Python计时 - 必须有更好的方法!

Question

我希望有人可以帮助我解决这个问题.我想测量排序算法.这是我目前的做法:

M = 1000 # number of executions
N = [1000, 2000, 4000, 16000] # size of the list
L = [100, 1000, 2000,16000] # max element of the list

# timing:
print 'Number of executions: %i' % (M)
print '-'*80
print '\tL\N\t|\t%i\t|\t%i\t|\t%i\t|\t%i' % (N[0], N[1], N[2], N[3])
print '-'*80
for l in L:
    print '\t%i\t' % l,
    for n in N: 
        t = 0
        for m in xrange(M):
            A = [random.randint(0,l-1) for r in xrange(n)] # generates an n long random list
            t0 = time.clock()
            pass # sort function call goes here
            t1 = time.clock()
            t += (t1-t0)
        print '|\t%0.3f\t' % ((t*1000.0)/M ), # avg time
    print
print '-'*80

这个空测试大约需要4分钟.我很感激有关如何加快速度的建议.

干杯

编辑: 在Rafe Kettler的暗示之后,我想出了这个:

def sorting(LST):
    pass

if __name__ == "__main__" :
    M = 1000
    N = [1000, 2000, 4000, 16000]
    L = [100, 1000, 2000,16000]

    print 'Number of executions: %i' % (M)
    print '-'*80
    print '\tL\N\t|\t%i\t|\t%i\t|\t%i\t|\t%i' % (N[0], N[1], N[2], N[3])
    print '-'*80
    for l in L:
        print '\t%i\t' % l,
        for n in N:
            #------------------------
            t = timeit.Timer('sorting([random.randint(0,l-1) for r in xrange(n)])', 'from __main__ import sorting, n, l, random')
            #------------------------
            print '|\t%0.3f\t' % (t.timeit(M)/M ), # avg time
        print
    print '-'*80

不幸的是它变慢了.我究竟做错了什么？

Answer 1

时间.期间Python的最佳时间.将算法重构为函数并用于timeit测试执行时间.