In [1]:
from dsc80_utils import *
import lec15_util as util

Lecture 15 – Pipelines, Multicollinearity, and Generalization¶

DSC 80, Spring 2024¶

Announcements 📣¶

  • Final Project Checkpoint 1 due tomorrow.
  • Lab 8 due next Wednesday.

Agenda 📆¶

  • Pipelines.
  • Multicollinearity.
  • Generalization.
    • Bias and variance.
    • Train-test splits.

Question 🤔 (Answer at q.dsc80.com)

Remember, you can always ask questions at q.dsc80.com!

Pipelines¶

No description has been provided for this image

So far, we've used transformers for feature engineering and models for prediction. We can combine these steps into a single Pipeline.

Pipelines in sklearn¶

From sklearn's documentation:

Pipeline allows you to sequentially apply a list of transformers to preprocess the data and, if desired, conclude the sequence with a final predictor for predictive modeling.

Intermediate steps of the pipeline must be "transforms", that is, they must implement fit and transform methods. The final estimator only needs to implement fit.

  • General template: pl = Pipeline([trans_1, trans_2, ..., model])
    • Note that the model is optional.
  • Once a Pipeline is instantiated, you can fit all steps (transformers and model) using pl.fit(X, y).
  • To make predictions using raw, untransformed data, use pl.predict(X).
  • The actual list we provide Pipeline with must be a list of tuples, where
    • The first element is a "name" (that we choose) for the step.
    • The second element is a transformer or estimator instance.

Our first Pipeline¶

Let's build a Pipeline that:

  • One hot encodes the categorical features in tips.
  • Fits a regression model on the one hot encoded data.
In [3]:
tips = px.data.tips()
In [4]:
tips_cat = tips[['sex', 'smoker', 'day', 'time']]
tips_cat.head()
Out[4]:
sex smoker day time
0 Female No Sun Dinner
1 Male No Sun Dinner
2 Male No Sun Dinner
3 Male No Sun Dinner
4 Female No Sun Dinner
In [5]:
from sklearn.pipeline import Pipeline
In [8]:
from sklearn.preprocessing import OneHotEncoder
from sklearn.linear_model import LinearRegression

pl = Pipeline([
    ('one-hot', OneHotEncoder()),
    ('lin-reg', LinearRegression())
])

Now that pl is instantiated, we fit it the same way we would fit the individual steps.

In [9]:
pl.fit(tips_cat, tips['tip'])
Out[9]:
Pipeline(steps=[('one-hot', OneHotEncoder()), ('lin-reg', LinearRegression())])

Now, to make predictions using raw data, all we need to do is use pl.predict:

In [10]:
pl.predict(tips_cat.iloc[:5])
Out[10]:
array([3.1 , 3.27, 3.27, 3.27, 3.1 ])

pl performs both feature transformation and prediction with just a single call to predict!

We can access individual "steps" of a Pipeline through the named_steps attribute:

In [11]:
pl.named_steps
Out[11]:
{'one-hot': OneHotEncoder(), 'lin-reg': LinearRegression()}
In [12]:
pl.named_steps['one-hot'].transform(tips_cat).toarray()
Out[12]:
array([[1., 0., 1., ..., 0., 1., 0.],
       [0., 1., 1., ..., 0., 1., 0.],
       [0., 1., 1., ..., 0., 1., 0.],
       ...,
       [0., 1., 0., ..., 0., 1., 0.],
       [0., 1., 1., ..., 0., 1., 0.],
       [1., 0., 1., ..., 1., 1., 0.]])
In [13]:
pl.named_steps['one-hot'].get_feature_names_out()
Out[13]:
array(['sex_Female', 'sex_Male', 'smoker_No', 'smoker_Yes', 'day_Fri',
       'day_Sat', 'day_Sun', 'day_Thur', 'time_Dinner', 'time_Lunch'],
      dtype=object)
In [14]:
pl.named_steps['lin-reg'].coef_
Out[14]:
array([-0.09,  0.09, -0.04,  0.04, -0.2 , -0.13,  0.14,  0.19,  0.25,
       -0.25])

pl also has a score method, the same way a fit LinearRegression instance does:

In [15]:
# Why is this so low?
pl.score(tips_cat, tips['tip'])
Out[15]:
0.027496790201475663

More sophisticated Pipelines¶

  • In the previous example, we one hot encoded every input column. What if we want to perform different transformations on different columns?
  • Solution: Use a ColumnTransformer.
    • Instantiate a ColumnTransformer using a list of tuples, where:
      • The first element is a "name" we choose for the transformer.
      • The second element is a transformer instance (e.g. OneHotEncoder()).
      • The third element is a list of relevant column names.

Planning our first ColumnTransformer¶

In [16]:
from sklearn.compose import ColumnTransformer

Let's perform different transformations on the quantitative and categorical features of tips (note that we are not transforming 'tip').

