The estimator API

Skill positionModule 1 of 4 · Topic 1 of 5

1.1 The estimator API

—Lab score— / 100Total XP—Streak—

A model is an object with a contract

Everything in scikit-learn is an estimator, and every estimator follows the same contract. You create it with its settings, called hyperparameters, for example the strength of a penalty. You call $\mathrm{fit}(X,y)$ so that it learns from data. What it learned is stored in attributes whose names end with an underscore, such as the mean or the coefficients. Then you use it: a preprocessor offers $\mathrm{transform}(X)$ , a model offers $\mathrm{predict}(X)$ . The call $\mathrm{fit\_transform}$ is fit followed by transform. The table $X$ has one row per sample and one column per feature, and $y$ is the target. Because the contract is the same everywhere, you can swap a model for another and keep the rest of your code.

z=\frac{x-\hat\mu}{\hat\sigma},\qquad \hat y=\hat a+\hat b\,x

(1)

The golden rule is that $\mathrm{fit}$ may only see training data. The numbers a preprocessor learns, such as $\hat\mu$ and $\hat\sigma$ , are then applied unchanged to validation and test data by $\mathrm{transform}$ . If you fitted the scaler on all days, the test days would have shaped the scaling, which is a quiet leak of the future. Settings such as $\mathrm{random\_state}$ make runs repeatable, and $\mathrm{clone}$ builds a fresh unfitted copy of an estimator, which is what cross-validation does for every fold. In the market-data labs of this skill, rows are stock-months and columns are features such as momentum and volatility.

Two tiny fits by handFit a standard scaler on the training values

2,4,6,8

. It learns the mean

5

and the standard deviation

\sqrt5=2.236

. A new value

7

becomes

(7-5)/2.236=0.894

. Now fit a linear regression on

(x,y)=(1,2),(2,4),(3,7)

. It learns the slope

2.5

and the intercept

-0.667

, so a prediction for

x=4

-0.667+2.5\times4=9.333

Settings in, numbers learned: hyperparameters go into the constructor, learned numbers end with an underscore.
Fit on training only: transform and predict reuse the learned numbers on any data.
Same contract everywhere: swapping a model for another keeps the rest of your code.

Lab · what's graded

What the scaler learns: the mean10 pts
What the scaler learns: the spread10 pts
Transform a test value10 pts
What the regression learns: the slope15 pts
A prediction10 pts
Which attributes are learned10 pts
A safe workflow10 pts
Match the vocabulary10 pts
Why fit the scaler on training data only15 pts

0/ 9 answered

1Numeric answer10 ptsNot answered

A standard scaler is fitted on the training values $2,4,6,8$ . What is the learned mean?

m =

accepted: ± 0.01

2Numeric answer10 ptsNot answered

The values $2,4,6,8$ have mean $5$ and squared deviations $9,1,1,9$ . The scaler uses the population standard deviation (divide by $n=4$ ). What is it? Give three decimals.

s =

accepted: ± 0.003

3Numeric answer10 ptsNot answered

The scaler learned mean $5$ and standard deviation $2.236$ . A test value $7$ arrives. What does transform return, $(7-5)/2.236$ ? Give three decimals.

z =

accepted: ± 0.003

4Numeric answer15 ptsNot answered

A linear regression is fitted on $(x,y)=(1,2),(2,4),(3,7)$ . The slope is $\sum(x-\bar x)(y-\bar y)/\sum(x-\bar x)^{2}$ with $\bar x=2$ , $\bar y=4.333$ . What is the learned slope? Give one decimal.

b =

accepted: ± 0.05

5Numeric answer10 ptsNot answered

The regression learned slope $2.5$ and intercept $-0.667$ . What does predict return for $x=4$ ? Give three decimals.

y =

accepted: ± 0.005

6Choice10 ptsNot answered

Which of these are numbers learned by fit rather than settings you choose? Pick all that apply.

7Order the steps10 ptsNot answered

Put the steps of a safe modelling workflow in order.

Drag to reorder, or focus an item, press Space to lift it, the arrow keys to move it and Space to drop it.

1Fit the model on the transformed training data
2Split the days into training and test by time
3Predict on the transformed test data and score
4Apply transform to the training and the test data
5Fit the scaler on the training data only

8Match10 ptsNot answered

Match each call with what it does.

