# Logistic Regression Loss Function: Scikit Learn vs Glmnet

The loss function in sklearn is

$$\min_{w,c}{\frac{1}{2}w^Tw+C\sum_{i=1}^N{\log(\exp(-y_i(X_i^Tw+c))+1)}}$$

Whereas the loss function in glmnet is

$$\min_{\beta,\beta_0}{-\bigg[\frac{1}{N} \sum_{i=1}^N{y_i(\beta_0 + x_i^T\beta) - \log(1+e^{\beta_0 + x_i^T\beta})}\bigg] + \lambda[(1-\alpha)||\beta||_2^2/2+\alpha||\beta||_1]}$$

When setting $$\alpha=0$$, this post states that they differ by a factor of $$\lambda$$ if $$C=\frac{1}{N\lambda}$$, but I can't seem to work this out. From the loss function in glmnet and changing the variable names and notations to match that of scikit-learn, I get

\begin{align} &\min_{w,c}{-\bigg[\frac{1}{N} \sum_{i=1}^N{y_i(X_i^Tw+c) - \log(1+\exp({X_i^Tw+c}))}\bigg] + \lambda \frac{1}{2}w^Tw} \\ = &\min_{w,c}{\lambda \frac{1}{2}w^Tw -\bigg[\frac{1}{N} \sum_{i=1}^N{y_i(X_i^Tw+c) - \log(1+\exp({X_i^Tw+c}))}\bigg]} \\ = &\min_{w,c}{\lambda \frac{1}{2}w^Tw +\bigg[\frac{1}{N} \sum_{i=1}^N{\log(1+\exp({X_i^Tw+c}))-y_i(X_i^Tw+c)}\bigg]}\\ = &\min_{w,c}{\lambda \frac{1}{2}w^Tw +\bigg[\frac{1}{N} \sum_{i=1}^N{\log(1+\exp({X_i^Tw+c}))-\log(\exp(y_i(X_i^Tw+c)))}\bigg]} \\ = &\min_{w,c}{\lambda \frac{1}{2}w^Tw +\frac{1}{N} \sum_{i=1}^N{\log\bigg(\frac{1+\exp(X_i^Tw+c)}{\exp(y_i(X_i^Tw+c))}\bigg)}} \\ = &\min_{w,c}{\lambda \frac{1}{2}w^Tw +\frac{1}{N} \sum_{i=1}^N{\log\bigg(\frac{1}{\exp(y_i(X_i^Tw+c))} + \frac{\exp(X_i^Tw+c)}{\exp(y_i(X_i^Tw+c))}\bigg)}}\\ = &\min_{w,c}{\lambda \frac{1}{2}w^Tw +\frac{1}{N} \sum_{i=1}^N{\log\bigg(\exp(-y_i(X_i^Tw+c)) + \frac{\exp(X_i^Tw+c)}{\exp(y_i(X_i^Tw+c))}\bigg)}} \end{align}

which is only true if $$\frac{\exp(X_i^Tw+c)}{\exp(y_i(X_i^Tw+c))}=1$$ but I can't reason for that case. Which step is wrong here?

Inspired from this answer, I think the reason they are different is that they work on different domains. For sklearn, $$y_{sklearn}\in\{-1, 1\}$$; for glmnet, $$y_{glmnet}\in\{0,1\}$$. If we plug in $$y=-1$$ in the sklearn loss function and $$y=0$$ in that of glmnet, we get the same result. Same for $$y=1$$.