How to modify one column of a selected row from a numpy structured array

Question

I am looking for an easy way to modify one field of a numpy structured array of a selected line of it. Here is my SWE :

import numpy as np
dt=np.dtype([('name',np.unicode,80),('x',np.float),('y',np.float)])
a=np.array( [('a',0.,0.),('b',0.,0.),('c',0.,0.) ],dtype=dt)
b=a.copy()
a[a['name']=='a']['x']=1
print a==b # return [ True  True  True]

In this example, the a==b results should return [False True True].Actually, I would like to selected the line of my array from the 'name' field and modify the value of one field of it (here 'x').

It's time to go to pandas ;). df=pandas.Dataframe(a) ; df.loc[df.name=='a','x']=1. — B. M.
– B. M., Commented Apr 18, 2016 at 17:05

Sigmun · Accepted Answer · 2016-04-18 12:04:11Z

2

I found the answer... The point is the position of the field and of the mask. You need to apply the mask to the field column, not looking for the field of the masked array :

a['x'][a['name']=='a']=1
print a==b # returns [False  True  True]

answered Apr 18, 2016 at 12:04

Sigmun

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1 Comment

unutbu Over a year ago

The order of indexing matters because "advanced indexing", such as a[a['name']=='a'], always returns a copy, while "basic indexing" such as a['x'] always returns view. Modifying the view with a['x'][a['name']=='a']=1 affects the original array a, while modifying the copy with a[a['name']=='a']['x']=1 does not affect a.

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