thanks. I am also getting an error in ndi.mean Were you getting the error "RuntimeError: data type not supported"?
-Mathew On Wed, Jun 2, 2010 at 9:40 AM, Wes McKinney <wesmck...@gmail.com> wrote: > On Wed, Jun 2, 2010 at 3:41 AM, Vincent Schut <sc...@sarvision.nl> wrote: > > On 06/02/2010 04:52 AM, josef.p...@gmail.com wrote: > >> On Tue, Jun 1, 2010 at 9:57 PM, Zachary Pincus<zachary.pin...@yale.edu> > wrote: > >>>> 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.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? > >> > >> If the lat lon can be converted to a 1d label as Wes suggested, then > >> in a similar timing exercise ndimage was the fastest. > >> http://mail.scipy.org/pipermail/scipy-user/2009-February/019850.html > > > > And as you said your lats and lons are integers, you could simply do > > > > ll = lat*1000 + lon > > > > to get unique 'hashes' or '1d labels' for you latlon pairs, as a lat or > > lon will never exceed 360 (degrees). > > > > After that, either use the ndimage approach, or you could use > > histogramming with weighting by data values and divide by histogram > > withouth weighting, or just loop. > > > > Vincent > > > >> > >> (this was for python 2.4, also later I found np.bincount which > >> requires that the labels are consecutive integers, but is as fast as > >> ndimage) > >> > >> I don't know how it would compare to the new suggestions. > >> > >> Josef > >> > >> > >> > >>>> > >>>> Zach > >>>> _______________________________________________ > >>>> NumPy-Discussion mailing list > >>>> NumPy-Discussion@scipy.org > >>>> http://mail.scipy.org/mailman/listinfo/numpy-discussion > >>>> > >>>> _______________________________________________ > >>>> NumPy-Discussion mailing list > >>>> NumPy-Discussion@scipy.org > >>>> http://mail.scipy.org/mailman/listinfo/numpy-discussion > >>> > >>> _______________________________________________ > >>> NumPy-Discussion mailing list > >>> NumPy-Discussion@scipy.org > >>> http://mail.scipy.org/mailman/listinfo/numpy-discussion > >>> > > > > _______________________________________________ > > NumPy-Discussion mailing list > > NumPy-Discussion@scipy.org > > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > > > I was curious about how fast ndimage was for this operation so here's > the complete function. > > import scipy.ndimage as ndi > > N = 10000 > > lat = np.random.randint(0, 360, N) > lon = np.random.randint(0, 360, N) > data = np.random.randn(N) > > def group_mean(lat, lon, data): > indexer = np.lexsort((lon, lat)) > lat = lat.take(indexer) > lon = lon.take(indexer) > sorted_data = data.take(indexer) > > keys = 1000 * lat + lon > unique_keys = np.unique(keys) > > result = ndi.mean(sorted_data, labels=keys, index=unique_keys) > decoder = keys.searchsorted(unique_keys) > > return dict(zip(zip(lat.take(decoder), lon.take(decoder)), result)) > > Appears to be about 13x faster (and could be made faster still) than > the naive version on my machine: > > def group_mean_naive(lat, lon, data): > grouped = collections.defaultdict(list) > for lt, ln, da in zip(lat, lon, data): > grouped[(lt, ln)].append(da) > > averaged = dict((ltln, np.mean(da)) for ltln, da in grouped.items()) > > return averaged > > I had to get the latest scipy trunk to not get an error from ndimage.mean > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
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