[Numpy-discussion] Standard functions (z-score) on nan (again)
I've bodged my way through my median problems (see previous postings). Now I need to take a z-score of an array that might contain nans. At the moment, if the array, which is 7000 elements, contains 1 nan or more, all the results come out as nan. My other problem is that my array is indexed from somewhere else - position 0 holds the value for a particular id, position 1 holds the value for a particular id and so on, so if I remove the nans, all the indices are messed up. Can anybody suggest a sensible fix for this problem? I know I should really be using masked arrays, but I can't seem to find a masked-array version of zs. Is there one? Peter ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Standard functions (z-score) on nan (again)
2008/9/25 Peter Saffrey [EMAIL PROTECTED]: I've bodged my way through my median problems (see previous postings). Now I need to take a z-score of an array that might contain nans. At the moment, if the array, which is 7000 elements, contains 1 nan or more, all the results come out as nan. My other problem is that my array is indexed from somewhere else - position 0 holds the value for a particular id, position 1 holds the value for a particular id and so on, so if I remove the nans, all the indices are messed up. Can anybody suggest a sensible fix for this problem? I know I should really be using masked arrays, but I can't seem to find a masked-array version of zs. Is there one? Just keep an array of indices: c = ~np.isnan(A) Aclean = A[c] indices = np.arange(len(A))[c] Then the original index of Aclean[i] is indices[i]. Anne ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
