> You might want to try isfinite() to first remove nan, +/- infinity > before doing that. > numpy.median(a[numpy.isfinite(a)])
We just had this discussion a month or two ago, I think even on this list, and continued it at the SciPy conference. The problem with numpy.median(a[numpy.isfinite(a)]) is that it breaks when you have a multi-dimensional array, such as an array of 5000x3 as in this case, and take median down an axis. The example above flattens the array and eliminates the possibility of taking the median down an axis in a single call, as the poster desires. Currently the only way you can handle NaNs is by using masked arrays. Create a mask by doing isfinite(a), then call the masked array median(). There's an example here: http://sd-2116.dedibox.fr/pydocweb/doc/numpy.ma/ Note that our competitor language IDL does have a /nan flag to its single median routine, making this common task much easier in that language than ours. --jh-- _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
