2014-02-11 14:55 GMT+01:00 Andreas Hilboll <li...@hilboll.de>: > On 11.02.2014 14:47, Daniele Nicolodi wrote: > > On 11/02/2014 14:41, Andreas Hilboll wrote: > >> On 11.02.2014 14:22, Daniele Nicolodi wrote: > >>> On 11/02/2014 14:10, Andreas Hilboll wrote: > >>>> On 11.02.2014 14:08, Daniele Nicolodi wrote: > >>>>> Hello, > >>>>> > >>>>> I have two time series (2xN dimensional arrays) recorded on the same > >>>>> time basis, but each with it's own dead times (and start and end > >>>>> recording times). I would like to obtain two time series containing > >>>>> only the time overlapping segments of the data. > >>>>> > >>>>> Does numpy or scipy offer something that may help in this? > >>>>> > >>>>> I can imagine strategies about how to approach the problem, but none > >>>>> that would be efficient. Ideas? > >>>> > >>>> Take a look at pandas. It has built-in time series functionality. > >>> > >>> Even using Pandas (and I would like to avoid to have to depend on it) > it > >>> is not clear to me how I would achieve what I want. Am I missing > something? > >> > >> If the two time series are pandas.Series objects and are called s1 and > s2: > >> > >> new1 = s1.ix[s2.dropna().index].dropna() > >> new2 = s2.ix[s1.dropna().index].dropna() > >> new1 = new1.ix[s2.dropna().index].dropna() > >> > >> Looks hackish, so there might be a more elegant solution. For further > >> questions about how to use pandas, please look at the pydata mailing > >> list or stackoverflow. > > > > Correct me if I'm wrong, but this assumes that missing data points are > > represented with Nan. In my case missing data points are just missing. > > pandas doesn't care. > > In pandas, you could simply do something like this (assuming the time is set as the index):
pd.concat([s1, s2], axis=1) and then remove the nan's (where the index was not overlapping) or use `join='inner'` Joris > Andreas. > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
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