将numpy数组列表合并为一个数组(快速)

Question

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？

我尝试了两种方法:

• merged_array = array(list_of_arrays)从Python化的方式来从numpy的阵列的列表创建numpy的阵列和

• vstack

您可以看到vstack更快,但由于某种原因,第一次运行比第二次运行时间长三倍.我假设这是由(缺少)预分配引起的.那么我将如何为数组预分配vstack？或者你知道更快的方法吗？

谢谢!

[UPDATE]

我(25280, 320)不想要(80, 320, 320)哪种方式,不merged_array = array(list_of_arrays)适合我.谢谢Joris指出这个!

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

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"

Answer 1

你有80个阵列320x320？所以你可能想要使用dstack:

first3 = numpy.dstack(firstmatrices)

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

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

如果要使用vstack,它将返回25600x320阵列:

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