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

On Wed, Feb 13, 2019 at 3:45 PM Mauro Cavalcanti wrote: > Chuck, > > I attempted to find such a list from the Numpy website. A complete list > like yours should be quite handy for users if available there. > > In ipython In [1]: numpy.lib.nanfunctions? will give it to you. But it looks like a m

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

Thanks! I didn't heard of this package and will look into it. Best regards, Em Qua, 13 de fev de 2019 22:10, Robert McLeod The `bottleneck` library is a very good package if there's some function > in NumPy that you want to handle `nan`s in reductions without exploding. > > https://github.com/kw

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

The `bottleneck` library is a very good package if there's some function in NumPy that you want to handle `nan`s in reductions without exploding. https://github.com/kwgoodman/bottleneck On Wed, Feb 13, 2019 at 12:35 PM Mauro Cavalcanti wrote: > Dear ALL, > > In the process of porting an existin

Re: [Numpy-discussion] Why slicing Pandas column and then subtract gives NaN?

This is more a question for the pandas list, but since i'm here i'll take a crack. - numpy aligns arrays by position. - pandas aligns by label. So what you did in pandas is roughly equivalent to the following: a = pandas.Series([85, 86, 87, 86], name='a').iloc[1:4].to_frame() b = pandas.S

[Numpy-discussion] Why slicing Pandas column and then subtract gives NaN?

Dear list, I have the following to Pandas Series: a, b. I want to slice and then subtract. Like this: a[1:4] - b[0:3]. Why does it give me NaN? But it works in Numpy. Example 1: did not work >>>a = pd.Series([85, 86, 87, 86]) >>>b = pd.Series([15, 72, 2, 3]) >>> a[1:4]-b[0:3] 0 NaN 1 14.0 2

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

Chuck, I attempted to find such a list from the Numpy website. A complete list like yours should be quite handy for users if available there. Best regards, Em qua, 13 de fev de 2019 às 20:10, Charles R Harris < charlesr.har...@gmail.com> escreveu: > > > On Wed, Feb 13, 2019 at 2:14 PM Mauro Cav

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

On Wed, Feb 13, 2019 at 2:14 PM Mauro Cavalcanti wrote: > 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 Cav

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.har...@gmail.com> escreveu: > > > On Wed, Feb 13, 2019 at 1:35 PM Mauro Cavalcanti > wrote: > >> Dear ALL, >> >> In the process of porting an existing (

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

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 normTran

[Numpy-discussion] 'nansqrt' function?

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

Re: [Numpy-discussion] Porting code for Numpy 1.13+ to get rid of "boolean index did not match indexed array along dimension 1" error

Eric, Implementing either of your suggestions (swapping the lines or using an intermediate variable) worked fine under the latest Numpy (v1.16.1)! Thanks a lot for your help! Best regards, Em ter, 12 de fev de 2019 às 23:06, Eric Wieser escreveu: > It looks like your code is wrong, and numpy