from dsc80_utils import *

Lecture 17 – Random Forests

diabetes = pd.read_csv(Path("data") / "diabetes.csv")

from sklearn.model_selection import train_test_split

X_train, X_test, y_train, y_test = train_test_split(
    diabetes[["Glucose", "BMI"]], diabetes["Outcome"], random_state=1
)

from sklearn.tree import DecisionTreeClassifier

Grid search

Grid search

GridSearchCV takes in:

• an un-fit instance of an estimator, and
• a dictionary of hyperparameter values to try,

and performs $k$-fold cross-validation to find the combination of hyperparameters with the best average validation performance.

from sklearn.model_selection import GridSearchCV

The following dictionary contains the values we're considering for each hyperparameter. (We're using GridSearchCV with 3 hyperparameters, but we could use it with even just a single hyperparameter.)

hyperparameters = {
    "max_depth": [2, 3, 4, 5, 7, 10, 13, 15, 18, None],
    "min_samples_split": [2, 5, 10, 20, 50, 100, 200],
    "criterion": ["gini", "entropy"],
}

Note that there are 140 combinations of hyperparameters we need to try. We need to find the best combination of hyperparameters, not the best value for each hyperparameter individually.

len(hyperparameters["max_depth"]) * len(
    hyperparameters["min_samples_split"]
) * len(hyperparameters["criterion"])

140

GridSearchCV needs to be instantiated and fit.

searcher = GridSearchCV(DecisionTreeClassifier(), hyperparameters, cv=5)

%%time
searcher.fit(X_train, y_train)

CPU times: user 780 ms, sys: 16.8 ms, total: 797 ms
Wall time: 797 ms

GridSearchCV(cv=5, estimator=DecisionTreeClassifier(),
             param_grid={'criterion': ['gini', 'entropy'],
                         'max_depth': [2, 3, 4, 5, 7, 10, 13, 15, 18, None],
                         'min_samples_split': [2, 5, 10, 20, 50, 100, 200]})

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After being fit, the best_params_ attribute provides us with the best combination of hyperparameters to use.

searcher.best_params_

{'criterion': 'gini', 'max_depth': 4, 'min_samples_split': 50}

All of the intermediate results – validation accuracies for each fold, mean validation accuaries, etc. – are stored in the cv_results_ attribute:

searcher.cv_results_["mean_test_score"]  # Array of length 140.

array([0.73, 0.73, 0.73, ..., 0.75, 0.74, 0.72])

# Rows correspond to folds, columns correspond to hyperparameter combinations.
pd.DataFrame(
    np.vstack([searcher.cv_results_[f"split{i}_test_score"] for i in range(5)])
)

	0	1	2	3	...	136	137	138	139
0	0.71	0.71	0.71	0.71	...	0.70	0.70	0.71	0.73
1	0.77	0.77	0.77	0.77	...	0.82	0.83	0.77	0.76
2	0.74	0.74	0.74	0.74	...	0.68	0.72	0.74	0.73
3	0.70	0.70	0.70	0.70	...	0.77	0.79	0.76	0.70
4	0.72	0.72	0.72	0.72	...	0.70	0.71	0.72	0.70

5 rows × 140 columns

Note that the above DataFrame tells us that 5 * 140 = 700 models were trained in total!

Now that we've found the best combination of hyperparameters, we should fit a decision tree instance using those hyperparameters on our entire training set.

searcher.best_params_

{'criterion': 'gini', 'max_depth': 4, 'min_samples_split': 50}

final_tree = DecisionTreeClassifier(**searcher.best_params_)
final_tree

DecisionTreeClassifier(max_depth=4, min_samples_split=50)

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final_tree.fit(X_train, y_train)

DecisionTreeClassifier(max_depth=4, min_samples_split=50)

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# Training accuracy.
final_tree.score(X_train, y_train)

0.7881944444444444

# Testing accuracy.
# A bit lower than the `dt` tree we fit above!
final_tree.score(X_test, y_test)

0.765625

Remember, searcher itself is a model object (we had to fit it). After performing $k$-fold cross-validation, behind the scenes, searcher is trained on the entire training set using the optimal combination of hyperparameters.

In other words, searcher makes the same predictions that final_tree does!

searcher.score(X_train, y_train)

0.7881944444444444

searcher.score(X_test, y_test)

0.765625

Choosing possible hyperparameter values

• A full grid search can take a long time.

