kennard-stone 1.0.4

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

pip install 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 you include them in the argument, you will not get an error, but they have no effect, so be careful.

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/