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up vote 2 down vote favorite

Having a large array in which column[0] corresponds with the day, col[1]=month, col[2]=year and col[3]=hours (the latter is a float and also contains info on minutes and seconds in the fraction), what is the most efficient way to convert these columns into an array of datetimes?

update below: I tinkered with the dt.datetime function so it handles array input as well as fractional years, months whatever. I haven't tested this thoroughly yet and there are probably more elegant ways to do it but here goes.

from __future__import division

def getrem(input):
    "this function yields the value behind the decimal point"
    import numpy as np
    output=abs(input-np.fix(input))
    return output

def datenum(Yr,Mo=1,Da=1,Hr=0,Mi=0,Se=0,Ms=0):
    "this function works as regular datetime.datetime, but allows for float input"
    import numpy as np    
    import datetime as dt
    import calendar

    #correct faulty zero input
    if Mo<1:
        Mo+=1
    if Da<1:
        Da+=1        

    #distribute the year fraction over days    
    if  getrem(Yr)>0:
        if calendar.isleap(np.floor(Yr)):
            fac=366       
        else:
            fac=365               
        Da=Da+getrem(Yr)*fac
        Yr=int(Yr)
    #if months exceeds 12, pump to years         
    while int(Mo)>12:
        Yr=Yr+1
        Mo=Mo-12
    #distribute fractional months to days              
    if getrem(Mo)>0:
        Da=Da+getrem(Mo)*calendar.monthrange(Yr,int(Mo))[1]
        Mo=int(Mo)
    #datetime input for 28 days always works excess is pumped to timedelta    
    if Da>28:
        extraDa=Da-28
        Da=28
    else:
        extraDa=0 
    # sometimes input is such that you get 0 day or month values, this fixes this anomaly           
    if int(Da)==0:
       Da+=1
    if int(Mo)==0:
       Mo+=1

    #datetime calculation           
    mytime=dt.datetime(int(Yr),int(Mo),int(Da))+dt.timedelta(days=extraDa+getrem(Da),hours=Hr,minutes=Mi,seconds=Se,microseconds=Ms)
    return mytime    

def araydatenum(*args):
    mydatetimes=[datenum(*[a.squeeze()[x] for a in args]) for x in range(len(args[0].squeeze()))]
    return mydatetimes
share|improve this question
4  
By datetimes do you mean Python datetime.datetime objects (for which you'll need an array of dtype='object'), or do you mean NumPy datetime64 objects introduced in NumPy 1.7? –  unutbu Nov 4 '13 at 14:22
    
either format will do –  tragewombat Nov 6 '13 at 8:36
1  
Do you really have fractional years and months? Sounds like a pain. :-) BTW, timedelta will handle the extra hours, minutes, and seconds, saving you at least the part of pushing up the remainders. –  ratatoskr Nov 7 '13 at 16:47
    
Fractional years is a common occurrence, as is days and sometimes hours. Months is pretty rare. You're right about timedelta, I might tweak it a bit further still. –  tragewombat Nov 8 '13 at 9:23

1 Answer 1

up vote 2 down vote accepted

Can't speak to the most efficient, but it can be done easily like this:

import datetime as dt
mydatetimes = [dt.datetime(x[2], x[1], x[0]) + dt.timedelta(hours=x[3]) for x in myarray]

This creates a regular python list, not a numpy array. Just add numpy.array( ... ) around the right hand side to make it an array with dtype=object.

share|improve this answer
1  
A variation would be np.apply_along_axis(foo,1,myarray), where foo is a simple function wrapper around your dt.datetime.... But it's actually slower than your np.numpy([foo(x) for x in myarray]). For custom actions like this, straightforward Python is often the best starting point. –  hpaulj Nov 4 '13 at 23:19
    
This works (see edit update above), but I still wonder if a for loop (or apply_along_axis) cannot be avoided. –  tragewombat Nov 7 '13 at 15:31

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