We often work with high-dimensional data that contain many columns/features. Given all these dimensions, this data can be difficult to visualize and model. However, not all the data in this high-dimensional space is useful — there could be repeated features or outliers that make the data seem more complex than it really is. The most concise representation of high-dimensional data is its intrinsic dimension. Our goal with this lecture is to use dimensionality reduction to find the intrinsic dimension of a high-dimensional dataset. In other words, we want to find a smaller set of new features/columns that approximates the original data well without losing that much information. This is especially useful because this smaller set of features allows us to better visualize the data and do EDA to understand which modeling techniques would fit the data well. Let’s treat this as a matrix factorization problem, and review what we covered in the previous lecture.
In order to find the intrinsic dimension of a high-dimensional dataset, we’ll use techniques from linear algebra. Suppose we have a high-dimensional dataset, \(X\), that has \(n\) rows and \(d\) columns. We want to factor (split) \(X\) into two matrices, \(Z\) and \(W\). \(Z\) has \(n\) rows and \(k\) columns; \(W\) has \(k\) rows and \(d\) columns.
\[ X \approx ZW\]
We can reframe this problem as a loss function: in other words, if we want \(X\) to roughly equal \(ZW\), their difference should be as small as possible, ideally 0. This difference becomes our loss function, \(L(Z, W)\):
\[L(Z, W) = \frac{1}{n}\sum_{i=1}^{n}||X_i - Z_iW||^2\]
Breaking down the variables in this formula:
Using calculus and optimization techniques (take EECS 127 if you’re interested!), we find that this loss is minimized when \[Z = XW^T\] The proof for this is out of scope for Data 100, but for those who are interested, we:
Take a look at the previous lecture for the full proof. This gives us a very cool result of
\[\Sigma w^T = \lambda w^T\]
\(\Sigma\) is the covariance matrix of \(X\). The equation above implies that:
This tells us that the principal components (rows of \(W\)) are the eigenvectors with the largest eigenvalues of the covariance matrix \(\Sigma\). They represent the directions of maximum variance in the data. We can construct the latent factors, or the \(Z\) matrix, by projecting the centered data \(X\) onto the principal component vectors, \(W^T\).
But how do we compute the eigenvectors of \(\Sigma\)? Let’s dive into SVD to answer this question.
Singular value decomposition (SVD) is an important concept in linear algebra. Since this class requires a linear algebra course (MATH 54, MATH 56, or EECS 16A) as a pre/co-requisite, we assume you have taken or are taking a linear algebra course, so we won’t explain SVD in its entirety. In particular, we will not prove why SVD is a valid decomposition of rectangular matrices.
We will not dive deep into the theory and details of SVD. Instead, we will only cover what is needed for a data science interpretation. If you’d like more information, check out EECS 16B Note 14 or EECS 16B Note 15.
Orthonormal is a combination of two words: orthogonal and normal.
When we say the columns of a matrix are orthonormal, we know that:
Orthonormal matrices have a few important properties:
Diagonal matrices are square matrices with non-zero values on the diagonal axis and zeros everywhere else.
Right-multiplied diagonal matrices scale each column up or down by a constant factor. Geometrically, this transformation can be viewed as scaling the coordinate system.
Singular value decomposition (SVD) describes a matrix \(X\)’s decomposition into three matrices: \[ X = U S V^T \]
Let’s break down each of these terms one by one.
Here is a proof that \(U\) consists of the eigenvectors of \(XX^T\): \[\begin{align} XX^T &= USV^T (USV^T)^T \\ &= USV^T V S^T U^T \\ &= USS^T U^T \end{align}\]
Now right multiply by \(U\): \[\begin{align} XX^T U &= US S^T U^T U \\ &= US S^T \end{align}\]
Just like before, here is a proof that \(V\) consists of the eigenvectors of \(X^TX\): \[\begin{align} X^TX &= (USV^T)^T USV^T \\ &= V S^T U^T USV^T \\ &= VS^TS V^T \end{align}\]
Now right multiply by \(V\): \[\begin{align} X^TX V &= VS^T S V^T V \\ &= VS^T S \end{align}\]
NumPyFor this demo, we’ll work with a rectangular dataset containing \(n=100\) rows and \(d=4\) columns.
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
import numpy as np
np.random.seed(23) # kallisti
plt.rcParams["figure.figsize"] = (4, 4)
plt.rcParams["figure.dpi"] = 150
sns.set()
rectangle = pd.read_csv("data/rectangle_data.csv")
rectangle.head(5)| width | height | area | perimeter | |
|---|---|---|---|---|
| 0 | 8 | 6 | 48 | 28 |
| 1 | 2 | 4 | 8 | 12 |
| 2 | 1 | 3 | 3 | 8 |
| 3 | 9 | 3 | 27 | 24 |
| 4 | 9 | 8 | 72 | 34 |
In NumPy, the SVD decomposition function can be called with np.linalg.svd (documentation). There are multiple versions of SVD; to get the version that we will follow, we need to set the full_matrices parameter to False.
