Machine Learning Fundamentals in Python
A GradientBoostingClassifier model is fitted on the training data X_train and y_train, and stored as model. Use model and the X_test feature data to predict values for the response variable, and store it in y_pred.
from sklearn.ensemble import GradientBoostingClassifier
from sklearn.metrics import accuracy_score
model = GradientBoostingClassifier(n_estimators=300, max_depth=1, random_state=1)
model.fit(X_train, y_train)
y_pred = model.predict(X_test)
print(accuracy_score(y_test, y_pred))
0.95
0.95
You are supplied with 2 sets of variables; y_pred are predicted using the model supplied and y_test are the actual response values.
Print the main classification metrics for the model.
from sklearn.tree import DecisionTreeClassifier
from sklearn import metrics
model = DecisionTreeClassifier(max_depth=4, random_state=42)
model.fit(X_train, y_train)
y_pred = model.predict(X_test)
print(metrics.classification_report(y_test, y_pred ))
precision recall f1-score support
0 0.93 0.94 0.94 54
1 0.97 0.96 0.96 89
accuracy 0.95 143
macro avg 0.95 0.95 0.95 143
weighted avg 0.95 0.95 0.95 143
precision recall f1-score support
0 0.93 0.94 0.94 54
1 0.97 0.96 0.96 89
accuracy 0.95 143
macro avg 0.95 0.95 0.95 143
weighted avg 0.95 0.95 0.95 143
3 out of 15
To understand the impact of weekly income on the amount a family spends each week on groceries, fit a suitable model using the weekly_spend data. What is the value of the intercept of the model?
spend income children car
0 20 19549 0 0
1 52 95248 1 1
2 18 27693 0 1
3 37 50788 1 1
4 46 50312 0 1
import numpy as np
from sklearn.linear_model import LinearRegression
X=np.array(weekly_spend['income']).reshape(-1,1)
mod=LinearRegression()
mod=LinearRegression(fit_intercept=True)
mod.fit(X, y)
mod.fit(X, weekly_spend["spend"])
print(mod.intercept_.round(2))
14.39
Traceback (most recent call last):
File "<stdin>", line 8, in <module>
mod.fit(X, y)
NameError: name 'y' is not defined
4 out of 15
A random forest model has been fitted to train data.
Use the results from a random forest classifier to determine the importance of each feature in determining whether a patient does or does not have heart disease.
import matplotlib.pyplot as plt
import seaborn as sns
from sklearn.ensemble import RandomForestClassifier
model = RandomForestClassifier(n_estimators=15,
random_state=1)
model.fit(X_train, y_train)
# Create a DataFrame with the feature importances
feature_importances = pd.DataFrame(
{"feature": list(X.columns), "importance": model.feature_importances_}
).sort_values("importance", ascending=False)
sns.barplot(data=feature_importances, x="importance", y="feature")
plt.show()
5 out of 15
Available in your working session is the dataset scaled_samples. Instantiate a principal component analysis model object with 2 components, and fit the model to the scaled_samples object.
from sklearn.decomposition import PCA
pca = PCA(n_components=2)
pca.fit(scaled_samples)
pca_features = pca.transform(scaled_samples)
print(pca_features.shape)
(85, 2)
6 out of 15
As part of a brainstorming session for a medical startup idea that predicts the chance of diabetes based on a few key readings, you have a sample DataFrame named df with key readings for diabetes:
Glucose BloodPressure Insulin
0 148 72 0
1 85 66 0
2 183 64 0
3 89 66 94
4 78 50 168
Apply Min-Max scaling so that all numeric columns are between 0 and 1.
from sklearn.preprocessing import MinMaxScaler
min_max = MinMaxScaler()
df_scaled = pd.DataFrame(min_max.fit_transform(df), columns=df.columns)
print(df_scaled.head())
BloodPressure Glucose Insulin
0 1.000000 0.666667 0.000000
1 0.727273 0.066667 0.000000
2 0.636364 1.000000 0.000000
3 0.727273 0.104762 0.559524
4 0.000000 0.000000 1.000000
7 out of 15
Create and fit random forest classifier with 15 trees to predict whether a patient has or does not have heart disease. The data has already been split into train and test sets (X_train, X_test, y_train, y_test).
from sklearn.ensemble import RandomForestClassifier
from sklearn.metrics import accuracy_score
model = RandomForestClassifier(n_estimators=15,
random_state=1)
model.fit(X_train, y_train)
y_pred = model.predict(X_test)
print(accuracy_score(y_test, y_pred))
0.8032786885245902
8 out of 15
The scatterplot shows data for a sample of 14 biofuels. Fit a linear regression model and print the intercept and coefficient.
from sklearn.linear_model import LinearRegression
model = LinearRegression(fit_intercept=True)
model.fit(x, y)
print("Regression coefficients: {}".format(model.coef_))
print("Regression intercept: {}".format(model.intercept_))
Regression coefficients: [[-0.20938742]]
Regression intercept: [75.21243193]
9 out of 15
Consider the first five rows of the data frame df shown below. Apply a pre-processing step to standardize all numeric features.
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1 5.1 3.5 1.4 0.2 setosa
2 4.9 3.0 1.4 0.2 setosa
3 4.7 3.2 1.3 0.2 setosa
4 4.6 3.1 1.5 0.2 setosa
5 5.0 3.6 1.4 0.2 setosa
6 5.4 3.9 1.7 0.4 setosa
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
df_scaled = pd.DataFrame(scaler.fit_transform(df), columns=df.columns)
print(df_scaled.head())
sepal length (cm) sepal width (cm) petal length (cm) petal width (cm)
0 -0.900681 1.019004 -1.340227 -1.315444
1 -1.143017 -0.131979 -1.340227 -1.315444
2 -1.385353 0.328414 -1.397064 -1.315444
3 -1.506521 0.098217 -1.283389 -1.315444
4 -1.021849 1.249201 -1.340227 -1.315444
10 out of 15
Using the original data, df, training (X_train, y_train) and test (X_test, y_test) sets have been created. Complete the code using the data sets in the appropriate places.
from sklearn.linear_model import LinearRegression
from sklearn.metrics import mean_squared_error
lin_reg = LinearRegression()
lin_reg.fit(X_train,y_train)
predictions = lin_reg.predict(X_test)
print("Mean squared error: %.2f" % mean_squared_error(y_test, predictions))
Mean squared error: 8.47
11 out of 15
A dataset has been prepared for you and split into test and training sets (X_train, X_test, y_train, y_test).
Use sklearn to fit a classification gradient boosting model on the training data with 300 estimators and 0.01 learning rate
from sklearn.ensemble import GradientBoostingClassifier
model = GradientBoostingClassifier(n_estimators=300, learning_rate=0.01, random_state=42)
model.fit(X_train, y_train)
y_pred = model.predict(X_test)
print(model.score(X_test, y_test))
0.8888888888888888
12 out of 15
A dataset has been prepared for you and fed into a random forest model.
Use sklearn to show the predicted probabilities of a new data point belonging to each class.
>>> print(new)
Alcohol Malic.acid Phenols Flavanoids
13.64 3.10 2.70 3.01
from sklearn.ensemble import RandomForestClassifier
model = RandomForestClassifier(random_state=42)
model.fit(X_train, y_train)
print(model.predict_proba(new))
[[0. 1. 0.]]
13 out of 15
Consider the data frame df below that shows the total number of observations per month. Fit a suitable imputer to fill the missing values.
Date Ozone Solar Wind
1976-05-31 26 27 31
1976-06-30 9 30 30
1976-07-31 26 31 31
1976-08-31 26 28 31
1976-09-30 29 30 31
from sklearn.impute import SimpleImputer
imputer = SimpleImputer(strategy='median')
print(imputer.fit(df))
SimpleImputer(add_indicator=False, copy=True, fill_value=None,
missing_values=nan, strategy='median', verbose=0)
14 out of 15
A linear regression model has been fitted to X_train with respect to y_train.
Test values are provided in the arrays X_test and y_test.
Diagnose potential problems by plotting the residuals against the fitted values, including a lowess smoother.
from matplotlib import pyplot as plt
import seaborn as sns
model = LinearRegression().fit(X_train,y_train)
y_fitted = model.predict(X_test)
sns.residplot(y_test - y_fitted, y_fitted)
plt.show()
15 out of 15
Consider the variable x in the Pandas DataFrame df shown in the plot below. Note that the data contains positive and negative values. Apply a suitable transformation to the data.
from sklearn.preprocessing import PowerTransformer
log = PowerTransformer(method='yeo-johnson')
df['log_x'] = log.fit_transform(df[['x']])
print(df['log_x'].head())
0 -0.319518
1 1.714791
2 0.823256
3 0.414669
4 -1.054036
Name: log_x, dtype: float64
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ScoreCompareApr 23rd, 20212:48pmJul 4th, 20213:53pmNoviceIntermediateAdvanced+29 overall134
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