我最近问a question on the optimization of a mask function in Matlab。我得到了两个对我有帮助的答案，但似乎根据我的时间，所有Matlab解决方案看起来都比Numpy解决方案慢得多。不同功能的代码可以在我之前的问题中找到，但为了了解我在做什么，我给出了Numpy“循环”解决方案，当然这不是最快的解决方案，但可能是最简单的解读：
def dealiasing2d(where, data):
nk, n0, n1 = data.shape
for i0 in xrange(n0):
for i1 in xrange(n1):
if where[i0, i1]:
data[:, i0, i1] = 0.
我获得（使用Matlab R2014b和与Blas和Lapack相关联的“基本”Numpy 1.9.1）（n0 = n1 = N）：
N = 500 ; nk = 500:
Method | time (s) | normalized
----------------|----------|------------
Numpy | 0.05 | 1.0
Numpy loop | 0.05 | 1.0
Matlab find | 0.74 | 14.8
Matlab bsxfun2 | 0.76 | 15.2
Matlab bsxfun | 0.78 | 15.6
Matlab loop | 0.78 | 15.6
Matlab repmat | 0.89 | 17.8
N = 500 ; nk = 100:
Method | time (s) | normalized
----------------|----------|------------
Numpy | 0.01 | 1.0
Numpy loop | 0.03 | 3.0
Matlab find | 0.15 | 13.6
Matlab bsxfun2 | 0.15 | 13.6
Matlab bsxfun | 0.16 | 14.5
Matlab loop | 0.16 | 14.5
Matlab repmat | 0.18 | 16.4
N = 2000 ; nk = 10:
Method | time (s) | normalized
----------------|----------|------------
Numpy | 0.02 | 1.0
Matlab find | 0.23 | 13.8
Matlab bsxfun2 | 0.23 | 13.8
Matlab bsxfun | 0.26 | 15.6
Matlab repmat | 0.28 | 16.8
Matlab loop | 0.34 | 20.4
Numpy loop | 0.42 | 25.1
在我看来，这些结果非常奇怪。对我来说，Numpy和Matlab在科学计算方面非常相似所以性能应该相似，而这里有10倍以上！所以我的第一个猜测是我比较两种语言的方式有问题。另一种可能是我的Matlab设置有问题，但我不明白为什么。或者是Matlab和Numpy之间真正的深刻区别？
有人可以用这些功能验证这些结果吗？你知道为什么Matlab在这个简单的例子中看起来比Python慢得多吗？
为了对Matlab函数进行计时，我使用了一个文件：
N = 500;
n0 = N;
n1 = N;
nk = 500;
disp(['N = ', num2str(N), ' ; nk = ', num2str(nk)])
where = false([n1, n0]);
where(1:100, 1:100) = 1;
data = (5.+1i)*ones([n1, n0, nk]);
disp('time dealiasing2d_loops:')
time = timeit(@() dealiasing2d_loops(where, data));
disp([' ', num2str(time), ' s'])
disp('time dealiasing2d_find:')
time = timeit(@() dealiasing2d_find(where, data));
disp([' ', num2str(time), ' s'])
disp('time dealiasing2d_bsxfun:')
time = timeit(@() dealiasing2d_bsxfun(where, data));
disp([' ', num2str(time), ' s'])
disp('time dealiasing2d_bsxfun2:')
time = timeit(@() dealiasing2d_bsxfun2(where, data));
disp([' ', num2str(time), ' s'])
disp('time dealiasing2d_repmat:')
time = timeit(@() dealiasing2d_repmat(where, data));
disp([' ', num2str(time), ' s'])
我用
衡量Python函数的性能from __future__ import print_function
import numpy as np
from timeit import timeit, repeat
import_lines = {
'numpy_bad': 'from dealiasing_numpy_bad import dealiasing2d as dealiasing',
'numpy': 'from dealiasing_numpy import dealiasing'}
tools = import_lines.keys()
time_approx_one_bench = 5.
setup = """
import numpy as np
N = 500
n0, n1 = N, N
nk = 500
where = np.zeros((n0, n1), dtype=np.uint8)
where[0:100, 0:100] = 1
data = (5.+1j)*np.ones((nk, n0, n1), dtype=np.complex128)
"""
exec(setup)
print('n0 = n1 = {}, nk = {}'.format(N, nk))
print(13*' ' + 'min mean')
durations = np.empty([len(tools)])
for it, tool in enumerate(tools):
setup_tool = import_lines[tool] + setup
duration = timeit(setup_tool + 'dealiasing(where, data)', number=1)
nb_repeat = int(round((time_approx_one_bench - duration)/duration))
nb_repeat = max(1, nb_repeat)
ds = np.array(repeat('dealiasing(where, data)',
setup=setup_tool, number=1, repeat=nb_repeat))
duration = ds.min()
print('{:11s} {:8.2e} s ; {:8.2e} s'.format(
tool.capitalize() + ':', duration, ds.mean()))
durations[it] = duration
fastest = tools[durations.argmin()].capitalize()
durations = durations / durations.min()
print('Durations normalized by the fastest method (' + fastest + '):')
for it, tool in enumerate(tools):
print('{:11s} {:8.3f}'.format(tool.capitalize() + ':', durations[it]))
答案 0 :(得分：3)
我认为这主要与制作data变量的副本有关。如果您安排的事情使MATLAB的写时复制行为对您有利，那么您可以获得相当不错的时间。我用线性索引编写了一个简单的版本
function data = dealiasing2d2(where_dealiased, data)
[n1, n2, nk] = size(data);
where_li = find(where_dealiased);
for idx = 1:nk
offset = n1 * n2 * (idx-1);
data(where_li + offset) = 0;
end
我这样运行（请注意，timed是功能非常重要，而不是允许重复使用data的脚本。
function timed
N = 2000;
nk = 10;
where = false([N, N]);
where(1:100, 1:100) = 1;
data = (5.+1j)*ones([N, N, nk]);
tic, data = dealiasing2d2(where,data); toc
在运行R2014b的GLNXA64机器上运行0.00437秒。