In [17]:
tips_features = tips.drop('tip', axis=1)
tips_features.head()
Out[17]:
total_bill sex smoker day time size
0 16.99 Female No Sun Dinner 2
1 10.34 Male No Sun Dinner 3
2 21.01 Male No Sun Dinner 3
3 23.68 Male No Sun Dinner 2
4 24.59 Female No Sun Dinner 4
  • We will leave the 'total_bill' column untouched.
  • To the 'size' column, we will apply the Binarizer transformer with a threshold of 2 (big tables vs. small tables).
  • To the categorical columns, we will apply the OneHotEncoder transformer.
  • In essence, we will create a transformer that reproduces the following DataFrame:
size x0_Female x0_Male x1_No x1_Yes x2_Fri x2_Sat x2_Sun x2_Thur x3_Dinner x3_Lunch total_bill
0 0 1.0 0.0 1.0 0.0 0.0 0.0 1.0 0.0 1.0 0.0 16.99
1 1 0.0 1.0 1.0 0.0 0.0 0.0 1.0 0.0 1.0 0.0 10.34
2 1 0.0 1.0 1.0 0.0 0.0 0.0 1.0 0.0 1.0 0.0 21.01
3 0 0.0 1.0 1.0 0.0 0.0 0.0 1.0 0.0 1.0 0.0 23.68
4 1 1.0 0.0 1.0 0.0 0.0 0.0 1.0 0.0 1.0 0.0 24.59

Building a Pipeline using a ColumnTransformer¶

Let's start by creating our ColumnTransformer.

In [19]:
from sklearn.preprocessing import Binarizer

preproc = ColumnTransformer(
    transformers=[
        ('size', Binarizer(threshold=2), ['size']),
        ('categorical_cols', OneHotEncoder(), ['sex', 'smoker', 'day', 'time'])
    ],
    remainder='passthrough' # Specify what to do with all other columns ('total_bill' here) – drop or passthrough.
)

Now, let's create a Pipeline using preproc as a transformer, and fit it:

In [20]:
pl = Pipeline([
    ('preprocessor', preproc), 
    ('lin-reg', LinearRegression())
])
In [21]:
pl.fit(tips_features, tips['tip'])
Out[21]:
Pipeline(steps=[('preprocessor',
                 ColumnTransformer(remainder='passthrough',
                                   transformers=[('size',
                                                  Binarizer(threshold=2),
                                                  ['size']),
                                                 ('categorical_cols',
                                                  OneHotEncoder(),
                                                  ['sex', 'smoker', 'day',
                                                   'time'])])),
                ('lin-reg', LinearRegression())])

Prediction is as easy as calling predict:

In [22]:
tips_features.head()
Out[22]:
total_bill sex smoker day time size
0 16.99 Female No Sun Dinner 2
1 10.34 Male No Sun Dinner 3
2 21.01 Male No Sun Dinner 3
3 23.68 Male No Sun Dinner 2
4 24.59 Female No Sun Dinner 4
In [23]:
# Note that we fit the Pipeline using tips_features, not tips_features.head()!
pl.predict(tips_features.head())
Out[23]:
array([2.74, 2.32, 3.37, 3.37, 3.75])

Aside: FunctionTransformer¶

A transformer you'll often use as part of a ColumnTransformer is the FunctionTransformer, which enables you to use your own functions on entire columns. Think of it as the sklearn equivalent of apply.

In [24]:
from sklearn.preprocessing import FunctionTransformer
In [25]:
f = FunctionTransformer(np.sqrt)
f.transform([1, 2, 3])
Out[25]:
array([1.  , 1.41, 1.73])

💡 Pro-Tip: Using make_pipeline and make_column_transformer¶

Instead of using Pipeline and ColumnTransformer classes directly, scikit-learn provides nifty shortcut methods called make_pipeline and make_column_transformer:

In [26]:
# Old code

preproc = ColumnTransformer(
    transformers=[
        ('size', Binarizer(threshold=2), ['size']),
        ('categorical_cols', OneHotEncoder(), ['sex', 'smoker', 'day', 'time'])
    ],
    remainder='passthrough' 
)

pl = Pipeline([
    ('preprocessor', preproc), 
    ('lin-reg', LinearRegression())
])
In [30]:
from sklearn.pipeline import make_pipeline
from sklearn.compose import make_column_transformer

preproc = make_column_transformer(
    (Binarizer(threshold=2), ['size']),
    (OneHotEncoder(), ['sex', 'smoker', 'day', 'time']),
    remainder='passthrough',
)

pl = make_pipeline(preproc, LinearRegression())
# Notice that the steps in the pipeline and column transformer are
# automatically named
pl
Out[30]:
Pipeline(steps=[('columntransformer',
                 ColumnTransformer(remainder='passthrough',
                                   transformers=[('binarizer',
                                                  Binarizer(threshold=2),
                                                  ['size']),
                                                 ('onehotencoder',
                                                  OneHotEncoder(),
                                                  ['sex', 'smoker', 'day',
                                                   'time'])])),
                ('linearregression', LinearRegression())])

