Chuck, Sure, using numpy.sqrt works fine.
Thank you very much. Best regards, Em qua, 13 de fev de 2019 às 19:09, Charles R Harris < charlesr.har...@gmail.com> escreveu: > > > On Wed, Feb 13, 2019 at 1:35 PM Mauro Cavalcanti <mauro...@gmail.com> > wrote: > >> Dear ALL, >> >> In the process of porting an existing (but abandoned) package to the >> latest version of Numpy, I stumbled upon a call to a 'numpy.nansqrt' >> function, which seems not to exist. >> >> Here is the specific code: >> >> def normTrans(y): >> denom = np.nansqrt(np.nansum(y**2)) >> return y/denom >> >> As far as I could find, there is no such 'nansqrt' function in the >> current version of Numpy, so I suspect that the above code has not been >> properly tested. >> >> Am I right, or that function had existed in some past version of Numpy? >> >> Thanks in advance for any comments or suggestions. >> >> > I don't recall any such function, but nansum will not result in any nans, > so plain old sqrt should work. > > Chuck > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@python.org > https://mail.python.org/mailman/listinfo/numpy-discussion > -- Dr. Mauro J. Cavalcanti E-mail: mauro...@gmail.com Web: http://sites.google.com/site/maurobio "Life is complex. It consists of real and imaginary parts."
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