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I'm trying to bruteforce through some list iterations in python (I'm a pynoob) and I can't seem to understand why this is such an issue.

My data structure looks like this:

pprint.pprint(list)

[[1355759452000L, 1],
 [1355759191000L, 1],
 [1355758983000L, 1],
 [1355758939000L, 1],
 ... items removed for brevity...
 [1355742844000L, 1],
 [1355742833000L, 1],
 [1355742558000L, 1]]

I want to iterate over this list, however, the only way I was able to get the timestamp from this was to perform the following (seems wrong):

startEpoch = 0
endEpoch = ...some future date...
newList = []
while currentTime <= endEpoch:
        for i,l in enumerate(list):
            for epoch in enumerate(l):
                if epoch[1] >= currentTime and epoch[1] <= (currentTime + 7200):
                    newList.append(currentTime)
        currentTime += 7200

The goal of this is to iterate over 'list' and add up ever entry that falls with a 2 hour range. So, if the start is 0, count each entry between 0 and 7200, then count each entry between 7200 and 14200, etc.

Ideally, I want newList to be something like:

[0][12]
[7200][11]

[the time stamp][the count]

For whatever reason, my bad habits in other languages and my lack of understanding in python, this is becoming error prone and way more difficult than it should be.

Any help and guidance is appreciated.

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2 Answers 2

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2

How about something like this:

In [17]: l = [[0, 1], [3, 1], [200, 1], [8000, 1], [9000, 1], [20000, 1]]

In [24]: [(k,len(list(g))) for k,g in itertools.groupby(l, lambda x:x[0]-x[0]%7200)]
Out[24]: [(0, 3), (7200, 2), (14400, 1)]

This assumes that that the timestamps are arranged chronologically. If they aren't, you could use collections.Counter:

In [26]: sorted(collections.Counter(x[0]-x[0]%7200 for x in l).items())
Out[26]: [(0, 3), (7200, 2), (14400, 1)]

is there any way to get this to include a 0 for that 7200 time group? So if I have a time segment of 7200 with no matching entries in the original list, can I have it place a zero. (I'm graphing this data)

In [29]: c = collections.Counter(x[0]-x[0]%7200 for x in l)

In [30]: [(t, c.get(t, 0)) for t in range(0, 72000, 7200)]
Out[30]: 
[(0, 3),
 (7200, 2),
 (14400, 1),
 (21600, 0),
 (28800, 0),
 (36000, 0),
 (43200, 0),
 (50400, 0),
 (57600, 0),
 (64800, 0)]

Here, t iterates over the timestamps you wish to plot; c.get(t, 0) gets the count if there is data and returns 0 if there isn't.

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3 Comments

This is pretty sick, in a good way. I wish I had an understanding of python like this - it's really clear the power over a language like PHP. The real question is where can I learn all those cool tricks?
The downside is I don't really understand this and HATE using it ... I'll have to take some time to pick it apart and REALLY understand it.
Works great - is there any way to get this to include a 0 for that 7200 time group? So if I have a time segment of 7200 with no matching entries in the original list, can I have it place a zero. (I'm graphing this data).
1

Similar to NPE's answer,

for k, g in itertools.groupby(items, lambda x: x[0] / 7200000):
    print k,
    print sum([x[1] for x in g])

I am assuming (possibly incorrectly) that the second item in the sublist is a count of that timestamp. If not, then taking the length of the list (as in NPE's answer) is the right way to do it.

NOTE
You probably want to use 7200000 instead of 7200 since it appears your timestamps are in milliseconds.

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1 Comment

Yeah, you are right about the 7200000/milliseconds, thanks. I'm probably going to use NPE's answer since I started messing with it first, but +1 to you and thanks. Like I said above (in NPEs thread), I need to learn python a LOT better to make use of these features.

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