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 !

• Cheezey: 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?
• R.Bahl: 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.

Answer 1

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

1. x[:,1] = 1
2. x[:,2] *= x[:,1]
3. x[:,1] = np.where( x[:,0] > 5, 0, 4 )

• R.Bahl: Jon Clements: Thanks for this ! I was indexing in the wrong way, something like x[:][1] .. Thanks again for this !
• Sven Marnach: Number 2 can be improved by using x[:,2] *= x[:,1] – this saves the creation of a temporary array.
• Jon Clements: @Sven good point - not sure why I didn't write that - thanks - edited