# Accessing vector fields

**URL:** <https://discourse.julialang.org/t/accessing-vector-fields/61857>\
**Category:** New to Julia\
**Tags:** question\
**Created:** [May 26, 2021, 1:53pm UTC](https://discourse.julialang.org/t/accessing-vector-fields/61857 "2021-05-26T13:53:14Z")\
**Posts on this page:** 10\
**Page:** 1

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**Author:** ![francesco123212](https://avatars.discourse-cdn.com/v4/letter/f/bbe5ce/32.png) [@francesco123212](https://discourse.julialang.org/u/francesco123212)\
**Post date:** [May 26, 2021, 1:53pm UTC](https://discourse.julialang.org/t/accessing-vector-fields/61857/1 "2021-05-26T13:53:14Z")

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Hi everyone, hope you are all healthy.

I’m a mechanical engineering student and I come from MATLAB, so I am sorry for my heresies.

I currently am working on a NASTRAN .hdf5 output. There is a [package](https://github.com/JuliaIO/HDF5.jl) to open such files, that stores informations in a vector with several fields (hope it is the correct word!) as shown in the following piece of code:

```julia
GRIDcard
111403-element Array{HDF5.HDF5Compound{7},1}:
 HDF5.HDF5Compound{7}((567377, 0, [-76.9957, -13.1615, -82.5], 0, 0, 0, 1), ("ID", "CP", "X", "CD", "PS", "SEID", "DOMAIN_ID"), (Int64, Int64, HDF5.FixedArray{Float64,(3,)}, Int64, Int64, Int64, Int64))
 HDF5.HDF5Compound{7}((567378, 0, [-74.1493, -12.1665, -82.5], 0, 0, 0, 1), ("ID", "CP", "X", "CD", "PS", "SEID", "DOMAIN_ID"), (Int64, Int64, HDF5.FixedArray{Float64,(3,)}, Int64, Int64, Int64, Int64))

```

I would like to save all X entries in a single matrix defined as:

```julia
GRID = [
X[1]'
X[2]'
...
X[n]'
]

```

Is it possible to do it without a for loop that reads all entries as:

```julia
GRID = zeros(n,3)
for i = 1:n
    GRID[i,:] = GRIDcard[i].data[3]
end

```

Thank you all in advance, and I hope I haven’t broken too many rules!

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**Author:** ![jlchan](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/jlchan/32/10958_2.png) [@jlchan](https://discourse.julialang.org/u/jlchan)\
**Post date:** [May 26, 2021, 1:58pm UTC](https://discourse.julialang.org/t/accessing-vector-fields/61857/2 "2021-05-26T13:58:00Z")

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Welcome to the Julia discourse!

Loops are generally fast in Julia, but if you want to do it without a loop, you could use a comprehension

```julia
GRID = vcat([GRIDcard[i].data[3] for i = 1:n]...).

```

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**Author:** ![rdeits](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/rdeits/32/286_2.png) [@rdeits](https://discourse.julialang.org/u/rdeits)\
**Post date:** [May 26, 2021, 2:08pm UTC](https://discourse.julialang.org/t/accessing-vector-fields/61857/3 "2021-05-26T14:08:03Z")

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Note that using `vcat` in this way is unnecessarily slow both at compile time (because you end up calling a function with `n` arguments, and `n` can be very large), and at run time (because Julia isn’t well-optimized to handle function calls with huge numbers or arguments).

(I’m demonstrating here with `hcat` instead, since that seems closer to the loop result in the original code, but the result applies to vcat or hcat).

For example:

```julia
julia> data = [rand(3) for _ in 1:1000000];

julia> @btime hcat($data...);
  38.808 ms (7 allocations: 38.15 MiB)

```

You’ll get the same result with better performance from `reduce(hcat, ...)`, like this:

```julia
julia> @btime reduce(hcat, $data);
  13.763 ms (2 allocations: 22.89 MiB)

```

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**Author:** ![francesco123212](https://avatars.discourse-cdn.com/v4/letter/f/bbe5ce/32.png) [@francesco123212](https://discourse.julialang.org/u/francesco123212)\
**Post date:** [May 26, 2021, 2:09pm UTC](https://discourse.julialang.org/t/accessing-vector-fields/61857/4 "2021-05-26T14:09:19Z")

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Thanks a lot!