U, S, Vt = np.linalg.svd(rectangle, full_matrices=False)First, let’s examine U. As we can see, it’s dimensions are \(n \times d\).
U.shape(100, 4)The first 5 rows of U are shown below.
pd.DataFrame(U).head(5)| 0 | 1 | 2 | 3 | |
|---|---|---|---|---|
| 0 | -0.155151 | 0.064830 | -0.029935 | 0.934418 |
| 1 | -0.038370 | -0.089155 | 0.062019 | -0.299462 |
| 2 | -0.020357 | -0.081138 | 0.058997 | 0.006852 |
| 3 | -0.101519 | -0.076203 | -0.148160 | -0.011848 |
| 4 | -0.218973 | 0.206423 | 0.007274 | -0.056580 |
\(S\) is a little different in NumPy. Since the only useful values in the diagonal matrix \(S\) are the singular values on the diagonal axis, only those values are returned and they are stored in an array.
Our rectangle_data has a rank of \(3\), so we should have 3 non-zero singular values, sorted from largest to smallest.
Sarray([3.62932568e+02, 6.29904732e+01, 2.56544651e+01, 2.56364534e-14])It seems like we have 4 non-zero values instead of 3, but notice that the last value is so small (\(10^{-15}\)) that it’s practically \(0\). Hence, we can round the values to get 3 singular values.
np.round(S)array([363., 63., 26., 0.])To get S in matrix format, we use np.diag.
Sm = np.diag(S)
Smarray([[3.62932568e+02, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00],
[0.00000000e+00, 6.29904732e+01, 0.00000000e+00, 0.00000000e+00],
[0.00000000e+00, 0.00000000e+00, 2.56544651e+01, 0.00000000e+00],
[0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 2.56364534e-14]])Finally, we can see that Vt is indeed a \(d \times d\) matrix.
Vt.shape(4, 4)pd.DataFrame(Vt)| 0 | 1 | 2 | 3 | |
|---|---|---|---|---|
| 0 | -0.146436 | -0.129942 | -8.100201e-01 | -0.552756 |
| 1 | -0.192736 | -0.189128 | 5.863482e-01 | -0.763727 |
| 2 | -0.704957 | 0.709155 | 7.951614e-03 | 0.008396 |
| 3 | -0.666667 | -0.666667 | 9.775109e-17 | 0.333333 |
To check that this SVD is a valid decomposition, we can reverse it and see if it matches our original table (it does, yay!).
pd.DataFrame(U @ Sm @ Vt).head(5)| 0 | 1 | 2 | 3 | |
|---|---|---|---|---|
| 0 | 8.0 | 6.0 | 48.0 | 28.0 |
| 1 | 2.0 | 4.0 | 8.0 | 12.0 |
| 2 | 1.0 | 3.0 | 3.0 | 8.0 |
| 3 | 9.0 | 3.0 | 27.0 | 24.0 |
| 4 | 9.0 | 8.0 | 72.0 | 34.0 |
Principal Component Analysis (PCA) and Singular Value Decomposition (SVD) can be easily mixed up, especially when you have to keep track of so many acronyms. Here is a quick summary:
After centering the original data matrix \(X\) so that each column has a mean of 0, we find its SVD: \[ X = U S V^T \]
Because \(X\) is centered, the covariance matrix of \(X\), \(\Sigma\), is equal to \(X^T X\). Rearranging this equation, we get
\[ \begin{align} \Sigma &= X^T X \\ &= (U S V^T)^T U S V^T \\ &= V S^T U^T U S V^T & \text{U is orthonormal, so $U^T U = I$} \\ &= V S^2 V^T \end{align} \]
Multiplying both sides by \(V\), we get
\[ \begin{align} \Sigma V &= VS^2 V^T V \\ &= V S^2 \end{align} \]
This shows that the columns of \(V\) are the eigenvectors of the covariance matrix \(\Sigma\) and, therefore, the principal components. Additionally, the squared singular values \(S^2\) are the eigenvalues of \(\Sigma\).
We’ve now shown that the first \(k\) columns of \(V\) (equivalently, the first \(k\) rows of \(V^{T}\)) are the first k principal components of \(X\). We can use them to construct the latent vector representation of \(X\), \(Z\), by projecting \(X\) onto the principal components.
