Question

我正在研究分类问题我有一个ndarray形状(604329, 33)，其中有32个要素和一个标签列：

>>> n_data.shape   
(604329, 33)

此ndarray的第三列是带有0和1的标签我需要将第三列作为最后一列移动，以便在需要切片时更容易使用。

问题：
有没有办法重建ndarray我们可以将第三列作为最后一列移动？

Answer 1

如果我理解正确，你想做：

my_array = numpy.roll(my_array,-3,axis=1)

Answer 2

以下将会这样做：

x = np.hstack((x[:,:3],x[:,4:],x[:,3:4]))

其中x是您的ndarray。

Answer 3

作为aix解决方案的替代方案，您可以直接对数组进行切片，而不使用hstack。

>>> a = numpy.array([range(33) for _ in range(4)])
>>> indices = range(33)
>>> indices.append(indices.pop(3))
>>> a[:,indices]
array([[ 0,  1,  2,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17,
        18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32,  3],
       [ 0,  1,  2,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17,
        18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32,  3],
       [ 0,  1,  2,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17,
        18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32,  3],
       [ 0,  1,  2,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17,
        18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32,  3]])

小阵列的速度要快一些：

>>> %timeit numpy.hstack((a[:,:3], a[:,4:], a[:, 3:4]))
100000 loops, best of 3: 19.1 us per loop
>>> %timeit indices = range(33); indices.append(indices.pop(3)); a[:,indices]
100000 loops, best of 3: 14 us per loop

但实际上，对于较大的数组，它的速度较慢。

>>> a = numpy.array([range(33) for _ in range(600000)])
>>> %timeit numpy.hstack((a[:,:3], a[:,4:], a[:, 3:4]))
1 loops, best of 3: 385 ms per loop
>>> %timeit indices = range(33); indices.append(indices.pop(3)); a[:,indices]
1 loops, best of 3: 670 ms per loop

如果您不需要保留列的顺序（例如，如果您可以使用roll）那么Mr. E的解决方案对于大a来说是最快的：

>>> %timeit numpy.roll(a, -3, axis=1)
10 loops, best of 3: 120 ms per loop

Numpy / Scipy：如何重建一个ndarray？

3 个答案: