Is there a better way of making numpy.argmin() ignore NaN values

Question

I want to get the index of the min value of a numpy array that contains NaNs and I want them ignored

>>> a = array([ nan,   2.5,   3.,  nan,   4.,   5.])   >>> a   array([ NaN,  2.5,  3. ,  NaN,  4. ,  5. ])   

if I run argmin, it returns the index of the first NaN

>>> a.argmin()   0   

I substitute NaNs with Infs and then run argmin

>>> a[isnan(a)] = Inf   >>> a   array([ Inf,  2.5,  3. ,  Inf,  4. ,  5. ])   >>> a.argmin()   1   

My dilemma is the following: I'd rather not change NaNs to Infs and then back after I'm done with argmin (since NaNs have a meaning later on in the code). Is there a better way to do this?

There is also a question of what should the result be if all of the original values of a are NaN? In my implementation the answer is 0

Answer 1

Sure! Use nanargmin:

import numpy as np a = np.array([ np.nan,   2.5,   3.,  np.nan,   4.,   5.]) print(np.nanargmin(a)) # 1 

There is also nansum, nanmax, nanargmax, and nanmin,

In scipy.stats, there is nanmean and nanmedian.

For more ways to ignore nans, check out masked arrays.