We also have a second way of computing \(Z\) as follows:
\[ \begin{align} Z &= X V \\ &= USV^T V \\ &= U S \end{align} \]
\[Z = XV = US\]
In other words, we can construct \(X\)’s’ latent vector representation \(Z\) through:
Using \(Z\), we can approximately recover the centered \(X\) matrix by multiplying \(Z\) by \(V^T\): \[ Z V^T = XV V^T = USV^T = X\]
If we choose a k that is less than the rank of X then we will only recover X approximately. Note that to recover the original (uncentered) \(X\) matrix, we would also need to add back the mean.
As we discussed above, when conducting PCA, we first center the data matrix \(X\) and then rotate it such that the direction with the most variation (i.e., the direction that is most spread out) aligns with the x-axis.
In particular, the elements of each principal component or column of \(V\) (row of \(V^{T}\)) rotate the original feature vectors, projecting \(X\) onto the principal components. The first column of \(V\) indicates how each feature contributes (e.g. positive, negative, etc.) to the first latent factor; it essentially assigns “weights” to each feature.
Coupled together, this interpretation also allows us to understand that:
Let’s summarize the steps to obtain Principal Components via SVD:
X = X - np.mean(X, axis = 0)X = (X - np.mean(X, axis = 0)) / np.std(X, axis = 0)Let’s now walk through an example where we compute PCA using SVD. In order to get the first \(k\) principal components from an \(n \times d\) matrix \(X\), we:
axis=0 so that the mean is computed per column.centered_df = rectangle - np.mean(rectangle, axis=0)
centered_df.head(5)| width | height | area | perimeter | |
|---|---|---|---|---|
| 0 | 2.97 | 1.35 | 24.78 | 8.64 |
| 1 | -3.03 | -0.65 | -15.22 | -7.36 |
| 2 | -4.03 | -1.65 | -20.22 | -11.36 |
| 3 | 3.97 | -1.65 | 3.78 | 4.64 |
| 4 | 3.97 | 3.35 | 48.78 | 14.64 |
U, S, Vt = np.linalg.svd(centered_df, full_matrices=False)
Sm = pd.DataFrame(np.diag(np.round(S, 1)))two_PCs = Vt.T[:, :2]
pd.DataFrame(two_PCs).head()| 0 | 1 | |
|---|---|---|
| 0 | -0.098631 | 0.668460 |
| 1 | -0.072956 | -0.374186 |
| 2 | -0.931226 | -0.258375 |
| 3 | -0.343173 | 0.588548 |
We define the total variance of a data matrix as the sum of variances of attributes. The principal components are a low-dimension representation that capture as much of the original data’s total variance as possible. Formally, the \(i\)-th singular value tells us the component score, or how much of the data variance is captured by the \(i\)-th principal component. Assuming the number of datapoints is \(n\):
\[\text{i-th component score} = \frac{(\text{i-th singular value}^2)}{n}\]
Summing up the component scores is equivalent to computing the total variance if we center our data.
If you want to dive deeper into PCA, Steve Brunton’s SVD Video Series is a great resource.
We often plot the first two principal components using a scatter plot, with PC1 on the \(x\)-axis and PC2 on the \(y\)-axis. This is often called a PCA plot.
If the first two singular values are large and all others are small, then two dimensions are enough to describe most of what distinguishes one observation from another. If not, a PCA plot omits a lot of information.
PCA plots help us assess similarities between our data points and if there are any clusters in our dataset. Here is one example:
A scree plot shows the variance ratio captured by each principal component, with the largest variance ratio first. Recall that we could use each singular value to determine the component score:
\[\text{i-th component score} = \frac{(\text{i-th singular value}^2)}{n} = \frac{{s_i}^2}{N}\]
Therefore, the total variance is the sum of all component scores: \[\text{Total Variance} = \sum_{i=1}^{k} \frac{{s_i}^2}{N}\]
Then the variance ratio, or the proportion of fraction of total variance represented by a single principal component is the following: \[\text{Variance Ratio of Principal Component j} = \frac{\frac{{s_j}^2}{N}}{\sum_{i=1}^{k} \frac{{s_i}^2}{N}} = \frac{{s_j}^2}{\sum_{i=1}^{k} {s_i}^2}\]
In other words, the variance ratio is a principal component’s singular value squared divided by the sum of all singular values squared. Scree plots can help us visually determine the number of dimensions needed to describe the data reasonably. The singular values that fall in the region of the plot after a large drop-off correspond to principal components that are not needed to describe the data since they explain a relatively low proportion of the total variance of the data. This point where adding more principal components results in diminishing returns is called the “elbow” and is the point just before the line flattens out. Using this “elbow method”, we can see that the elbow is at the second principal component, so two dimensions are enough to describe most of what distinguishes one observation from another. If this was not the case, then a PCA scatter plot is omitting lots of information.
