On Tue, Jun 1, 2010 at 4:49 PM, Zachary Pincus <zachary.pin...@yale.edu> wrote: >> Hi >> Can anyone think of a clever (non-lopping) solution to the following? >> >> A have a list of latitudes, a list of longitudes, and list of data >> values. All lists are the same length. >> >> I want to compute an average of data values for each lat/lon pair. >> e.g. if lat[1001] lon[1001] = lat[2001] [lon [2001] then >> data[1001] = (data[1001] + data[2001])/2 >> >> Looping is going to take wayyyy to long. > > As a start, are the "equal" lat/lon pairs exactly equal (i.e. either > not floating-point, or floats that will always compare equal, that is, > the floating-point bit-patterns will be guaranteed to be identical) or > approximately equal to float tolerance? > > If you're in the approx-equal case, then look at the KD-tree in scipy > for doing near-neighbors queries. > > If you're in the exact-equal case, you could consider hashing the lat/ > lon pairs or something. At least then the looping is O(N) and not > O(N^2): > > import collections > grouped = collections.defaultdict(list) > for lt, ln, da in zip(lat, lon, data): > grouped[(lt, ln)].append(da) > > averaged = dict((ltln, numpy.mean(da)) for ltln, da in grouped.items()) > > Is that fast enough? > > Zach > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
This is a pretty good example of the "group-by" problem that will hopefully work its way into a future edition of NumPy. Given that, a good approach would be to produce a unique key from the lat and lon vectors, and pass that off to the groupby routine (when it exists). _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