An example Pipeline¶

One of the transformers we used was the StandardScaler transformer, which standardizes columns.

$$z(x_i) = \frac{x_i - \text{mean of } x}{\text{SD of } x}$$

Let's build a Pipeline that:

  • Takes in the 'total_bill' and 'size' features of tips.
  • Standardizes those features.
  • Uses the resulting standardized features to fit a linear model that predicts 'tip'.
In [31]:
# Let's define these once, since we'll use them repeatedly.
X = tips[['total_bill', 'size']]
y = tips['tip']
In [33]:
from sklearn.preprocessing import StandardScaler

model_with_std = make_pipeline(
    StandardScaler(),
    LinearRegression(),
)


model_with_std.fit(X, y)
Out[33]:
Pipeline(steps=[('standardscaler', StandardScaler()),
                ('linearregression', LinearRegression())])

How well does our model do? We can compute its $R^2$ and RMSE.

In [34]:
model_with_std.score(X, y)
Out[34]:
0.46786930879612587
In [36]:
from sklearn.metrics import mean_squared_error

mean_squared_error(y, model_with_std.predict(X), squared=False)
Out[36]:
1.007256127114662

Does this model perform any better than one that doesn't standardize its features? Let's find out.

In [37]:
model_without_std = LinearRegression()
model_without_std.fit(X, y)
Out[37]:
LinearRegression()
In [38]:
model_without_std.score(X, y)
Out[38]:
0.46786930879612587
In [39]:
mean_squared_error(y, model_without_std.predict(X), squared=False)
Out[39]:
1.007256127114662

No!

The purpose of standardizing features¶

If you're performing "vanilla" linear regression – that is, using the LinearRegression object – then standardizing your features will not change your model's error.

  • There are other models where standardizing your features will improve performance, because the methods assume features are standardized.
    • Regularized linear regression (see DSC 140A).
    • PCA (assumes centered data, not necessarily standardized: see DSC 140B).
    • Clustering algorithms, e.g. $k$-means clustering (saw in DSC 40A!).
  • There is a benefit to standardizing features when performing vanilla linear regression, as we saw in DSC 40A: the features are brought to the same scale, so the coefficients can be compared directly.
In [40]:
# Total bill, table size.
model_without_std.coef_
Out[40]:
array([0.09, 0.19])
In [42]:
# Total bill, table size.
model_with_std.named_steps['linearregression'].coef_
Out[42]:
array([0.82, 0.18])

Aside: Pipelines of just transformers¶

If you want to apply multiple transformations to the same column in a dataset, you can create a Pipeline just for that column.

For example, suppose we want to:

  • One hot encode the 'sex', 'smoker', and 'time' columns.
  • One hot encode the 'day' column, but as either 'Weekday', 'Sat', or 'Sun'.
  • Binarize the 'size' column.

Here's how we might do that:

In [58]:
def is_weekend(s):
    # The input to is_weekend is a Series!
    return s.replace({'Thur': 'Weekday', 'Fri': 'Weekday'})
In [60]:
pl_day = make_pipeline(
    FunctionTransformer(is_weekend),
    OneHotEncoder(),
)
In [61]:
col_trans = make_column_transformer(
    (pl_day, ['day']),
    (OneHotEncoder(drop='first'), ['sex', 'smoker', 'time']),
    (Binarizer(threshold=2), ['size']),
    remainder='passthrough')
In [63]:
pl = make_pipeline(
    col_trans,
    LinearRegression(),
)

pl.fit(tips.drop('tip', axis=1), tips['tip'])
Out[63]:
Pipeline(steps=[('columntransformer',
                 ColumnTransformer(remainder='passthrough',
                                   transformers=[('pipeline',
                                                  Pipeline(steps=[('functiontransformer',
                                                                   FunctionTransformer(func=<function is_weekend at 0x7f96590a9ee0>)),
                                                                  ('onehotencoder',
                                                                   OneHotEncoder())]),
                                                  ['day']),
                                                 ('onehotencoder',
                                                  OneHotEncoder(drop='first'),
                                                  ['sex', 'smoker', 'time']),
                                                 ('binarizer',
                                                  Binarizer(threshold=2),
                                                  ['size'])])),
                ('linearregression', LinearRegression())])

Question 🤔 (Answer at q.dsc80.com)

How many weights does this linear model have?

In [ ]:

Multicollinearity¶

Heights and weights¶

We have a dataset containing the weights and heights of 25,000 18 year olds, taken from here.

