In the following code, I can’t understand why sum(a[1,:],dims=2) gives a different answer then the first row of sum(a[1:2,:],dims=2). julia> a = [1 2 3 4; 5 6 7 8; 9 10 11 12]; julia> sum(a, dims=2) 3×1 Array{Int64,2}: 10 26 42 julia> sum(a[1:2,:], dims=2) 2×1 Array{Int64,2}: 10 26 julia> s…

When tackling things like this, I always recommend simplifying things as much as possible to figure it out. In this case, removing the sum does the trick: [image] kw_martin: julia> a = [1 2 3 4; 5 6 7 8; 9 10 11 12]; julia> a[1:2,:] 2×4 Array{Int64,2}: 1 2 3 4 5 6 7 8 julia> a[1,:] …

mbauman, thank you. I could only find two examples in indexing that show this but I couldn’t find any discussions; I scanned through it twice, but didn’t read word for word so I might have missed it. I come from a Matlab background where the shapes and dimensions of matrices are not changed. Indeed,…

The reason Julia drops dimensions is for consistency. If A is a matrix, most people want A[i,j] to be a scalar. In matlab, scalars don’t exist – this is produces a 1x1 matrix. This sucks for speed and efficiency and also just common sense. The generalization of this rule is that every scalar index …

[image] kw_martin: a = [1 2 3 4; 5 6 7 8; 9 10 11 12]; Oscar, thank you for taking the time to reply. I’m certain this can’t be changed at this time, but I don’t think I agree with everything you are saying. “The reason Julia drops dimensions is for consistency. If A is a matrix, most people…

[image] kw_martin: I certainly agree that indexing a single element should produce a scalar. But I don’t agree that indexing multiple elements should always drop the dimensions to be consistent. I see these as being different operations and being consistent here achieves little functionality. …

[image] kw_martin: Do you know why vectors are actually needed? I will leave the other questions aside, but I have to say that the absence of proper vectors in Matlab is a huge annoyance. It’s causing me endless trouble almost every time I use Matlab (which is at least 5 days a week.) You rea…

The reason scalar indexing drops dimensions is a little more subtle than I explained initially. The idea is that A[i,:][j]=A[i,j] is an identity that feels like it should hold. Since A[i,j] is a scalar, A[i,:] needs to be a type which when indexed by 1 index gives a scalar. The fact that Julia has N…

We could indeed use a bit more in the documentation on this rule — right now it’s very tersely described as: X = A[I_1, I_2, ..., I_n] […] X is an array with the same number of dimensions as the sum of the dimensionalities of all the indices. […] If I_1 is changed to a two-dimensional matri…

Trouble Understanding Slicing

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Trouble Understanding Slicing

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