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up vote 1 down vote favorite

Say I make a weird little array:

>>> a = np.array([[[1,2,3],4],[[4,5,6],5]])
>>> a
array([[[1, 2, 3], 4],
       [[4, 5, 6], 5]], dtype=object)

And then take a the first column as a slice:

>>> b = a[:,0]
>>> b
array([[1, 2, 3], [4, 5, 6]], dtype=object)
>>> b.shape
(2,)

Say I now want to reshape b so that its shape is (2,3):

>>> b.reshape((-1,3))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: total size of new array must be unchanged

I presume that numpy is treating each array in b as an object rather than an array in and of itself. The question is, is there a good way of doing the desired resize?

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2 Answers

up vote 1 down vote accepted

In your particular example, you could use numpy.vstack : 

import numpy as np


a = np.array([[[1,2,3],4],[[4,5,6],5]])
b = a[:,0]

c = np.vstack(b)
print c.shape # (2,3)

EDIT : Since your array a is not a real matrix but a collection of arrays (as pointed by wim ), you can also do the following : 

   b = np.array([ line for line in a[:,0]])
   print b.shape #(2,3)
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up vote 1 down vote

You can not change the shape of b in place, but you create a copy of the desired shape with np.vstack(b). I guess you probably already knew this much though.

Note that you did not make an array in the first column of a, if you examine type(a[0,0]) you will see that you actually have a list there. i.e. your slice a[:,0] is actually a column vector of two list objects, it isn't (and was never) an array in and of itself.

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You're right. However, it doesn't make a difference whether I put an array in a[0,0]. The result of the a[:,0] slice still returns an array of objects. Using vstack does do the trick though. Upvote for you and accept to georgesl for immediately pointing out the desired resulting shape after the vstack (which is what I was looking for). –  juniper- Oct 24 '12 at 13:15
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