将numpy数组列表合并为一个数组（快速）

what would be the fastest way to merge a list of numpy arrays into one array if one knows the length of the list and the size of the arrays, which is the same for all?

如果知道列表的长度和数组的大小，那么将numpy数组列表合并到一个数组中的最快方法是什么？

I tried two approaches:

我尝试了两种方法：

merged_array = array(list_of_arrays) from Pythonic way to create a numpy array from a list of numpy arrays and

mergethar = array（list_of_arrays）来自Pythonic方式，从numpy数组列表中创建一个numpy数组
vstack

vstack

A you can see vstack is faster, but for some reason the first run takes three times longer than the second. I assume this caused by (missing) preallocation. So how would I preallocate an array for vstack? Or do you know a faster methode?

你可以看到vstack更快，但由于某种原因，第一次运行比第二次运行长三倍。我假设这是由（缺少）预分配引起的。那么我如何为vstack预分配一个数组呢？或者你知道更快的方法吗？

Thanks!

谢谢！

[UPDATE]

[UPDATE]

I want (25280, 320) not (80, 320, 320) which means, merged_array = array(list_of_arrays) wont work for me. Thanks Joris for pointing that out!!!

我想（25280,320）不是（80,320,320）这意味着，merged_array = array（list_of_arrays）对我不起作用。谢谢Joris指出这个！

Output:

输出：

0.547468900681 s merged_array = array(first_list_of_arrays)
0.547191858292 s merged_array = array(second_list_of_arrays)
0.656183958054 s vstack first
0.236850976944 s vstack second

Code:

码：

import numpy
import time
width = 320
height = 320
n_matrices=80

secondmatrices = list()
for i in range(n_matrices):
    temp = numpy.random.rand(height, width).astype(numpy.float32)
    secondmatrices.append(numpy.round(temp*9))

firstmatrices = list()
for i in range(n_matrices):
    temp = numpy.random.rand(height, width).astype(numpy.float32)
    firstmatrices.append(numpy.round(temp*9))


t1 = time.time()
first1=numpy.array(firstmatrices)
print time.time() - t1, "s merged_array = array(first_list_of_arrays)"

t1 = time.time()
second1=numpy.array(secondmatrices)
print time.time() - t1, "s merged_array = array(second_list_of_arrays)"

t1 = time.time()
first2 = firstmatrices.pop()
for i in range(len(firstmatrices)):
    first2 = numpy.vstack((firstmatrices.pop(),first2))
print time.time() - t1, "s vstack first"

t1 = time.time()
second2 = secondmatrices.pop()
for i in range(len(secondmatrices)):
    second2 = numpy.vstack((secondmatrices.pop(),second2))

print time.time() - t1, "s vstack second"

1 个解决方案

#1

You have 80 arrays 320x320? So you probably want to use dstack:

你有80个阵列320x320？所以你可能想要使用dstack：

first3 = numpy.dstack(firstmatrices)

This returns one 80x320x320 array just like numpy.array(firstmatrices) does:

这将返回一个80x320x320数组，就像numpy.array（firstmatrices）一样：

timeit numpy.dstack(firstmatrices)
10 loops, best of 3: 47.1 ms per loop


timeit numpy.array(firstmatrices)
1 loops, best of 3: 750 ms per loop

If you want to use vstack, it will return a 25600x320 array:

如果你想使用vstack，它将返回一个25600x320数组：

timeit numpy.vstack(firstmatrices)
100 loops, best of 3: 18.2 ms per loop

#1