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The procedure that you can use is the following. First cluster your data with gaussian mixture models. This method should also work with multiple lines with different slopes. It should be able to deal with intersections as points near an intersection can belong to both clusters and a wrong classification will not lead to huge differences in the results of ...


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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 $w_0=c$. ...


1

If I understand your question right the answer is yes. For example in tensorflow functional API could you have as your final layers: out = tf.keras.ayers.Dense(10*10)(previous_layer) out = tf.keras.layers.Lambda(lambda t: tf reshape(t, [..., 10, 10]))(out) out = tf.keras.Lambda(lambda t: tf.nn.softmax(t, axis=-1)) Your output would have shape [batch, 10, 10]...


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If your goal is to set the optimum value of hyperparameters (weather it be learning rate, no of layers, activation function etc.) you should look into Keras Tuner. The reason as to why the learning curve is oscillating is not clear to me without seeing your code/data. But if you want to set an optimum value of learning rate then look into Kera Tuner provided ...


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Instead of using the estimator attribute you should be using the best_estimator attribute, after which you can access the underlying estimators of the MultiOutputRegressor using the estimators_ attribute. You can then access the coefficients as follows: coefficients = [estimator.coef_ for estimator in best_model.best_estimator_.estimators_] # [array([-0. ...


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Provided you have the data already, and the data is labelled (i.e., split into the two classes $A$ and $B$), it makes sense to produce a number of visualisations to gauge what the model output would be. If you start with traditional classification algorithms like logistic regression, then the model output is going to be the probability of belonging to a ...


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You could simulate data and fit a model to it as if it were real data. there are packages and functions in R and Python to do this. You'd have to be very clear that the data is faked. You could then examine the model and produce graphs as if it were a real one. This has the downside that it involves writing all the code and writing code to sim data, which ...


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Look at your past experience. Even though you're a novice, you were hired as a data scientist, so you'll probably have some experience with data science projects. A simple binary classification problem with a few hundred datapoints can be solved in a productive afternoon, whereas a large project that requires significant upfront engineering for the ...


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These string data, called categorical data can be converted to numerical data using many Categorical Encoding Techniques. Encoding categorical data is a process of converting categorical data into integer format so that the data with converted categorical values can be provided to the different models. Types of Categorical Techniques: Backward Difference ...


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You can multiple things here : Converting them to numerical introduces some sense of ordering For example if you say slovenia is 1 and USA is 2 ans ordering is introduced instead you can use one hot encoding. Pandas getdummies function will do it for you If one of your string has a lot of values say 1000 one hot encoding does not makes sense. In those ...


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I think the issue is mostly with your network architecture. You are using only one convolutional layers and you are using all sigmoid activiations. Adding more convolutional layers, changing the activations from sigmoid to relu, and changing the optimizer to Adam gives me a loss below 5 after 30 epochs: model = tf.keras.Sequential([ tf.keras.layers.Conv2D(...


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I assume the "dose" $y$ is limited to $y \in [0,1]$. So in the moment you have "bunching" in your target value $y$ which you try to remove. In this case, a linear regression could lead to "overshooting" (see here for more details). So it could be beneficial to use some estimator which "restrics" $\hat{y} \in [0,1]$ as ...


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