Building a Time series prediction model on web login timestamp

Question

So I'm trying to build a time series prediction model. All I have is a sequence of timestamps of a user when he logs in to a site.

Here is the first few rows of the data. This is a Panda series I've got this in

0   2012-03-01 00:05:55
1   2012-03-01 00:06:23
2   2012-03-01 00:06:52
3   2012-03-01 00:11:23
4   2012-03-01 00:12:47
5   2012-03-01 00:12:54
6   2012-03-01 00:16:14
7   2012-03-01 00:17:31
8   2012-03-01 00:21:23
9   2012-03-01 00:21:26

Now the questions I have are;

1). How to Graph the user behavior on an hourly basis when all I have is timestamps and no Y values or any other features

2). Build a model which fits this time series and predict for the next two weeks.

Comments

I'm dealing with the same problem statement right now, But i need to predict next login date and time both ....whenever it is... Did u find any solution for this. If so, can u help me with a brief regarding how u dealt this?

Answer 1

I had done something similar and ran into the same problem.

So, what I did was that I grouped the time series using epoch and loaded it into a dictionary. 

From there I could work on the time series in hour chunks. (data source is json) Then you can convert it to a panda DataFrame and chart directly using matplotlib. Since your data is already in panda, you could skip the data pull and edit the initial loop to process your raw data. I hope this helps.

for key in responseJson['All'].keys():
        t = time.strftime('%Y,%m,%d %H:00:00', time.gmtime(float(key) / 1000.0))
        h = responseJson['All'][key]
        word = t
        epochkey = int(time.mktime(time.strptime(t, '%Y,%m,%d %H:00:00')))

        if word not in dict:
            dict[word] = h
            epochdict[epochkey] = h
        else:
            dict[word] += h
            epochdict[epochkey] += h

Then I converted it to a panda DataFrame:

for row in epochdict:
        if(row[0] not in data):
            data[row[0]]={}
        data[str(row[0])][str(row[2])]=round(row[3],3)

            df=DataFrame(data).T.fillna(0)