[Numpy-discussion] 2D binning
Zachary Pincus
zachary.pincus@yale....
Tue Jun 1 20:57:35 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)
You could (in c or cython) try a brain-dead "hashtable" with no
collision detection:
for lat, long, data in dataset:
bin = (lat ^ long) % num_bins
hashtable[bin] = update_incremental_mean(hashtable[bin], data)
you'll of course want to do some experiments to see if your data are
sufficiently sparse and/or you can afford a large enough hashtable
array that you won't get spurious hash collisions. Adding error-
checking to ensure that there are no collisions would be pretty
trivial (just keep a table of the lat/long for each hash value, which
you'll need anyway, and check that different lat/long pairs don't get
assigned the same bin).
Zach
> -Mathew
>
> On Tue, Jun 1, 2010 at 1:49 PM, Zachary Pincus <zachary.pincus@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
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