Biplots superimpose the directions onto the plot of PC1 vs. PC2, where vector \(j\) corresponds to the direction for feature \(j\) (e.g., \(v_{1j}, v_{2j}\)). There are several ways to scale biplot vectors — in this course, we plot the direction itself. For other scalings, which can lead to more interpretable directions/loadings, see SAS biplots.
Through biplots, we can interpret how features correlate with the principal components shown: positively, negatively, or not much at all.
The directions of the arrow are (\(v_1\), \(v_2\)), where \(v_1\) and \(v_2\) are how that specific feature column contributes to PC1 and PC2, respectively. \(v_1\) and \(v_2\) are elements of the first and second columns of \(V\), respectively (i.e., the first two rows of \(V^T\)).
Say we were considering feature 3, and say that was the purple arrow labeled “520” here (pointing bottom right).
Let’s examine how the House of Representatives (of the 116th Congress, 1st session) voted in the month of September 2019.
Specifically, we’ll look at the records of Roll call votes. From the U.S. Senate (link): roll call votes occur when a representative or senator votes “yea” or “nay” so that the names of members voting on each side are recorded. A voice vote is a vote in which those in favor or against a measure say “yea” or “nay,” respectively, without the names or tallies of members voting on each side being recorded.
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
import numpy as np
import yaml
from datetime import datetime
import plotly.express as px
import plotly.graph_objects as go
votes = pd.read_csv("data/votes.csv")
votes = votes.astype({"roll call": str})
votes.head()| chamber | session | roll call | member | vote | |
|---|---|---|---|---|---|
| 0 | House | 1 | 555 | A000374 | Not Voting |
| 1 | House | 1 | 555 | A000370 | Yes |
| 2 | House | 1 | 555 | A000055 | No |
| 3 | House | 1 | 555 | A000371 | Yes |
| 4 | House | 1 | 555 | A000372 | No |
Suppose we pivot this table to group each legislator and their voting pattern across every (roll call) vote in this month. We mark 1 if the legislator voted Yes (“yea”), and 0 otherwise (“No”, “nay”, no vote, speaker, etc.).
def was_yes(s):
return 1 if s.iloc[0] == "Yes" else 0
vote_pivot = votes.pivot_table(
index="member", columns="roll call", values="vote", aggfunc=was_yes, fill_value=0
)
print(vote_pivot.shape)
vote_pivot.head()(441, 41)| roll call | 515 | 516 | 517 | 518 | 519 | 520 | 521 | 522 | 523 | 524 | ... | 546 | 547 | 548 | 549 | 550 | 551 | 552 | 553 | 554 | 555 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| member | |||||||||||||||||||||
| A000055 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | ... | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
| A000367 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | ... | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 |
| A000369 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | ... | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
| A000370 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | ... | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
| A000371 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | ... | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
5 rows × 41 columns
Do legislators’ roll call votes show a relationship with their political party?
While we could consider loading information about the legislator, such as their party, and see how this relates to their voting pattern, it turns out that we can do a lot with PCA to cluster legislators by how they vote. Let’s calculate the principal components using the SVD method.
vote_pivot_centered = vote_pivot - np.mean(vote_pivot, axis=0)
u, s, vt = np.linalg.svd(vote_pivot_centered, full_matrices=False) # SVDWe can use the singular values in s to construct a scree plot:
fig = px.line(y=s**2 / sum(s**2), title='Variance Explained', width=700, height=600, markers=True)
fig.update_xaxes(title_text='Principal Component i')
fig.update_yaxes(title_text='Proportion of Variance Explained')It looks like this graph plateaus after the third principal component, so our “elbow” is at PC3, and most of the variance is captured by just the first three principal components. Let’s use these PCs to visualize the latent vector representation of \(X\)!
# Calculate the latent vector representation (US or XV)
# using the first 3 principal components
vote_2d = pd.DataFrame(index=vote_pivot_centered.index)
vote_2d[["z1", "z2", "z3"]] = (u * s)[:, :3]
# Plot the latent vector representation
fig = px.scatter_3d(vote_2d, x='z1', y='z2', z='z3', title='Vote Data', width=800, height=600)
fig.update_traces(marker=dict(size=5))Baesd on the plot above, it looks like there are two clusters of datapoints. What do you think this corresponds to?
