On Sat, Jan 5, 2013 at 2:15 PM, Eric Emsellem <eric.emsel...@eso.org> wrote: > Dear all, > > I have a code using lots of "numpy.where" to make some constrained > calculations as in: > > data = arange(10) > result = np.where(data == 0, 0., 1./data) > > # or > data1 = arange(10) > data2 = arange(10)+1.0 > result = np.where(data1 > data2, np.sqrt(data1-data2), np.sqrt(data2-data2)) > > which then produces warnings like: > /usr/bin/ipython:1: RuntimeWarning: invalid value encountered in sqrt > > or for the first example: > > /usr/bin/ipython:1: RuntimeWarning: divide by zero encountered in divide > > How do I avoid these messages to appear? > > I know that I could in principle use numpy.seterr. However, I do NOT > want to remove these warnings for other potential divide/multiply/sqrt > etc errors. Only when I am using a "where", to in fact avoid such > warnings! Note that the warnings only happen once, but since I am going > to release that code, I would like to avoid the user to get such > messages which are irrelevant here (because I am testing, with the > where, when NOT to divide by zero or take a sqrt of a negative number).
You can't avoid it while using np.where like this, because the warning is being issued before np.where is even called. It's basically doing: # Calculate all possible sqrts tmp1 = np.sqrt(data1-data2) tmp2 = np.sqrt(data2-data2) # let's pretend this isn't just all zeros... # Use np.where to pick out the useful ones and put them together into one array mashed_up = np.where(data1 > data2, tmp1, tmp2) So you need to somehow apply the indexing while doing the sqrt. In this case the easiest way would just be np.sqrt(np.where(data1 > data2, data1 - data2, data2 - data2)) Or, slightly faster (avoiding some temporaries): np.sqrt(np.where(data1 > data2, data1, data2) - data2) If your operation doesn't factor like this though then you can always use something more cumbersome like result = np.empty_like(data) mask = (data == 0) result[mask] = 0 result[~mask] = 1.0/data[~mask] Or in 1.7 this could be written result = np.zeros_like(data) np.divide(1.0, data, where=(data != 0), out=result) -n _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
