我有一個如下所示的 3-d 矩陣，并希望沿軸 1 取最大值，并將所有非最大值保持為零。A = np.random.rand(3,3,2) [[[0.34444547, 0.50260393], [0.93374423, 0.39021899], [0.94485653, 0.9264881 ]], [[0.95446736, 0.335068 ], [0.35971558, 0.11732342], [0.72065402, 0.36436023]], [[0.56911013, 0.04456443], [0.17239996, 0.96278067], [0.26004909, 0.06767436]]]想要的結(jié)果： [[0 , 0 ], [0 , 0 ], [0.94485653, 0.9264881]], [[0.95446736, 0 ], [0 , 0 ], [0 , 0.36436023]], [[0.56911013, 0 ], [0 , 0.96278067], [0 , 0 ]]])我試過了：B = np.zeros_like(A) #return matrix of zero with same shape as Amax_idx = np.argmax(A, axis=1) #index along axis 1 with max value array([[2, 0], [2, 2], [0, 2], [0, 1]])C = np.max(A, axis=1, keepdims = True) #gives a (4,1,2) matrix of max value along axis 1 array([[[0.95377958, 0.92940525]], [[0.94485653, 0.9264881 ]], [[0.95446736, 0.36436023]], [[0.56911013, 0.96278067]]])但我不知道如何將這些想法結(jié)合在一起以獲得我想要的輸出。請幫忙！！
1 回答
Cats萌萌
TA貢獻(xiàn)1805條經(jīng)驗(yàn) 獲得超9個贊
你可以得到3維指數(shù)的最大值從max_idx。中的值max_idx是最大值沿軸 1 的索引。有六個值，因?yàn)槟钠渌S是 3 和 2 (3 x 2 = 6)。您只需要了解 numpy 通過它們獲取其他每個軸的索引的順序。您首先遍歷最后一個軸：
d0, d1, d2 = A.shape
a0 = [i for i in range(d0) for _ in range(d2)] # [0, 0, 1, 1, 2, 2]
a1 = max_idx.flatten() # [2, 2, 0, 2, 0, 1]
a2 = [k for _ in range(d0) for k in range(d2)] # [0, 1, 0, 1, 0, 1]
B[a0, a1, a2] = A[a0, a1, a2]
輸出：
array([[[0. , 0. ],
[0. , 0. ],
[0.94485653, 0.9264881 ]],
[[0.95446736, 0. ],
[0. , 0. ],
[0. , 0.36436023]],
[[0.56911013, 0. ],
[0. , 0.96278067],
[0. , 0. ]]])
添加回答
舉報(bào)
0/150
提交
取消