Authors: [Maria Sumarokova](https://www.linkedin.com/in/mariya-sumarokova-230b4054/), and [Yury Kashnitsky](https://www.linkedin.com/in/festline/). Translated and edited by Gleb Filatov, Aleksey Kiselev, [Anastasia Manokhina](https://www.linkedin.com/in/anastasiamanokhina/), [Egor Polusmak](https://www.linkedin.com/in/egor-polusmak/), and [Yuanyuan Pao](https://www.linkedin.com/in/yuanyuanpao/). All content is distributed under the [Creative Commons CC BY-NC-SA 4.0](https://creativecommons.org/licenses/by-nc-sa/4.0/) license. Same assignment as a [Kaggle Kernel](https://www.kaggle.com/kashnitsky/a3-demo-decision-trees) + [solution](https://www.kaggle.com/kashnitsky/a3-demo-decision-trees-solution). Fill in the answers in the [web-form](https://docs.google.com/forms/d/1wfWYYoqXTkZNOPy1wpewACXaj2MZjBdLOL58htGWYBA/edit). Let's start by loading all necessary libraries: ```{code-cell} ipython3 %matplotlib inline from matplotlib import pyplot as plt plt.rcParams["figure.figsize"] = (10, 8) import collections import numpy as np import pandas as pd from sklearn.ensemble import RandomForestClassifier from sklearn.metrics import accuracy_score from sklearn.model_selection import GridSearchCV, cross_val_score from sklearn.preprocessing import LabelEncoder from sklearn.tree import DecisionTreeClassifier, plot_tree ``` ## Part 1. Toy dataset "Will They? Won't They?" Your goal is to figure out how decision trees work by walking through a toy problem. While a single decision tree does not yield outstanding results, other performant algorithms like gradient boosting and random forests are based on the same idea. That is why knowing how decision trees work might be useful. We'll go through a toy example of binary classification - Person A is deciding whether they will go on a second date with Person B. It will depend on their looks, eloquence, alcohol consumption (only for example), and how much money was spent on the first date. ### Creating the dataset ```{code-cell} ipython3 # Create dataframe with dummy variables def create_df(dic, feature_list): out = pd.DataFrame(dic) out = pd.concat([out, pd.get_dummies(out[feature_list])], axis=1) out.drop(feature_list, axis=1, inplace=True) return out # Some feature values are present in train and absent in test and vice-versa. def intersect_features(train, test): common_feat = list(set(train.keys()) & set(test.keys())) return train[common_feat], test[common_feat] ``` ```{code-cell} ipython3 features = ["Looks", "Alcoholic_beverage", "Eloquence", "Money_spent"] ``` ### Training data ```{code-cell} ipython3 df_train = {} df_train["Looks"] = [ "handsome", "handsome", "handsome", "repulsive", "repulsive", "repulsive", "handsome", ] df_train["Alcoholic_beverage"] = ["yes", "yes", "no", "no", "yes", "yes", "yes"] df_train["Eloquence"] = ["high", "low", "average", "average", "low", "high", "average"] df_train["Money_spent"] = ["lots", "little", "lots", "little", "lots", "lots", "lots"] df_train["Will_go"] = LabelEncoder().fit_transform(["+", "-", "+", "-", "-", "+", "+"]) df_train = create_df(df_train, features) df_train ``` ### Test data ```{code-cell} ipython3 df_test = {} df_test["Looks"] = ["handsome", "handsome", "repulsive"] df_test["Alcoholic_beverage"] = ["no", "yes", "yes"] df_test["Eloquence"] = ["average", "high", "average"] df_test["Money_spent"] = ["lots", "little", "lots"] df_test = create_df(df_test, features) df_test ``` ```{code-cell} ipython3 # Some feature values are present in train and absent in test and vice-versa. y = df_train["Will_go"] df_train, df_test = intersect_features(train=df_train, test=df_test) df_train ``` ```{code-cell} ipython3 df_test ``` ### Draw a decision tree (by hand or in any graphics editor) for this dataset. Optionally you can also implement tree construction and draw it here. 1\. What is the entropy $S_0$ of the initial system? By system states, we mean values of the binary feature "Will_go" - 0 or 1 - two states in total. Answer: $S_0 = -\frac{3}{7}\log_2{\frac{3}{7}}-\frac{4}{7}\log_2{\frac{4}{7}} = 0.985$. 