list calculation in python

Question

I have a list like this:

[(ip1, video1, 12345.00000),(ip1, video1, 12346.12362),(ip1, video1, 12347.12684),(ip1, video2,12367.12567),(ip2, video1, 14899.93736), (ip2,video1, 24566.12345).....]

It records the video id and the time when the video was requested by each user.

Now I want to go through the list, and calculate the time interval between the first and the last request for each video,my list has already sorted by ip address.

The result I want to get is like:

ip1, video1, 2.12684
ip1, video2, 0

0 means, the requests never repeats.

Can anyone help?

The following is the code I create the dictionary:

for line in fd_in.readlines():
    (time, addr, iptp, userag, usertp, hash, vlanid) = line.split()

    if addr not in client_dict:
        client_dict[addr] = {}

    hash_dict = client_dict[addr]

    if hash not in hash_dict:
        hash_dict[hash] = []

    hash_dict[hash].append((float(time), addr, iptp, userag, usertp, hash, vlanid))


for addr, hash_dict in client_dict.items():
    for hash, hits_list in hash_dict.items():
        hits_list_sorted = sorted(hits_list, key=lambda item: item[0])

        for (time, addr, iptp, userag,usertp,hash,vlanid) in hits_list_sorted:

                xxxxxxxx[Dont know how to do the calculation]

                fd_out.write("%f\t%s\t%s\t%s\n" % (addr, hash, timeinternal))

The client_dict can be filled with client_dict.setdefault(addr, {}).setdefault(hash, []).append((float(time), addr, iptp, userag, usertp, hash, vlanid)). Also, hash is a function, so don't overwrite it with your variables. — eumiro
– eumiro, Commented Feb 20, 2012 at 9:31

DrTyrsa · Accepted Answer · 2012-02-20 09:33:06Z

3

Somethig like this

from itertools import groupby

for video, group in groupby(sorted(data, key=lambda x: x[1]), key=lambda x: x[1]):
    times = [x[2] for x in group]
    print 'Video: %s, interval: %f' % (video, max(times) - min(times))

answered Feb 20, 2012 at 9:33

DrTyrsa

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