  • In our previous example, we tried 140 combinations of hyperparameters.
  • Since we performed 5-fold cross-validation, we trained 700 decision trees under the hood.

• Question: How do we pick the possible hyperparameter values to try?

• Answer: Trial and error.

  • If the "best" choice of a hyperparameter was at an extreme, try increasing the range.
  • For instance, if you try max_depths from 32 to 128, and 32 was the best, try including max_depths under 32.

Key takeaways

• Decision trees are trained by finding the best questions to ask using the features in the training data. A good question is one that isolates classes as much as possible.
• Decision trees have a tendency to overfit to training data. One way to mitigate this is by restricting the maximum depth of the tree.
• To efficiently find hyperparameters through cross-validation, use GridSearchCV.
  • Specify which values to try for each hyperparameter, and GridSearchCV will try all unique combinations of hyperparameters and return the combination with the best average validation performance.
  • GridSearchCV is not the only solution – see RandomizedSearchCV if you're curious.

Decision tree pros and cons

Pros:

• Fast to fit (usually log-linear time w.r.t. size of design matrix)
• Fast to predict (usually log time)
• Interpretable
• Robust to irrelevant features (think about why!)
• Linear transformations on features don't affect predictions (think about why!)

Cons:

• High variance: complete tree (no depth limit) will almost always overfit!
• Aren't the best at prediction in general.

Random Forests

Another idea:

Train a bunch of decision trees, then have them vote on a prediction!

• Problem: If you use the same training data, you will always get the same tree.
• Solution: Introduce randomness into training procedure to get different trees.

Idea 1: Bootstrap the training data

• We can bootstrap the training data $T$ times, then train one tree on each resample.
• Also known as bagging (Bootstrap AGgregating). In general, combining different predictors together is a useful technique called ensemble learning.
• For decision trees though, doesn't make trees different enough from each other (e.g. if you have one really strong predictor, it'll always be the first split).

Idea 2: Only use a subset of features

• At each split, take a random subset of $ m $ features instead of choosing from all $ d $ of them.

• Rule of thumb: $ m \approx \sqrt d $ seems to work well.

• Key idea: For ensemble learning, you want the individual predictors to have low bias, high variance, and be uncorrelated with each other. That way, when you average them together, you have low bias AND low variance.

• Random forest algorithm: Fit $ T $ trees by using bagging and a random subset of features at each split. Predict by taking a vote from the $ T $ trees.

Question 🤔

How will increasing $ m $ affect the bias / variance of each decision tree?

Example

# Let's use more features for prediction
from sklearn.model_selection import train_test_split

X_train, X_test, y_train, y_test = train_test_split(
    diabetes.drop(columns=["Outcome"]), diabetes["Outcome"], random_state=1
)

from sklearn.ensemble import RandomForestClassifier

clf = RandomForestClassifier()
clf.fit(X_train, y_train)
clf.score(X_train, y_train)

1.0

clf.score(X_test, y_test)

0.8177083333333334

Compared to our previous best decision tree with depth 4:

dt = DecisionTreeClassifier(max_depth=4, criterion="entropy")
dt.fit(X_train, y_train)
dt.score(X_train, y_train)

0.7829861111111112

dt.score(X_test, y_test)

0.734375

Example: Modeling using text features

Example: Fake news

We have a dataset containing news articles and labels for whether the article was deemed "fake" or "real". Credit to https://github.com/KaiDMML/FakeNewsNet.

news = pd.read_csv("data/fake_news_training.csv")
news

	baseurl	content	label
0	twitter.com	\njavascript is not available.\n\nwe’ve detect...	real
1	whitehouse.gov	remarks by the president at campaign event -- ...	real
2	web.archive.org	the committee on energy and commerce\nbarton: ...	real
...	...	...	...
658	politico.com	full text: jeff flake on trump speech transcri...	fake
659	pol.moveon.org	moveon.org political action: 10 things to know...	real
660	uspostman.com	uspostman.com is for sale\nyes, you can transf...	fake

661 rows × 3 columns

Goal: Use an article's content to predict its label.

news["label"].value_counts(normalize=True)

label
real    0.55
fake    0.45
Name: proportion, dtype: float64

Question: What is the worst possible accuracy we should expect from a classifier, given the above distribution?

Aside: CountVectorizer

Entries in the 'content' column are not currently quantitative! We can use the bag of words encoding to create quantitative features out of each 'content'.

