2018-02-06 11:32:46 +00:00
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using Base.Test
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using Flux: onehotbatch, mse, crossentropy, logitcrossentropy,
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σ, binarycrossentropy, logitbinarycrossentropy
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2017-12-05 23:38:15 +00:00
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@testset "losses" begin
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# First, regression-style y's
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y = [1, 1, 0, 0]
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2018-02-06 11:32:46 +00:00
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ŷ = [.9, .1, .1, .9]
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2017-12-05 23:38:15 +00:00
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@testset "mse" begin
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2018-02-06 11:32:46 +00:00
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@test mse(ŷ, y) ≈ (.1^2 + .9^2)/2
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2017-12-05 23:38:15 +00:00
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end
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# Now onehot y's
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y = onehotbatch([1, 1, 0, 0], 0:1)
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2018-02-06 11:32:46 +00:00
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ŷ = [.1 .9; .9 .1; .9 .1; .1 .9]'
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v = log(.1 / .9)
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logŷ = [v 0.0; 0.0 v; 0.0 v; v 0.0]'
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lossvalue = 1.203972804325936
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2017-12-05 23:38:15 +00:00
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@testset "crossentropy" begin
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2018-02-06 11:32:46 +00:00
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@test crossentropy(ŷ, y) ≈ lossvalue
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end
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@testset "logitcrossentropy" begin
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@test logitcrossentropy(logŷ, y) ≈ lossvalue
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2017-12-05 23:38:15 +00:00
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end
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@testset "weighted_crossentropy" begin
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2018-02-06 11:32:46 +00:00
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@test crossentropy(ŷ, y, weight = ones(2)) ≈ lossvalue
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@test crossentropy(ŷ, y, weight = [.5, .5]) ≈ lossvalue/2
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@test crossentropy(ŷ, y, weight = [2, .5]) ≈ 1.5049660054074199
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end
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@testset "weighted_logitcrossentropy" begin
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@test logitcrossentropy(logŷ, y, weight = ones(2)) ≈ lossvalue
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@test logitcrossentropy(logŷ, y, weight = [.5, .5]) ≈ lossvalue/2
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@test logitcrossentropy(logŷ, y, weight = [2, .5]) ≈ 1.5049660054074199
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end
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logŷ, y = randn(3), rand(3)
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@testset "binarycrossentropy" begin
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@test binarycrossentropy.(σ.(logŷ), y) ≈ -y.*log.(σ.(logŷ)) - (1 - y).*log.(1 - σ.(logŷ))
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end
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@testset "logitbinarycrossentropy" begin
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@test logitbinarycrossentropy.(logŷ, y) ≈ binarycrossentropy.(σ.(logŷ), y)
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2017-12-05 23:38:15 +00:00
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end
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end
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