Reputation: 163
I've run into this interaction with arrays that I'm a little confused. I can work around it, but for my own understanding, I'd like to know what is going on.
Essentially, I have a datafile that I'm trying to tailor so I can run this as an input for some code I've already written. This involves some calculations on some columns, rows, etc. In particular, I also need to rearrange some elements, where the original array isn't being modified as I expect it would.
import numpy as np
ex_data = np.arange(12).reshape(4,3)
ex_data[2,0] = 0 #Constructing some fake data
ex_data[ex_data[:,0] == 0][:,1] = 3
print ex_data
Basically, I look in a column of interest, collect all the rows where that column contains a parameter value of interest and just reassigning values.
With the snippet of code above, I would expect ex_data to have it's column 1 elements, conditional if it's column 0 element is equal to 0, to be assigned a value of 3. However what I'm seeing is that there is no effect at all.
>>> ex_data
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 0, 7, 8],
[ 9, 10, 11]])
In another case, if I don't 'slice', my 'sliced' data file, then the reassignment goes on as normal.
ex_data[ex_data[:,0] == 0] = 3
print ex_data
Here I'd expect my entire row, conditional to where column 0 is equal to 0, be populated with 3. This is what you see.
>>> ex_data
array([[ 3, 3, 3],
[ 3, 4, 5],
[ 3, 3, 3],
[ 9, 10, 11]])
Can anyone explain the interaction?
Upvotes: 1
Views: 1511
Reputation: 231385
In [368]: ex_data
Out[368]:
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 0, 7, 8],
[ 9, 10, 11]])
The column 0 test:
In [369]: ex_data[:,0]==0
Out[369]: array([ True, False, True, False])
That boolean mask can be applied to the rows as:
In [370]: ex_data[ex_data[:,0]==0,0]
Out[370]: array([0, 0]) # the 0's you expected
In [371]: ex_data[ex_data[:,0]==0,1]
Out[371]: array([1, 7]) # the col 1 values you want to replace
In [372]: ex_data[ex_data[:,0]==0,1] = 3
In [373]: ex_data
Out[373]:
array([[ 0, 3, 2],
[ 3, 4, 5],
[ 0, 3, 8],
[ 9, 10, 11]])
The indexing you tried:
In [374]: ex_data[ex_data[:,0]==0]
Out[374]:
array([[0, 3, 2],
[0, 3, 8]])
produces a copy. Assigning ...[:,1]=3
just changes that copy, not the original array. Fortunately in this case, it is easy to use
ex_data[ex_data[:,0]==0,1]
instead of
ex_data[ex_data[:,0]==0][:,1]
Upvotes: 3