# Grouping by Split, apply, combine

**URL:** <https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417>\
**Category:** General Usage\
**Tags:** question, package, tuple, splitapplycombine\
**Created:** [July 18, 2022, 5:21pm UTC](https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417 "2022-07-18T17:21:23Z")\
**Posts on this page:** 10\
**Page:** 1

<div class="post-metadata">

**Author:** ![Optimization](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/optimization/32/32462_2.png) [@Optimization](https://discourse.julialang.org/u/Optimization)\
**Post date:** [July 18, 2022, 5:21pm UTC](https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417/1 "2022-07-18T17:21:23Z")

</div>

I have a big set of 4-tuples and would like to group them based on their third position. As an example:

```julia
using SplitApplyCombine
A = [(1,5,2,5), (,12,5,9,6), (7,5,2,6), (11,5,2,5), (8,5,9,6), (3,5,3,6), (7,5,9,6), (1,5,3,6)]
Third = [2,9,3] 

```

I used `group(3, A)` and `group(Third, A)` but both returns error. How to group them?

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**Author:** ![pdeffebach](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/pdeffebach/32/10320_2.png) [@pdeffebach](https://discourse.julialang.org/u/pdeffebach)\
**Post date:** [July 18, 2022, 5:44pm UTC](https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417/2 "2022-07-18T17:44:32Z")

</div>

```julia
julia> group(t -> t[3], A)
3-element Dictionaries.Dictionary{Int64, Vector{NTuple{4, Int64}}}
 2 │ [(1, 5, 2, 5), (7, 5, 2, 6), (11, 5, 2, 5)]
 9 │ [(12, 5, 9, 6), (8, 5, 9, 6), (7, 5, 9, 6)]
 3 │ [(3, 5, 3, 6), (1, 5, 3, 6)]

```

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<div class="post-metadata">

**Author:** ![Optimization](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/optimization/32/32462_2.png) [@Optimization](https://discourse.julialang.org/u/Optimization)\
**Post date:** [July 18, 2022, 6:31pm UTC](https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417/3 "2022-07-18T18:31:11Z")

</div>

Hi @pdeffebach Thank you! It’s great. Is there any way to impose extra condition via group??

Like filtering all element in each of the keys, if the first element in in certain list? In other words, if `grouped = group(t -> t[3], A)` now that we know grouped(2) has `3-element Vector{Tuple{Int64, Int64, Int64}}` is it possible to further filter those whos second element is `11`. Or in general, if associated with each keys we have want to filter those whose let’s say are in `[11, 8, 3]`. So it would return

```julia
2 │ [(11, 5, 2, 5)]
 9 │ [(8, 5, 9, 6)]
 3 │ [(3, 5, 3, 6)]

```

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<div class="post-metadata">

**Author:** ![pdeffebach](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/pdeffebach/32/10320_2.png) [@pdeffebach](https://discourse.julialang.org/u/pdeffebach)\
**Post date:** [July 18, 2022, 7:21pm UTC](https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417/4 "2022-07-18T19:21:00Z")

</div>

Read the documentation of `group`. The key is just to write a function that returns a unique value for each set of requirements. And group based on that function.

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<div class="post-metadata">

**Author:** ![Optimization](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/optimization/32/32462_2.png) [@Optimization](https://discourse.julialang.org/u/Optimization)\
**Post date:** [July 18, 2022, 7:40pm UTC](https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417/5 "2022-07-18T19:40:47Z")

</div>

@pdeffebach Yes, I’m going to reread it. I wish Julia’s documetation were a bit longer.

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<div class="post-metadata">

**Author:** ![rocco\_sprmnt21](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/rocco_sprmnt21/32/20127_2.png) [@rocco\_sprmnt21](https://discourse.julialang.org/u/rocco_sprmnt21)\
**Post date:** [July 18, 2022, 7:49pm UTC](https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417/6 "2022-07-18T19:49:37Z")

</div>

You could filter before or after the group.

There is the possibility of transforming the elements of the groups through a function.  
A solution that comes close to what you are asking could be this:

```julia
group(t->t[3], t->t[1] ∉ [11,8,3] ? () : t, A )

```

But if you really want a function that groups and filters at the same time, you have to build it.  
One could be the following

```julia

function groupbynth(A,n, sieve)
    d=Dict{Int64, Vector{Tuple}}()
    foreach(tA-> tA[1] ∈ sieve ? push!(get!(()->Vector{Tuple}(), d, tA[n]),tA) : nothing, A)
    d
end

```

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<div class="post-metadata">

**Author:** ![Optimization](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/optimization/32/32462_2.png) [@Optimization](https://discourse.julialang.org/u/Optimization)\
**Post date:** [July 18, 2022, 9:10pm UTC](https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417/7 "2022-07-18T21:10:21Z")

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> [@rocco\_sprmnt21](#):
>
> `group(t->t[3], t->t[1] ∉ [11,8,3] ? () : t, A )`

Thanks a lot @rocco_sprmnt21 it was very very helpful!

