This is the loveliest of all solutions: c[isfinite(c)].mean()
You are all very helpful and funny. I am sure most of you spend more than 16 hours a day in front of or by your screens :) On Tue, Aug 4, 2009 at 11:46 AM, Gökhan Sever <[email protected]> wrote: > Hello, > > I know this has to have a very simple answer, but stuck at this very moment > and can't get a meaningful result out of np.mean() > > > In [121]: a = array([NaN, 4, NaN, 12]) > > In [122]: b = array([NaN, 2, NaN, 3]) > > In [123]: c = a/b > > In [124]: mean(c) > Out[124]: nan > > In [125]: mean a > --------> mean(a) > Out[125]: nan > > Further when I tried: > > In [138]: c > Out[138]: array([ NaN, 2., NaN, 4.]) > > In [139]: np.where(c==NaN) > Out[139]: (array([], dtype=int32),) > > > In [141]: mask = [c != NaN] > > In [142]: mask > Out[142]: [array([ True, True, True, True], dtype=bool)] > > > Any ideas? > > -- > Gökhan > -- Gökhan
_______________________________________________ NumPy-Discussion mailing list [email protected] http://mail.scipy.org/mailman/listinfo/numpy-discussion