Instead of performing a bag of words encoding manually as we did before, we can rely on sklearn's CountVectorizer. (There is also a TfidfVectorizer.)

from sklearn.feature_extraction.text import CountVectorizer

example_corp = [
    "hey hey hey my name is billy",
    "hey billy how is your dog billy",
]

count_vec = CountVectorizer()
count_vec.fit(example_corp)

CountVectorizer()

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count_vec learned a vocabulary from the corpus we fit it on.

count_vec.vocabulary_

{'hey': 2,
 'my': 5,
 'name': 6,
 'is': 4,
 'billy': 0,
 'how': 3,
 'your': 7,
 'dog': 1}

count_vec.transform(example_corp).toarray()

array([[1, 0, 3, 0, 1, 1, 1, 0],
       [2, 1, 1, 1, 1, 0, 0, 1]])

Note that the values in count_vec.vocabulary_ correspond to the positions of the columns in count_vec.transform(example_corp).toarray(), i.e. 'billy' is the first column and 'your' is the last column.

example_corp

['hey hey hey my name is billy', 'hey billy how is your dog billy']

pd.DataFrame(
    count_vec.transform(example_corp).toarray(),
    columns=pd.Series(count_vec.vocabulary_).sort_values().index,
)

	billy	dog	hey	how	is	my	name	your
0	1	0	3	0	1	1	1	0
1	2	1	1	1	1	0	0	1

Creating an initial Pipeline

Let's build a Pipeline that takes in summaries and overall ratings and:

• Uses CountVectorizer to quantitatively encode summaries.

• Fits a RandomForestClassifier to the data.

But first, a train-test split (like always).

from sklearn.ensemble import RandomForestClassifier
from sklearn.model_selection import train_test_split
from sklearn.pipeline import make_pipeline

X = news["content"]
y = news["label"]
X_train, X_test, y_train, y_test = train_test_split(X, y)

To start, we'll create a random forest with 100 trees (n_estimators) each of which has a maximum depth of 3 (max_depth).

pl = make_pipeline(
    CountVectorizer(),
    RandomForestClassifier(
        max_depth=3,
        n_estimators=100,  # Uses 100 separate decision trees!
        random_state=42,
    ), 
)

pl.fit(X_train, y_train)

Pipeline(steps=[('countvectorizer', CountVectorizer()),
                ('randomforestclassifier',
                 RandomForestClassifier(max_depth=3, random_state=42))])

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# Training accuracy.
pl.score(X_train, y_train)

0.7515151515151515

# Testing accuracy.
pl.score(X_test, y_test)

0.7469879518072289

The accuracy of our random forest is just under 70%, on the test set. How much better does it do compared to a classifier that predicts "real" every time?

y_train.value_counts(normalize=True)

label
real    0.55
fake    0.45
Name: proportion, dtype: float64

# Distribution of predicted ys in the training set:

# stops scientific notation for pandas
pd.set_option("display.float_format", "{:.3f}".format)
pd.Series(pl.predict(X_train)).value_counts(normalize=True)

fake   0.642
real   0.358
Name: proportion, dtype: float64

pl.named_steps

{'countvectorizer': CountVectorizer(),
 'randomforestclassifier': RandomForestClassifier(max_depth=3, random_state=42)}

len(pl.named_steps["countvectorizer"].vocabulary_)  # Lots of features!

26096

Choosing tree depth via GridSearchCV

We arbitrarily chose max_depth=3 before, but it seems like that isn't working well. Let's perform a grid search to find the max_depth with the best generalization performance.

# Note that we've used the key randomforestclassifier__max_depth,
# not max_depth because max_depth is a hyperparameter of randomforestclassifier,
# not of pl.

hyperparameters = {"randomforestclassifier__max_depth": np.arange(2, 200, 20)}

Note that while pl has already been fit, we can still give it to GridSearchCV, which will repeatedly re-fit it during cross-validation.

