Hi Mukulikaa,
these all sound great to me! I have used numpy for a while, and it took me a long time to figure out how to index arrays efficiently (tbh, I'd rather say I am still in the process of it). Could I help you with writing this how-doc?

Hi @AnjaTRPES, I would love to use your help! I'm curious to know if you referred to any specific resources or used special tricks to learn how to index arrays efficiently. I will make a draft PR outlining the doc soon (to make it easier to collaborate) and I'd appreciate your inputs for best practices there.

Nice @Mukulikaa - it might be useful to ask in the mailing list as well and see if there are any extra suggestions there.

I can instantly think of two different patterns that are hard to figure out on your own, and confusing when you see them for the first time, but fairly easy to understand and learn to use if you've seen them before (and an "indexing howto" might be a potential place for this!).

The first pattern that can confound new users while being a fairly basic use case for broadcasting and fancy indexing is

I have a 2d array of shape (n, m), and for each row I have k column indices. How do I get the (n, k)-shaped values I'm looking for?

(Note sure if this is included in your "Indexing with index arrays of shapes different from the ndarray" bullet point.) The solution being a range in the remaining leading dimension(s):

arr = np.arange(3*4).reshape(3, 4)
column_indices = [[1, 3], [0, 2], [2, 2]]
print(arr[np.arange(arr.shape[0])[:,None], column_indices])

More generally you'd need an mgrid (ogrid) in every leading dimension, with an additional trailing singleton dimension.

The second pattern is only a partial match for the "indexing" topic, because there are multiple ways to solve the problem. This is

If I give you a permutation index array, can you tell me the inverse?

I.e. if you have b = a[inds], how can you get a = b[inverse_inds]? I know two solutions off the top of my head:

a = np.arange(20, 10, -1)
rng = np.random.default_rng()
inds = rng.permutation(a.size)
b = a[inds]

# solution 1: argsort the indices
inverse_inds = inds.argsort()

# solution 2: index into the left-hand side
inverse_inds2 = np.empty_like(inds)
inverse_inds2[inds] = np.arange(inds.size)

print(np.array_equal(inverse_inds, inverse_inds2))  # True
print(np.array_equal(a, b[inverse_inds]))  # True

@Mukulikaa your suggested list of topics in the OP seems like a great start for a how-to. Keep in mind that the document doesn't have to be completely comprehensive, so I would recommend not to worry about getting all of the bullets in the first version of the document. I think this will be a very nice addition to the docs!

Hey @adeak, thanks for your helpful inputs!

And, @rossbar - will definitely keep your point in mind 🙂

I believe there's nothing else to do here so closing. Thanks everyone!