[Numpy-discussion] Help with np.where and datetime functions

Pierre GM pgmdevlist@gmail....
Wed Jul 8 13:44:50 CDT 2009


On Jul 8, 2009, at 7:03 AM, John [H2O] wrote:

>
> Hello,
>
> I have several issues which require me to iterate through a fairly  
> large
> array (300000+ records).
>
> The first case is calculating and hourly average from non-regularly  
> sampled
> data.

Would you like to give the scikits.timeseries package a try ? It's  
available at pytseries.sourceforge.net.
Calculatng the hourly average should be straightforward.


> The second is screening one array, based on data in the second array.
> The functions are defined below, but inherent to each is the following
> snippet:
>
>    ind = np.where( (t1 < X[:,0]) & (X[:,0] < t2) )
>
>    where X is a (n,2) array and X[:,0] = a vector of datetime objects.
>
> What I am trying to do (obviously?) is find all the values of X that  
> fall
> within a time range.

Well, timeseries could help you on this one as well.

>
> Specifically, one point I do not understand is why the following two  
> methods
> fail:
>
> --> 196         ind = np.where( (t1 < Y[:,0] < t2) ) #same result
> with/without inner parens
> TypeError: can't compare datetime.datetime to numpy.ndarray
>
>
> OR trying the 'and' method:
>
> --> 196         ind = np.where( (Y[:,0]>t1) and (Y[:,0]<t2) )
> ValueError: The truth value of an array with more than one element is
> ambiguous. Use a.any() or a.all()

As mentioned by Neil, you need to use & instead of and in your  
expression. You can also use the lengthier np.logical_and(Y[:,0]>t1,  
Y[:,0]<t2)

Also, using np.where without additional inputs  
(np.where(condition,x,y)) is equivalent to .nonzero


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