By incorporating member information (source), we can augment our graph with biographic data like each member’s party and gender.
legislators_data = yaml.safe_load(open("data/legislators-2019.yaml"))
def to_date(s):
return datetime.strptime(s, "%Y-%m-%d")
legs = pd.DataFrame(
columns=[
"leg_id",
"first",
"last",
"gender",
"state",
"chamber",
"party",
"birthday",
],
data=[
[
x["id"]["bioguide"],
x["name"]["first"],
x["name"]["last"],
x["bio"]["gender"],
x["terms"][-1]["state"],
x["terms"][-1]["type"],
x["terms"][-1]["party"],
to_date(x["bio"]["birthday"]),
]
for x in legislators_data
],
)
legs["age"] = 2024 - legs["birthday"].dt.year
legs.set_index("leg_id")
legs.sort_index()
vote_2d = vote_2d.join(legs.set_index("leg_id")).dropna()
np.random.seed(42)
vote_2d["z1_jittered"] = vote_2d["z1"] + np.random.normal(0, 0.1, len(vote_2d))
vote_2d["z2_jittered"] = vote_2d["z2"] + np.random.normal(0, 0.1, len(vote_2d))
vote_2d["z3_jittered"] = vote_2d["z3"] + np.random.normal(0, 0.1, len(vote_2d))
px.scatter_3d(vote_2d, x='z1_jittered', y='z2_jittered', z='z3_jittered', color='party', symbol="gender", size='age',
title='Vote Data', width=800, height=600, size_max=10,
opacity = 0.7,
color_discrete_map={'Democrat':'blue', 'Republican':'red', "Independent": "green"},
hover_data=['first', 'last', 'state', 'party', 'gender', 'age'])Using SVD and PCA, we can clearly see a separation between the red dots (Republican) and blue dots (Deomcrat).
We can also look at \(V^T\) directly to try to gain insight into why each component is as it is.
fig_eig = px.bar(x=vote_pivot_centered.columns, y=vt[0, :])
# extract the trace from the figure
fig_eig.show()We have the party affiliation labels so we can see if this eigenvector aligns with one of the parties.
party_line_votes = (
vote_pivot_centered.join(legs.set_index("leg_id")["party"])
.groupby("party")
.mean()
.T.reset_index()
.rename(columns={"index": "call"})
.melt("call")
)
fig = px.bar(
party_line_votes,
x="call", y="value", facet_row = "party", color="party",
color_discrete_map={'Democrat':'blue', 'Republican':'red', "Independent": "green"})
fig.for_each_annotation(lambda a: a.update(text=a.text.split("=")[-1]))loadings = pd.DataFrame(
{"pc1": np.sqrt(s[0]) * vt[0, :], "pc2": np.sqrt(s[1]) * vt[1, :]},
index=vote_pivot_centered.columns,
)
vote_2d["num votes"] = votes[votes["vote"].isin(["Yes", "No"])].groupby("member").size()
vote_2d.dropna(inplace=True)
fig = px.scatter(
vote_2d,
x='z1_jittered',
y='z2_jittered',
color='party',
symbol="gender",
size='num votes',
title='Biplot',
width=800,
height=600,
size_max=10,
opacity = 0.7,
color_discrete_map={'Democrat':'blue', 'Republican':'red', "Independent": "green"},
hover_data=['first', 'last', 'state', 'party', 'gender', 'age'])
for (call, pc1, pc2) in loadings.head(20).itertuples():
fig.add_scatter(x=[0,pc1], y=[0,pc2], name=call,
mode='lines+markers', textposition='top right',
marker= dict(size=10,symbol= "arrow-bar-up", angleref="previous"))
figEach roll call from the 116th Congress - 1st Session: https://clerk.house.gov/evs/2019/ROLL_500.asp
As shown in the demo, the primary goal of PCA is to transform observations from high-dimensional data down to low dimensions through linear transformations.
In machine learning, PCA is often used as a preprocessing step prior to training a supervised model.
Let’s explore how PCA is useful for building an image classification model based on the Fashion-MNIST dataset, a dataset containing images of articles of clothing; these images are gray scale with a size of 28 by 28 pixels. The copyright for Fashion-MNIST is held by Zalando SE. Fashion-MNIST is licensed under the MIT license.
First, we’ll load in the data.
import requests
from pathlib import Path
import time
import gzip
import os
import numpy as np
import plotly.express as px
def fetch_and_cache(data_url, file, data_dir="data", force=False):
"""
Download and cache a url and return the file object.
data_url: the web address to download
file: the file in which to save the results.
data_dir: (default="data") the location to save the data
force: if true the file is always re-downloaded
return: The pathlib.Path object representing the file.