In [45]:
people_path = Path('data') / 'SOCR-HeightWeight.csv'
people = pd.read_csv(people_path).drop(columns=['Index'])
people.head()
Out[45]:
Height (Inches) Weight (Pounds)
0 65.78 112.99
1 71.52 136.49
2 69.40 153.03
3 68.22 142.34
4 67.79 144.30
In [46]:
people.plot(kind='scatter', x='Height (Inches)', y='Weight (Pounds)', 
            title='Weight vs. Height for 25,000 18 Year Olds')

Motivating example¶

Suppose we fit a simple linear regression model that uses height in inches to predict weight in pounds.

$$\text{predicted weight (pounds)} = w_0 + w_1 \cdot \text{height (inches)}$$
In [47]:
X = people[['Height (Inches)']]
y = people['Weight (Pounds)']
In [48]:
lr_one_feat = LinearRegression()
lr_one_feat.fit(X, y)
Out[48]:
LinearRegression()

$w_0^*$ and $w_1^*$ are shown below, along with the model's training set RMSE.

In [49]:
lr_one_feat.intercept_, lr_one_feat.coef_
Out[49]:
(-82.57574306454099, array([3.08]))
In [50]:
mean_squared_error(y, lr_one_feat.predict(X), squared=False)
Out[50]:
10.079113675632819

Now, suppose we fit another regression model, that uses height in inches AND height in centimeters to predict weight.

$$\text{predicted weight (pounds)} = w_0 + w_1 \cdot \text{height (inches)} + w_2 \cdot \text{height (cm)}$$
In [51]:
people['Height (cm)'] = people['Height (Inches)'] * 2.54 # 1 inch = 2.54 cm
In [52]:
X2 = people[['Height (Inches)', 'Height (cm)']]
In [53]:
lr_two_feat = LinearRegression()
lr_two_feat.fit(X2, y)
Out[53]:
LinearRegression()

What are $w_0^*$, $w_1^*$, $w_2^*$, and the model's test RMSE?

In [54]:
lr_two_feat.intercept_, lr_two_feat.coef_
Out[54]:
(-82.57585227669999, array([-2.46e+10,  9.67e+09]))
In [55]:
mean_squared_error(y, lr_two_feat.predict(X2), squared=False)
Out[55]:
10.079113005376787

Observation: The intercept is the same as before (roughly -82.57), as is the RMSE. However, the coefficients on 'Height (Inches)' and 'Height (cm)' are massive in size!

What's going on?

Redundant features¶

Let's use simpler numbers for illustration. Suppose in the first model, $w_0^* = -80$ and $w_1^* = 3$.

$$\text{predicted weight (pounds)} = -80 + 3 \cdot \text{height (inches)}$$

In the second model, we have:

$$\begin{align*}\text{predicted weight (pounds)} &= w_0^* + w_1^* \cdot \text{height (inches)} + w_2^* \cdot \text{height (cm)} \\ &= w_0^* + w_1^* \cdot \text{height (inches)} + w_2^* \cdot \big( 2.54^* \cdot \text{height (inches)} \big) \\ &= w_0^* + \left(w_1^* + 2.54 \cdot w_2^* \right) \cdot \text{height (inches)} \end{align*}$$

In the first model, we already found the "best" intercept ($-80$) and slope ($3$) in a linear model that uses height in inches to predict weight.

So, as long as $w_1^* + 2.54 \cdot w_2^* = 3$ in the second model, the second model's training predictions will be the same as the first, and hence they will also minimize RMSE.

Infinitely many parameter choices¶

Issue: There are an infinite number of $w_1^*$ and $w_2^*$ that satisfy $w_1^* + 2.54 \cdot w_2^* = 3$!

$$\text{predicted weight} = -80 - 10 \cdot \text{height (inches)} + \frac{13}{2.54} \cdot \text{height (cm)}$$
$$\text{predicted weight} = -80 + 10 \cdot \text{height (inches)} - \frac{7}{2.54} \cdot \text{height (cm)}$$
  • Both prediction rules look very different, but actually make the same predictions.
  • lr.coef_ could return either set of coefficients, or any other of the infinitely many options.
  • But neither set of coefficients is has any meaning!
In [56]:
(-80 - 10 * people.iloc[:, 0] + (13 / 2.54) * people.iloc[:, 2]).head()
Out[56]:
0    117.35
1    134.55
2    128.20
3    124.65
4    123.36
dtype: float64
In [57]:
(-80 + 10 * people.iloc[:, 0] - (7 / 2.54) * people.iloc[:, 2]).head()
Out[57]:
0    117.35
1    134.55
2    128.20
3    124.65
4    123.36
dtype: float64

Multicollinearity¶

  • Multicollinearity occurs when features in a regression model are highly correlated with one another.
    • In other words, multicollinearity occurs when a feature can be predicted using a linear combination of other features, fairly accurately.
  • When multicollinearity is present in the features, the coefficients in the model are uninterpretable – they have no meaning.
    • A "slope" represents "the rate of change of $y$ with respect to a feature", when all other features are held constant – but if there's multicollinearity, you can't hold other features constant.
  • Note: Multicollinearity doesn't impact a model's predictions!
    • It doesn't impact a model's ability to generalize to unseen data.
    • If features are multicollinear in the training data, they will probably be multicollinear in the test data too.
  • Solutions:
    • Manually remove highly correlated features.
    • Use a dimensionality reduction technique (such as PCA) to automatically reduce dimensions.

