计算重复的URL,尽可能快的方法

我仍在使用这个庞大的URL列表,我收到的所有帮助都很棒.

目前我的列表看起来像这样(但是17000个URL):

http://www.domain.com/page?CONTENT_ITEM_ID=1
http://www.domain.com/page?CONTENT_ITEM_ID=3
http://www.domain.com/page?CONTENT_ITEM_ID=2
http:// www .domain.com/page？CONTENT_ITEM_ID = 1
http://www.domain.com/page?CONTENT_ITEM_ID=2
http://www.domain.com/page?CONTENT_ITEM_ID=3
http://www.domain.com/页面？CONTENT_ITEM_ID = 3

我可以过滤掉重复项没有问题,有几种方法,awk等.我真正想做的是取出重复的URL,但同时计算列表中存在的URL数和打印次数带管道分隔符的URL旁边的计数.处理完列表后,它应如下所示:

网址| 计算
http://www.domain.com/page?CONTENT_ITEM_ID=1 | 2
http://www.domain.com/page?CONTENT_ITEM_ID=2 | 2
http://www.domain.com/page?CONTENT_ITEM_ID=3 | 3

实现这一目标的最快方法是什么方法？

干杯

这可能与您无需编写代码即可获得的速度一样快.

    $ cat foo.txt
    http://www.domain.com/page?CONTENT_ITEM_ID=1
    http://www.domain.com/page?CONTENT_ITEM_ID=3
    http://www.domain.com/page?CONTENT_ITEM_ID=2
    http://www.domain.com/page?CONTENT_ITEM_ID=1
    http://www.domain.com/page?CONTENT_ITEM_ID=2
    http://www.domain.com/page?CONTENT_ITEM_ID=3
    http://www.domain.com/page?CONTENT_ITEM_ID=3
    $ sort foo.txt | uniq -c
          2 http://www.domain.com/page?CONTENT_ITEM_ID=1
          2 http://www.domain.com/page?CONTENT_ITEM_ID=2
          3 http://www.domain.com/page?CONTENT_ITEM_ID=3

做了一些测试,并没有特别快,虽然对于17k它只需要1秒钟(在装载的P4 2.8Ghz机器上)

$ wc -l foo.txt
174955 foo.txt
vinko@mithril:~/i3media/2008/product/Pending$ time sort foo.txt | uniq -c
  54482 http://www.domain.com/page?CONTENT_ITEM_ID=1
  48212 http://www.domain.com/page?CONTENT_ITEM_ID=2
  72261 http://www.domain.com/page?CONTENT_ITEM_ID=3

real    0m23.534s
user    0m16.817s
sys     0m0.084s

$ wc -l foo.txt
14955 foo.txt
$ time sort foo.txt | uniq -c
   4233 http://www.domain.com/page?CONTENT_ITEM_ID=1
   4290 http://www.domain.com/page?CONTENT_ITEM_ID=2
   6432 http://www.domain.com/page?CONTENT_ITEM_ID=3

real    0m1.349s
user    0m1.216s
sys     0m0.012s

虽然O()像往常一样赢得比赛.测试了S.Lott的解决方案和

$ cat pythoncount.py
from collections import defaultdict
myFile = open( "foo.txt", "ru" )
fq= defaultdict( int )
for n in myFile:
    fq[n] += 1
for n in fq.items():
    print "%s|%s" % (n[0].strip(),n[1])

$ wc -l foo.txt
14955 foo.txt

$ time python pythoncount.py
http://www.domain.com/page?CONTENT_ITEM_ID=2|4290
http://www.domain.com/page?CONTENT_ITEM_ID=1|4233
http://www.domain.com/page?CONTENT_ITEM_ID=3|6432

real    0m0.072s
user    0m0.028s
sys     0m0.012s

$ wc -l foo.txt
1778955 foo.txt

$ time python pythoncount.py
http://www.domain.com/page?CONTENT_ITEM_ID=2|504762
http://www.domain.com/page?CONTENT_ITEM_ID=1|517557
http://www.domain.com/page?CONTENT_ITEM_ID=3|756636

real    0m2.718s
user    0m2.440s
sys     0m0.072s