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You need to set self.theta to be an array, not a scalar (at least in this specific problem).

In your case, (intercepted-augmented) X is a '3 by n' array, so try self.theta = [0, 0, 0] for example. This will correct the specific error 'bool' object has no attribute 'mean'. Still, this will just produce preds as a zero vector; you haven't fit the model yet.

To let you know how I approached the error, I first went to the exact line the error message was pointing to, and put print(preds == y) before the line, and it printed out False. I guess what you expected was a vector of True and Falses. Your y seemed okay; it was a vector (a list to be specific). So I tried print(pred), which showed me a '3 by n' array, which is weird. Now going up from that line, I found out that pred comes from predict_prob(), especially np.dot(X, self.theta). Here, when X is a '3 by n' array and self.theta is a scalar, numpy seems to multiply the scalar to each item in the array and return the array (having the same dimension as the original array), instead of doing matrix multiplication! So you need to explicitly provide self.theta as an array (conforming to the dimension of X).

Hope the answer and the reasoning behind it helped.

As for the red line you mentioned in the comment, I guess it is also because you are not fitting the model. (To see the problem, put print(probs) before plt.countour(...). You'll see an array with 0.5 only.)

So try putting model.fit(X, y) before preds = model.predict(X). (You'll also need to put self.verbose = verbose in the __init__().)

After that, I get the following:

Like in mathematics, you start with the inner most parenthesis or brackets..

You can vectorize the calculation using numpy:

Use time

The problem with result_df.astype(timedelta_df.dtypes.to_dict()) resulting in wrong values is that the datatype of the days column is timedelta64[ns], i.e. it expects nanoseconds, whereas to_json defaults to serializing timedeltas as milliseconds.

One simple way to fix this would therefore be to explicitly serialize it as nanoseconds: timedelta_df.to_json(date_unit="ns").

Try something like this

The actual problem is really silly, but since it took quite some time to track down, I'll share an outline of my debugging.

First, rather than plotting, print the actual values of the decision_function: you'll find that the first one comes out unique, but that after that everything is constant. Running the same on various slices of the dataset, this pattern persists. So I thought perhaps some values were being overwritten, and I dug into the SVC code a bit. That lead to some useful internal functions/attributes, like ._BaseLibSVM__Xfit containing the training data, _decision_function and _dense_decision_function, and _compute_kernel. But none of the code indicated a problem, and running them just showed the same problem. Running _compute_kernel gave results that were all-zeros past the first row, and then coming back to your code, running linear_kernel does that already. So, finally, it comes back to your linear_kernel function.

You return inside the outer for loop, so you only ever use the first row of X, never computing the rest of the matrix. (This brings up a surprise: why did the predictions look good? That seems to have been a fluke. Changing the definition for f_lin, to change the classes, the model still learns the slope-1 line.)

Think of y as a two-dimensional matrix, although it has only one column. But the fit method expects y to be a flat array. That's why you have to use ravel, to convert the 2d to a 1d array.

It's common in machine learning papers and textbooks to write y as a matrix, because it can simplify the notation when matrices are multiplied. But you could also write it as a simple one-dimensional vector. You could say it makes no difference, because it really only has one dimension in either case, but mathematically and in the Python implementation, the matrix and the vector are two different objects.