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Prevent overflow in mean(::AbstractRange) and relax type constraint #115

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Prevent overflow in mean(::AbstractRange) and relax type constraint #115

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12 changes: 6 additions & 6 deletions src/Statistics.jl
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Original file line number Diff line number Diff line change
Expand Up @@ -182,9 +182,9 @@ function _mean(f, A::AbstractArray, dims::Dims=:) where Dims
end
end

function mean(r::AbstractRange{<:Real})
isempty(r) && return oftype((first(r) + last(r)) / 2, NaN)
(first(r) + last(r)) / 2
function mean(r::AbstractRange{T}) where T
isempty(r) && return zero(T)/0
return first(r)/2 + last(r)/2
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Suggested change
return first(r)/2 + last(r)/2
return middle(r)

This method is already defined appropriately for ranges and behaves as desired when the input is non-empty:

julia> middle(typemax(Int):typemax(Int)) ≈ typemax(Int)
true

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It would be less performant to call middle because it currently doesn't elide bounds checks and a[end] is unfortunately slow for StepRange (it calls lastindex, which calls length, which requires a division). In contrast, first and last simply accesses fields of the range struct (except for StepRangeLen).

I've just opened a PR (#116) to have middle call mean instead. Would you mind taking a look?

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Given that both mean here and middle at #116 check that the input is not empty, better not have one call the other and instead duplicate the (very short) computation?

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Sure, sounds good to me. I'll update #116 to not call mean.

end

median(r::AbstractRange{<:Real}) = mean(r)
Expand Down Expand Up @@ -997,9 +997,9 @@ end
require_one_based_indexing(v)

n = length(v)

@assert n > 0 # this case should never happen here

m = alpha + p * (one(alpha) - alpha - beta)
aleph = n*p + oftype(p, m)
j = clamp(trunc(Int, aleph), 1, n-1)
Expand All @@ -1012,7 +1012,7 @@ end
a = v[j]
b = v[j + 1]
end

if isfinite(a) && isfinite(b)
return a + γ*(b-a)
else
Expand Down
20 changes: 14 additions & 6 deletions test/runtests.jl
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Expand Up @@ -171,8 +171,16 @@ end
@test f(2:0.1:n) ≈ f([2:0.1:n;])
end
end
@test mean(2:1) === NaN
@test mean(big(2):1) isa BigFloat
@test mean(2:0.1:4) === 3.0 # N.B. mean([2:0.1:4;]) != 3
@test mean(LinRange(2im, 4im, 21)) === 3.0im
@test mean(2:1//10:4) === 3//1
@test isnan(@inferred(mean(2:1))::Float64)
@test isnan(@inferred(mean(big(2):1))::BigFloat)
z = @inferred(mean(LinRange(2im, 1im, 0)))::ComplexF64
@test isnan(real(z)) & isnan(imag(z))
@test_throws DivideError mean(2//1:1)
@test mean(typemax(Int):typemax(Int)) === float(typemax(Int))
@test mean(prevfloat(Inf):prevfloat(Inf)) === prevfloat(Inf)
end

@testset "var & std" begin
Expand Down Expand Up @@ -547,16 +555,16 @@ end
@test cor(tmp, tmp) <= 1.0
@test cor(tmp, tmp2) <= 1.0
end

@test cor(Int[]) === 1.0
@test cor([im]) === 1.0 + 0.0im
@test_throws MethodError cor([])
@test_throws MethodError cor(Any[1.0])

@test cor([1, missing]) === 1.0
@test ismissing(cor([missing]))
@test_throws MethodError cor(Any[1.0, missing])

@test Statistics.corm([true], 1.0) === 1.0
@test_throws MethodError Statistics.corm(Any[0.0, 1.0], 0.5)
@test Statistics.corzm([true]) === 1.0
Expand Down Expand Up @@ -958,4 +966,4 @@ end
@test isequal(cor(mx, Int[]), fill(NaN, 2, 1))
@test isequal(cov(Int[], my), fill(-0.0, 1, 3))
@test isequal(cor(Int[], my), fill(NaN, 1, 3))
end
end