NumPy MemoryError when using array constructors

Question

I'm creating a couple of arrays using numpy and list constructors, and I can't figure out why this is failing. My code is:

import numpy as np
A = np.ndarray( [i for i in range(10)] ) # works fine
B = np.ndarray( [i**2 for i in range(10)] ) # fails, with MemoryError

I've also tried just B = [i**2 for i in range(10)] which works, but I need it to be an ndarray. I don't see why the normal constructor would work but calling a function wouldn't. As far as I understand, the ndarray constructor shouldn't even see the inside of that, it should get a length 10 list with ints in it for both.

Use np.array([...]). np.ndarray is an advanced constructor with different parameters. np.array is the one we normally use. — hpaulj
– hpaulj, Commented Sep 17, 2016 at 19:26
Or use np.zeros (or ones or empty) if trying to create an array with a list of dimensions. — hpaulj
– hpaulj, Commented Sep 17, 2016 at 19:30
As @hpaulj suggests, normally you should use np.array() to create a new array from a list. The first argument of ndarray is the shape of the array, not the data to put in the array. So you are requesting an array with shape [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]. It is interesting that it generates a MemoryError. With that 0 in there, the total size of the array would actually be 0. — Warren Weckesser
– Warren Weckesser, Commented Sep 17, 2016 at 19:35

hpaulj · Accepted Answer · 2016-09-17 19:36:18Z

2

http://docs.scipy.org/doc/numpy/reference/generated/numpy.ndarray.html is a low-level method that we normally don't use. Its first argument is the shape.

In [98]: np.ndarray([x for x in range(3)])
Out[98]: array([], shape=(0, 1, 2), dtype=float64)
In [99]: np.ndarray([x**2 for x in range(3)])
Out[99]: array([], shape=(0, 1, 4), dtype=float64)

Normally use zeros or ones to construct a blank array of a given shape:

In [100]: np.zeros([x**2 for x in range(3)])
Out[100]: array([], shape=(0, 1, 4), dtype=float64)

Use np.array if you want to turn a list into an array:

In [101]: np.array([x for x in range(3)])
Out[101]: array([0, 1, 2])
In [102]: np.array([x**2 for x in range(3)])
Out[102]: array([0, 1, 4])

You can generate the range numbers, and then perform the math on the whole array (without iteration):

In [103]: np.arange(3)**2
Out[103]: array([0, 1, 4])

answered Sep 17, 2016 at 19:36

hpaulj

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1 Comment

user5563109 Over a year ago

Thank you! I guess I knew the type was ndarray so I assumed that would be the constructor.

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