New answers tagged regularization
1
In a comment you say
I just need an average so that in the example Company B will rank
above Company A when creating a Top 10 or Something like that.
So, if that is the goal, you can use the number of reviews to calculate a standard error of the mean (arithmetic mean) of the ratings. Using a normal distribution as an approximation, you can choose the top ...
1
I know it's an oldie, but I put an example here of how to do this in Python either by using ElasticNet (approximation), or via nnls().
Basically, for ElasticNet, you can use:
from sklearn import linear_model as lm
eln = lm.ElasticNet(l1_ratio=0, fit_intercept=False)
act_alphas, coefs, dual_gaps = eln.path(X, y, alphas=alphas, positive=True)
but be ...
21
No difference. It's just a symbol. Sometimes mathematics uses symbols by convention, but there's no rule or requirement that you must use a certain symbol for a concept.
In this particular case, the word lambda is reserved by the Python language, so alpha avoids overlapping with that word.
As an aside, one sharp corner in sklearn is that sklearn.linear_model....
0
It is a decent question, asked a decade ago. The chat is long gone.
Question
Given a random forest:
That has bias in the outputs
That is decently trained with decent parameters and is otherwise non-pathological
has some similarity to this (link)
Find:
Elastic net regression, or other regularized regression, to calibrate the output
Logistic regression ...
0
$N(\lambda, a)$ in general is not a good function for tuning the regularization parameter $\lambda$ without more constraint. In general, $L(y,f_L(x))=0$ for $f_L(x)=y$ and $J(f)=0$ for some $f=f_J$. Then $\ln(L(y,f_L))=\ln(J(f_J))=-\infty$. Obviously $f_{\lambda=0}=f_L\,\bigvee\, f_{\lambda=1}=f_J\implies N(\lambda,a)=-\infty$. Thus there is no ...
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Your evaluation should always be performed on a holdout set. In the simplest setups, this reduces to train & test sets. So, you should split. Ridge regression has regularisation mechanics but it may not save you from overfitting.
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Writing the values for observation $i$ as $\mathbf{a}_i = (a_{i1},a_{i2},\ldots, a_{im}),$ $\mathbf{b}_i=(b_{i1},b_{i2},\ldots, b_{in})$, and $y_i$ ($i=1,2,\ldots, N$), this model can be expressed as
$$E[y_i] = \sum_{j=1}^m\sum_{k=1}^n a_{ij}X_{jk}b_{ik} = \sum_l \alpha_{il} X_l $$
where $l$ indexes the $mn$ ordered pairs $(j,k)$ and $\alpha_{il} = a_{ij}b_{...
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