計算機視覺學習——李飛飛斯坦福大學計算機視覺公開課 Assignment 1 Q1代碼整理

Assignment1作業詳情頁面:http://cs231n.github.io/assignments2017/assignment1/ 

使用juputer notebook分別依次運行一下的每一個文件的所有cell,某些代碼塊的運行時間可能需要一到兩分鐘,等吧……

knn.py 完整實現代碼:

# Load the raw CIFAR-10 data.
cifar10_dir = 'cs231n/datasets/cifar-10-batches-py'
X_train, y_train, X_test, y_test = load_CIFAR10(cifar10_dir)

# As a sanity check, we print out the size of the training and test data.
print ('Training data shape: ', X_train.shape)
print ('Training labels shape: ', y_train.shape)
print ('Test data shape: ', X_test.shape)
print ('Test labels shape: ', y_test.shape)
# Run some setup code for this notebook.

import random
import numpy as np
from cs231n.data_utils import load_CIFAR10
import matplotlib.pyplot as plt
from __future__ import print_function
from past.builtins import xrange

# This is a bit of magic to make matplotlib figures appear inline in the notebook
# rather than in a new window.
%matplotlib inline
plt.rcParams['figure.figsize'] = (10.0, 8.0) # set default size of plots
plt.rcParams['image.interpolation'] = 'nearest'
plt.rcParams['image.cmap'] = 'gray'

# Some more magic so that the notebook will reload external python modules;
# see http://stackoverflow.com/questions/1907993/autoreload-of-modules-in-ipython
%load_ext autoreload
%autoreload 2

 

# Visualize some examples from the dataset.
# We show a few examples of training images from each class.
classes = ['plane', 'car', 'bird', 'cat', 'deer', 'dog', 'frog', 'horse', 'ship', 'truck']
num_classes = len(classes)
samples_per_class = 7
for y, cls in enumerate(classes):
    idxs = np.flatnonzero(y_train == y)
    idxs = np.random.choice(idxs, samples_per_class, replace=False)
    for i, idx in enumerate(idxs):
        plt_idx = i * num_classes + y + 1
        plt.subplot(samples_per_class, num_classes, plt_idx)
        plt.imshow(X_train[idx].astype('uint8'))
        plt.axis('off')
        if i == 0:
            plt.title(cls)
plt.show()
# Subsample the data for more efficient code execution in this exercise
num_training = 5000
mask = range(num_training)
X_train = X_train[mask]
y_train = y_train[mask]

num_test = 500
mask = range(num_test)
X_test = X_test[mask]
y_test = y_test[mask]
# Reshape the image data into rows

X_train = np.reshape(X_train, (X_train.shape[0], -1))
X_test = np.reshape(X_test, (X_test.shape[0], -1))
print (X_train.shape, X_test.shape)

 

from cs231n.classifiers import KNearestNeighbor

# Create a kNN classifier instance. 
# Remember that training a kNN classifier is a noop: 
# the Classifier simply remembers the data and does no further processing 
classifier = KNearestNeighbor()
classifier.train(X_train, y_train)
# Open cs231n/classifiers/k_nearest_neighbor.py and implement
# compute_distances_two_loops.

# Test your implementation:
dists = classifier.compute_distances_two_loops(X_test)
print(dists.shape)
# We can visualize the distance matrix: each row is a single test example and
# its distances to training examples
plt.imshow(dists, interpolation='none')
plt.show()
# Now implement the function predict_labels and run the code below:
# We use k = 1 (which is Nearest Neighbor).
y_test_pred = classifier.predict_labels(dists, k=1)

# Compute and print the fraction of correctly predicted examples
num_correct = np.sum(y_test_pred == y_test)
accuracy = float(num_correct) / num_test
print( 'Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy))
y_test_pred = classifier.predict_labels(dists, k=5)
num_correct = np.sum(y_test_pred == y_test)
accuracy = float(num_correct) / num_test
print('Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy))
# Now lets speed up distance matrix computation by using partial vectorization
# with one loop. Implement the function compute_distances_one_loop and run the
# code below:
dists_one = classifier.compute_distances_one_loop(X_test)

# To ensure that our vectorized implementation is correct, we make sure that it
# agrees with the naive implementation. There are many ways to decide whether
# two matrices are similar; one of the simplest is the Frobenius norm. In case
# you haven't seen it before, the Frobenius norm of two matrices is the square
# root of the squared sum of differences of all elements; in other words, reshape
# the matrices into vectors and compute the Euclidean distance between them.
difference = np.linalg.norm(dists - dists_one, ord='fro')
print('Difference was: %f' % (difference, ))
if difference < 0.001:
  print('Good! The distance matrices are the same')
else:
  print('Uh-oh! The distance matrices are different')
# Now implement the fully vectorized version inside compute_distances_no_loops
# and run the code
dists_two = classifier.compute_distances_no_loops(X_test)