Example: One hot encoding¶

A one hot encoding will result in multicollinearity unless you drop one of the one hot encoded features.

Suppose we have the following fitted model:

$$ \begin{aligned} H(x) = 1 + 2 \cdot (\text{smoker==Yes}) - 2 \cdot (\text{smoker==No}) \end{aligned} $$

This is equivalent to:

$$ \begin{aligned} H(x) = 10 - 7 \cdot (\text{smoker==Yes}) - 11 \cdot (\text{smoker==No}) \end{aligned} $$

Solution: Drop one of the one hot encoded columns. sklearn.preprocessing.OneHotEncoder has an option to do this.

Key takeaways¶

  • Multicollinearity is present in a linear model when one feature can be accurately predicted using one or more other features.
    • In other words, it is present when a feature is redundant.
  • Multicollinearity doesn't pose an issue for prediction; it doesn't hinder a model's ability to generalize. Instead, it renders the coefficients of a linear model meaningless.

Question 🤔 (Answer at q.dsc80.com)

(Wi23 Final Q9)

One piece of information that may be useful as a feature is the proportion of SAT test takers in a state in a given year that qualify for free lunches in school. The Series lunch_props contains 8 values, each of which are either "low", "medium", or "high". Since we can’t use strings as features in a model, we decide to encode these strings using the following Pipeline:

# Note: The FunctionTransformer is only needed to change the result
# of the OneHotEncoder from a "sparse" matrix to a regular matrix
# so that it can be used with StandardScaler;
# it doesn't change anything mathematically.
pl = Pipeline([
    ("ohe", OneHotEncoder(drop="first")),
    ("ft", FunctionTransformer(lambda X: X.toarray())),
    ("ss", StandardScaler())
])

After calling pl.fit(lunch_props), pl.transform(lunch_props) evaluates to the following array:

array([[ 1.29099445, -0.37796447],
       [-0.77459667, -0.37796447],
       [-0.77459667, -0.37796447],
       [-0.77459667,  2.64575131],
       [ 1.29099445, -0.37796447],
       [ 1.29099445, -0.37796447],
       [-0.77459667, -0.37796447],
       [-0.77459667, -0.37796447]])

and pl.named_steps["ohe"].get_feature_names() evaluates to the following array:

array(["x0_low", "x0_med"], dtype=object)

Fill in the blanks: Given the above information, we can conclude that lunch_props has ____________ value(s) equal to "low", ____________ value(s) equal to "medium", and _____________ value(s) equal to "high". (Note: You should write one positive integer in each box such that the numbers add up to 8.)

In [ ]:

Generalization¶

Motivation¶

  • You and Billy are studying for an upcoming exam. You both decide to test your understanding by taking a practice exam.
    • Your logic: If you do well on the practice exam, you should do well on the real exam.
  • You each take the practice exam once and look at the solutions afterwards.
  • Your strategy: Memorize the answers to all practice exam questions, e.g. "Question 1: A; Question 2: C; Question 3: A."
  • Billy's strategy: Learn high-level concepts from the solutions, e.g. "data are NMAR if the likelihood of missingness depends on the missing values themselves."
  • Who will do better on the practice exam? Who will probably do better on the real exam? 🧐

Evaluating the quality of a model¶

  • So far, we've computed the RMSE (and $R^2$) of our fit regression models on the data that we used to fit them, i.e. the training data.
  • We've said that Model A is better than Model B if Model A's RMSE is lower than Model B's RMSE.
    • Remember, our training data is a sample from some population.
    • Just because a model fits the training data well doesn't mean it will generalize and work well on similar, unseen samples from the same population!

Example: Overfitting and underfitting¶

Let's collect two samples $\{(x_i, y_i)\}$ from the same population.

In [66]:
np.random.seed(23) # For reproducibility.

def sample_from_pop(n=100):
    x = np.linspace(-2, 3, n)
    y = x ** 3 + (np.random.normal(0, 3, size=n))
    return pd.DataFrame({'x': x, 'y': y})

sample_1 = sample_from_pop()
sample_2 = sample_from_pop()

For now, let's just look at Sample 1. The relationship between $x$ and $y$ is roughly cubic; that is, $y \approx x^3$ (remember, in reality, you won't get to see the population).

In [67]:
px.scatter(sample_1, x='x', y='y', title='Sample 1')

Polynomial regression¶

Let's fit three polynomial models on Sample 1:

  • Degree 1.
  • Degree 3.
  • Degree 25.

The PolynomialFeatures transformer will be helpful here.

In [68]:
from sklearn.preprocessing import PolynomialFeatures
In [69]:
# fit_transform fits and transforms the same input.
d2 = PolynomialFeatures(3)
d2.fit_transform(np.array([1, 2, 3, 4, -2]).reshape(-1, 1))
Out[69]:
array([[ 1.,  1.,  1.,  1.],
       [ 1.,  2.,  4.,  8.],
       [ 1.,  3.,  9., 27.],
       [ 1.,  4., 16., 64.],
       [ 1., -2.,  4., -8.]])

