Python中数组中的非零不同元素和索引

Question

我有一个shape (3,3)的数组A 。 我想识别所有非零不同元素及其索引。 我提出了预期的输出。

import numpy as np
A=np.array([[10,2,0],[2,20,1.3],[0,1.3,30]])

预期的输出是

Distinct=[10,2,20,1.3,30]
Indices=[[(0,0)],[(0,1),(1,0)],[(1,1)],[(1,2),(2,1)],[(2,2)]]

Answer 1

也许不是最漂亮的选择，但这确实有效。

import numpy as np
A=np.array([[10,2,0],[2,20,1.3],[0,1.3,30]])

Distinct=sorted(set(list(np.reshape(A,A.shape[0]*A.shape[1]))))
Distinct = [x for x in Distinct if x!=0]
Indices = [[] for x in Distinct]

for i,x in enumerate(Distinct):
    for j in range(A.shape[0]):
        for k in range(A.shape[1]):
            if x==A[j,k]:
                Indices[i].append((j,k))

或者一个 numpy 解决方案，从 Kilian 的解决方案中汲取灵感：

import numpy as np
A=np.array([[10,2,0],[2,20,1.3],[0,1.3,30]])
Distinct = np.unique(A[A!=0])
Indices = [np.argwhere(A==x) for x in Distinct]

Answer 2

干得好！

import numpy as np

A = np.array([[10,2,0], [2,20,1.3], [0,1.3,30]])
indices = np.argwhere(A!=0)
distinct = np.unique(A[A!=0])

print(indices)
print(distinct)