Conversion from Matlab Timestamp Date to Python Timestamp Date

Question

I have a timestamp Matrix with (72500, 1) dimension in MATLAB. I have written such a statement in MATLAB:

returnMatrix = datestr(M(:,1)/86400 + datenum(1970,1,1)- 4/24);

And I successfully get the output date matrix back as a return Matrix.

returnMatrix = ['29/06/2015',...,'06/07/2015']

In order to get the same output, I have written such a statement in Python:

returnMatrix = (M[:, 0] / 86400 + date.toordinal(date(1970, 1, 1)) + 366 - (4/24))
returnMatrix = np.apply_along_axis((lambda x:[date.fromtimestamp(x[0]).strftime("%d/%m/%y")]),1,returnMatrix)

But as an output, I get this result:

returnMatrix = ['09/01/1970',...,'09/01/1970']

How should I write the statement in order to get the output as the one in MATLAB?

Sharad · Accepted Answer · 2015-10-11 14:54:49Z

1

What I understand from this question is that you have data in seconds which you want to add to the base date which is 1/1/1970 and get the new date. Python 'timedelta' method from 'datetime' module will be handy to solve this problem.

Here I present my solution to your problem

    import datetime
    base_date = datetime.datetime(1970,1,1,0,0,0)
    return_matrix = []
    for time in M:
        return_matrix.append(base_date + datetime.timedelta(seconds=time))

The return matrix will contain all the dates. No need for additional calculation.

I assume the data is saved in list M. I don't understand the way you are using python list. The way you have accessed the list i.e M[:0] will actually split the list from zeroth element to zeroth element excluding the zeroth. If it is a one dimensional data then you can directly use M the way I have used. I think you got confused there coming from a MATLAB background.

answered Oct 11, 2015 at 14:54

Sharad

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5 Comments

yusuf Over a year ago

Thank you for your answer Sharad. I will let you know after trying your solution. But there is no other way to write it by using lambda? Because for loop takes so much time to process.

Sharad Over a year ago

I can't say with surety about lambda because I haven't used it.

Sharad Over a year ago

list comprehension can be another solution for the problem. you can create a list in this way: return_matrix = [base_date + datetime.timedelta(seconds=time) for time in M]

chthonicdaemon Over a year ago

@Sharad It seems like OP is using a two-dimensional NumPy array for M by analogy to how it would work in Matlab. The indexing in his example is correct.

Sharad Over a year ago

Thanks for the info. I haven't used numpy so had no idea

chthonicdaemon · Accepted Answer · 2015-10-11 16:32:13Z

0

Given M is a numpy array containing seconds, you can do this, if you can live with having your date strings in ISO8601 format

dates = (M*numpy.timedelta64(1, 's') + 
         numpy.datetime64('1970-01-01T00:00') -
         numpy.timedelta64(4, 'h').astype('timedelta64[s]')).astype('datetime64[D]')

(that .astype is there to squelch a warning about implicit conversion).

edited Oct 11, 2015 at 16:32

answered Oct 11, 2015 at 16:21

chthonicdaemon

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