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pythonnumpyflatten
IssamLaradji
IssamLaradji

Reputation: 6855

How to flatten only some dimensions of a numpy array

Is there a quick way to "sub-flatten" or flatten only some of the first dimensions in a numpy array?

For example, given a numpy array of dimensions (50,100,25), the resultant dimensions would be (5000,25)

Upvotes: 250

Views: 247977

Answers (5)

Sherman
Sherman

Reputation: 555

numpy.vstack is perfect for this situation

import numpy as np
arr = np.ones((50,100,25))
np.vstack(arr).shape
> (5000, 25)

I prefer to use stack, vstack or hstack over reshape because reshape just scans through the data and seems to brute-force it into the desired shape. This can be problematic if you are e.g. going to take column averages.

Here's an illustration of what I mean. Suppose we have the following array

>>> arr.shape
(2, 3, 4)
>>> arr 
array([[[1, 2, 3, 4],
        [1, 2, 3, 4],
        [1, 2, 3, 4]],

       [[7, 7, 7, 7],
        [7, 7, 7, 7],
        [7, 7, 7, 7]]])

We apply both methods to get an array of shape (3,8)

>>> arr.reshape((3,8)).shape
(3, 8)
>>> np.hstack(arr).shape 
(3, 8)

However if we look at how they have been reshaped in each case, the hstack would allow us to take column sums that we could also have calculated from the original array. With reshape this isn't possible.

>>> arr.reshape((3,8))
array([[1, 2, 3, 4, 1, 2, 3, 4],
       [1, 2, 3, 4, 7, 7, 7, 7],
       [7, 7, 7, 7, 7, 7, 7, 7]])
>>> np.hstack(arr)
array([[1, 2, 3, 4, 7, 7, 7, 7],
       [1, 2, 3, 4, 7, 7, 7, 7],
       [1, 2, 3, 4, 7, 7, 7, 7]])

Upvotes: 19

kmario23
kmario23

Reputation: 61305

An alternative approach is to use numpy.resize() as in:

In [37]: shp = (50,100,25)
In [38]: arr = np.random.random_sample(shp)
In [45]: resized_arr = np.resize(arr, (np.prod(shp[:2]), shp[-1]))
In [46]: resized_arr.shape
Out[46]: (5000, 25)

# sanity check with other solutions
In [47]: resized = np.reshape(arr, (-1, shp[-1]))
In [48]: np.allclose(resized_arr, resized)
Out[48]: True

Upvotes: 6

KeithWM
KeithWM

Reputation: 1345

A slight generalization to Peter's answer -- you can specify a range over the original array's shape if you want to go beyond three dimensional arrays.

e.g. to flatten all but the last two dimensions:

arr = numpy.zeros((3, 4, 5, 6))
new_arr = arr.reshape(-1, *arr.shape[-2:])
new_arr.shape
# (12, 5, 6)

EDIT: A slight generalization to my earlier answer -- you can, of course, also specify a range at the beginning of the of the reshape too:

arr = numpy.zeros((3, 4, 5, 6, 7, 8))
new_arr = arr.reshape(*arr.shape[:2], -1, *arr.shape[-2:])
new_arr.shape
# (3, 4, 30, 7, 8)

Upvotes: 77

Alexander
Alexander

Reputation: 12775

Take a look at numpy.reshape .

>>> arr = numpy.zeros((50,100,25))
>>> arr.shape
# (50, 100, 25)

>>> new_arr = arr.reshape(5000,25)
>>> new_arr.shape   
# (5000, 25)

# One shape dimension can be -1. 
# In this case, the value is inferred from 
# the length of the array and remaining dimensions.
>>> another_arr = arr.reshape(-1, arr.shape[-1])
>>> another_arr.shape
# (5000, 25)

Upvotes: 219

Peter
Peter

Reputation: 13475

A slight generalization to Alexander's answer - np.reshape can take -1 as an argument, meaning "total array size divided by product of all other listed dimensions":

e.g. to flatten all but the last dimension:

>>> arr = numpy.zeros((50,100,25))
>>> new_arr = arr.reshape(-1, arr.shape[-1])
>>> new_arr.shape
# (5000, 25)

Upvotes: 123

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