2\. Let's split the data by the feature "Looks_handsome". What is the entropy $S_1$ of the left group - the one with "Looks_handsome". What is the entropy $S_2$ in the opposite group? What is the information gain (IG) if we consider such a split? Answer: $S_1 = -\frac{1}{4}\log_2{\frac{1}{4}}-\frac{3}{4}\log_2{\frac{3}{4}} = 0.811$, $S_2 = -\frac{2}{3}\log_2{\frac{2}{3}}-\frac{1}{3}\log_2{\frac{1}{3}} = 0.918$, $IG = S_0-\frac{4}{7}S_1-\frac{3}{7}S_2 = 0.128$. ### Train a decision tree using sklearn on the training data. You may choose any depth for the tree. ```{code-cell} ipython3 dt = DecisionTreeClassifier(criterion="entropy", random_state=17) dt.fit(df_train, y); ``` ### Additional: display the resulting tree using graphviz. ```{code-cell} ipython3 plot_tree( dt, feature_names=df_train.columns, filled=True, class_names=["Won't go", "Will go"] ); ``` ## Part 2. Functions for calculating entropy and information gain. Consider the following warm-up example: we have 9 blue balls and 11 yellow balls. Let ball have label **1** if it is blue, **0** otherwise. ```{code-cell} ipython3 balls = [1 for i in range(9)] + [0 for i in range(11)] ``` Next split the balls into two groups: ```{code-cell} ipython3 # two groups balls_left = [1 for i in range(8)] + [0 for i in range(5)] # 8 blue and 5 yellow balls_right = [1 for i in range(1)] + [0 for i in range(6)] # 1 blue and 6 yellow ``` ### Implement a function to calculate the Shannon Entropy ```{code-cell} ipython3 from math import log def entropy(a_list): lst = list(a_list) size = len(lst) entropy = 0 set_elements = len(set(lst)) if set_elements in [0, 1]: return 0 for i in set(lst): occ = lst.count(i) entropy -= occ / size * log(occ / size, 2) return entropy ``` Tests ```{code-cell} ipython3 print(entropy(balls)) # 9 blue and 11 yellow ones print(entropy(balls_left)) # 8 blue and 5 yellow ones print(entropy(balls_right)) # 1 blue and 6 yellow ones print(entropy([1, 2, 3, 4, 5, 6])) # entropy of a fair 6-sided die ``` 3\. What is the entropy of the state given by the list **balls_left**? Answer: 0.961 4\. What is the entropy of a fair dice? (where we look at a dice as a system with 6 equally probable states)? Answer: 2.585 ```{code-cell} ipython3 # information gain calculation def information_gain(root, left, right): """ root - initial data, left and right - two partitions of initial data""" return ( entropy(root) - 1.0 * len(left) / len(root) * entropy(left) - 1.0 * len(right) / len(root) * entropy(right) ) ``` ```{code-cell} ipython3 print(information_gain(balls, balls_left, balls_right)) ``` 5\. What is the information gain from splitting the initial dataset into **balls_left** and **balls_right** ? Answer: 0.161 ```{code-cell} ipython3 def information_gains(X, y): """Outputs information gain when splitting with each feature""" out = [] for i in X.columns: out.append(information_gain(y, y[X[i] == 0], y[X[i] == 1])) return out ``` ### Optional: - Implement a decision tree building algorithm by calling `information_gains` recursively - Plot the resulting tree ```{code-cell} ipython3 information_gains(df_train, y) ``` ```{code-cell} ipython3 def btree(X, y, feature_names): clf = information_gains(X, y) best_feat_id = clf.index(max(clf)) best_feature = feature_names[best_feat_id] print(f"Best feature to split: {best_feature}") x_left = X[X.iloc[:, best_feat_id] == 0] x_right = X[X.iloc[:, best_feat_id] == 1] print(f"Samples: {len(x_left)} (left) and {len(x_right)} (right)") y_left = y[X.iloc[:, best_feat_id] == 0] y_right = y[X.iloc[:, best_feat_id] == 1] entropy_left = entropy(y_left) entropy_right = entropy(y_right) print(f"Entropy: {entropy_left} (left) and {entropy_right} (right)") print("_" * 30 + "\n") if entropy_left != 0: print(f"Splitting the left group with {len(x_left)} samples:") btree(x_left, y_left, feature_names) if entropy_right != 0: print(f"Splitting the right group with {len(x_right)} samples:") btree(x_right, y_right, feature_names) ``` ```{code-cell} ipython3 btree(df_train, y, df_train.columns) ``` This visualization is far from perfect, but it's easy to grasp if you compare it to the normal tree visualization (by sklearn) done above. ## Part 3. The "Adult" dataset **Dataset description:** [Dataset](http://archive.ics.uci.edu/ml/machine-learning-databases/adult) UCI Adult (no need to download it, we have a copy in the course repository): classify people using demographical data - whether they earn more than \$50,000 per year or not. Feature descriptions: - **Age** – continuous feature - **Workclass** – continuous feature - **fnlwgt** – final weight