On Tue, Aug 4, 2009 at 1:53 PM, Bruce Southey<[email protected]> wrote: > On Tue, Aug 4, 2009 at 1:40 PM, Gökhan Sever<[email protected]> wrote: >> This is the loveliest of all solutions: >> >> c[isfinite(c)].mean() > > This handling of nonfinite elements has come up before. > Please remember that this only for 1d or flatten array so it not work > in general especially along an axis.
If you don't want to use nanmean from scipy.stats you could use: np.nansum(c, axis=0) / (~np.isnan(c)).sum(axis=0) or np.nansum(c, axis=0) / (c == c).sum(axis=0) But if c contains ints then you'll run into trouble with the division, so you'll need to protect against that. _______________________________________________ NumPy-Discussion mailing list [email protected] http://mail.scipy.org/mailman/listinfo/numpy-discussion