Isn’t it possible to do something like GRIDcard[:].data[3]?

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**Author:** ![jlchan](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/jlchan/32/10958_2.png) [@jlchan](https://discourse.julialang.org/u/jlchan)\
**Post date:** [May 26, 2021, 2:09pm UTC](https://discourse.julialang.org/t/accessing-vector-fields/61857/5 "2021-05-26T14:09:42Z")

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Feels like I learn a new trick every day on this site 😄

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**Author:** ![francesco123212](https://avatars.discourse-cdn.com/v4/letter/f/bbe5ce/32.png) [@francesco123212](https://discourse.julialang.org/u/francesco123212)\
**Post date:** [May 26, 2021, 2:11pm UTC](https://discourse.julialang.org/t/accessing-vector-fields/61857/6 "2021-05-26T14:11:01Z")

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> [@francesco123212](#):
>
> Isn’t it possible to do something like GRIDcard[:].data[3]?

or would be anyway as slow as a for or hcat?

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**Author:** ![rdeits](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/rdeits/32/286_2.png) [@rdeits](https://discourse.julialang.org/u/rdeits)\
**Post date:** [May 26, 2021, 2:16pm UTC](https://discourse.julialang.org/t/accessing-vector-fields/61857/7 "2021-05-26T14:16:35Z")

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> [@francesco123212](#):
>
> Isn’t it possible to do something like GRIDcard[:].data[3]?

I can see how that would be useful, but no–it’s not a supported syntax in Julia. It is the kind of thing that could probably be implemented in a macro (kind of like how [GitHub - mcabbott/Tullio.jl: ⅀](https://github.com/mcabbott/Tullio.jl) implements Einstein summation), but I don’t know of an existing implementation.

I think the primary thing to take away here is not the particular trick of how to use `hcat` most efficiently. The most important thing is what @jlchan said at the very beginning:

> [@jlchan](#):
>
> Loops are generally fast in Julia

It’s hard to overstate how important this is. Your initial post included a loop which was easy to write and easy to understand, and simple loops like that are often _the fastest_ way to do things in Julia. If you can describe what you want in a loop, then you can often just declare victory and move on to the next problem.

By the way, if you do want to improve performance a bit, you might try transposing the way you store your matrix. Julia matrices are column-major (this is like fortran and Matlab and unlike numpy and C), so it’s usually better to store data that are accessed together as columns instead of rows. That would mean making `GRID` into a `3xn` matrix instead of `nx3` and accessing it as `GRID[:, i]`.

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<div class="post-metadata">

**Author:** ![francesco123212](https://avatars.discourse-cdn.com/v4/letter/f/bbe5ce/32.png) [@francesco123212](https://discourse.julialang.org/u/francesco123212)\
**Post date:** [May 26, 2021, 2:24pm UTC](https://discourse.julialang.org/t/accessing-vector-fields/61857/8 "2021-05-26T14:24:16Z")

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> [@rdeits](#):
>
> Julia matrices are column-major

didn’t know that, thanks a lot, now I see why the NASTRAN’s (written in fortran) output is `nx3`!

The best way would be to modify the package I guess to get exactly what I want from the .h5 file, right?

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<div class="post-metadata">

**Author:** ![rdeits](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/rdeits/32/286_2.png) [@rdeits](https://discourse.julialang.org/u/rdeits)\
**Post date:** [May 26, 2021, 2:26pm UTC](https://discourse.julialang.org/t/accessing-vector-fields/61857/9 "2021-05-26T14:26:08Z")

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If you think the modification would be of use to other users of the package, then sure! But there’s no performance difference between package code or user code in Julia, so you might find it easier to just write a function to do whatever data manipulation you need and keep that function in your own code.

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<div class="post-metadata">

**Author:** ![francesco123212](https://avatars.discourse-cdn.com/v4/letter/f/bbe5ce/32.png) [@francesco123212](https://discourse.julialang.org/u/francesco123212)\
**Post date:** [May 26, 2021, 2:27pm UTC](https://discourse.julialang.org/t/accessing-vector-fields/61857/10 "2021-05-26T14:27:08Z")

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Okay, thank you very much, have a nice day!