CODEWITHSUNDEEP
pythonarraysnumpyscipy
Nick Alger
Nick Alger

Reputation: 1104

Numpy reshape acts different on copied vs. uncopied arrays

I've come across seemingly inconsistent results when flattening certain numpy arrays with numpy.reshape. Sometimes if I reshape an array it returns a 2D array with one row, whereas if I first copy the array then do the exact same operation, it returns a 1D array.

This seems to happen primarily when combining numpy arrays with scipy arrays, and creates alignment problems when I want to later multiply the flattened array by a matrix.

For example, consider the following code:

import numpy as np
import scipy.sparse as sps

n = 10
A = np.random.randn(n,n)
I = sps.eye(n)
X = I+A

x1 = np.reshape(X, -1)
x2 = np.reshape(np.copy(X), -1)

print 'x1.shape=', x1.shape
print 'x2.shape=', x2.shape

When run it prints:

x1.shape= (1, 100)
x2.shape= (100,)

The same thing happens with numpy.flatten(). What is going on here? Is this behavior intentional?

Upvotes: 3

Views: 189

Answers (1)

user2357112
user2357112

Reputation: 281252

You added together a sparse matrix object and a normal ndarray:

X = I+A

The result is a dense matrix object, an instance of np.matrix, not a normal ndarray.

This:

np.reshape(X, -1)

ends up returning a matrix, which can't be less than 2D.

This:

np.reshape(np.copy(X), -1)

makes a normal ndarray in np.copy(X), so you get a 1D output from np.reshape.

Be very careful at all times about whether you're dealing with sparse matrices, dense matrices, or standard ndarrays. Avoid np.matrix whenever possible.

Upvotes: 5

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