Flattening a list of NumPy arrays?

Question

It appears that I have data in the format of a list of NumPy arrays (type() = np.ndarray):

[array([[ 0.00353654]]), array([[ 0.00353654]]), arra         


        

Answer 1

You could use numpy.concatenate, which as the name suggests, basically concatenates all the elements of such an input list into a single NumPy array, like so -

import numpy as np
out = np.concatenate(input_list).ravel()

If you wish the final output to be a list, you can extend the solution, like so -

out = np.concatenate(input_list).ravel().tolist()

Sample run -

In [24]: input_list
Out[24]: 
[array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]])]

In [25]: np.concatenate(input_list).ravel()
Out[25]: 
array([ 0.00353654,  0.00353654,  0.00353654,  0.00353654,  0.00353654,
        0.00353654,  0.00353654,  0.00353654,  0.00353654,  0.00353654,
        0.00353654,  0.00353654,  0.00353654])

Convert to list -

In [26]: np.concatenate(input_list).ravel().tolist()
Out[26]: 
[0.00353654,
 0.00353654,
 0.00353654,
 0.00353654,
 0.00353654,
 0.00353654,
 0.00353654,
 0.00353654,
 0.00353654,
 0.00353654,
 0.00353654,
 0.00353654,
 0.00353654]

Answer 2

I came across this same issue and found a solution that combines 1-D numpy arrays of variable length:

np.column_stack(input_list).ravel()

See numpy.column_stack for more info.

Example with variable-length arrays with your example data:

In [135]: input_list
Out[135]: 
[array([[ 0.00353654,  0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654]]),
 array([[ 0.00353654,  0.00353654,  0.00353654]])]

In [136]: [i.size for i in input_list]    # variable size arrays
Out[136]: [2, 1, 1, 3]

In [137]: np.column_stack(input_list).ravel()
Out[137]: 
array([ 0.00353654,  0.00353654,  0.00353654,  0.00353654,  0.00353654,
        0.00353654,  0.00353654])

Note: Only tested on Python 2.7.12

Answer 3

Another simple approach would be to use numpy.hstack() followed by removing the singleton dimension using squeeze() as in:

In [61]: np.hstack(list_of_arrs).squeeze()
Out[61]: 
array([0.00353654, 0.00353654, 0.00353654, 0.00353654, 0.00353654,
       0.00353654, 0.00353654, 0.00353654, 0.00353654, 0.00353654,
       0.00353654, 0.00353654, 0.00353654])

Answer 4

Can also be done by

np.array(list_of_arrays).flatten().tolist()

resulting in

[0.00353654, 0.00353654, 0.00353654, 0.00353654, 0.00353654, 0.00353654, 0.00353654, 0.00353654, 0.00353654, 0.00353654, 0.00353654, 0.00353654, 0.00353654]

---

Update

As @aydow points out in the comments, using numpy.ndarray.ravel can be faster if one doesn't care about getting a copy or a view

np.array(list_of_arrays).ravel()

Although, according to docs

When a view is desired in as many cases as possible, arr.reshape(-1) may be preferable.

In other words

np.array(list_of_arrays).reshape(-1)

The initial suggestion of mine was to use numpy.ndarray.flatten that returns a copy every time which affects performance.

Let's now see how the time complexity of the above-listed solutions compares using perfplot package for a setup similar to the one of the OP

import perfplot

perfplot.show(
    setup=lambda n: np.random.rand(n, 2),
    kernels=[lambda a: a.ravel(),
             lambda a: a.flatten(),
             lambda a: a.reshape(-1)],
    labels=['ravel', 'flatten', 'reshape'],
    n_range=[2**k for k in range(16)],
    xlabel='N')

Here flatten demonstrates piecewise linear complexity which can be reasonably explained by it making a copy of the initial array compare to constant complexities of ravel and reshape that return a view.

It's also worth noting that, quite predictably, converting the outputs .tolist() evens out the performance of all three to equally linear.

Answer 5

Another way using itertools for flattening the array:

import itertools

# Recreating array from question
a = [np.array([[0.00353654]])] * 13

# Make an iterator to yield items of the flattened list and create a list from that iterator
flattened = list(itertools.chain.from_iterable(a))

This solution should be very fast, see https://stackoverflow.com/a/408281/5993892 for more explanation.

If the resulting data structure should be a numpy array instead, use numpy.fromiter() to exhaust the iterator into an array:

# Make an iterator to yield items of the flattened list and create a numpy array from that iterator
flattened_array = np.fromiter(itertools.chain.from_iterable(a), float)

Docs for itertools.chain.from_iterable(): https://docs.python.org/3/library/itertools.html#itertools.chain.from_iterable

Docs for numpy.fromiter(): https://docs.scipy.org/doc/numpy/reference/generated/numpy.fromiter.html