πŸ“ƒ Solution for Exercise M3.02

πŸ“ƒ Solution for Exercise M3.02#

The goal is to find the best set of hyperparameters which maximize the generalization performance on a training set.

from sklearn.datasets import fetch_california_housing
from sklearn.model_selection import train_test_split

data, target = fetch_california_housing(return_X_y=True, as_frame=True)
target *= 100  # rescale the target in k$

data_train, data_test, target_train, target_test = train_test_split(
    data, target, random_state=42
)

In this exercise, we progressively define the regression pipeline and later tune its hyperparameters.

Start by defining a pipeline that:

  • uses a StandardScaler to normalize the numerical data;

  • uses a sklearn.neighbors.KNeighborsRegressor as a predictive model.

# solution
from sklearn.pipeline import make_pipeline
from sklearn.preprocessing import StandardScaler
from sklearn.neighbors import KNeighborsRegressor

scaler = StandardScaler()
model = make_pipeline(scaler, KNeighborsRegressor())

Use RandomizedSearchCV with n_iter=20 to find the best set of hyperparameters by tuning the following parameters of the model:

  • the parameter n_neighbors of the KNeighborsRegressor with values np.logspace(0, 3, num=10).astype(np.int32);

  • the parameter with_mean of the StandardScaler with possible values True or False;

  • the parameter with_std of the StandardScaler with possible values True or False.

Notice that in the notebook β€œHyperparameter tuning by randomized-search” we pass distributions to be sampled by the RandomizedSearchCV. In this case we define a fixed grid of hyperparameters to be explored. Using a GridSearchCV instead would explore all the possible combinations on the grid, which can be costly to compute for large grids, whereas the parameter n_iter of the RandomizedSearchCV controls the number of different random combination that are evaluated. Notice that setting n_iter larger than the number of possible combinations in a grid (in this case 10 x 2 x 2 = 40) would lead to repeating already-explored combinations.

Once the computation has completed, print the best combination of parameters stored in the best_params_ attribute.

# solution
import numpy as np
from sklearn.model_selection import RandomizedSearchCV

param_distributions = {
    "kneighborsregressor__n_neighbors": np.logspace(0, 3, num=10).astype(
        np.int32
    ),
    "standardscaler__with_mean": [True, False],
    "standardscaler__with_std": [True, False],
}

model_random_search = RandomizedSearchCV(
    model,
    param_distributions=param_distributions,
    n_iter=20,
    n_jobs=2,
    verbose=1,
    random_state=1,
)
model_random_search.fit(data_train, target_train)
model_random_search.best_params_

Fitting 5 folds for each of 20 candidates, totalling 100 fits

{'standardscaler__with_std': True,
 'standardscaler__with_mean': False,
 'kneighborsregressor__n_neighbors': np.int32(10)}

So the best hyperparameters give a model where the features are scaled but not centered.

Getting the best parameter combinations is the main outcome of the hyper-parameter optimization procedure. However it is also interesting to assess the sensitivity of the best models to the choice of those parameters. The following code, not required to answer the quiz question shows how to conduct such an interactive analysis for this this pipeline using a parallel coordinate plot using the plotly library.

We could use cv_results = model_random_search.cv_results_ to make a parallel coordinate plot as we did in the previous notebook (you are more than welcome to try!).

import pandas as pd

cv_results = pd.DataFrame(model_random_search.cv_results_)

To simplify the axis of the plot, we rename the column of the dataframe and only select the mean test score and the value of the hyperparameters.

column_name_mapping = {
    "param_kneighborsregressor__n_neighbors": "n_neighbors",
    "param_standardscaler__with_mean": "centering",
    "param_standardscaler__with_std": "scaling",
    "mean_test_score": "mean test score",
}

cv_results = cv_results.rename(columns=column_name_mapping)
cv_results = cv_results[column_name_mapping.values()].sort_values(
    "mean test score", ascending=False
)

In addition, the parallel coordinate plot from plotly expects all data to be numeric. Thus, we convert the boolean indicator informing whether or not the data were centered or scaled into an integer, where True is mapped to 1 and False is mapped to 0. As n_neighbors has dtype=object, we also convert it explicitly to an integer.

column_scaler = ["centering", "scaling"]
cv_results[column_scaler] = cv_results[column_scaler].astype(np.int64)
cv_results["n_neighbors"] = cv_results["n_neighbors"].astype(np.int64)
cv_results
n_neighbors centering scaling mean test score
17 10 0 1 0.687926
18 4 0 1 0.674812
6 46 0 1 0.668778
9 100 0 1 0.648317
16 2 1 1 0.629772
12 215 0 1 0.617295
15 215 1 1 0.617295
10 464 1 1 0.567164
0 1 0 1 0.508809
13 1000 1 1 0.486503
11 21 1 0 0.103390
8 21 0 0 0.103390
3 46 1 0 0.061394
4 100 0 0 0.033122
5 215 1 0 0.017583
1 215 0 0 0.017583
19 464 0 0 0.007987
14 464 1 0 0.007987
7 1000 0 0 0.002900
2 1 0 0 -0.238830 
import plotly.express as px

fig = px.parallel_coordinates(
    cv_results,
    color="mean test score",
    dimensions=["n_neighbors", "centering", "scaling", "mean test score"],
    color_continuous_scale=px.colors.diverging.Tealrose,
)
fig.show()

We recall that it is possible to select a range of results by clicking and holding on any axis of the parallel coordinate plot. You can then slide (move) the range selection and cross two selections to see the intersections.

Selecting the best performing models (i.e. above an accuracy of ~0.68), we observe that in this case:

  • scaling the data is important. All the best performing models use scaled features;

  • centering the data does not have a strong impact. Both approaches, centering and not centering, can lead to good models;

  • using some neighbors is fine but using too many is a problem. In particular no pipeline with n_neighbors=1 can be found among the best models. However, scaling features has an even stronger impact than the choice of n_neighbors in this problem.

The reason is that fitting scaled data leads to a completely different KNeighbors model: if you have two variables A and B where A has values which vary between 0 and 10,000 (e.g. the variable "Population") and B is a feature that varies between 1 and 10 (e.g. the variable "AveRooms"), then distances between samples (rows of the dataframe) are mostly impacted by differences in values of the column A, while values of the column B are comparatively ignored. If one applies StandardScaler to such a database, both the values of A and B will be approximately between -3 and 3 and the neighbor structure will be impacted more or less equivalently by both variables.

Note that in this case the models with scaled features perform better than the models with non-scaled features because all the variables are expected to be predictive and we rather avoid some of them being comparatively ignored.

If the variables in lower scales were not predictive one may experience a decrease of the performance after scaling the features: noisy features would contribute more to the prediction after scaling and therefore scaling would increase overfitting.