Below, we look at our three models' predictions on Sample 1 (which they were trained on).

In [70]:
# Look at the definition of train_and_plot in lec15_util.py if you're curious as to how the plotting works.
fig = util.train_and_plot(train_sample=sample_1, test_sample=sample_1, degs=[1, 3, 25], data_name='Sample 1')
fig.update_layout(title='Trained on Sample 1, Performance on Sample 1')

The degree 25 polynomial has the lowest RMSE on Sample 1.

How do the same fit polynomials look on Sample 2?

In [71]:
fig = util.train_and_plot(train_sample=sample_1, test_sample=sample_2, degs=[1, 3, 25], data_name='Sample 2')
fig.update_layout(title='Trained on Sample 1, Performance on Sample 2')
  • The degree 3 polynomial has the lowest RMSE on Sample 2.
  • Note that we didn't get to see Sample 2 when fitting our models!
  • As such, it seems that the degree 3 polynomial generalizes better to unseen data than the degree 25 polynomial does.

What if we fit a degree 1, degree 3, and degree 25 polynomial on Sample 2 as well?

In [72]:
util.plot_multiple_models(sample_1, sample_2, degs=[1, 3, 25])

Key idea: Degree 25 polynomials seem to vary more when trained on different samples than degree 3 and 1 polynomials do.

Bias and variance¶

The training data we have access to is a sample from the population. We are concerned with our model's ability to generalize and work well on different datasets drawn from the same population.

Suppose we fit a model $H$ (e.g. a degree 3 polynomial) on several different datasets from a population. There are three sources of error that arise:

  • ⭐️ Bias: The expected deviation between a predicted value and an actual value.
    • In other words, for a given $x_i$, how far is $H(x_i)$ from the true $y_i$, on average?
    • Low bias is good! ✅
    • High bias is a sign of underfitting, i.e. that our model is too basic to capture the relationship between our features and response.
  • ⭐️ Model variance ("variance"): The variance of a model's predictions.
    • In other words, for a given $x_i$, how much does $H(x_i)$ vary across all datasets?
    • Low model variance is good! ✅
    • High model variance is a sign of overfitting, i.e. that our model is too complicated and is prone to fitting to the noise in our training data.
  • Observation error: The error due to the random noise in the process we are trying to model (e.g. measurement error). We can't control this, without collecting more data!

Here, suppose:

  • The red bulls-eye represents your true weight and height 🧍.
  • The dark blue darts represent predictions of your weight and height using different models that were fit on the same DGP.

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We'd like our models to be in the top left, but in practice that's hard to achieve!

Risk vs. empirical risk¶

  • In DSC 40A, we started using empirical risk minimization to find optimal model parameters $w^*$:
$$ \text{choose the $w$ such that } \frac{1}{n} \sum_{i = 1}^n \left( y_i - H(x_i) \right)^2 \text{ is minimized}$$
or, equivalently:
$$w^* = \underset{w}{\text{argmin}} \frac{1}{n} \sum_{i = 1}^n \left( y_i - H(x_i) \right)^2$$
  • Key idea: A model that works well on past data should work well on future data, if future data looks like past data.
  • What we really want is for the expected loss for a new data point $(x_{\text{new}}, y_{\text{new}})$, drawn from the same population as the training set, to be small. That is, we want $$\mathbb{E}[y_{\text{new}} - H(x_{\text{new}})]^2$$ to be minimized. The quantity above is called risk.
  • What's that fancy $\mathbb{E}$? It is the expectation operator of a random variable: it computes the average value of the random variable across its entire distribution.
    • (From Math 183/180A) If $X \sim \text{Binomial}(n, p)$, then $\mathbb{E}[X] = np$.
  • In general, we don't know the entire population distribution of $x$s and $y$s, so we can't compute risk exactly. That's why we compute empirical risk!
$$\mathbb{E}[y_{\text{new}} - H(x_{\text{new}})]^2 \approx \frac{1}{n} \sum_{i = 1}^n \left( y_i - H(x_i) \right)^2$$

The bias-variance decomposition¶

Risk can be decomposed as follows:

$$\mathbb{E}[y_{\text{new}} - H(x_{\text{new}})]^2 = \text{model bias}^2 + \text{model variance} + \text{observation error}$$
  • Remember, this expectation $\mathbb{E}$ is over the entire population of $x$s and $y$s: in real life, we don't know what this population distribution is, so we can't put actual numbers to this.
  • If $H$ is too simple to capture the relationship between $x$s and $y$s in the population, $H$ will underfit to training sets and have high bias.
  • If $H$ is overly complex, $H$ will overfit to training sets and have high variance, meaning it will change significantly from one training set to the next.
  • Generally:
    • Training error reflects bias, not variance.
    • Test error reflects both bias and variance.