# check that the distance matrix agrees with the one we computed before:
difference = np.linalg.norm(dists - dists_two, ord='fro')
print('Difference was: %f' % (difference, ))
if difference < 0.001:
  print( 'Good! The distance matrices are the same')
else:
  print( 'Uh-oh! The distance matrices are different')
# Let's compare how fast the implementations are
def time_function(f, *args):
  """
  Call a function f with args and return the time (in seconds) that it took to execute.
  """
  import time
  tic = time.time()
  f(*args)
  toc = time.time()
  return toc - tic

two_loop_time = time_function(classifier.compute_distances_two_loops, X_test)
print('Two loop version took %f seconds' % two_loop_time)

one_loop_time = time_function(classifier.compute_distances_one_loop, X_test)
print('One loop version took %f seconds' % one_loop_time)

no_loop_time = time_function(classifier.compute_distances_no_loops, X_test)
print('No loop version took %f seconds' % no_loop_time)

# you should see significantly faster performance with the fully vectorized implementation
num_folds = 5
k_choices = [1, 3, 5, 8, 10, 12, 15, 20, 50, 100]

X_train_folds = []
y_train_folds = []
################################################################################
# TODO:                                                                        #
# Split up the training data into folds. After splitting, X_train_folds and    #
# y_train_folds should each be lists of length num_folds, where                #
# y_train_folds[i] is the label vector for the points in X_train_folds[i].     #
# Hint: Look up the numpy array_split function.                                #
################################################################################
X_train_folds = np.array_split(X_train,num_folds)
y_train_folds = np.array_split(y_train,num_folds)
################################################################################
#                                 END OF YOUR CODE                             #
################################################################################

# A dictionary holding the accuracies for different values of k that we find
# when running cross-validation. After running cross-validation,
# k_to_accuracies[k] should be a list of length num_folds giving the different
# accuracy values that we found when using that value of k.
k_to_accuracies = {}


################################################################################
# TODO:                                                                        #
# Perform k-fold cross validation to find the best value of k. For each        #
# possible value of k, run the k-nearest-neighbor algorithm num_folds times,   #
# where in each case you use all but one of the folds as training data and the #
# last fold as a validation set. Store the accuracies for all fold and all     #
# values of k in the k_to_accuracies dictionary.                               #
################################################################################
for k in k_choices:
    classifier = KNearestNeighbor()
    k_to_accuracies[k]=[]
    for i in range(num_folds):
        tmp=list(range(i))+list(range(i+1,num_folds))
        train_data = np.concatenate([X_train_folds[j] for j in tmp])
        train_label = np.concatenate([y_train_folds[j] for j in tmp])
        classifier.train(train_data,train_label)
        test = X_train_folds[i]
        dists = classifier.compute_distances_no_loops(test)
        y_test_pred = classifier.predict_labels(dists, k)
        num_correct = np.sum(y_test_pred == y_train_folds[i])
        num_test_train=np.shape(X_train_folds[i])[0]
        accuracy = float(num_correct) / num_test_train
        k_to_accuracies[k].append(accuracy)

################################################################################
#                                 END OF YOUR CODE                             #
################################################################################

# Print out the computed accuracies
for k in sorted(k_to_accuracies):
    for accuracy in k_to_accuracies[k]:
        print('k = %d, accuracy = %f' % (k, accuracy))
# plot the raw observations
for k in k_choices:
  accuracies = k_to_accuracies[k]
  plt.scatter([k] * len(accuracies), accuracies)

# plot the trend line with error bars that correspond to standard deviation
accuracies_mean = np.array([np.mean(v) for k,v in sorted(k_to_accuracies.items())])
accuracies_std = np.array([np.std(v) for k,v in sorted(k_to_accuracies.items())])
plt.errorbar(k_choices, accuracies_mean, yerr=accuracies_std)
plt.title('Cross-validation on k')
plt.xlabel('k')
plt.ylabel('Cross-validation accuracy')
plt.show()
# Based on the cross-validation results above, choose the best value for k,   
# retrain the classifier using all the training data, and test it on the test
# data. You should be able to get above 28% accuracy on the test data.

best_k = 10

classifier = KNearestNeighbor()
classifier.train(X_train, y_train)
y_test_pred = classifier.predict(X_test, k=best_k)

# Compute and display the accuracy

num_correct = np.sum(y_test_pred == y_test)
accuracy = float(num_correct) / num_test
print('Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy))

k_nearest_neighbor.py 完整代碼:

import numpy as np

class KNearestNeighbor(object):
  """ a kNN classifier with L2 distance """

  def __init__(self):
    pass

  def train(self, X, y):
    """
    Train the classifier. For k-nearest neighbors this is just 
    memorizing the training data.

    Inputs:
    - X: A numpy array of shape (num_train, D) containing the training data
      consisting of num_train samples each of dimension D.
    - y: A numpy array of shape (N,) containing the training labels, where
         y[i] is the label for X[i].
    """
    self.X_train = X
    self.y_train = y
    
  def predict(self, X, k=1, num_loops=0):
    """
    Predict labels for test data using this classifier.

    Inputs:
    - X: A numpy array of shape (num_test, D) containing test data consisting
         of num_test samples each of dimension D.
    - k: The number of nearest neighbors that vote for the predicted labels.
    - num_loops: Determines which implementation to use to compute distances
      between training points and testing points.

