NumPy resize method

Question

Can anyone explain this to me? (Python 3.3.2, numpy 1.7.1):

>>> a = np.array([[1,2],[3,4]])
>>> a    # just a peek
array([[1, 2],
       [3, 4]])
>>> a.resize(3,2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: cannot resize an array references or is referenced
by another array in this way.  Use the resize function
>>> a = np.array([[1,2],[3,4]])
>>> a.resize(3,2)
>>> a
array([[1, 2],
       [3, 4],
       [0, 0]])
>>> a = np.array([[1,2],[3,4]])
>>> print(a)   # look properly this time
[[1 2]
 [3 4]]
>>> a.resize(3,2)
>>> a
array([[1, 2],
       [3, 4],
       [0, 0]])

Why does taking a peek at the array create a reference to it? (or, at least, why does that reference persist after I'm done looking?) Also, is it just me or does that Exception need a bit of a rewrite?

It's not just you. The exception message is missing a "which" in front of "references" and could use a healthy dose of punctuation and capitalization. — Neil Traft
– Neil Traft, Commented Feb 10, 2014 at 7:04

jonrsharpe · Accepted Answer · 2013-12-22 14:42:39Z

16

From the documentation (emphasis mine):

The purpose of the reference count check is to make sure you do not use this array as a buffer for another Python object and then reallocate the memory. However, reference counts can increase in other ways so if you are sure that you have not shared the memory for this array with another Python object, then you may safely set refcheck to False.

Your "peek", unlike print, does not decrement the reference count afterwards. This is because, in the interpreter, the result of the last computation is assigned to _. Try:

print(_) # shows array
a.resize((3, 2), refcheck=False) # works

Alternatively, if you do any other computation (e.g. just 1 + 2) in-between, this will dereference your array from _.

edited Dec 22, 2013 at 14:42

answered Dec 22, 2013 at 14:34

jonrsharpe

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3 Comments

xnx Over a year ago

Make sense (more so than the error message). Perhaps it shouldn't, though... I understood that creating a view on an array will create a reference - what other operations do so?

jonrsharpe Over a year ago

In this case, it's only because you're using the interpreter, which assigns the result of the last computation to _. I agree that the error message isn't terribly helpful!

xnx Over a year ago

Got it! I actually didn't know about _ until now. Sneaky... Thanks!

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