假設(shè)我有 2 個 numpy 數(shù)組；arr1 = np.array([ [1, 2, 3, 4], [2, 3, 1, 4], [2, 4, 1, 5], ...)arr2 = np.array([ [2, 4, 1, 5], [2, 1, 3, 5], [1, 2, 3, 4], ...)arr1我想獲得和的行的交集arr2。我努力了：intersect = np.intersect1d(arr1, arr2)它返回array([1, 2, 3, 4, 5])這意味著按元素相交。我想按行完成這個。它應(yīng)該返回；array( [2, 4, 1, 5], [1, 2, 3, 5])
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TA貢獻(xiàn)1856條經(jīng)驗(yàn) 獲得超11個贊
In [3]: arr1 = np.array([
...: [1, 2, 3, 4],
...: [2, 3, 1, 4],
...: [2, 4, 1, 5],
...: ])
...:
...: arr2 = np.array([
...: [2, 4, 1, 5],
...: [2, 1, 3, 5],
...: [1, 2, 3, 4],
...: ])
廣播平等后跟all和的適當(dāng)組合any：
In [8]: (arr1[:,None,:]==arr2[None,:,:]).shape
Out[8]: (3, 3, 4)
In [9]: (arr1[:,None,:]==arr2[None,:,:]).all(axis=2)
Out[9]:
array([[False, False, True],
[False, False, False],
[ True, False, False]])
In [10]: (arr1[:,None,:]==arr2[None,:,:]).all(axis=2).any(axis=0)
Out[10]: array([ True, False, True])
In [12]: arr1[_]
Out[12]:
array([[1, 2, 3, 4],
[2, 4, 1, 5]])
帶套
In [19]: set([tuple(row) for row in arr1])
Out[19]: {(1, 2, 3, 4), (2, 3, 1, 4), (2, 4, 1, 5)}
In [20]: set([tuple(row) for row in arr2])
Out[20]: {(1, 2, 3, 4), (2, 1, 3, 5), (2, 4, 1, 5)}
In [21]: _19.intersection(_20)
Out[21]: {(1, 2, 3, 4), (2, 4, 1, 5)}
===
如果我擴(kuò)展arr2到 4 行：
...: arr2 = np.array([
...: [2, 4, 1, 5],
...: [2, 1, 3, 5],
...: [1, 2, 3, 4],
...: [1, 1, 1, 1],
...: ])
In [34]: (arr1[:,None,:]==arr2[None,:,:]).all(axis=2).any(axis=0)
Out[34]: array([ True, False, True, False])
anyon 0 產(chǎn)生一個 4 元素數(shù)組，它必須用于索引arr2（不像arr1我原來那樣）：
In [35]: arr2[_]
Out[35]:
array([[2, 4, 1, 5],
[1, 2, 3, 4]])
或者any沿著另一個軸：
In [36]: (arr1[:,None,:]==arr2[None,:,:]).all(axis=2).any(axis=1)
Out[36]: array([ True, False, True])
In [37]: arr1[_]
Out[37]:
array([[1, 2, 3, 4],
[2, 4, 1, 5]])
產(chǎn)生all（在本例中）一個 (3,4) 數(shù)組：
In [38]: (arr1[:,None,:]==arr2[None,:,:]).all(axis=2)
Out[38]:
array([[False, False, True, False],
[False, False, False, False],
[ True, False, False, False]])
any可以減少行或列。
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