How can I tell if NumPy creates a view or a copy?

Question

For a minimal working example, let’s digitize a 2D array. numpy.digitize requires a 1D array:

import numpy as np
N = 200
A = np.random.random((N, N))
X = np.linspace(0, 1, 20)
print np.digitize(A.ravel(), X).reshape((N, N))

Now the documentation says:

… A copy is made only if needed.

How do I know if the ravel copy it is “needed” in this case? In general – is there a way I can determine if a particular operation creates a copy or a view?

Asked By: Hooked

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Answer 1

This question is very similar to a question that I asked a while back:

You can check the base attribute.

a = np.arange(50)
b = a.reshape((5, 10))
print (b.base is a)

However, that’s not perfect. You can also check to see if they share memory using np.may_share_memory.

print (np.may_share_memory(a, b))

There’s also the flags attribute that you can check:

print (b.flags['OWNDATA'])  #False -- apparently this is a view
e = np.ravel(b[:, 2])
print (e.flags['OWNDATA'])  #True -- Apparently this is a new numpy object.

But this last one seems a little fishy to me, although I can’t quite put my finger on why…

Answered By: mgilson

Answer 2

In the documentation for reshape there is some information about how to ensure an exception if a view cannot be made:

It is not always possible to change the shape of an array without copying the data. If you want an error to be raised if the data is copied, you should assign the new shape to the shape attribute of the array:

>>> a = np.zeros((10, 2))
# A transpose make the array non-contiguous
>>> b = a.T
# Taking a view makes it possible to modify the shape without modiying the
# initial object.
>>> c = b.view()
>>> c.shape = (20)
AttributeError: incompatible shape for a non-contiguous array

This is not exactly an answer to your question, but in certain cases it may be just as useful.

Answered By: amicitas

How can I tell if NumPy creates a view or a copy?

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