[Numpy-discussion] 2D binning
Tue Jun 1 17:43:02 CDT 2010
I guess it's as fast as I'm going to get. I don't really see any other way.
BTW, the lat/lons are integers)
On Tue, Jun 1, 2010 at 1:49 PM, Zachary Pincus <email@example.com>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 lon = lat [lon  then
> > data = (data + data)/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
> 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?
> NumPy-Discussion mailing list
-------------- next part --------------
An HTML attachment was scrubbed...
More information about the NumPy-Discussion