"""
data_dir = Path(data_dir)
data_dir.mkdir(exist_ok=True)
file_path = data_dir / Path(file)
# If the file already exists and we want to force a download then
# delete the file first so that the creation date is correct.
if force and file_path.exists():
file_path.unlink()
if force or not file_path.exists():
print("Downloading...", end=" ")
resp = requests.get(data_url)
with file_path.open("wb") as f:
f.write(resp.content)
print("Done!")
last_modified_time = time.ctime(file_path.stat().st_mtime)
else:
last_modified_time = time.ctime(file_path.stat().st_mtime)
print("Using cached version that was downloaded (UTC):", last_modified_time)
return file_path
def head(filename, lines=5):
"""
Returns the first few lines of a file.
filename: the name of the file to open
lines: the number of lines to include
return: A list of the first few lines from the file.
"""
from itertools import islice
with open(filename, "r") as f:
return list(islice(f, lines))
def load_data():
"""
Loads the Fashion-MNIST dataset.
This is a dataset of 60,000 28x28 grayscale images of 10 fashion categories,
along with a test set of 10,000 images. This dataset can be used as
a drop-in replacement for MNIST.
The classes are:
| Label | Description |
|:-----:|-------------|
| 0 | T-shirt/top |
| 1 | Trouser |
| 2 | Pullover |
| 3 | Dress |
| 4 | Coat |
| 5 | Sandal |
| 6 | Shirt |
| 7 | Sneaker |
| 8 | Bag |
| 9 | Ankle boot |
Returns:
Tuple of NumPy arrays: `(x_train, y_train), (x_test, y_test)`.
**x_train**: uint8 NumPy array of grayscale image data with shapes
`(60000, 28, 28)`, containing the training data.
**y_train**: uint8 NumPy array of labels (integers in range 0-9)
with shape `(60000,)` for the training data.
**x_test**: uint8 NumPy array of grayscale image data with shapes
(10000, 28, 28), containing the test data.
**y_test**: uint8 NumPy array of labels (integers in range 0-9)
with shape `(10000,)` for the test data.
Example:
(x_train, y_train), (x_test, y_test) = fashion_mnist.load_data()
assert x_train.shape == (60000, 28, 28)
assert x_test.shape == (10000, 28, 28)
assert y_train.shape == (60000,)
assert y_test.shape == (10000,)
License:
The copyright for Fashion-MNIST is held by Zalando SE.
Fashion-MNIST is licensed under the [MIT license](
https://github.com/zalandoresearch/fashion-mnist/blob/master/LICENSE).
"""
dirname = os.path.join("datasets", "fashion-mnist")
base = "https://storage.googleapis.com/tensorflow/tf-keras-datasets/"
files = [
"train-labels-idx1-ubyte.gz",
"train-images-idx3-ubyte.gz",
"t10k-labels-idx1-ubyte.gz",
"t10k-images-idx3-ubyte.gz",
]
paths = []
for fname in files:
paths.append(fetch_and_cache(base + fname, fname))
# paths.append(get_file(fname, origin=base + fname, cache_subdir=dirname))
with gzip.open(paths[0], "rb") as lbpath:
y_train = np.frombuffer(lbpath.read(), np.uint8, offset=8)
with gzip.open(paths[1], "rb") as imgpath:
x_train = np.frombuffer(imgpath.read(), np.uint8, offset=16).reshape(
len(y_train), 28, 28
)
with gzip.open(paths[2], "rb") as lbpath:
y_test = np.frombuffer(lbpath.read(), np.uint8, offset=8)
with gzip.open(paths[3], "rb") as imgpath:
x_test = np.frombuffer(imgpath.read(), np.uint8, offset=16).reshape(
len(y_test), 28, 28
)
return (x_train, y_train), (x_test, y_test)class_names = [
"T-shirt/top",
"Trouser",
"Pullover",
"Dress",
"Coat",
"Sandal",
"Shirt",
"Sneaker",
"Bag",
"Ankle boot",
]
class_dict = {i: class_name for i, class_name in enumerate(class_names)}
(train_images, train_labels), (test_images, test_labels) = load_data()
print("Training images", train_images.shape)
print("Test images", test_images.shape)
rng = np.random.default_rng(42)
n = 5000
sample_idx = rng.choice(np.arange(len(train_images)), size=n, replace=False)
# Invert and normalize the images so they look better
img_mat = -1 * train_images[sample_idx].astype(np.int16)
img_mat = (img_mat - img_mat.min()) / (img_mat.max() - img_mat.min())
images = pd.DataFrame(
{
"images": img_mat.tolist(),
"labels": train_labels[sample_idx],
"class": [class_dict[x] for x in train_labels[sample_idx]],
}
)Using cached version that was downloaded (UTC): Fri Sep 27 15:27:33 2024
Using cached version that was downloaded (UTC): Fri Sep 27 15:27:33 2024
Using cached version that was downloaded (UTC): Fri Sep 27 15:27:33 2024
Using cached version that was downloaded (UTC): Fri Sep 27 15:27:33 2024
Training images (60000, 28, 28)
Test images (10000, 28, 28)Let’s see what some of the images contained in this dataset look like.
