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I think of an acquisition function as describing the utility of the point to be evaluated next in the Bayesian optimization framework. To give more details, let's think about the general concept of Bayesian Optimization and the setting in which it is usually applied. Consider a black-box function $f$ which is expensive to evaluate and we want to find the ...


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We can extend the reasoning presented in Can a random forest be used for feature selection in multiple linear regression? to this context. The data in this figure are obviously separated by the circle $1=x_1^2 + x_2^2$, which is a nonlinear boundary in $x_1, x_2$. Because the relationship is pretty obvious, it shouldn't be surprising that random forest ...


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To add to what @Sycorax commented: If you're satisfied with merely being a consumer of scientific software, you can skate by without much knowledge at all. But if you're curious and interested in really understanding what's going on when your script runs, knowledge of statistics, linear algebra, calculus and numerical optimization are essential; ...


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One of the advantages of standardization is making the values unitless, so that we can compare/sum different quantities with each other. Your scaled y is unit-free, because scaling divides your target (- mean) by its deviation, which is still of unit u. When we divide a quantity of unit u, with another quantity of unit u, the resulting quantity become unit-...


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It's a simple result of Bayes rule. If the outcome variable is conditionally dependent on any of the inputs, then the imputation model for the inputs will not be correct unless it includes the outcome as a feature. To impute missing data in a new data frame, you have to train an imputation model which includes missing outcome as a missingness pattern. In ...


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Several considerations here pertain to the background upon which the question is based rather than the text of the question itself. As the say in Maine, "You can't get there from here." In simplest terms, one must first have shape information in order to test for shape, and there isn't any shape information to test, as follows. 1) Concentrations are not ...


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I think linear algebra is absolutely essential to the data scientist. A card carrying data scientist should have more than just a comprehensive knowledge of what tools are out there, but he or she should also have the ability to compare them, to develop new tools, and, when a job simply can't be solved by any tool, be able to explain why. Linear algebra (...


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I worked with data scientists who do not know linear algebra. The field, which pretentiosly has "science" in its name, is so vast that there's something to do for everyone willing. It is somewhat similar to programmers not knowing electronics, and most of them have no clue. You can survive and even prosper without linear algebra but you will not be able to ...


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What you can do is something called nested cross validation. Instead of creating one train-test split you create K train-test splits (as you would in k-fold cross validation). For each of your K training sets you perform cross-validation to select hyper-parameters (this is called the inner CV loop) and then test on the test dataset (outer loop). You will ...


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To me, this sounds like a problem where quantile regression could come in handy. The reason for this is that you need to compare distributions of model performance metrics (e.g., AUC) across levels of a factor such as country. To achieve this comparison, you could choose to compare certain quantiles of these distributions across the levels of this factor, ...


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Typically a dense layer follows the LSTM/RNN layer(s), because the output of the RNN cell is of dimension of your choice, i.e. latent space dimension. Since you've three outputs, the final Dense layer will have three neurons, aiming to regress genre points. The RNN layer's duty is to figure out a compressed, latent representation of your series going back a ...


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You are training a machine learner to identify patterns in your (training) data. If you later incorporate data it has never encountered before it has no idea what to do with it (it literally is seeing this information for the first time). Think of it as reading Chinese, while you've only learned English. Sure there's a lot of info in the Chinese text but ...


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Ultimately, yes, if you are representing individual cards as 17-dimensional one-hot vectors and passing all four hands into the network, then you'll need a total of 544 inputs. (An alternative might be to ignore suit in the representation, represent individual cards as 52-dimensional one-hot vectors, and represent hands as sums of the one-hot card vectors. ...


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The number of distinct subsets (including the empty set) of a set with $p$ elements is $$\sum_{k=0}^n \binom pk = (1+1)^p = 2^p$$ To each subset (of variables) corresponds a model, so there are $2^p$ models.


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You have a vector that describes your model: $$ v = \overbrace{ v_1v_2v_3\dots v_p }^\text{variables}$$ where $v_i=1$ if the variable $i$ is included into a model, or 0 if it's skipped. How may combinations of variables you can get? It's $2^p$


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