scikit-learn | scikit-learn : machine learning in Python | Machine Learning library

I'm pretty sure that this is due to bias correction that generally takes place when you have MSE. You can read about the formula for bias correction that is used in the references they provided in ?sae::meanSFH. In one of the articles, they provided a case study where the average MSE is negative. (I found this in Molina et al., 2009. They identify the bias correction in a few places, but it's very clear on pp. 452-453.)

You can visualize the errors and see how very close they are to zero.

It happened the same to me last friday. I think it has something to do with Cuda instalation in Google Colab but I don't know exactly the reason

The immediate error is that the build is generating a Python 3.10 version, but when testing Conda doesn't recognize any constraint on the Python version, and creates a Python 3.9 environment.

I think the main issue is that python >=3.5 is only a valid constraint when doing noarch builds, which this is not. That is, once a package builds with a given Python version, the version must be constrained to exactly that version (up through minor). So, in this case, the package is built with Python 3.10, but it reports in its metadata that it is compatible with all versions of Python 3.5+, which simply isn't true because Conda Python packages install the modules into Python-version-specific site-packages (e.g., lib/python-3.10/site-packages/jive).

Typically, Python versions are controlled by either the --python argument given to conda-build or a matrix supplied by the conda_build_config.yaml file (see documentation on "Build variants").

Try adjusting the meta.yaml to something like

I got the conda working by modifying the script as below, emr python versions were colliding with the conda version.:

When you set Lasso(..normalize=True) the normalization is different from that in StandardScaler(). It divides by the l2-norm instead of the standard deviation. If you read the help page:

normalize bool, default=False This parameter is ignored when fit_intercept is set to False. If True, the regressors X will be normalized before regression by subtracting the mean and dividing by the l2-norm. If you wish to standardize, please use StandardScaler before calling fit on an estimator with normalize=False.

Deprecated since version 1.0: normalize was deprecated in version 1.0 and will be removed in 1.2.

It is also touched upon in this post. Since it will be deprecated, I think it's better to just use the StandardScaler normalization. You can see it's reproducible as long as you scale it in the same way:

Streamlit share runs the app in a linux environment meaning there is no pywin32 because this is for windows.

Delete the pywin32 from the requirements file and also the pywinpty==1.1.6 for the same reason.

After deleting these requirements re-deploy your app and it will work.

In your example, you're using a DecisionTreeClassifier which by default support targets of dimension (n, m) where m > 1.

However if you want to have as result the marginal probability of each class then use the OneVsRestClassifier.

Notice that CalibratedClassifierCV expects target to be 1d so the "trick" is to extend it to support Multilabel Classification with MultiOutputClassifier.

Full Example

For me, you can actually use predict_proba() after calibration to apply a different cutoff.

What happens within class CalibratedClassifierCV (as you noticed) is effectively that the output of predict() is based on the output of predict_proba() (see here for reference), i.e. np.argmax(self.predict_proba(X), axis=1) == self.predict(X).

On the other side, for the non-calibrated classifier that you're passing to CalibratedClassifierCV (depending on whether it is a probabilistic classifier or not) the above equality may or may not hold (e.g. it does not for an SVC() classifier - see here, for instance, for some other details on this).

Though very late, I'll try to give my answer. According to the doc, here's the considered primal optimization problem for LinearSVC: ,phi being the Identity matrix, given that LinearSVC only solves linear problems.

Effectively, this is just one of the possible problems that LinearSVC admits (it is the L2-regularized, L1-loss in the terms of the LIBLINEAR paper) and not the default one (which is the L2-regularized, L2-loss). The LIBLINEAR paper gives a more general formulation for what concerns what's referred to as loss in Chapter 2, then it further elaborates also on what's referred to as penalty within the Appendix (A2+A4).

Basically, it states that LIBLINEAR is meant to solve the following unconstrained optimization pb with different loss functions xi(w;x,y) (which are hinge and squared_hinge); the default setting of the model in LIBLINEAR does not consider the bias term, that's why you won't see any reference to b from now on (there are many posts on SO on this).

• , hinge or L1-loss
• , squared_hinge or L2-loss.

For what concerns the penalty, basically this represents the norm of the vector w used. The appendix elaborates on the different problems:

• L2-regularized, L1-loss (penalty='l2', loss='hinge'):
• L2-regularized, L2-loss (penalty='l2', loss='squared_hinge'), default in LinearSVC:
• L1-regularized, L2-loss (penalty='l1', loss='squared_hinge'):

Instead, as stated within the documentation, LinearSVC does not support the combination of penalty='l1' and loss='hinge'. As far as I see the paper does not specify why, but I found a possible answer here (within the answer by Arun Iyer).

Eventually, effectively the combination of penalty='l2', loss='hinge', dual=False is not supported as specified in here (it is just not implemented in LIBLINEAR) or here; not sure whether that's the case, but within the LIBLINEAR paper from Appendix B onwards it is specified the optimization pb that's solved (which in the case of L2-regularized, L1-loss seems to be the dual).

For a theoretical discussion on SVC pbs in general, I found that chapter really useful; it shows how the minimization of the norm of w relates to the idea of the maximum-margin.