Is there any way to use max() and min() based on a function?

General Usage
1

This is easily done in, for example, Python, and can be very useful. I have spent quite a bit of time searching documentation and reading forum questions, without finding anything.

Here’s a small example of what I would like to do:

arr = [1,2,3,4,-5]
sqmax = max(arr, by = x -> x^2)

I know this can be done as follows, but it seems like it would make sense for max() to have a by option.

sqmax = sort(arr, by = x -> x^2)[end]
2

You are looking for what is called maximum in Julia. Take a look at the documentation on max and maximum for more examples.

Welcome to the forum. A question like this is better suited for the “Usage” category.

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3

But still there is no built in maxiumum function offering whats requested.

I’d go with:
findmax(map(x->x^2,arr))

Update: this is only needed if you want the index, which I initially thought was asked.

4

maximum does take an optional function as its first argument. Try maximum(x -> x^2, arr).

6 Likes
5

why not:

maximum(arr.^2)
6

The problem with that is that it creates a temporary array, which can be avoided by passing the function as an argument.

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7

That allocates an intermediate array for the squared array whereas the version from simon does not. It would be nice if we could reduce over a broadcasted array without allocs one day but we’re not there yet.

8

thanks for the answer, but maximum(f, arr) will return the maximal element in f(arr) rather than in arr.
Is there any keyword of maximum or other functons to “find x in arr s.t. f(x) is maximal”

9

Example:

xs = [1, 3, 2]; i = argmax([x^2 for x in xs]); println(xs[i])
10

You can also pass the function to argmax, to avoid the allocation of the intermediate array:

julia> x = [1,3,2]
3-element Vector{Int64}:
 1
 3
 2

julia> argmax(x -> -x^2, x)
1

julia> argmax(x -> x^2, x)
3
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11 
julia> findmax(abs2, -3:2)
(9, 1)

the second returned value can be used to index the arr

or just use:

julia> argmax(abs2, -3:2)
-3

which does the indexing for you under the hood

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12 
foldl((x,y)->x^2>y^2 ? x : y, arr)

or

foldl((x,y)->ifelse(x^2>y^2 , x , y), arr)

maxby(f,arr)=foldl((x,y)->f(x)>f(y) ? x : y, arr)
1 Like
13

Yes, for those wondering, this feature was added in Julia v1.7. Quite useful/convenient on occasion.

14

Open issue up for grabs: Add a `by` keyword to `maximum` and `minimum` · Issue #28210 · JuliaLang/julia · GitHub

(Of course, you could just write your own loop for this, which is perfectly fast.)

1 Like
15

Thanks! I found out this implementation is new in Julia 1.7. For 1.6 or lower version, it still can be done by a less direct way

data[argmax(f.data)]