Array indexing in numpy

Question

Is there a way in numpy to retrieve all items in an array except the item of the index provided.

 x = 
 array([[[4, 2, 3],
    [2, 0, 1],
    [1, 3, 4]],

   [[2, 1, 2],
    [3, 2, 3],
    [3, 4, 2]],

   [[2, 4, 1],
    [0, 2, 2],
    [4, 0, 0]]])

and by asking for

x[not 1,:,:]

you will get

array([[[4, 2, 3],
    [2, 0, 1],
    [1, 3, 4]],

   [[2, 4, 1],
    [0, 2, 2],
    [4, 0, 0]]])

Thanks

so what if x = np.random.ranint(10,size=(1000,3,3)) and I am trying to get a view of the matrix without the ith 3X3 matrix. Your approach cant be generalized :) — JustInTime
– JustInTime, Commented Jan 3, 2012 at 12:43
For the general case, I think indexing with a tuple would be easiest.. although maybe there is some numpy magic that I'm not aware of! — wim
– wim, Commented Jan 3, 2012 at 12:46

unutbu · Accepted Answer · 2012-01-03 12:44:56Z

6

In [42]: x[np.arange(x.shape[0])!=1,:,:]
Out[42]: 
array([[[4, 2, 3],
        [2, 0, 1],
        [1, 3, 4]],

       [[2, 4, 1],
        [0, 2, 2],
        [4, 0, 0]]])

answered Jan 3, 2012 at 12:44

unutbu

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3 Comments

eumiro Over a year ago

+1 This is just beautiful! With x[np.arange(x.shape[0])!=1,:,:] it would be even perfect :-)

unutbu Over a year ago

Thanks for the improvement, eurimo.

wim Over a year ago

I knew unutbu would chime in with a nice answer :) even x[np.arange(x.shape[0]) != 1] will work, the other dims will take : by default

jcollado · Accepted Answer · 2012-01-03 12:40:29Z

2

Have you tried this?

a[(0,2), :, :]

Instead of blacklisting what you don't want to get, you can try to whitelist what you need.

If you need to blacklist anyway, you can do something like this:

a[[i for i in range(a.shape[0]) if i != 1], :, :]

Basically you just create a list with all possible indexes (range(a.shape[0])) and filter out those that you don't want to get displayed (if i != 1).

answered Jan 3, 2012 at 12:40

jcollado

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2 Comments

JustInTime Over a year ago

What if I have 1000 2D matrices ?? should it be (0,2,3,4,...).

jcollado Over a year ago

@JustInTime I'm not sure I understand, but the [i for i in range(a.shape[0]) if i != 1] expression will let you create a mask for all the indexes you need except 1 regardless of number of matrices because of a.shape[0] usage.

eumiro · Accepted Answer · 2012-01-03 12:43:48Z

2

This is quite a generic solution:

x[range(0,i)+range(i+1,x.shape[0]),:,:]

answered Jan 3, 2012 at 12:43

eumiro

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