range(i+1,len(K)):  #travel down each other rows
        if np.array_equal(K[i],K[j]) and np.any(K[i] != 0, axis=1) and np.any(K[j] != 0, axis=1):
        myList.append(K[i])
        print ('indices of similar-non-zeros rows are\n',(i, j)),
        elif not np.array_equal(K[i],K[j]) and np.any(K[i] != 0,axis=1) and np.any(K[j] != 0, axis=1): 
        myList.append(K[i])
        print ('indices of non-similar-non-zeros rows are\n',(i, j)),
        else: 
            continue

new_K = np.asmatrix(np.asarray(myList))
new_new_K = np.unique(new_K,axis=0)
print('Now K is \n',new_new_K)

答案是：

    new_new_K = [[-1.  1.  0.]
                 [ 0. -1.  1.]]

问题1：如何以pythonic方式进行。以上是具有矩阵存储限制的替代解决方案，但更优选地将索引存储在阵列中。

Answer 1

您可以使用简单的rating_position_recalculate_service: class: App\Service\RatingPositionRecalculateService autowire: false arguments: $entityManager: '@doctrine.orm.entity_manager' $producer: '@old_sound_rabbit_mq.rating_position_producer'循环for来实现此目的。

enumerate

<强>结果

import numpy as np

A = np.array([[-1,  1,  0],
              [ 0,  0,  0],
              [ 0, -1,  1],
              [ 0,  0,  0],
              [ 0, -1,  1],
              [ 0,  0,  0]])

seen = {(0, 0, 0)}
res = []

for idx, row in enumerate(map(tuple, A)):
    if row not in seen:
        res.append(idx)
        seen.add(row)

示例＃2

print(A[res])

[[-1  1  0]
 [ 0 -1  1]]

Answer 2

您可以使用np.unique及其axis参数来获取起始唯一行索引，然后过滤掉其对应行为全零的唯一一行索引，如下所示 -

def unq_row_indices_wozeros(a):
    # Get unique rows and their first occuring indices
    unq, idx = np.unique(a, axis=0, return_index=1)

    # Filter out the index, the corresponding row of which is ALL 0s
    return idx[(unq!=0).any(1)]

示例运行 -

In [53]: # Setup input array with few all zero rows and duplicates
    ...: np.random.seed(0)
    ...: a = np.random.randint(0,9,(10,3))
    ...: a[[2,5,7]] = 0
    ...: a[4] = a[1]
    ...: a[8] = a[3]

In [54]: a
Out[54]: 
array([[5, 0, 3],
       [3, 7, 3],
       [0, 0, 0],
       [7, 6, 8],
       [3, 7, 3],
       [0, 0, 0],
       [1, 5, 8],
       [0, 0, 0],
       [7, 6, 8],
       [2, 3, 8]])

In [55]: unq_row_indices_wozeros(a)
Out[55]: array([6, 9, 1, 0, 3])

# Sort those indices if needed
In [56]: np.sort(unq_row_indices_wozeros(a))
Out[56]: array([0, 1, 3, 6, 9])

Python：在将每行与矩阵中的每隔一行进行比较后，存储非零唯一行的索引

2 个答案: