I noticed the following oddity println(findmin(identity, x for x in [3, 2, 1])) println(findmin(identity, x for x in [3, 2, 1] if x != 1)) showing (1, 3) ERROR: MethodError: no method matching keys(::Base.Iterators.Filter{var"#9#10", Vector{Int64}}) Do you know an easy fix for this?

This seems to fix the issue: julia> import Base: keys julia> keys(itr::Base.Iterators.Filter) = Iterators.map(first, enumerate(itr)) julia> println(findmin(identity, x for x in [3, 2, 1] if x != 1)) (2, 2) Might be a good idea to make an appropriate change in Base. Not sure this fix is ideal (as…

This update, moves the fix up the call chain and prevents the double cycling of iterator. On the other hand, I’m not familiar enough with other side-effects this fix might have. julia> import Base: pairs julia> pairs(g::Base.Generator{T,I}) where {T<:Base.Iterators.Filter, I} = Iterators.…

adding parens works julia> findmin(identity, (x for x in [3, 2, 1] if x != 1)) (2, 2)

You shouldn’t be using findmin here, but instead minimum. findmin returns a tuple of the minimum value and its index in the collection or iterator. If you pass to it a filtered iterator, the index is meaningless, so it rightly throws a fit. Thus, to get it to work, you’re forced to collect all the …

@uniment is right. It is important to get the semantics accurately. Perhaps what we are looking for is: julia> minimum(x => x * x for x in (6,5,4) if x != 4) 5 => 25 The confusing bit is that above expression calculates the mathematical argmin of the expression x^2 in the set. The Julia argmin fu…

[image] Dan: Perhaps what we are looking for is: julia> minimum(x => x * x for x in (6,5,4) if x != 4) 5 => 25 The confusing bit is that above expression calculates the mathematical argmin of the expression x^2 in the set. The Julia argmin function calculates the minimum and its index, which i…

If you’re using findmin because you’re hoping to find both the minimum squared value and its index, for any value not equal to 1, then this is likely what you’re looking for: julia> findmin(x==1 ? Inf : x^2 for x ∈ (4, 3, 2, 1)) (4, 3)

Again @uniment , you are right. A Pair is ordered by key first and value later. Somehow I had assumed it was ordered by value first. To get the result I wanted: julia> minimum((x * x, x) for x in (6,5,4) if x != 4) (25, 5) which is the same output, but gets things right when function is not order …

julia> findmin(x==1 ? Inf : x^2 for x ∈ (4, 3, 2, 1)) (4, 3) This is: a. not type stable. b. the second tuple number is somewhat useless, you would want the value of x for an argmin of the function x^2.

Findmin/max and generators

General Usage

goerch October 11, 2022, 2:19am 5

@Dan: to make this work for

@btime findmin(x * x for x in (3, 2, 1) if x != 1)

I had to use

Base.pairs(g::Base.Generator{I,F}) where {I<:Base.Iterators.Filter, F<: Function} = Iterators.map(=>, 1:typemax(Int), g)

To all: thanks for your help!

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Findmin/max and generators

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