将可变大小的子列表的嵌套列表展平为SciPy数组

Question

How can I use numpy/scipy to flatten a nested list with sublists of different sizes? Speed is very important and the lists are large.

 lst = [[1, 2, 3, 4],[2, 3],[1, 2, 3, 4, 5],[4, 1, 2]]

Is anything faster than this?

 vec = sp.array(list(*chain(lst)))

Answer 1

How about np.fromiter :

In [49]: %timeit np.hstack(lst*1000)
10 loops, best of 3: 25.2 ms per loop

In [50]: %timeit np.array(list(chain.from_iterable(lst*1000)))
1000 loops, best of 3: 1.81 ms per loop

In [52]: %timeit np.fromiter(chain.from_iterable(lst*1000), dtype='int')
1000 loops, best of 3: 1 ms per loop

Answer 2

You can try numpy.hstack

>>> lst = [[1, 2, 3, 4],[2, 3],[1, 2, 3, 4, 5],[4, 1, 2]]
>>> np.hstack(lst)
array([1, 2, 3, 4, 2, 3, 1, 2, 3, 4, 5, 4, 1, 2])

Answer 3

The fastest way to create a numpy array from an iterator is to use numpy.fromiter :

>>> %timeit numpy.fromiter(itertools.chain.from_iterable(lst), numpy.int64)
100000 loops, best of 3: 3.76 us per loop
>>> %timeit numpy.array(list(itertools.chain.from_iterable(lst)))
100000 loops, best of 3: 14.5 us per loop
>>> %timeit numpy.hstack(lst)
10000 loops, best of 3: 57.7 us per loop

As you can see, this is faster than converting to a list, and much faster than hstack .

Answer 4

尝试怎么样：

np.hstack(lst)

Answer 5

Use chain.from_iterable :

vec = sp.array(list(chain.from_iterable(lst)))

This avoids using * which is quite expensive to handle if the iterable has many sublists.

An other option might be to sum the lists:

vec = sp.array(sum(lst, []))

Note however that this will cause quadratic reallocation . Something like this performs much better:

def sum_lists(lst):
    if len(lst) < 2:
        return sum(lst, [])
    else:
        half_length = len(lst) // 2
        return sum_lists(lst[:half_length]) + sum_lists(lst[half_length:])

On my machine I get:

>>> L = [[random.randint(0, 500) for _ in range(x)] for x in range(10, 510)]
>>> timeit.timeit('sum(L, [])', 'from __main__ import L', number=1000)
168.3029818534851
>>> timeit.timeit('sum_lists(L)', 'from __main__ import L,sum_lists', number=1000)
10.248489141464233
>>> 168.3029818534851 / 10.248489141464233
16.422223757114615

As you can see, a 16x speed-up. The chain.from_iterable is even faster:

>>> timeit.timeit('list(itertools.chain.from_iterable(L))', 'import itertools; from __main__ import L', number=1000)
1.905594825744629
>>> 10.248489141464233 / 1.905594825744629
5.378105042586658

An other 6x speed-up.

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I looked for a "pure-python" solution, not knowing numpy. I believe Abhijit unutbu/senderle's solution is the way to go in your case.

Answer 6

Use a function to flatten the list

>>> flatten = lambda x: [y for l in x for y in flatten(l)] if type(x) is list else [x]
>>> flatten(lst)