Corrections for Julia 0.6.0 #13
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src/performance_profiles.jl
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| @@ -13,21 +13,22 @@ function performance_ratios(T :: Array{Float64,2}; logscale :: Bool=true) | |||
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| (np, ns) = size(T); # Number of problems and number of solvers. | |||
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| T[isinf(T)] = NaN; | |||
| T[T .< 0] = NaN; | |||
| T[isinf.(T)] = Inf; | |||
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Is this one really necessary then? Does it do anything?!
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I think is should be T[isnan.(T)] = Inf? This way, maximum is also kept.
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You're right it should be isnan instead of is inf. But I don't see why maximum should be kept. At least on my tests with isnan the maximum function still produce bad performance profiles
src/performance_profiles.jl
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| logscale && (r = log2(r)); | ||
| max_ratio = maximum(r); | ||
| logscale && (r = log2.(r)); | ||
| max_ratio = findmax(r)[1] |
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In Julia 0.5.2 maximum([2.0, NaN]) returned 2.0 and findmax([2.0, NaN])[1] returned 2.0 as well. But in Julia 0.6.0 maximum([2.0, NaN]) returns NaN and findmax([2.0, NaN])[1] returns 2.0. That's why I changed it.
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I think that's a bug in Julia. See
- Inconsistency between
findmaxandmaximumwith respect toNaNJuliaLang/julia#22337 and - https://discourse.julialang.org/t/inconsistency-between-findmax-and-maximum-with-respect-to-nan/4214/3.
It's time to use NaNMath.jl:
julia> NaNMath.max(2, NaN)
2.0
julia> NaNMath.maximum([2, NaN])
2.0
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Thanks. We also need to add |
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I've added NaNMath in REQUIRE! |
REQUIRE
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| @@ -1,2 +1,5 @@ | |||
| julia 0.5 | |||
| Plots | |||
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| julia 0.6 | |||
| Plots, NaNMath | |||
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Just add NaNMath to the list on line 3.
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ps: the failure on Windows 32 bit is known and fixed in 0.7-dev : JuliaPlots/Plots.jl#968. It may be backported to Julia 0.6?! |
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Thank you! |
I've made some changes so that the performance profiles appear correctly.
Basically, NaN have been changed to Inf and then to NaN again. Now it outputs the same things that it did in Julia 0.5.2.