Re: [Numpy-discussion] 'nansqrt' function?

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.harris@gmail.com> escreveu:

On Wed, Feb 13, 2019 at 1:35 PM Mauro Cavalcanti 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

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