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The lambda parameter in ridge regression penalizes larger coefficients and pushes the model to balance the trade-off between fitting the data the best it can while taking into account the size of the coefficient. As a result coefficients are generally pushed closer to zero, which a larger amount of shrinkage for larger values of lambda.

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This sounds as a supervised change point detection task. I believe your approach with the 0/1 dummy variable is fine. Alternatively to the neural network approach you could try out some classification algorithms. You might find helpful this review on change point detection (A Survey of Methods for Time Series Change Point Detection), where some alternative ...

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I'm not sure if I understand your problem. However, if you want to try a linear regression on this you can simply arrange all the $x$ in a matrix. Say you have (I try to immitate your notation): $$80.3 = 3 x_2 + 9 x_3 \\ 72.5 = 5 x_1 + 7 x_3 \\ 68.8 = 2x_2 + 9x_3$$ This would yield a $y$ vector and $X$ matrix like: $\left( \begin{array}{rrr} 80.3 \\ 72.5 \... 2 It is possible to constrain to linear regression in scikit-learn to only positive coefficients. The sklearn.linear_model.LinearRegression has an option for positive=True which: When set to True, forces the coefficients to be positive. This option is only supported for dense arrays. The positive=True option is not available for ridge regression in scikit-... 3 I am assuming a linear regression of the form $$y = w_0x_0 + w_1x_1+ \ldots w_px_p + \varepsilon.$$ If we combine all output observations into a single vector$\mathbf{y}$and represent the data matrix with an 1-column from left as$\mathbf{X}$, then we can express the linear regression $$\mathbf{y} = \mathbf{X}\mathbf{w} + \mathbf{\varepsilon},$$ in which$\...

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