torch.unique¶
- torch.unique(input, sorted=True, return_inverse=False, return_counts=False, dim=None) Tuple[Tensor, Tensor, Tensor] [來源]¶
返回輸入張量中的唯一元素。
注意
此函數不同於
torch.unique_consecutive()
的地方在於,此函數也會消除非連續的重複值。注意
目前在 CUDA 實現和 CPU 實現中,無論 sort 參數為何,torch.unique 始終會在開始時對張量進行排序。 排序可能會很慢,因此如果您的輸入張量已經排序,建議使用
torch.unique_consecutive()
,它可以避免排序。- 參數
- 返回
包含以下內容的張量或張量元組
output (Tensor):唯一純量元素的輸出列表。
inverse_indices (Tensor):(可選)如果
return_inverse
為 True,則會多返回一個張量(與輸入的形狀相同),表示原始輸入中的元素映射到輸出中的索引; 否則,此函數只會返回單個張量。counts (Tensor):(可選)如果
return_counts
為 True,則會多返回一個張量(與輸出的形狀相同,如果指定了 dim,則與 output.size(dim) 相同),表示每個唯一值或張量的出現次數。
- 返回類型
範例
>>> output = torch.unique(torch.tensor([1, 3, 2, 3], dtype=torch.long)) >>> output tensor([1, 2, 3]) >>> output, inverse_indices = torch.unique( ... torch.tensor([1, 3, 2, 3], dtype=torch.long), sorted=True, return_inverse=True) >>> output tensor([1, 2, 3]) >>> inverse_indices tensor([0, 2, 1, 2]) >>> output, inverse_indices = torch.unique( ... torch.tensor([[1, 3], [2, 3]], dtype=torch.long), sorted=True, return_inverse=True) >>> output tensor([1, 2, 3]) >>> inverse_indices tensor([[0, 2], [1, 2]]) >>> a = torch.tensor([ ... [ ... [1, 1, 0, 0], ... [1, 1, 0, 0], ... [0, 0, 1, 1], ... ], ... [ ... [0, 0, 1, 1], ... [0, 0, 1, 1], ... [1, 1, 1, 1], ... ], ... [ ... [1, 1, 0, 0], ... [1, 1, 0, 0], ... [0, 0, 1, 1], ... ], ... ]) >>> # If we call `torch.unique(a, dim=0)`, each of the tensors `a[idx, :, :]` >>> # will be compared. We can see that `a[0, :, :]` and `a[2, :, :]` match >>> # each other, so one of them will be removed. >>> (a[0, :, :] == a[2, :, :]).all() tensor(True) >>> a_unique_dim0 = torch.unique(a, dim=0) >>> a_unique_dim0 tensor([[[0, 0, 1, 1], [0, 0, 1, 1], [1, 1, 1, 1]], [[1, 1, 0, 0], [1, 1, 0, 0], [0, 0, 1, 1]]]) >>> # Notice which sub-tensors from `a` match with the sub-tensors from >>> # `a_unique_dim0`: >>> (a_unique_dim0[0, :, :] == a[1, :, :]).all() tensor(True) >>> (a_unique_dim0[1, :, :] == a[0, :, :]).all() tensor(True) >>> # For `torch.unique(a, dim=1)`, each of the tensors `a[:, idx, :]` are >>> # compared. `a[:, 0, :]` and `a[:, 1, :]` match each other, so one of >>> # them will be removed. >>> (a[:, 0, :] == a[:, 1, :]).all() tensor(True) >>> torch.unique(a, dim=1) tensor([[[0, 0, 1, 1], [1, 1, 0, 0]], [[1, 1, 1, 1], [0, 0, 1, 1]], [[0, 0, 1, 1], [1, 1, 0, 0]]]) >>> # For `torch.unique(a, dim=2)`, the tensors `a[:, :, idx]` are compared. >>> # `a[:, :, 0]` and `a[:, :, 1]` match each other. Also, `a[:, :, 2]` and >>> # `a[:, :, 3]` match each other as well. So in this case, two of the >>> # sub-tensors will be removed. >>> (a[:, :, 0] == a[:, :, 1]).all() tensor(True) >>> (a[:, :, 2] == a[:, :, 3]).all() tensor(True) >>> torch.unique(a, dim=2) tensor([[[0, 1], [0, 1], [1, 0]], [[1, 0], [1, 0], [1, 1]], [[0, 1], [0, 1], [1, 0]]])