%%time

# Takes a few seconds to run – how many trees are being trained?
from sklearn.model_selection import GridSearchCV

grids = GridSearchCV(
    pl,
    n_jobs=-1,  # Use multiple processors to parallelize
    param_grid=hyperparameters,
    return_train_score=True,
)
grids.fit(X_train, y_train)

CPU times: user 1.04 s, sys: 247 ms, total: 1.29 s
Wall time: 6.15 s

GridSearchCV(estimator=Pipeline(steps=[('countvectorizer', CountVectorizer()),
                                       ('randomforestclassifier',
                                        RandomForestClassifier(max_depth=3,
                                                               random_state=42))]),
             n_jobs=-1,
             param_grid={'randomforestclassifier__max_depth': array([  2,  22,  42,  62,  82, 102, 122, 142, 162, 182])},
             return_train_score=True)

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grids.best_params_

{'randomforestclassifier__max_depth': np.int64(42)}

Recall, fit GridSearchCV objects are estimators on their own as well. This means we can compute the training and testing accuracies of the "best" random forest directly:

# Training accuracy.
grids.score(X_train, y_train)

0.9919191919191919

# Testing accuracy.
grids.score(X_test, y_test)

0.8674698795180723

~10% better test set error!

Training and validation accuracy vs. depth

Below, we plot how training and validation accuracy varied with tree depth. Note that the $y$-axis here is accuracy, and that larger accuracies are better (unlike with RMSE, where smaller was better).

index = grids.param_grid["randomforestclassifier__max_depth"]
train = grids.cv_results_["mean_train_score"]
valid = grids.cv_results_["mean_test_score"]

pd.DataFrame(
    {"train": train, "valid": valid}, index=index
).plot().update_layout(xaxis_title="max_depth", yaxis_title="Accuracy")

Question 🤔

Suppose we write the following code:

hyperparameters = {
    'n_estimators': [10, 100, 1000], # number of trees per forest
    'max_depth': [None, 100, 10]     # max depth of each tree
}
grids = GridSearchCV(
    RandomForestClassifier(), param_grid=hyperparameters,
    cv=3, # 3-fold cross-validation
)
grids.fit(X_train, y_train)

Answer the following questions with a single number.

1. How many random forests are fit in total?
2. How many decision trees are fit in total?
3. How many times in total is the first point in X_train used to train a decision tree?

Classifier Evaluation

Accuracy isn't everything!

$$ \text{accuracy} = \frac{\text{\# data points classified correctly}}{\text{\# data points}} $$

• Accuracy is defined as the proportion of predictions that are correct.

• It weighs all correct predictions the same, and weighs all incorrect predictions the same.

• But some incorrect predictions may be worse than others!

  • Example: Suppose you take a COVID test 🦠. Which is worse:
    • The test saying you have COVID, when you really don't, or
    • The test saying you don't have COVID, when you really do?

The Boy Who Cried Wolf 👦😭🐺

(source)

A shepherd boy gets bored tending the town's flock. To have some fun, he cries out, "Wolf!" even though no wolf is in sight. The villagers run to protect the flock, but then get really mad when they realize the boy was playing a joke on them.

Repeat the previous paragraph many, many times.

One night, the shepherd boy sees a real wolf approaching the flock and calls out, "Wolf!" The villagers refuse to be fooled again and stay in their houses. The hungry wolf turns the flock into lamb chops. The town goes hungry. Panic ensues.

The wolf classifier

• Predictor: Shepherd boy.
• Positive prediction: "There is a wolf."
• Negative prediction: "There is no wolf."

Some questions to think about:

• What is an example of an incorrect, positive prediction?

• Was there a correct, negative prediction?

• There are four possibilities. What are the consequences of each?
  • (predict yes, predict no) x (actually yes, actually no).

The wolf classifier

Below, we present a confusion matrix, which summarizes the four possible outcomes of the wolf classifier.

Outcomes in binary classification

When performing binary classification, there are four possible outcomes.

(Note: A "positive prediction" is a prediction of 1, and a "negative prediction" is a prediction of 0.)

Outcome of Prediction	Definition	True Class
True positive (TP) ✅	The predictor correctly predicts the positive class.	P
False negative (FN) ❌	The predictor incorrectly predicts the negative class.	P
True negative (TN) ✅	The predictor correctly predicts the negative class.	N
False positive (FP) ❌	The predictor incorrectly predicts the positive class.	N

⬇️

	Predicted Negative	Predicted Positive
Actually Negative	TN ✅	FP ❌
Actually Positive	FN ❌	TP ✅

The confusion matrix above is organized the same way that sklearn's confusion matrices are (but differently than in the wolf example).

Note that in the four acronyms – TP, FN, TN, FP – the first letter is whether the prediction is correct, and the second letter is what the prediction is.