    Returns:
    - y: A numpy array of shape (num_test,) containing predicted labels for the
      test data, where y[i] is the predicted label for the test point X[i].  
    """
    if num_loops == 0:
      dists = self.compute_distances_no_loops(X)
    elif num_loops == 1:
      dists = self.compute_distances_one_loop(X)
    elif num_loops == 2:
      dists = self.compute_distances_two_loops(X)
    else:
      raise ValueError('Invalid value %d for num_loops' % num_loops)

    return self.predict_labels(dists, k=k)

  def compute_distances_two_loops(self, X):
    """
    Compute the distance between each test point in X and each training point
    in self.X_train using a nested loop over both the training data and the 
    test data.

    Inputs:
    - X: A numpy array of shape (num_test, D) containing test data.

    Returns:
    - dists: A numpy array of shape (num_test, num_train) where dists[i, j]
      is the Euclidean distance between the ith test point and the jth training
      point.
    """
    num_test = X.shape[0]
    num_train = self.X_train.shape[0]
    dists = np.zeros((num_test, num_train))
    for i in range(num_test):
      for j in range(num_train):
        #####################################################################
        # TODO:                                                             #
        # Compute the l2 distance between the ith test point and the jth    #
        # training point, and store the result in dists[i, j]. You should   #
        # not use a loop over dimension.                                    #
        #####################################################################
        dists[i][j] = np.sqrt(np.sum(np.square(self.X_train[j,:] - X[i,:])))
#         pass
        #####################################################################
        #                       END OF YOUR CODE                            #
        #####################################################################
    return dists

  def compute_distances_one_loop(self, X):
    """
    Compute the distance between each test point in X and each training point
    in self.X_train using a single loop over the test data.

    Input / Output: Same as compute_distances_two_loops
    """
    num_test = X.shape[0]
    num_train = self.X_train.shape[0]
    dists = np.zeros((num_test, num_train))
    for i in range(num_test):
        for j in range(num_train):
      #######################################################################
      # TODO:                                                               #
      # Compute the l2 distance between the ith test point and all training #
      # points, and store the result in dists[i, :].                        #
      #######################################################################
            dists[i][j] = np.sqrt(np.sum(np.square(X[i,:] - self.X_train[j,:])))
      #######################################################################
      #                         END OF YOUR CODE                            #
      #######################################################################
    return dists

  def compute_distances_no_loops(self, X):
    """
    Compute the distance between each test point in X and each training point
    in self.X_train using no explicit loops.

    Input / Output: Same as compute_distances_two_loops
    """
    num_test = X.shape[0]
    num_train = self.X_train.shape[0]
    dists = np.zeros((num_test, num_train)) 
    #########################################################################
    # TODO:                                                                 #
    # Compute the l2 distance between all test points and all training      #
    # points without using any explicit loops, and store the result in      #
    # dists.                                                                #
    #                                                                       #
    # You should implement this function using only basic array operations; #
    # in particular you should not use functions from scipy.                #
    #                                                                       #
    # HINT: Try to formulate the l2 distance using matrix multiplication    #
    #       and two broadcast sums.                                         #
    #########################################################################
    dists = np.multiply(np.dot(X,self.X_train.T),-2)
    sq1 = np.sum(np.square(X),axis=1,keepdims = True)
    sq2 = np.sum(np.square(self.X_train),axis=1)
    dists = np.add(dists,sq1)
    dists = np.add(dists,sq2)
    dists = np.sqrt(dists)
    #########################################################################
    #                         END OF YOUR CODE                              #
    #########################################################################
    return dists

  def predict_labels(self, dists, k=1):
    """
    Given a matrix of distances between test points and training points,
    predict a label for each test point.

    Inputs:
    - dists: A numpy array of shape (num_test, num_train) where dists[i, j]
      gives the distance betwen the ith test point and the jth training point.

    Returns:
    - y: A numpy array of shape (num_test,) containing predicted labels for the
      test data, where y[i] is the predicted label for the test point X[i].  
    """
    num_test = dists.shape[0]
    y_pred = np.zeros(num_test)
    for i in range(num_test):
      # A list of length k storing the labels of the k nearest neighbors to
      # the ith test point.
        closest_y = []
      #########################################################################
      # TODO:                                                                 #
      # Use the distance matrix to find the k nearest neighbors of the ith    #
      # testing point, and use self.y_train to find the labels of these       #
      # neighbors. Store these labels in closest_y.                           #
      # Hint: Look up the function numpy.argsort.                             #
      #########################################################################
        closest_y = self.y_train[np.argsort(dists[i])[:k]]
      #########################################################################
      # TODO:                                                                 #
      # Now that you have found the labels of the k nearest neighbors, you    #
      # need to find the most common label in the list closest_y of labels.   #
      # Store this label in y_pred[i]. Break ties by choosing the smaller     #
      # label.                                                                #
      #########################################################################
        y_pred[i] = np.argmax(np.bincount(closest_y))
      #########################################################################
      #                           END OF YOUR CODE                            # 
      #########################################################################

    return y_pred

最終結果:

當k=10時得到的數據準確率最高,

Got 141 / 500 correct => accuracy: 0.282000
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