[Solved] DATA201 – Assignment 5

Start by downloading the data and loading it into a dataframe.In [2]: df = pd.read_csv(‘landmark_faces.txt’, sep=” “, header=None) In [3]:

Out[3]:0 1 2 3 4 5 6 7 8 9 … 128 129 130

0 1_0_2_20161219140530307.jpg -4 71 -4 96 -3 120 -1 144 9 … 136 130 1351 1_0_2_20161219140525218.jpg 13 76 13 96 15 117 18 137 25 … 137 121 1412 1_0_2_20161219140540938.jpg 11 62 14 84 18 105 23 127 33 … 135 135 1363 6_1_2_20161219140554092.jpg 36 66 34 86 38 108 45 129 54 … 140 120 1544 1_1_2_20161219140604000.jpg -3 60 -3 85 -1 110 3 134 12 … 137 126 1415 rows × 138 columns

For now, we will simply extract the locations for each landmark. They are stored as x1y1x2y2…x68y68 in columns 1 to 136 of the dataframe. The following code will extract them into a data array, remove the first and last columns (the filename and the last column, which is blank), and make sure that numpy knows that the data are numbers.In [4]:

Get the size of the data array, and store the relevant information in variables nfaces and npoints In [5]:

9780 136To plot a face, take one row of the data matrix, and plot the x coordinates (even-numbered indices) against y coordinates (odd-numbered ones). It is helpful to use ‘.-‘ to plot a line as well as the points. At first this will be upside down. One way to turn it the right way up is to use 1-data on the y coordinates.In [6]:face = data[4]pl.plot(face[::2], 1-(face[1::2]),’-‘, color=’#38ada9′, linewidth=3) pl.plot(face[::2], 1-(face[1::2]),’o’, color = ‘#3c6382’, markersize=6)Out[6]:[<matplotlib.lines.Line2D at 0x1517906a940>]

The points are all meant to be matched on each face (the jargon for this is that they are ‘corresponding points’). This means that to find the mean face (the average face, not a nasty-looking face) you just compute the mean of each datapoint. Compute this and plot it.mean_points = np.mean(data, axis=0)pl.plot(mean_points[0::2], 1-mean_points[1::2], ‘-‘, color=’#b71540’, linewidth=3) pl.plot(mean_points[0::2], 1-mean_points[1::2], ‘o’, color = ‘#0c2461’, markersize=6)Out[7]:[<matplotlib.lines.Line2D at 0x15179039cc0>]

We’ve made the PCA algorithm a lot. Here it is again for your use.

Out[8]:array([[ 1.51075747e+01, 4.45351127e+01, -2.70876248e+01, …, -8.70254115e-02, 5.19442691e-03, 1.85535201e-01],[-2.68442070e+01, 4.80654703e+01, -3.19136447e+01, …, 1.31049520e-01, -3.85654002e-01, 3.36220834e-02],[-4.43991541e+01, -2.00963925e+00, -2.51369001e+01, …,-4.19986364e-01, 2.58111825e-01, 5.36453208e-01],…,[-4.13029519e+00, -2.56470272e+01, -1.84527494e+01, …, 2.66547407e-01, 2.90493378e-01, -1.24834955e-01],[-2.06792748e+01, -2.52999978e+01, 2.30844453e+01, …, 1.42926961e-01, -1.42641704e-01, -2.55764792e-01],[ 2.81662919e+01, -2.06048621e+01, -1.29233591e+00, …, -3.67411483e-01, -5.13240937e-02, -1.29443573e-02]])My own implementation of the PCA algorithm:

[[ 1.51075747e+01 4.45351127e+01 -2.70876248e+01 … -8.70254115e-02 5.19442691e-03 1.85535201e-01][-2.68442070e+01 4.80654703e+01 -3.19136447e+01 … 1.31049520e-01 -3.85654002e-01 3.36220834e-02][-4.43991541e+01 -2.00963925e+00 -2.51369001e+01 … -4.19986364e-01 2.58111825e-01 5.36453208e-01]…[-4.13029519e+00 -2.56470272e+01 -1.84527494e+01 … 2.66547407e-01 2.90493378e-01 -1.24834955e-01][-2.06792748e+01 -2.52999978e+01 2.30844453e+01 … 1.42926961e-01 -1.42641704e-01 -2.55764792e-01][ 2.81662919e+01 -2.06048621e+01 -1.29233591e+00 … -3.67411483e-01-5.13240937e-02 -1.29443573e-02]]Out[9]:TrueCompute the scree plot of the principal components and see how many you think might be useful.

Reconstruct a face from the dataset using different numbers of dimensions, and plot both the original and reconstructed face. Use this to decide how many principal components you actually need. As a hint, here is the code to reconstruct the original data. You simply need to reduce the size of the eigenvector matrix appropriately in order to reconstruct lower-dimensional versions of it.