def show_images(images, ncols=5, max_images=30):
# conver the subset of images into a n,28,28 matrix for facet visualization
img_mat = np.array(images.head(max_images)["images"].to_list())
fig = px.imshow(
img_mat,
color_continuous_scale="gray",
facet_col=0,
facet_col_wrap=ncols,
height=220 * int(np.ceil(len(images) / ncols)),
)
fig.update_layout(coloraxis_showscale=False)
# Extract the facet number and convert it back to the class label.
fig.for_each_annotation(
lambda a: a.update(text=images.iloc[int(a.text.split("=")[-1])]["class"])
)
return fig
fig = show_images(images.groupby("class", as_index=False).sample(2), ncols=6)
fig.show()Let’s break this down further and look at it by class, or the category of clothing:
print(class_dict)
show_images(images.groupby('class',as_index=False).sample(2), ncols=6){0: 'T-shirt/top', 1: 'Trouser', 2: 'Pullover', 3: 'Dress', 4: 'Coat', 5: 'Sandal', 6: 'Shirt', 7: 'Sneaker', 8: 'Bag', 9: 'Ankle boot'}As we can see, each 28x28 pixel image is labelled by the category of clothing it belongs to. Us humans can very easily look at these images and identify the type of clothing being displayed, even if the image is a little blurry. However, this task is less intuitive for machine learning models. To illustrate this, let’s take a small sample of the training data to see how the images above are represented in their raw format:
images.head()| images | labels | class | |
|---|---|---|---|
| 0 | [[1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0,... | 3 | Dress |
| 1 | [[1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0,... | 4 | Coat |
| 2 | [[1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0,... | 0 | T-shirt/top |
| 3 | [[1.0, 1.0, 1.0, 1.0, 1.0, 0.996078431372549, ... | 2 | Pullover |
| 4 | [[1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0,... | 1 | Trouser |
Each row represents one image. Every image belongs to a "class" of clothing with it’s enumerated "label". In place of a typically displayed image, the raw data contains a 28x28 2D array of pixel values; each pixel value is a float between 0 and 1. If we just focus on the images, we get a 3D matrix. You can think of this as a matrix containing 2D images.
X = np.array(images["images"].to_list())
X.shape (5000, 28, 28)However, we’re not used to working with 3D matrices for our training data X. Typical training data expects a vector of features for each datapoint, not a matrix per datapoint. We can reshape our 3D matrix so that it fits our typical training data by “unrolling” the the 28x28 pixels into a single row vector containing 28*28 = 784 dimensions.
X = X.reshape(X.shape[0], -1)
X.shape(5000, 784)What we have now is 5000 datapoints that each have 784 features. That’s a lot of features! Not only would training a model on this data take a very long time, it’s also very likely that our matrix is linearly independent. PCA is a very good strategy to use in situations like these when there are lots of features, but we want to remove redundant information.
sklearnTo perform PCA, let’s begin by centering our data.
X = X - X.mean(axis=0)We can run PCA using sklearn’s PCA package.
from sklearn.decomposition import PCA
n_comps = 50
pca = PCA(n_components=n_comps)
pca.fit(X)PCA(n_components=50)In a Jupyter environment, please rerun this cell to show the HTML representation or trust the notebook.
PCA(n_components=50)
Now that sklearn helped us find the principal components, let’s visualize a scree plot.
# Make a line plot and show markers
fig = px.line(y=pca.explained_variance_ratio_ * 100, markers=True)
fig.show()We can see that the line starts flattening out around 2 or 3, which suggests that most of the data is explained by just the first two or three dimensions. To illustrate this, let’s plot the first three principal components and the datapoints’ corresponding classes. Can you identify any patterns?
images[['z1', 'z2', 'z3']] = pca.transform(X)[:, :3]
fig = px.scatter_3d(images, x='z1', y='z2', z='z3', color='class', hover_data=['labels'],
width=1000, height=800)
# set marker size to 5
fig.update_traces(marker=dict(size=5))As we saw in the demos, we often perform PCA during the Exploratory Data Analysis (EDA) stage of our data science lifecycle (if we already know what to model, we probably don’t need PCA!). It helps us with:
PCA is commonly used in biomedical contexts, which have many named variables! It can be used to:
Suppose we know the child mortality rate of a given country. Linear regression tries to predict the fertility rate from the mortality rate; for example, if the mortality is 6, we might guess the fertility is near 4. The regression line tells us the “best” prediction of fertility given all possible mortality values by minimizing the root mean squared error. See the vertical red lines (note that only some are shown).
