我是靠谱客的博主 无限香水,这篇文章主要介绍softmax求导,现在分享给大家,希望可以做个参考。

softmax函数:
S i = e a i ∑ j e a j S_{i}=frac{e^{a_{i}}}{sum_{j} e^{a_{j}}} Si​=∑j​eaj​eai​​

softmax求导

∂ S i ∂ a j = ∂ e a i ∂ a j ⋅ Σ − ∂ Σ ∂ a j ⋅ e a i ∑ 2 frac{partial S_{i}}{partial a_{j}}=frac{frac{partial e^{a_{i}}}{partial a_{j}} cdot Sigma-frac{partial Sigma}{partial a_{j}} cdot e^{a_{i}}}{sum^{2}} ∂aj​∂Si​​=∑2∂aj​∂eai​​⋅Σ−∂aj​∂Σ​⋅eai​​

当 i = j i = j i=j时:
∂ S i ∂ a j = e a i ⋅ Σ − e a j e a i Σ 2 = e a i Σ ⋅ Σ − e a j Σ = S i ⋅ ( 1 − S j ) frac{partial S_{i}}{partial a_{j}}=frac{e^{a_{i}} cdot Sigma-e^{a_{j}} e^{a_{i}}}{Sigma^{2}} = frac{e^{a_{i}}}{Sigma} cdot frac{Sigma-e^{a_{j}}}{Sigma}=S_{i} cdotleft(1-S_{j}right) ∂aj​∂Si​​=Σ2eai​⋅Σ−eaj​eai​​=Σeai​​⋅ΣΣ−eaj​​=Si​⋅(1−Sj​)

当 i ≠ j i neq j i​=j时:
∂ S i ∂ a j = − e α j ⋅ e a i Σ 2 = − S i ⋅ S j frac{partial S_{i}}{partial a_{j}}=-frac{e^{alpha_{j}} cdot e^{a_{i}}}{Sigma^{2}}=-S_{i} cdot S_{j} ∂aj​∂Si​​=−Σ2eαj​⋅eai​​=−Si​⋅Sj​

交叉熵损失函数

L = − ∑ y i log ⁡ S i L=-sum y_{i} log S_{i} L=−∑yi​logSi​

∂ L ∂ S i = − y i ⋅ 1 S i frac{partial L}{partial S_{i}}=-y_{i} cdot frac{1}{S_{i}} ∂Si​∂L​=−yi​⋅Si​1​

对softmax的输入求导:

在这里插入图片描述

∂ L ∂ a i = ∑ j ∂ L ∂ S j ⋅ ∂ S j ∂ a i = ∂ L ∂ S i ⋅ ∂ S i ∂ a i + ∑ j ≠ i ∂ L ∂ S j ⋅ ∂ S j ∂ a i = − y i ⋅ 1 S i ⋅ S i ( 1 − S i ) + ∑ j ≠ i − y j S j ⋅ ( − 1 ) S i S j = − y i ( 1 − S i ) + ∑ j ≠ i y j ⋅ S i = − y i + y i S i + ∑ j ≠ i y j ⋅ S i = S i − y i frac{partial L}{partial a_{i}}=sum_{j} frac{partial L}{partial S_{j}} cdot frac{partial S_{j}}{partial a_{i}}=frac{partial L}{partial S_{i}} cdot frac{partial S_{i}}{partial a_{i}}+sum_{j neq i} frac{partial L}{partial S_{j}} cdot frac{partial S_{j}}{partial a_{i}} \ =-y_{i} cdot frac{1}{S_{i}} cdot S_{i}left(1-S_{i}right)+sum_{j neq i} frac{-y_{j}}{S_{j}} cdot(-1) S_{i} S_{j} \ =-y_{i}left(1-S_{i}right)+sum_{j neq i} y_{j} cdot S_{i} \ =-y_{i}+y_{i} S_{i}+sum_{j neq i} y_{j} cdot S_{i} \ =S_{i}-y_{i} ∂ai​∂L​=j∑​∂Sj​∂L​⋅∂ai​∂Sj​​=∂Si​∂L​⋅∂ai​∂Si​​+j​=i∑​∂Sj​∂L​⋅∂ai​∂Sj​​=−yi​⋅Si​1​⋅Si​(1−Si​)+j​=i∑​Sj​−yj​​⋅(−1)Si​Sj​=−yi​(1−Si​)+j​=i∑​yj​⋅Si​=−yi​+yi​Si​+j​=i∑​yj​⋅Si​=Si​−yi​

最后

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