Example: COVID testing 🦠

• UCSD Health administers hundreds of COVID tests a day. The tests are not fully accurate.

• Each test comes back either

  • positive, indicating that the individual has COVID, or
  • negative, indicating that the individual does not have COVID.

• Question: What is a TP in this scenario? FP? TN? FN?

• TP: The test predicted that the individual has COVID, and they do ✅.

• FP: The test predicted that the individual has COVID, but they don't ❌.

• TN: The test predicted that the individual doesn't have COVID, and they don't ✅.

• FN: The test predicted that the individual doesn't have COVID, but they do ❌.

Accuracy of COVID tests

The results of 100 UCSD Health COVID tests are given below.

	Predicted Negative	Predicted Positive
Actually Negative	TN = 90 ✅	FP = 1 ❌
Actually Positive	FN = 8 ❌	TP = 1 ✅

UCSD Health test results

🤔 Question: What is the accuracy of the test?

🙋 Answer: $$\text{accuracy} = \frac{TP + TN}{TP + FP + FN + TN} = \frac{1 + 90}{100} = 0.91$$

• Followup: At first, the test seems good. But, suppose we build a classifier that predicts that nobody has COVID. What would its accuracy be?

• Answer to followup: Also 0.91! There is severe class imbalance in the dataset, meaning that most of the data points are in the same class (no COVID). Accuracy doesn't tell the full story.

Recall

	Predicted Negative	Predicted Positive
Actually Negative	TN = 90 ✅	FP = 1 ❌
Actually Positive	FN = 8 ❌	TP = 1 ✅

UCSD Health test results

🤔 Question: What proportion of individuals who actually have COVID did the test identify?

🙋 Answer: $\frac{1}{1 + 8} = \frac{1}{9} \approx 0.11$

More generally, the recall of a binary classifier is the proportion of actually positive instances that are correctly classified. We'd like this number to be as close to 1 (100%) as possible.

$$\text{recall} = \frac{TP}{\text{\# actually positive}} = \frac{TP}{TP + FN}$$

To compute recall, look at the bottom (positive) row of the above confusion matrix.

Recall isn't everything, either!

$$\text{recall} = \frac{TP}{TP + FN}$$

🤔 Question: Can you design a "COVID test" with perfect recall?

🙋 Answer: Yes – just predict that everyone has COVID!

	Predicted Negative	Predicted Positive
Actually Negative	TN = 0 ✅	FP = 91 ❌
Actually Positive	FN = 0 ❌	TP = 9 ✅

everyone-has-COVID classifier $$\text{recall} = \frac{TP}{TP + FN} = \frac{9}{9 + 0} = 1$$

Like accuracy, recall on its own is not a perfect metric. Even though the classifier we just created has perfect recall, it has 91 false positives!

Precision

	Predicted Negative	Predicted Positive
Actually Negative	TN = 0 ✅	FP = 91 ❌
Actually Positive	FN = 0 ❌	TP = 9 ✅

everyone-has-COVID classifier

The precision of a binary classifier is the proportion of predicted positive instances that are correctly classified. We'd like this number to be as close to 1 (100%) as possible.

$$\text{precision} = \frac{TP}{\text{\# predicted positive}} = \frac{TP}{TP + FP}$$

To compute precision, look at the right (positive) column of the above confusion matrix.

• Tip: A good way to remember the difference between precision and recall is that in the denominator for 🅿️recision, both terms have 🅿️ in them (TP and FP).

• Note that the "everyone-has-COVID" classifier has perfect recall, but a precision of $\frac{9}{9 + 91} = 0.09$, which is quite low.

• 🚨 Key idea: There is a "tradeoff" between precision and recall. Ideally, you want both to be high. For a particular prediction task, one may be important than the other.

Precision and recall

(source)

Precision and recall

$$\text{precision} = \frac{TP}{TP + FP} \: \: \: \: \: \: \: \: \text{recall} = \frac{TP}{TP + FN}$$

Question 🤔

🤔 When might high precision be more important than high recall?

🤔 When might high recall be more important than high precision?

Question 🤔

Consider the confusion matrix shown below.

	Predicted Negative	Predicted Positive
Actually Negative	TN = 22 ✅	FP = 2 ❌
Actually Positive	FN = 23 ❌	TP = 18 ✅

What is the accuracy of the above classifier? The precision? The recall?