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<div class="post-metadata">

**Author:** ![rocco\_sprmnt21](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/rocco_sprmnt21/32/20127_2.png) [@rocco\_sprmnt21](https://discourse.julialang.org/u/rocco_sprmnt21)\
**Post date:** [July 18, 2022, 9:46pm UTC](https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417/8 "2022-07-18T21:46:05Z")

</div>

if you need performance, keep in mind that

```julia
using SplitApplyCombine, BenchmarkTools, Dictionaries

A=[Tuple(rand(1:100, 4)) for _ in 1:10^5]

function groupbynth(A,n, sieve)
    d=Dictionary{Int64, Vector{Tuple}}()
    foreach(tA-> tA[1] ∈ sieve ? push!(get!(()->Vector{Tuple}(), d, tA[n]),tA) : nothing, A)
    d
end

julia> @btime groupbynth(A,3, [11, 8, 3])
  893.400 μs (3354 allocations: 202.81 KiB)
100-element Dictionary{Int64, Vector{Tuple}}
   4 │ Tuple[(11, 35, 4, 28), (3, 93, 4, 98), (11, 82, 4, 97), (3, 75, 4, 76), (…
  38 │ Tuple[(8, 63, 38, 95), (11, 46, 38, 23), (3, 3, 38, 1), (3, 70, 38, 80), …
  19 │ Tuple[(8, 67, 19, 42), (11, 13, 19, 78), (11, 88, 19, 10), (3, 10, 19, 89…
  42 │ Tuple[(11, 57, 42, 78), (11, 27, 42, 97), (11, 6, 42, 16), (8, 97, 42, 52…
  99 │ Tuple[(11, 48, 99, 48), (8, 96, 99, 51), (11, 3, 99, 30), (8, 23, 99, 30)…

julia> @btime group(t->t[3], t->t[1] ∉ [11,8,3] ? () : t, A )
  5.605 ms (100637 allocations: 16.33 MiB)
100-element Dictionary{Int64, Vector{Union{Tuple{}, NTuple{4, Int64}}}}
  4 │ Union{Tuple{}, NTuple{4, Int64}}[(11, 35, 4, 28), (), (), (), (), (), (), …
 96 │ Union{Tuple{}, NTuple{4, Int64}}[(), (), (), (), (), (), (), (), (), () ……
 95 │ Union{Tuple{}, NTuple{4, Int64}}[(), (), (), (), (), (), (), (), (), () ……
 63 │ Union{Tuple{}, NTuple{4, Int64}}[(), (), (), (), (), (), (), (), (), () ……

```

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<div class="post-metadata">

**Author:** ![rocco\_sprmnt21](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/rocco_sprmnt21/32/20127_2.png) [@rocco\_sprmnt21](https://discourse.julialang.org/u/rocco_sprmnt21)\
**Post date:** [July 19, 2022, 9:24am UTC](https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417/9 "2022-07-19T09:24:18Z")

</div>

you could also try the `groupreduce` function In the following way( I have no way now to verify the correctness of the expressions, I just go by heart)

```julia
groupreduce(t->t[3],identity, (t1,t2)->t2[1] ∈ [3,8,11] ? push!(t1,t2) : t1, A; init=Tuple[])

```

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<div class="post-metadata">

**Author:** ![rafael.guerra](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/rafael.guerra/32/216610_2.png) [@rafael.guerra](https://discourse.julialang.org/u/rafael.guerra)\
**Post date:** [July 19, 2022, 10:06am UTC](https://discourse.julialang.org/t/grouping-by-split-apply-combine/84417/10 "2022-07-19T10:06:48Z")

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@Optimization, please check annotated code below, where the input was sligthly modified to try removing some ambiguity in the problem statement.

```julia
using SplitApplyCombine

A = [(1,5,2,5), (11,5,9,6), (7,5,2,6), (11,5,2,5), (8,5,9,6), (3,5,3,6), (7,5,9,6), (1,5,3,6)]
D = group(t -> t[3], A) # groups tuples by their 3rd element
b = (11, 8, 3) # tuple for filtering the last 3 groups computed, one value per group

[k => filter(x -> x[1] == v, d) for (v, (k,d)) in zip(b, pairs(D))] # matches first element of each tuple

3-element Vector{Pair{Int64, Vector{NTuple{4, Int64}}}}:
 2 => [(11, 5, 2, 5)]
 9 => [(8, 5, 9, 6)]
 3 => [(3, 5, 3, 6)]

```