of object, continuous feature - **Education** – categorical feature - **Education_Num** – number of years of education, continuous feature - **Martial_Status** – categorical feature - **Occupation** – categorical feature - **Relationship** – categorical feature - **Race** – categorical feature - **Sex** – categorical feature - **Capital_Gain** – continuous feature - **Capital_Loss** – continuous feature - **Hours_per_week** – continuous feature - **Country** – categorical feature **Target** – earnings level, categorical (binary) feature. **Reading train and test data** ```{code-cell} ipython3 # for Jupyter-book, we copy data from GitHub, locally, to save Internet traffic, # you can specify the data/ folder from the root of your cloned # https://github.com/Yorko/mlcourse.ai repo, to save Internet traffic DATA_PATH = "https://raw.githubusercontent.com/Yorko/mlcourse.ai/master/data/" ``` ```{code-cell} ipython3 data_train = pd.read_csv(DATA_PATH + "adult_train.csv", sep=";") ``` ```{code-cell} ipython3 data_train.tail() ``` ```{code-cell} ipython3 data_test = pd.read_csv(DATA_PATH + "adult_test.csv", sep=";") ``` ```{code-cell} ipython3 data_test.tail() ``` ```{code-cell} ipython3 # necessary to remove rows with incorrect labels in test dataset data_test = data_test[ (data_test["Target"] == " >50K.") | (data_test["Target"] == " <=50K.") ] # encode target variable as integer data_train.loc[data_train["Target"] == " <=50K", "Target"] = 0 data_train.loc[data_train["Target"] == " >50K", "Target"] = 1 data_test.loc[data_test["Target"] == " <=50K.", "Target"] = 0 data_test.loc[data_test["Target"] == " >50K.", "Target"] = 1 ``` **Primary data analysis** ```{code-cell} ipython3 data_test.describe(include="all").T ``` ```{code-cell} ipython3 data_train["Target"].value_counts() ``` ```{code-cell} ipython3 fig = plt.figure(figsize=(25, 15)) cols = 5 rows = int(data_train.shape[1] / cols) for i, column in enumerate(data_train.columns): ax = fig.add_subplot(rows, cols, i + 1) ax.set_title(column) if data_train.dtypes[column] == np.object: data_train[column].value_counts().plot(kind="bar", axes=ax) else: data_train[column].hist(axes=ax) plt.xticks(rotation="vertical") plt.subplots_adjust(hspace=0.7, wspace=0.2); ``` **Checking data types** ```{code-cell} ipython3 data_train.dtypes ``` ```{code-cell} ipython3 data_test.dtypes ``` As we see, in the test data, age is treated as type **object**. We need to fix this. ```{code-cell} ipython3 data_test["Age"] = data_test["Age"].astype(int) ``` Also we'll cast all **float** features to **int** type to keep types consistent between our train and test data. ```{code-cell} ipython3 data_test["fnlwgt"] = data_test["fnlwgt"].astype(int) data_test["Education_Num"] = data_test["Education_Num"].astype(int) data_test["Capital_Gain"] = data_test["Capital_Gain"].astype(int) data_test["Capital_Loss"] = data_test["Capital_Loss"].astype(int) data_test["Hours_per_week"] = data_test["Hours_per_week"].astype(int) ``` **Fill in missing data for continuous features with their median values, for categorical features with their mode.** ```{code-cell} ipython3 # we see some missing values data_train.info() ``` ```{code-cell} ipython3 # choose categorical and continuous features from data categorical_columns = [ c for c in data_train.columns if data_train[c].dtype.name == "object" ] numerical_columns = [ c for c in data_train.columns if data_train[c].dtype.name != "object" ] print("categorical_columns:", categorical_columns) print("numerical_columns:", numerical_columns) ``` ```{code-cell} ipython3 # fill missing data for c in categorical_columns: data_train[c].fillna(data_train[c].mode()[0], inplace=True) data_test[c].fillna(data_train[c].mode()[0], inplace=True) for c in numerical_columns: data_train[c].fillna(data_train[c].median(), inplace=True) data_test[c].fillna(data_train[c].median(), inplace=True) ``` ```{code-cell} ipython3 # no more missing values data_train.info() ``` We'll dummy code some categorical features: **Workclass**, **Education**, **Martial_Status**, **Occupation**, **Relationship**, **Race**, **Sex**, **Country**. It can be done via pandas method **get_dummies** ```{code-cell} ipython3 data_train = pd.concat( [data_train[numerical_columns], pd.get_dummies(data_train[categorical_columns])], axis=1, ) data_test = pd.concat( [data_test[numerical_columns], pd.get_dummies(data_test[categorical_columns])], axis=1, ) ``` ```{code-cell} ipython3 set(data_train.columns) - set(data_test.columns) ``` ```{code-cell} ipython3 data_train.shape, data_test.shape ``` **There is no Holland in the test data. Create new zero-valued feature.