4

I have a numpy array:

array([], shape=(0, 4), dtype=float64)

How can I remove this array in a multidimensional array? I tried 

import numpy as np

if array == []:
    np.delete(array)

But, the multidimensional array still has this empty array.

EDIT: The input is

new_array = [array([], shape=(0, 4), dtype=float64), 
   array([[-0.97,  0.99, -0.98, -0.93 ],
   [-0.97, -0.99,  0.59, -0.93 ],
   [-0.97,  0.99, -0.98, -0.93 ],
   [ 0.70 ,  1,  0.60,  0.65]]), array([[-0.82,  1,  0.61, -0.63],
   [ 0.92, -1,  0.77,  0.88],
   [ 0.92, -1,  0.77,  0.88],
   [ 0.65, -1,  0.73,  0.85]]), array([], shape=(0, 4), dtype=float64)]

The expected output after removing the empty arrays is:

new array = [array([[-0.97,  0.99, -0.98, -0.93 ],
   [-0.97, -0.99,  0.59, -0.93 ],
   [-0.97,  0.99, -0.98, -0.93 ],
   [ 0.70 ,  1,  0.60,  0.65]]), 
   array([[-0.82,  1,  0.61, -0.63],
   [ 0.92, -1,  0.77,  0.88],
   [ 0.92, -1,  0.77,  0.88],
   [ 0.65, -1,  0.73,  0.85]])]
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  • 1
    There is no shape keyword in numpy.array what function is that array ? Commented Jun 23, 2015 at 18:13
  • I just printed new_array which I constructed using new_array.append(array). And that's the output of the print. Commented Jun 23, 2015 at 18:16
  • 1
    @AnandSKumar: That's a NumPy ndarray. The shape and dtype are included in the repr when they can't be inferred (for example when there's a zero-length dimension, as here). np.zeros((0, 4), dtype=float) would have a repr like that, for example. Commented Jun 23, 2015 at 18:16

3 Answers 3

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4

new_array, as printed, looks like a list of arrays. And even if it were an array, it would be a 1d array of dtype=object.

==[] is not the way to check for an empty array:

In [10]: x=np.zeros((0,4),float)
In [11]: x
Out[11]: array([], shape=(0, 4), dtype=float64)
In [12]: x==[]
Out[12]: False
In [14]: 0 in x.shape  # check if there's a 0 in the shape
Out[14]: True

Check the syntax for np.delete. It requires an array, an index and an axis, and returns another array. It does not operate in place.

If new_array is a list, a list comprehension would do a nice job of removing the [] arrays:

In [33]: alist=[x, np.ones((2,3)), np.zeros((1,4)),x]

In [34]: alist
Out[34]: 
[array([], shape=(0, 4), dtype=float64), array([[ 1.,  1.,  1.],
        [ 1.,  1.,  1.]]), array([[ 0.,  0.,  0.,  0.]]), array([], shape=(0, 4), dtype=float64)]

In [35]: [y for y in alist if 0 not in y.shape]
Out[35]: 
[array([[ 1.,  1.,  1.],
        [ 1.,  1.,  1.]]), array([[ 0.,  0.,  0.,  0.]])]

It would also work if new_array was a 1d array:

new_array=np.array(alist)
newer_array = np.array([y for y in new_array if 0 not in y.shape])

To use np.delete with new_array, you have to specify which elements:

In [47]: np.delete(new_array,[0,3])
Out[47]: 
array([array([[ 1.,  1.,  1.],
       [ 1.,  1.,  1.]]),
       array([[ 0.,  0.,  0.,  0.]])], dtype=object)

to find [0,3] you could use np.where:

np.delete(new_array,np.where([y.size==0 for y in new_array]))

Better yet, skip the delete and where and go with a boolean mask

new_array[np.array([y.size>0 for y in new_array])]

I don't think there's a way of identifying these 'emtpy' arrays without a list comprehension, since you have to check the shape or size property, not the element's data. Also there's a limit as to what kinds of math you can do across elements of an object array. It's more like a list than a 2d array.

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3 Comments

I prefer x.size == 0 to 0 in x.shape: it seems more direct.
Interesting... it got me thinking about the 'empty list' Take ... >>> x = np.asanyarray( [] , dtype='float64') >>> y = a.reshape((0,4,4)) >>>print("\nx: {!r:} shape:{!r:}\ny: {!r:}".format(x,x.shape,y)) if you try to merge the first and the last array with the middle, you end up with only the middle as long as you reshape the original array derived from the empty list to match the shape of the middle array. I can find no documentation discussing combining arrays of 0 size, I know you can't if the shape differs.
np.concatenate([np.ones((2,4)),np.zeros((0,4))]) effectively removes the (0,4) array. As long as the other dimensions match, concatenate has no problems with 0 dimensions.
1

I had initially an array (3,11,11) and after a multprocessing using pool.map my array was transformed in a list like this: 

[array([], shape=(0, 11, 11), dtype=float64),
array([[[ 0.35318114,  0.36152024,  0.35572945,  0.34495254,  0.34169853,
       0.36553977,  0.34266126,  0.3492261 ,  0.3339431 ,  0.34759375,
       0.33490712],...

if a convert this list in an array the shape was (3,), so I used: 

myarray = np.vstack(mylist)

and this returned my first 3d array with the original shape (3,11,11).

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0

Delete takes the multidimensional array as a parameter. Then you need to specify the subarray to delete and the axis it's on. See http://docs.scipy.org/doc/numpy/reference/generated/numpy.delete.html

np.delete(new_array,<obj indicating subarray to delete (perhaps an array of integers in your case)>, 0)

Also, note that the deletion is not in-place.

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