We can also perform a regression in the reverse direction. That is, given fertility, we try to predict mortality. In this case, we get a different regression line that minimizes the root mean squared length of the horizontal lines.
The rank-1 approximation is close but not the same as the mortality regression line. Instead of minimizing horizontal or vertical error, our rank-1 approximation minimizes the error perpendicular to the subspace onto which we’re projecting. That is, SVD finds the line such that if we project our data onto that line, the error between the projection and our original data is minimized. The similarity of the rank-1 approximation and the fertility was just a coincidence. Looking at adiposity and bicep size from our body measurements dataset, we see the 1D subspace onto which we are projecting is between the two regression lines.
Even in higher dimensions, the idea behind principal components is the same! Suppose we have 30-dimensional data and decide to use the first 5 principal components. Our procedure minimizes the error between the original 30-dimensional data and the projection of that 30-dimensional data onto the “best” 5-dimensional subspace. See CS 189 Note 10 for more details.
One key fact to remember is that the decomposition is not arbitrary. The rank of a matrix limits how small our inner dimensions can be if we want to perfectly recreate our matrix. The proof for this is out of scope.
Even if we know we have to factorize our matrix using an inner dimension of \(R\), that still leaves a large space of solutions to traverse. What if we have a procedure to automatically factorize a rank \(R\) matrix into an \(R\)-dimensional representation with some transformation matrix?
What if we wanted a 2D representation? It’s valuable to compress all of the data that is relevant into as few dimensions as possible in order to plot it efficiently. Some 2D matrices yield better approximations than others. How well can we do?
The proof defining component score is out of scope for this class, but it is included below for your convenience.
Setup: Consider the design matrix \(X \in \mathbb{R}^{n \times d}\), where the \(j\)-th column (corresponding to the \(j\)-th feature) is \(x_j \in \mathbb{R}^n\) and the element in row \(i\), column \(j\) is \(x_{ij}\). Further, define \(\tilde{X}\) as the centered design matrix. The \(j\)-th column is \(\tilde{x}_j \in \mathbb{R}^n\) and the element in row \(i\), column \(j\) is \(\tilde{x}_{ij} = x_{ij} - \bar{x_j}\), where \(\bar{x_j}\) is the mean of the \(x_j\) column vector from the original \(X\).
Variance: Construct the covariance matrix: \(\frac{1}{n} \tilde{X}^T \tilde{X} \in \mathbb{R}^{d \times d}\). The \(j\)-th element along the diagonal is the variance of the \(j\)-th column of the original design matrix \(X\):
\[\left( \frac{1}{n} \tilde{X}^T \tilde{X} \right)_{jj} = \frac{1}{n} \tilde{x}_j ^T \tilde{x}_j = \frac{1}{n} \sum_{i=i}^n (\tilde{x}_{ij} )^2 = \frac{1}{n} \sum_{i=i}^n (x_{ij} - \bar{x_j})^2\]
SVD: Suppose singular value decomposition of the centered design matrix \(\tilde{X}\) yields \(\tilde{X} = U S V^T\), where \(U \in \mathbb{R}^{n \times d}\) and \(V \in \mathbb{R}^{d \times d}\) are matrices with orthonormal columns, and \(S \in \mathbb{R}^{d \times d}\) is a diagonal matrix with singular values of \(\tilde{X}\).
\[ \begin{aligned} \tilde{X}^T \tilde{X} &= (U S V^T )^T (U S V^T) \\ &= V S U^T U S V^T & (S^T = S) \\ &= V S^2 V^T & (U^T U = I) \\ \frac{1}{n} \tilde{X}^T \tilde{X} &= \frac{1}{n} V S V^T =V \left( \frac{1}{n} S \right) V^T \\ \frac{1}{n} \tilde{X}^T \tilde{X} V &= V \left( \frac{1}{n} S \right) V^T V = V \left( \frac{1}{n} S \right) & \text{(right multiply by }V \rightarrow V^T V = I \text{)} \\ V^T \frac{1}{n} \tilde{X}^T \tilde{X} V &= V^T V \left( \frac{1}{n} S \right) = \frac{1}{n} S & \text{(left multiply by }V^T \rightarrow V^T V = I \text{)} \\ \left( \frac{1}{n} \tilde{X}^T \tilde{X} \right)_{jj} &= \frac{1}{n}S_j^2 & \text{(Define }S_j\text{ as the} j\text{-th singular value)} \\ \frac{1}{n} S_j^2 &= \frac{1}{n} \sum_{i=i}^n (x_{ij} - \bar{x_j})^2 \end{aligned} \]
The last line defines the \(j\)-th component score.