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呢?
我尝试了两种方法:
merged_array = array(list_of_arrays)
vstack
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)preallocation引起的。那么我如何为vstack
预先分配一个数组呢?或者你知道更快的方法吗?
谢谢!
[UPDATE]
我希望(25280, 320)
不是(80, 320, 320)
,这意味着,merged_array = array(list_of_arrays)
不适合我。谢谢Joris指出这一点!!!
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
代码:
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"
答案 0 :(得分:21)
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)
那样:
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
如果您想使用vstack
,它将返回25600x320阵列:
timeit numpy.vstack(firstmatrices)
100 loops, best of 3: 18.2 ms per loop