Eigenvectors: (6, 136)Data: (9780, 6)Data X Eigenvectors: (9780, 136)Y: (9780, 136)

The next code computes some a set of shapes. Look at the code, and explain what it is doing in a couple of sentences below.

In [1]:print(“The code shifts the mean points by a magnitude determined by the eigenvalue in t he direction of it’s corresponding eigenvector. We know that an eigenvector is a specia l vector that doesn’t change direction when linearly transformed, it can only shrink or expand in a particular direction. This shrinkage or expansion is done via it’s magnitud e, in other words it’s eigenvalue. So essentially what this code is doing is looping th rough the first 3 principle components and offsetting the mean face by the square root of it’s eigenvalue(I’ll just say eigenvalue instead of squareroot of the eigenvalue fr om now on) in the direction of it’s corresponding eigenvector(each column is a vector). For each of the PCs(these are the different rows) it plots 3 different eigenvalues, the eigenvalue itself, the eigenvalue doubled and the eigenvalue tripled (these are the dif ferent columns). For each of the individual plots: the blue face is the mean face, thered face is the mean face shifted negatively and the green face is shifted positively.This not only shows us the differing principle components and how they compare but also the effect of the magnitude on the eigenvector.”) ### Copied in from emailThe code shifts the mean points by a magnitude determined by the eigenvalu e in the direction of it’s corresponding eigenvector. We know that an eige nvector is a special vector that doesn’t change direction when linearly tr ansformed, it can only shrink or expand in a particular direction. This sh rinkage or expansion is done via it’s magnitude, in other words it’s eigen value. So essentially what this code is doing is looping through the first 3 principle components and offsetting the mean face by the square root of it’s eigenvalue(I’ll just say eigenvalue instead of squareroot of the eige nvalue from now on) in the direction of it’s corresponding eigenvector(eac h column is a vector). For each of the PCs(these are the different rows) i t plots 3 different eigenvalues, the eigenvalue itself, the eigenvalue dou bled and the eigenvalue tripled (these are the different columns). For eac h of the individual plots: the blue face is the mean face, the red face is the mean face shifted negatively and the green face is shifted positively. This not only shows us the differing principle components and how they com pare but also the effect of the magnitude on the eigenvector.To get training labels, look at the names of the images (which are in df[:,0]). The names are age_gender_race_date.jpg. We are interested in the gender, which is 0 or 1 for male and female (I’m not sure which is which, and it doesn’t matter). Extract that digit ( stringname.find() might be helpful, where stringname is a variable) and store it as the target variable. There is one error in the data, which is index 8513. I think you should assign it to category 0.In [14]:

Out[14]:0 01 02 03 14 1 ..9775 09776 09777 09778 19779 0Name: 1, Length: 9780, dtype: int64Now use a classification algorithm such as Support Vector Classifier from sklearn to classify the faces based on (i) the original data, and (ii) some small number of principal components. You should compare both the accuracy and time to train. Don’t forget ot split the data into training and testing. You might find that very small numbers of principal components are good for this.In [15]:start = time.time()y_train, y_test, X_train, X_test = train_test_split(genders, data, test_size=0.2, rando m_state=42)knn = KNeighborsClassifier(n_neighbors=5) knn_model = knn.fit(X_train, y_train)knn_score = accuracy_score(y_train, knn_model.predict(X_train))end = time.time() og_time = (end-start)print(“Original Dataset Execution Time:
{:.4f}ms”.format(og_time)) print(“Original Dataset Accuracy Score:
{:.2f}%”.format(knn_score*100))print(“Original Dataset Confusion Matrix:
{}”.format(confusion_matrix(y_test, knn_mod el.predict(X_test))))Original Dataset Execution Time: 10.4520msOriginal Dataset Accuracy Score:80.02%Original Dataset Confusion Matrix:[[484 387][222 863]]In [16]:red_start = time.time()red_X_train, red_X_test = train_test_split(red_data, test_size=0.2, random_state=42) red_knn = KNeighborsClassifier(n_neighbors=5) red_knn_model = red_knn.fit(red_X_train, y_train)red_knn_score = accuracy_score(y_train, red_knn_model.predict(red_X_train))red_end = time.time() red_time = (red_end-red_start)print(“Reduced Dataset Execution Time:
{:.2f}ms”.format(red_time)) print(“Reduced Dataset Accuracy Score:
{:.2f}%”.format(red_knn_score*100)) print(“Reduced Dataset Confusion Matrix:
{}”.format(confusion_matrix(y_test, red_knn_ model.predict(red_X_test))))Reduced Dataset Execution Time: 0.30msReduced Dataset Accuracy Score:76.44%Reduced Dataset Confusion Matrix:[[464 407][272 813]]In [17]:print(“A PC of {} is {:.2f}ms faster than the data with all PCs but is less accurate by{:.2f}%”.format(ndim, og_time-red_time, (knn_score-red_knn_score)*100))A PC of 6 is 10.15ms faster than the data with all PCs but is less accurat e by 3.58% In [ ]:

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