Pick the partner of each item on the left.

fit(X, y)

transform(X)

predict(X)

clone(estimator)

9Explain in words15 ptsNot answered

In one or two sentences, explain why a scaler should be fitted on the training days only and not on all the days.

Graded on:Fitting on all days lets the test days, the future, influence the mean and spreadIn real use the future is not available when the numbers are chosen, so the test score would be too optimistic0 / 400 characters

Math modeKeyboard or LaTeX inputTask 1 of 9 · Ctrl ⏎ to check

The estimator API

Skill positionModule 1 of 4 · Topic 1 of 5

1.1 The estimator API

—Lab score— / 100Total XP—Streak—

A model is an object with a contract

z=\frac{x-\hat\mu}{\hat\sigma},\qquad \hat y=\hat a+\hat b\,x

(1)

Two tiny fits by handFit a standard scaler on the training values

2,4,6,8

. It learns the mean

5

and the standard deviation

\sqrt5=2.236

. A new value

7

becomes

(7-5)/2.236=0.894

. Now fit a linear regression on

(x,y)=(1,2),(2,4),(3,7)

. It learns the slope

2.5

and the intercept

-0.667

, so a prediction for

x=4

-0.667+2.5\times4=9.333

Settings in, numbers learned: hyperparameters go into the constructor, learned numbers end with an underscore.
Fit on training only: transform and predict reuse the learned numbers on any data.
Same contract everywhere: swapping a model for another keeps the rest of your code.

Lab · what's graded

What the scaler learns: the mean10 pts
What the scaler learns: the spread10 pts
Transform a test value10 pts
What the regression learns: the slope15 pts
A prediction10 pts
Which attributes are learned10 pts
A safe workflow10 pts
Match the vocabulary10 pts
Why fit the scaler on training data only15 pts

0/ 9 answered

1Numeric answer10 ptsNot answered

A standard scaler is fitted on the training values $2,4,6,8$ . What is the learned mean?

m =

accepted: ± 0.01

2Numeric answer10 ptsNot answered

The values $2,4,6,8$ have mean $5$ and squared deviations $9,1,1,9$ . The scaler uses the population standard deviation (divide by $n=4$ ). What is it? Give three decimals.

s =

accepted: ± 0.003

3Numeric answer10 ptsNot answered

The scaler learned mean $5$ and standard deviation $2.236$ . A test value $7$ arrives. What does transform return, $(7-5)/2.236$ ? Give three decimals.

z =

accepted: ± 0.003

4Numeric answer15 ptsNot answered

A linear regression is fitted on $(x,y)=(1,2),(2,4),(3,7)$ . The slope is $\sum(x-\bar x)(y-\bar y)/\sum(x-\bar x)^{2}$ with $\bar x=2$ , $\bar y=4.333$ . What is the learned slope? Give one decimal.

b =

accepted: ± 0.05

5Numeric answer10 ptsNot answered

The regression learned slope $2.5$ and intercept $-0.667$ . What does predict return for $x=4$ ? Give three decimals.

y =

accepted: ± 0.005

6Choice10 ptsNot answered

Which of these are numbers learned by fit rather than settings you choose? Pick all that apply.

7Order the steps10 ptsNot answered

Put the steps of a safe modelling workflow in order.

Drag to reorder, or focus an item, press Space to lift it, the arrow keys to move it and Space to drop it.

1Fit the model on the transformed training data
2Split the days into training and test by time
3Predict on the transformed test data and score
4Apply transform to the training and the test data
5Fit the scaler on the training data only

8Match10 ptsNot answered

Match each call with what it does.

Pick the partner of each item on the left.

fit(X, y)

transform(X)

predict(X)

clone(estimator)

9Explain in words15 ptsNot answered

In one or two sentences, explain why a scaler should be fitted on the training days only and not on all the days.

Math modeKeyboard or LaTeX inputTask 1 of 9 · Ctrl ⏎ to check

The estimator API

A model is an object with a contract

Lab · what's graded

Log in to run your code.

The estimator API

A model is an object with a contract

Lab · what's graded

Log in to run your code.