On Wed, Jun 2, 2010 at 1:23 PM, Mathew Yeates <mat.yea...@gmail.com> wrote: > 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 > > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion > >
That's the error I was getting. Depending on your OS upgrading to the scipy trunk should be the easiest fix. _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
