numpy数组过滤器并替换

Question

I have an array

a = np.array([1,2,3,4,np.nan])

I would like to replace anything which is less than 1.5 by np.nan , ie I would like

a = np.array([np.nan,2,3,4,np.nan])

how do I do that?

I did

 a[a<1.5] = np.nan

I got the following run time warning error in IPython (Py3.4) RuntimeWarning: invalid value encountered in less . Is this because my list had np.nan ? Is there anything I can do to prevent this?

Also is there a way doing this inline without assigning? Instead of doing

a[a<1.5]=np.nan 
return a 

I can just do

 return a... 

where that .... is something that need filling in.

Answer 1

Is this [ RuntimeWarning ] because my list had np.nan?

Yes.

Is there anything I can do to prevent this?

In your case, this warning can be safely ignored. So that you don't accidentally suppress unrelated warnings, please don't put anything else inside the context manager besides the one line shown .

>>> import numpy as np
>>> a = np.array([1,2,3,4,np.nan])
>>> with np.errstate(invalid='ignore'):
...     a[a<1.5] = np.nan
...     
>>> a
array([ nan,   2.,   3.,   4.,  nan])

This operates in-place, a copy is not created here. To return a copy, with the original a unmodified, prefer the masked array approach .

Answer 2

Another option that gets you to your return statement as desired:

mask = ~np.isnan(a)
mask[mask] &= a[mask] < 1.5
return np.where(mask, np.nan, a)

Example:

def ma_lessthan(arr, num):
    mask = ~np.isnan(arr)
    mask[mask] &= arr[mask] < num
    return np.where(mask, np.nan, arr)

print(ma_lessthan(a, 1.5))
[ nan   2.   3.   4.  nan]

mask credit to: @Jaime .