** ```{code-cell} ipython3 data_test["Country_ Holand-Netherlands"] = 0 ``` ```{code-cell} ipython3 set(data_train.columns) - set(data_test.columns) ``` ```{code-cell} ipython3 data_train.head(2) ``` ```{code-cell} ipython3 data_test.head(2) ``` ```{code-cell} ipython3 X_train = data_train.drop(["Target"], axis=1) y_train = data_train["Target"] X_test = data_test.drop(["Target"], axis=1) y_test = data_test["Target"] ``` ### 3.1 Decision tree without parameter tuning Train a decision tree **(DecisionTreeClassifier)** with a maximum depth of 3, and evaluate the accuracy metric on the test data. Use parameter **random_state = 17** for results reproducibility. ```{code-cell} ipython3 tree = DecisionTreeClassifier(max_depth=3, random_state=17) tree.fit(X_train, y_train) ``` Make a prediction with the trained model on the test data. ```{code-cell} ipython3 tree_predictions = tree.predict(X_test[X_train.columns]) ``` ```{code-cell} ipython3 accuracy_score(y_test, tree_predictions) ``` 6\. What is the test set accuracy of a decision tree with maximum tree depth of 3 and **random_state = 17**? ### 3.2 Decision tree with parameter tuning Train a decision tree **(DecisionTreeClassifier, random_state = 17).** Find the optimal maximum depth using 5-fold cross-validation **(GridSearchCV)**. ```{code-cell} ipython3 %%time tree_params = {"max_depth": range(2, 11)} locally_best_tree = GridSearchCV( DecisionTreeClassifier(random_state=17), tree_params, cv=5 ) locally_best_tree.fit(X_train, y_train) ``` ```{code-cell} ipython3 print("Best params:", locally_best_tree.best_params_) print("Best cross validaton score", locally_best_tree.best_score_) ``` Train a decision tree with maximum depth of 9 (it is the best **max_depth** in my case), and compute the test set accuracy. Use parameter **random_state = 17** for reproducibility. ```{code-cell} ipython3 tuned_tree = DecisionTreeClassifier(max_depth=9, random_state=17) tuned_tree.fit(X_train, y_train) tuned_tree_predictions = tuned_tree.predict(X_test) accuracy_score(y_test, tuned_tree_predictions) ``` 7\. What is the test set accuracy of a decision tree with maximum depth of 9 and **random_state = 17**? Answer: 0.848 ### 3.3 (Optional) Random forest without parameter tuning Let's take a sneak peek of upcoming lectures and try to use a random forest for our task. For now, you can imagine a random forest as a bunch of decision trees, trained on slightly different subsets of the training data. Train a random forest **(RandomForestClassifier)**. Set the number of trees to 100 and use **random_state = 17**. ```{code-cell} ipython3 rf = RandomForestClassifier(n_estimators=100, random_state=17) rf.fit(X_train, y_train) ``` Perform cross-validation. ```{code-cell} ipython3 %%time cv_scores = cross_val_score(rf, X_train, y_train, cv=3) ``` ```{code-cell} ipython3 cv_scores, cv_scores.mean() ``` Make predictions for the test data. ```{code-cell} ipython3 forest_predictions = rf.predict(X_test) ``` ```{code-cell} ipython3 accuracy_score(y_test, forest_predictions) ``` ### 3.4 (Optional) Random forest with parameter tuning Train a random forest **(RandomForestClassifier)** of 10 trees. Tune the maximum depth and maximum number of features for each tree using **GridSearchCV**. ```{code-cell} ipython3 forest_params = {"max_depth": range(10, 16), "max_features": range(5, 105, 20)} locally_best_forest = GridSearchCV( RandomForestClassifier(n_estimators=10, random_state=17, n_jobs=-1), forest_params, cv=3, verbose=1, ) locally_best_forest.fit(X_train, y_train) ``` ```{code-cell} ipython3 print("Best params:", locally_best_forest.best_params_) print("Best cross validaton score", locally_best_forest.best_score_) ``` Make predictions for the test data. ```{code-cell} ipython3 tuned_forest_predictions = locally_best_forest.predict(X_test) accuracy_score(y_test, tuned_forest_predictions) ``` Wow! Looks that with some tuning we made a forest of 10 trees perform better than a forest of 100 trees with default hyperparameter values.