kennard-stone 1.1.2

A method for selecting samples by spreading the training data evenly.

Kennard Stone

What is this?

This is an algorithm for evenly partitioning data in a scikit-learn-like interface. (See References for details of the algorithm.)

How to install

PyPI

pip install kennard-stone

Anaconda

conda install -c conda-forge kennard-stone

You need numpy, pandas and scikit-learn.

How to use

You can use them like scikit-learn.

See example for details.

In the following, X denotes an arbitrary explanatory variable and y an arbitrary objective variable. And, estimator indicates an arbitrary prediction model that conforms to scikit-learn.

train_test_split

kennard_stone

from kennard_stone import train_test_split

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.2)

scikit-learn

from sklearn.model_selection import train_test_split

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.2, random_state = 334)

KFold

kennard_stone

from kennard_stone import KFold

kf = KFold(n_splits = 5)
for i_train, i_test in kf.split(X, y):
    X_train = X[i_train]
    y_train = y[i_train]
    X_test = X[i_test]
    y_test = y[i_test]

scikit-learn

from sklearn.model_selection import KFold

kf = KFold(n_splits = 5, shuffle = True, random_state = 334)
for i_train, i_test in kf.split(X, y):
    X_train = X[i_train]
    y_train = y[i_train]
    X_test = X[i_test]
    y_test = y[i_test]

Others

If you ever specify cv in scikit-learn, you can assign KFold objects to it and apply it to various functions.

An example is cross_validate.

kennard_stone

from kennard_stone import KFold
from sklearn.model_selection import cross_validate

kf = KFold(n_splits = 5)
print(cross_validate(estimator, X, y, cv = kf))

scikit-learn

from sklearn.model_selection import KFold
from sklearn.model_selection import cross_validate

kf = KFold(n_splits = 5, shuffle = True, random_state = 334)
print(cross_validate(estimator, X, y, cv = kf))

OR

from sklearn.model_selection import cross_validate

print(cross_validate(estimator, X, y, cv = 5))

Points to note

There is no notion of random_state or shuffle because the partitioning is determined uniquely for the dataset. If these arguments are included, they do not cause an error. They simply have no effect on the result. Please be careful.

If you want to run the notebook in example directory, you will need to additionally download matplotlib, seaborn, tqdm, and jupyter other than the packages in requirements.txt.

LICENSE

MIT Licence

Copyright (c) 2021 yu9824

References

Papers

• R. W. Kennard & L. A. Stone (1969) Computer Aided Design of Experiments, Technometrics, 11:1, 137-148, DOI: 10.1080/00401706.1969.10490666

Sites

• https://datachemeng.com/trainingtestdivision/