Python: efficient operations on numpy arrays

Question

Say I have a numpy array x:

x = array([[  3,   2,   1],
           [  3,  25,  34],
           [ 33, 333,   3],
           [ 43,  32,   2]])

I want to carry out the following operations without explicitly writing a for loop i.e. say a method which uses automatic in built looping;

1) Replace the 2nd column by a column of all 1 i.e.

x = array([[  3,   1,   1],
           [  3,   1,  34],
           [ 33,   1,   3],
           [ 43,   1,   2]])

2) In the original array , replace 3rd column with the product of 2nd and 3rd i.e.

x = array([[  3,   2,   1*2],
           [  3,  25,  34*25],
           [ 33, 333,   3*333],
           [ 43,  32,   2*32]])

3) Finally, I would like to replace the 2nd column in the original array based on a condition i.e.

x[1] = 0  if x[0] > 5 else 4

i.e. the array now looks like:

x = array([[  3,   4,   1],
           [  3,   4,  34],
           [ 33,   0,   3],
           [ 43,   0,   2]])

Any suggestions ? Thanks !

You can do this using a while loop and an iterator variable that is incremented with every loop. May I ask why you can't use a for loop? — Cheezey
– Cheezey, Commented Jul 23, 2012 at 0:27
The array might get very tall, i.e say million elements, which might be repeated 100's of times. Using an explicit for loop is adding a lot of overhead. Therefore, I switched to numpy array in place of lists, hopping that there will be some kind of vectorized solution to this. — R.Bahl
– R.Bahl, Commented Jul 23, 2012 at 0:30

Jon Clements · Accepted Answer · 2012-07-23 00:40:13Z

6

The documentation on numpy is well worth reading as this is fairly basic stuff...

answered Jul 23, 2012 at 0:33

Jon Clements

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3 Comments

Jon Clements: Thanks for this ! I was indexing in the wrong way, something like x[:][1] .. Thanks again for this !

Number 2 can be improved by using x[:,2] *= x[:,1] – this saves the creation of a temporary array.

@Sven good point - not sure why I didn't write that - thanks - edited