Navigating the bias-variance tradeoff¶

$$\mathbb{E}[y_{\text{new}} - H(x_{\text{new}})]^2 = \text{model bias}^2 + \text{model variance} + \text{observation error}$$
  • As we collect more data points (i.e. as $n \uparrow$):
    • Model variance decreases.
    • If $H$ can exactly model the true population relationship between $x$ and $y$ (e.g. cubic), then model bias also decreases.
    • If $H$ can't exactly model the true population relationship between $x$ and $y$, then model bias will remain large.
  • As we add more features (i.e. as $d \uparrow$):
    • Model variance increases, whether or not the feature was useful.
    • Adding a useful feature decreases model bias.
    • Adding a useless feature doesn't change model bias.
  • Example: suppose the actual relationship between $x$ and $y$ in the population is linear, and we fit $H$ using simple linear regression.
    • Model bias = 0.
    • Model variance $\propto \frac{d}{n}$.
      • As $d \uparrow$, model variance $\uparrow$.
      • As $n \uparrow$, model variance $\downarrow$.

Read more here.

Question 🤔 (Answer at q.dsc80.com)

(Fa23 Final 9.4)

Determine how each change below affects model bias and variance compared to this model:

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For each change, choose all of the following that apply: increase bias, decrease bias, increase variance, decrease variance.

  1. Add degree 3 polynomial features.
  2. Add a feature of numbers chosen at random between 0 and 1.
  3. Collect 100 more points for the training set.
  4. Don’t use the 'veg' feature.

Train-test splits¶

Avoiding overfitting¶

  • We won't know whether our model has overfit to our sample (training data) unless we get to see how well it performs on a new sample from the same population.
  • 💡Idea: Split our sample into a training set and test set.
  • Use only the training set to fit the model (i.e. find $w^*$).
  • Use the test set to evaluate the model's error (RMSE, $R^2$).
  • The test set is like a new sample of data from the same population as the training data!
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Train-test split 🚆¶

sklearn.model_selection.train_test_split implements a train-test split for us! 🙏🏼

If X is an array/DataFrame of features and y is an array/Series of responses,

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.25)

randomly splits the features and responses into training and test sets, such that the test set contains 0.25 of the full dataset.

In [73]:
from sklearn.model_selection import train_test_split
In [74]:
# Read the documentation!
train_test_split?

Let's perform a train/test split on our tips dataset.

In [75]:
X = tips.drop('tip', axis=1)
y = tips['tip']
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2) # We don't have to choose 0.25.

Before proceeding, let's check the sizes of X_train and X_test.

In [76]:
print('Rows in X_train:', X_train.shape[0])
display(X_train.head())
print('Rows in X_test:', X_test.shape[0])
display(X_test.head())

Rows in X_train: 195
total_bill sex smoker day time size
4 24.59 Female No Sun Dinner 4
209 12.76 Female Yes Sat Dinner 2
178 9.60 Female Yes Sun Dinner 2
230 24.01 Male Yes Sat Dinner 4
5 25.29 Male No Sun Dinner 4 
Rows in X_test: 49
total_bill sex smoker day time size
146 18.64 Female No Thur Lunch 3
224 13.42 Male Yes Fri Lunch 2
134 18.26 Female No Thur Lunch 2
131 20.27 Female No Thur Lunch 2
147 11.87 Female No Thur Lunch 2
In [77]:
X_train.shape[0] / tips.shape[0]
Out[77]:
0.7991803278688525

Example train-test split¶

Steps:

  1. Fit a model on the training set.
  2. Evaluate the model on the test set.
In [78]:
tips.head()
Out[78]:
total_bill tip sex smoker day time size
0 16.99 1.01 Female No Sun Dinner 2
1 10.34 1.66 Male No Sun Dinner 3
2 21.01 3.50 Male No Sun Dinner 3
3 23.68 3.31 Male No Sun Dinner 2
4 24.59 3.61 Female No Sun Dinner 4
In [79]:
X = tips[['total_bill', 'size']] # For this example, we'll use just the already-quantitative columns in tips.
y = tips['tip']
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=1) # random_state is like np.random.seed.

Here, we'll use a stand-alone LinearRegression model without a Pipeline, but this process would work the same if we were using a Pipeline.

In [80]:
lr = LinearRegression()
lr.fit(X_train, y_train)
Out[80]:
LinearRegression()

Let's check our model's performance on the training set first.

In [81]:
pred_train = lr.predict(X_train)
rmse_train = mean_squared_error(y_train, pred_train, squared=False)
rmse_train
Out[81]:
0.9803205287924737

And the test set:

In [82]:
pred_test = lr.predict(X_test)
rmse_test = mean_squared_error(y_test, pred_test, squared=False)
rmse_test
Out[82]:
1.138177129113125

Since rmse_train and rmse_test are similar, it doesn't seem like our model is overfitting to the training data. If rmse_test was much larger than rmse_train, it would be evidence that our model is unable to generalize well.

