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To make it short the relation between R2 and MSE is the following : \begin{equation} \textrm{MSE}(y, \hat{y} ) = \frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{y_i})^{2} \end{equation} \begin{equation} \textrm{R}^2(y, \hat{y} ) = 1 - \frac{\sum_{i=1}^{n} (y_i - \hat{y_i})^{2}}{ \sum_{i=1}^{n} (y_i - \bar{y})^{2} } \end{equation} R2 is just MSE standatized between -1 ...


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First, T-test null hypothesis is that there are no differences between means of two samples. And p-value is the probability to observe the data, given that the null hypothesis is correct, so if p-value is small - you are likely to reject the null hypothesis. So in your case it is actually vice-versa to what you wrote: In case of StandardScaler your test ...


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One way to potentially do this is to choose peak widths such that those under a certain value are no longer detected as peaks and instead replaced with Median like Niels has suggested above.


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Not sure what is meant by a paper. Are you asking if there is mathematical proof that this is the best setting all of the time? There are some experiments pointed to from here, a text book quote in that answer, and in the link you posted. The answer is "it depends". You can tune this parameter for your data and problem. Perhaps n/3 is a good place ...


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I suggest to use R since it is open source and very powerful and thus is used by many companies and researchers. R does not only allow to deal with large amounts of data, it also allows to do state-of-art statistical analysis, including Tensorflow/Keras etc.


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I would go for R over SPSS any day. Anything you can do with SPSS can be done with R, the inverse is not necessarily true. The advantage of using R over SPSS in my opinion is that R is a full-fletched programming language, where SPSS is more a graphical user interface for doing statistics. In R you can never accidentally click the wrong option, and still get ...


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You could try to use the following method. $$y=a\sin \left[\dfrac{\pi}{12}x-b\right]+ c$$ $$=a\left[\sin \dfrac{\pi}{12}x \cos b - \sin b \cos \dfrac{\pi}{12}x \right] +c$$ $$=a\sin \dfrac{\pi}{12}x \cos b - a \sin b \cos \dfrac{\pi}{12}x +c$$ The last equation indicates that we can switch to new parameters $w_1 = a\cos b$, $w_2=-a\sin b$, and $w_0=c$. ...


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Box-Cox transformation cannot work with negative values. You can try feeding negative values to the box-cox transformation and it will give you an error.


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Yes you should keep the natural outliers in a dataset. They represent an extreme end of the data you have and contain useful info. They also help you with anomaly detection if you wish. But it also depends on the type of problem at hand. If for example in the case of Titanic dataset, where we are classifying who survived and who didn't. It is ok to remove ...


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These metrics are used to assess the performance of your model. If certain observations are not used by the model it would therefore not make sense to include them when calculating the model's metrics. You should therefore not take these type of records into account and only look at records that are processed by your model.


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I think I figured out a nice way on how to rank the datasets: The rank r(D) of dataset D is calculated by subtracting from the total number of datasets N_D the sum of the number of wins W_i of each observation i in D averaged over the total number of observations |D| in respect to the complete set of observations without |D|, which is N - |D|, multiplied ...


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Assuming by persistence you mean degree of autocorrelation. ACF is simply a function that can be derived for any stochastic process, whether stationary or not. But the estimation of ACF for non-stationary, which is what you are doing, is the problem. Generally, ACF is a function of both time, $t$ and lag, $h$: $$\gamma(t,t+h) \equiv Cov[X_t,X_{t+h}]$$ For ...


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