Numpy/Scipy: How to re-construct an ndarray?

Question

I am working on a classification problem.
I have a ndarray of shape (604329, 33) where there are 32 features and one column for label:

>>> n_data.shape   
(604329, 33)

The third column of this ndarray is a label with 0 and 1 .
I need to move this third column as the last column so that it is easier to work with when slicing is needed.

Question:
Is there a way to reconstruct the ndarray where we can move this third column as the last column?

Answer 1

如果我理解正确，您想执行以下操作：

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

Answer 2

The following will do it:

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

where x is your ndarray .

Answer 3

As an alternative to aix 's solution, you could slice the array directly, without 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]])

It's a bit faster for small arrays:

>>> %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

But actually, for larger arrays, it's slower.

>>> 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

If you don't need to preserve the order of the columns, (ie if you can use roll ) then Mr. E 's solution is fastest for large a :

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