Question

我正在努力争取时间二元搜索的不同大小。我只用这个程序获得2097152。我是python的新手。我知道缩进在python中至关重要。是缩进的吗？谢谢

import time
def bsearch(a,first,last,key):
    if first>last:
        #print "not found"
            return
    mid=(first+last)//2
    if key==a[mid]:
        #print "found"
            return
    elif key>a[mid]:
        bsearch(a,mid+1,last,key)
    else:
        bsearch(a,first,mid-1,key)

a=[1]*2097152
sizes=[128,512,2048,8192,32768,131072,524288,2097152]
for i in range(0,8):
    for j in range(0,sizes[i]):
        a[j]=j
start=time.time()
for k in range(0,20000):
    bsearch(a,0,j-1,j)
stop=time.time()
print ("time for size "+str(j)+" is: "+str((stop-start)*1000))

Answer 1

from timeit import timeit

for length in [128,512,2048,8192,32768,131072,524288,2097152]:
    l = list(range(length+1))
    command = 'bsearch(l, 0, length, length)'
    print('{:<23} seconds for {}'.format(timeit(command, globals=globals(), number=2000), length))

这是一个使用timeit.timeit来测量在给定大小的列表上执行bsearch 2000次所需的时间的版本。在我的机器上，结果是

0.005647999999382591    seconds for 128
0.007602000000588305    seconds for 512
0.009716999999909604    seconds for 2048
0.010829999999259599    seconds for 8192
0.012752000000546104    seconds for 32768
0.0143049999996947      seconds for 131072
0.01644899999973859     seconds for 524288
0.020023000000037428    seconds for 2097152

编辑：

如果您使用的是Python版本＆lt; 3.5，globals无法在timeit中使用import。您可以改为__main__

from timeit import timeit

for length in [128,512,2048,8192,32768,131072,524288,2097152]:
    l = list(range(length+1))
    command = 'bsearch(l, 0, length, length)'
    setup = 'from __main__ import bsearch, l, length'
    print('{:<23} seconds for {}'.format(timeit(command, setup=setup, number=2000), length))

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二进制搜索python

1 个答案: