In this notebook, we will implement k-nearest neighbors regression. You will:
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import os
import zipfile
import numpy as np
import pandas as pd
import matplotlib as mpl
import matplotlib.pyplot as plt
import seaborn as sns
sns.set_style('darkgrid')
%matplotlib inline
For this notebook, we use a subset of the King County housing dataset created by randomly selecting 40% of the houses in the full dataset.
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# Put files in current direction into a list
files_list = [f for f in os.listdir('.') if os.path.isfile(f)]
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# Filenames of unzipped files
unzip_files = ['kc_house_data_small.csv','kc_house_data_small_train.csv',
'kc_house_data_small_validation.csv', 'kc_house_data_small_test.csv' ]
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# If upzipped file not in files_list, unzip the file
for filename in unzip_files:
if filename not in files_list:
zip_file = filename + '.zip'
unzipping = zipfile.ZipFile(zip_file)
unzipping.extractall()
unzipping.close
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# Defining a dict w/ with the data type for each feature
dtype_dict = {'bathrooms':float, 'waterfront':int, 'sqft_above':int, 'sqft_living15':float,
'grade':int, 'yr_renovated':int, 'price':float, 'bedrooms':float,
'zipcode':str, 'long':float, 'sqft_lot15':float, 'sqft_living':float,
'floors':float, 'condition':int, 'lat':float, 'date':str, 'sqft_basement':int,
'yr_built':int, 'id':str, 'sqft_lot':int, 'view':int}
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sales = pd.read_csv('kc_house_data_small.csv', dtype=dtype_dict)
To efficiently compute pairwise distances among data points, we will convert the DataFrame into a 2D Numpy array. First import the numpy library and then copy and paste get_numpy_data() from the second notebook of Week 2.
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def get_numpy_data(input_df, features, output):
input_df['constant'] = 1.0 # Adding column 'constant' to input DataFrame with all values = 1.0
features = ['constant'] + features # Adding constant' to List of features
feature_matrix = input_df.as_matrix(columns=features) # Convert DataFrame w/ columns in features list in np.ndarray
output_array = input_df[output].values # Convert column with output feature into np.array
return(feature_matrix, output_array)
We will also need the normalize_features() function from Week 5 that normalizes all feature columns to unit norm. Paste this function below.
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def normalize_features(feature_matrix):
norms = np.linalg.norm(feature_matrix, axis=0)
normalized_features = feature_matrix/norms
return (normalized_features, norms)
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train_data = pd.read_csv('kc_house_data_small_train.csv', dtype=dtype_dict)
test_data = pd.read_csv('kc_house_data_small_test.csv', dtype=dtype_dict)
validation_data = pd.read_csv('kc_house_data_validation.csv', dtype=dtype_dict)
Using all of the numerical inputs listed in feature_list, transform the training, test, and validation DataFrames into Numpy arrays:
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feature_list = ['bedrooms',
'bathrooms',
'sqft_living',
'sqft_lot',
'floors',
'waterfront',
'view',
'condition',
'grade',
'sqft_above',
'sqft_basement',
'yr_built',
'yr_renovated',
'lat',
'long',
'sqft_living15',
'sqft_lot15']
features_train, output_train = get_numpy_data(train_data, feature_list, 'price')
features_test, output_test = get_numpy_data(test_data, feature_list, 'price')
features_valid, output_valid = get_numpy_data(validation_data, feature_list, 'price')
In computing distances, it is crucial to normalize features. Otherwise, for example, the sqft_living feature (typically on the order of thousands) would exert a much larger influence on distance than the bedrooms feature (typically on the order of ones). We divide each column of the training feature matrix by its 2-norm, so that the transformed column has unit norm.
IMPORTANT: Make sure to store the norms of the features in the training set. The features in the test and validation sets must be divided by these same norms, so that the training, test, and validation sets are normalized consistently.
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features_train, norms = normalize_features(features_train) # normalize training set features (columns)
features_test = features_test / norms # normalize test set by training set norms
features_valid = features_valid / norms # normalize validation set by training set norms
To start, let's just explore computing the "distance" between two given houses. We will take our query house to be the first house of the test set and look at the distance between this house and the 10th house of the training set.
To see the features associated with the query house, print the first row (index 0) of the test feature matrix. You should get an 18-dimensional vector whose components are between 0 and 1.
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print features_test[0]
print len(features_test[0])
Now print the 10th row (index 9) of the training feature matrix. Again, you get an 18-dimensional vector with components between 0 and 1.
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print features_train[9]
print len(features_train[9])
QUIZ QUESTION
What is the Euclidean distance between the query house and the 10th house of the training set?
Note: Do not use the np.linalg.norm function; use np.sqrt, np.sum, and the power operator (**) instead. The latter approach is more easily adapted to computing multiple distances at once.
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dist_euclid = np.sqrt( np.sum( (features_train[9] - features_test[0] )**2 ) )
print dist_euclid
Of course, to do nearest neighbor regression, we need to compute the distance between our query house and all houses in the training set.
To visualize this nearest-neighbor search, let's first compute the distance from our query house (features_test[0]) to the first 10 houses of the training set (features_train[0:10]) and then search for the nearest neighbor within this small set of houses. Through restricting ourselves to a small set of houses to begin with, we can visually scan the list of 10 distances to verify that our code for finding the nearest neighbor is working.
Write a loop to compute the Euclidean distance from the query house to each of the first 10 houses in the training set.
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# Setting the first house as the NN
min_euclid_dist = np.sqrt( np.sum( ( features_train[0] - features_test[0] )**2 ) )
min_house_index = 0
for i in range(1,10,1):
curr_euclid_dist = np.sqrt( np.sum( ( features_train[i] - features_test[0] )**2 ) )
# If distance of current house < current NN, update the NN
if curr_euclid_dist<min_euclid_dist:
min_euclid_dist = curr_euclid_dist
min_house_index = i
QUIZ QUESTION
Among the first 10 training houses, which house is the closest to the query house?
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print 'House', min_house_index + 1
It is computationally inefficient to loop over computing distances to all houses in our training dataset. Fortunately, many of the Numpy functions can be vectorized, applying the same operation over multiple values or vectors. We now walk through this process.
Consider the following loop that computes the element-wise difference between the features of the query house (features_test[0]) and the first 3 training houses (features_train[0:3]):
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for i in xrange(3):
print features_train[i]-features_test[0]
# should print 3 vectors of length 18
The subtraction operator (-) in Numpy is vectorized as follows:
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print features_train[0:3] - features_test[0]
Note that the output of this vectorized operation is identical to that of the loop above, which can be verified below:
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# verify that vectorization works
results = features_train[0:3] - features_test[0]
print results[0] - (features_train[0]-features_test[0])
# should print all 0's if results[0] == (features_train[0]-features_test[0])
print results[1] - (features_train[1]-features_test[0])
# should print all 0's if results[1] == (features_train[1]-features_test[0])
print results[2] - (features_train[2]-features_test[0])
# should print all 0's if results[2] == (features_train[2]-features_test[0])
Aside: it is a good idea to write tests like this cell whenever you are vectorizing a complicated operation.
Now that we have the element-wise differences, it is not too hard to compute the Euclidean distances between our query house and all of the training houses. First, write a single-line expression to define a variable diff such that diff[i] gives the element-wise difference between the features of the query house and the i-th training house.
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diff = features_train[:] - features_test[0]
To test the code above, run the following cell, which should output a value -0.0934339605842:
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print diff[-1].sum() # sum of the feature differences between the query and last training house
# should print -0.0934339605842
The next step in computing the Euclidean distances is to take these feature-by-feature differences in diff, square each, and take the sum over feature indices. That is, compute the sum of square feature differences for each training house (row in diff).
By default, np.sum sums up everything in the matrix and returns a single number. To instead sum only over a row or column, we need to specifiy the axis parameter described in the np.sum documentation. In particular, axis=1 computes the sum across each row.
Below, we compute this sum of square feature differences for all training houses and verify that the output for the 16th house in the training set is equivalent to having examined only the 16th row of diff and computing the sum of squares on that row alone.
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print np.sum(diff**2, axis=1)[15] # take sum of squares across each row, and print the 16th sum
print np.sum(diff[15]**2) # print the sum of squares for the 16th row -- should be same as above
With this result in mind, write a single-line expression to compute the Euclidean distances between the query house and all houses in the training set. Assign the result to a variable distances.
Hint: Do not forget to take the square root of the sum of squares.
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distances = np.sqrt( np.sum(diff**2, axis=1) )
To test the code above, run the following cell, which should output a value 0.0237082324496:
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print distances[100] # Euclidean distance between the query house and the 101th training house
# should print 0.0237082324496
Now you are ready to write a function that computes the distances from a query house to all training houses. The function should take two parameters: (i) the matrix of training features and (ii) the single feature vector associated with the query.
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def compute_distances(features_instances, features_query):
diff = features_instances[:] - features_query
distances = np.sqrt( np.sum(diff**2, axis=1) )
return distances
QUIZ QUESTIONS
Q1. Take the query house to be third house of the test set (features_test[2]). What is the index of the house in the training set that is closest to this query house?
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dist_Q1 = compute_distances(features_train, features_test[2])
index_NN = np.where(dist_Q1 == dist_Q1.min())[0][0]
print index_NN
Q2. What is the predicted value of the query house based on 1-nearest neighbor regression?
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print train_data['price'][index_NN]
For k-nearest neighbors, we need to find a set of k houses in the training set closest to a given query house. We then make predictions based on these k nearest neighbors.
Using the functions above, implement a function that takes in
and returns the indices of the k closest training houses. For instance, with 2-nearest neighbor, a return value of [5, 10] would indicate that the 6th and 11th training houses are closest to the query house.
Hint: Look at the documentation for np.argsort.
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def k_nearest_neighbors(k, feature_train, features_query):
distances = compute_distances(feature_train, features_query)
neighbors = np.argsort(distances)[0:k]
return neighbors
QUIZ QUESTION
Take the query house to be third house of the test set (features_test[2]). What are the indices of the 4 training houses closest to the query house?
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QQ_4NN = k_nearest_neighbors(4, features_train, features_test[2])
print QQ_4NN
Now that we know how to find the k-nearest neighbors, write a function that predicts the value of a given query house. For simplicity, take the average of the prices of the k nearest neighbors in the training set. The function should have the following parameters:
The function should return a predicted value of the query house.
Hint: You can extract multiple items from a Numpy array using a list of indices. For instance, output_train[[6, 10]] returns the prices of the 7th and 11th training houses.
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def predict_output_of_query(k, features_train, output_train, features_query):
kNN = k_nearest_neighbors(k, features_train, features_query)
prediction = np.average(output_train[kNN])
return prediction
QUIZ QUESTION
Again taking the query house to be third house of the test set (features_test[2]), predict the value of the query house using k-nearest neighbors with k=4 and the simple averaging method described and implemented above.
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QQ_pred = predict_output_of_query(4, features_train, train_data['price'].values, features_test[2])
print QQ_pred
Compare this predicted value using 4-nearest neighbors to the predicted value using 1-nearest neighbor computed earlier.
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print '1-NN prediction: ', train_data['price'][382]
print '4-NN prediction: ', QQ_pred
Write a function to predict the value of each and every house in a query set. (The query set can be any subset of the dataset, be it the test set or validation set.) The idea is to have a loop where we take each house in the query set as the query house and make a prediction for that specific house. The new function should take the following parameters:
The function should return a set of predicted values, one for each house in the query set.
Hint: To get the number of houses in the query set, use the .shape field of the query features matrix. See the documentation.
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def predict_output(k, features_train, output_train, features_query):
predictions = np.zeros(features_query.shape[0])
for i in range(len(predictions)):
predictions[i] = predict_output_of_query(k, features_train, output_train, features_query[i])
return predictions
QUIZ QUESTION
Make predictions for the first 10 houses in the test set using k-nearest neighbors with k=10.
Q1. What is the index of the house in this query set that has the lowest predicted value?
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QQ_10_preds = predict_output(10, features_train, train_data['price'].values, features_test[0:10])
index_low_pred = np.where(QQ_10_preds == QQ_10_preds.min())[0][0]
print index_low_pred
Q2. What is the predicted value of this house?
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print QQ_10_preds[index_low_pred]
There remains a question of choosing the value of k to use in making predictions. Here, we use a validation set to choose this value. Write a loop that does the following:
k in [1, 2, ..., 15]:rss_allk produced the lowest RSS on VALIDATION set.(Depending on your computing environment, this computation may take 10-15 minutes.)
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kvals = range(1, 16)
rss_all = np.zeros(len(kvals))
for i in range(len(kvals)):
pred_vals = predict_output(kvals[i], features_train, train_data['price'].values, features_valid)
rss_all[i] = sum( (pred_vals- validation_data['price'].values)**2 )
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index_min_rss = np.where(rss_all == rss_all.min())[0][0]
print 'Value of k which produces the lowest RSS on VALIDATION set: ', kvals[index_min_rss]
To visualize the performance as a function of k, plot the RSS on the VALIDATION set for each considered k value:
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plt.figure(figsize=(8,6))
plt.plot(kvals, rss_all,'bo-')
plt.xlabel('k-nearest neighbors used', fontsize=16)
plt.ylabel('Residual Sum of Squares', fontsize=16)
plt.title('k vs. RSS on Validation Dataset', fontsize=18)
plt.show()
QUIZ QUESTION
What is the RSS on the TEST data using the value of k found above? To be clear, sum over all houses in the TEST set.
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pred_vals_test = predict_output(8, features_train, train_data['price'].values, features_test)
rss_test = sum( (pred_vals_test- test_data['price'].values)**2 )
print '%.2e' % rss_test
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