Hyperparameters¶

Example: Polynomial regression¶

We recently looked at an example of polynomial regression.

In [83]:
fig = util.train_and_plot(train_sample=sample_1, test_sample=sample_2, degs=[1, 3, 25], data_name='Sample 2')
fig.update_layout(title='Trained on Sample 1, Performance on Sample 2')

When building these models:

  • We got to choose the degree of the polynomials (i.e. we chose 1, 3, and 25).
  • We didn't get to choose the exact formulas for the three polynomials – their formulas were learned from data.

Parameters vs. hyperparameters¶

  • A parameter defines the relationship between variables in a model.

    • We learn parameters from data.
    • For instance, suppose we fit a degree 3 polynomial to data, and end up with

    $$H(x) = 1 - 2x + 13x^2 - 4x^3$$

    • 1, -2, 13, and -4 are parameters.
  • A hyperparameter is a parameter that we get to choose before our model is fit to the data.
    • Think of hyperparameters as knobs 🎛 – we get to pick and tune them!
    • Polynomial degree was a hyperparameter in the previous example, and we tried three different values: 1, 3, and 25.
  • Question: How do we choose the "right" hyperparameter(s)?

Training error vs. test error¶

  • We know that a model's performance on a test set is a good estimate of its ability to generalize to unseen data.
  • We want to find the hyperparameter that leads to the best test set performance.
  • Idea:
    1. Come up with a list of hyperparameters to try.
    2. For each hyperparameter, train the model on the training set and compute its performance on the test set.
    3. Pick the hyperparameter with the best performance on the test set.

Training error vs. test error¶

  • Let's try this strategy on Sample 1 from our earlier example.

  • We'll try to fit a polynomial model on the dataset; we'll choose the polynomial's degree from the list [1, 2, ..., 25].

In [84]:
px.scatter(sample_1, x='x', y='y', title='Sample 1')

First, we perform a train-test split.

In [85]:
X = sample_1[['x']]
y = sample_1['y']

X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=100)

Polynomial degree vs. train/test error¶

Now, we'll create models with degree-1 through degree-25 polynomial features and compute their train and test errors.

In [88]:
train_errs = []
test_errs = []

for d in range(1, 26):
    pl = make_pipeline(PolynomialFeatures(d), LinearRegression())
    pl.fit(X_train, y_train)
    train_errs.append(mean_squared_error(y_train, pl.predict(X_train), squared=False))
    test_errs.append(mean_squared_error(y_test, pl.predict(X_test), squared=False))

errs = pd.DataFrame({'Train Error': train_errs, 'Test Error': test_errs})

Let's look at the plots of training error vs. degree and test error vs. degree.

In [89]:
fig = px.line(errs)
fig.update_layout(showlegend=True, xaxis_title='Polynomial Degree', yaxis_title='RMSE')
  • Training error appears to decrease as polynomial degree increases.
  • Test error appears to decrease until a "valley", and then increases again.
  • Here, we'd choose a degree of 3, since that degree has the lowest test error.

Training error vs. test error¶

The pattern we saw in the previous example is true more generally.

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We pick the hyperparameter(s) at the "valley" of test error.

Note that training error tends to underestimate test error, but it doesn't have to – i.e., it is possible for test error to be lower than training error (say, if the test set is "easier" to predict than the training set).

Conducting train-test splits¶

  • Recall, training data is used to fit our model, and test data is used to evaluate our model.
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  • Question: How should we split?
    • sklearn's train_test_split splits randomly, which usually works well.
    • However, if there is some element of time in the training data (say, when predicting the future price of a stock), a better split is "past" and "future".
  • Question: How large should the split be, e.g. 90%-10% vs. 75%-25%?
    • There's a tradeoff – a larger training set should lead to a "better" model, while a larger test set should lead to a better estimate of our model's ability to generalize.
    • There's no "right" choice, but we usually choose between 10% to 25% for the test set.

But wait...¶

  • With our current strategy, we are choosing the hyperparameter that creates the model that performs best on the test set.
  • As such, we are overfitting to the test set – the best hyperparameter for the test set might not be the best hyperparameter for a totally unseen dataset!
  • It seems like we need another split.

Summary, next time¶

Summary¶

  • We want to build models that generalize well to unseen data.
    • Models that have high bias are too simple to represent complex relationships in data, and underfit.
    • Models that have high variance are overly complex for the relationships in the data, and vary a lot when fit on different datasets. Such models overfit to the training data.
  • A model's training error tends to decrease as model complexity increases, while its test error tends to decrease, before reaching a "sweet spot" and increasing again.
  • A hyperparameter is a configuration that we choose before training a model; an important task in machine learning is selecting "good" hyperparameters.

Next time¶

We can't choose model hyperparameters just by using a train-test split, because this strategy overfits to the test data.

Is there a better way to choose model hyperparameters? Yes: cross-validation.