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Epsilon change in normalise for stability #2421
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Codecov ReportAll modified and coverable lines are covered by tests ✅
Additional details and impacted files@@ Coverage Diff @@
## master #2421 +/- ##
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+ Coverage 33.50% 60.45% +26.94%
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Files 31 31
Lines 1910 1942 +32
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+ Hits 640 1174 +534
+ Misses 1270 768 -502 ☔ View full report in Codecov by Sentry. |
I agree with this change and pytorch does the same thing. It should considered a breaking change thous, so let's wait for when we are near v01.5 before merging. |
Co-authored-by: Carlo Lucibello <[email protected]>
Can this have a test with input which triggers the NaN behaviour before? Ideally testing not just the function, but also LayerNorm, maybe BatchNorm, anything which uses this internally. Then if the implementation of these layers finally gets replaced, it will be harder to lose the change. |
Putting a backlink to #2096 because this work should close that. |
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Needs news entry?
@inline function normalise(x::AbstractArray; dims=ndims(x), eps=ofeltype(x, 1e-5)) | ||
@inline function normalise(x::AbstractArray; dims=ndims(x), eps=1f-5) |
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Why does this now assume Float32? Elsewhere we try to allow for Float16 too.
Normalise allows for an optional epsilon term aimed towards improving numerical stability. Previously the epsilon was added after computing the standard deviation of the input. The standard deviation computation involves a square root, leading to NaN's in gradients dependent on normalise when the variance is very low, and for instance LayerNorms applied to low variance inputs will result in NaN gradients. By first computing the variance and taking the square root after adding epsilon^2 (squaring to preserve scale), we prevent NaN's in gradients at low variance. See the following example with LayerNorm in the current patch.
We observe that while the gradients are fixed at low variance due to the epsilon addition in the denominator, this does prevent NaN's, due to the non-padded square root in the std computation. But, when using the updated normalise, these NaN's dissapear,
and remain fixed to the implicitly capped value. A simple test verifying this computation's equivalence with the previous one (modulo the differences at very low standard deviations) could be added if desired.