# 主成份分析

Principal component analysis

This work by Jephian Lin is licensed under a [Creative Commons Attribution 4.0 International License](http://creativecommons.org/licenses/by/4.0/).

$\newcommand{\trans}{^\top}
\newcommand{\adj}{^{\rm adj}}
\newcommand{\cof}{^{\rm cof}}
\newcommand{\inp}[2]{\left\langle#1,#2\right\rangle}
\newcommand{\dunion}{\mathbin{\dot\cup}}
\newcommand{\bzero}{\mathbf{0}}
\newcommand{\bone}{\mathbf{1}}
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\newcommand{\bq}{\mathbf{q}}
\newcommand{\br}{\mathbf{r}}
\newcommand{\bx}{\mathbf{x}}
\newcommand{\by}{\mathbf{y}}
\newcommand{\bz}{\mathbf{z}}
\newcommand{\bu}{\mathbf{u}}
\newcommand{\bv}{\mathbf{v}}
\newcommand{\bw}{\mathbf{w}}
\newcommand{\tr}{\operatorname{tr}}
\newcommand{\nul}{\operatorname{null}}
\newcommand{\rank}{\operatorname{rank}}
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from lingeo import random_int_list

## Main idea

Let $X$ be an $N\times d$ data matrix whose the sample vectors are centered at $\bzero\in\mathbb{R}^d$.
Let $C = \frac{1}{N-1}X\trans X$ be the covariance matrix.

Let $\bv$ be a unit vector in $\mathbb{R}^d$.
One may project all data points onto the direction of $\bv$.
Indeed, the projected data is $\bu = X\bv$.
Thus, the mean of $\bu$ is $\inp{\bu}{\bone} = \bone\trans X\bu = \bzero\trans\bu = 0$,
and the variance of $\bu$ is $\frac{1}{N-1}\inp{\bu}{\bu} = \frac{1}{N-1}\bu\trans X\trans X\bu = \bv\trans C\bv$.

It is natural to consider the optimization problem:

maximize $\bv\trans C\bv$,
subject to $\|\bv\| = 1$.

Let $\lambda_1\geq \cdots \geq \lambda_d$ be the eigenvalues of $C$ and $\{\bv_1,\ldots,\bv_d\}$ the corresponding orthonormal eigenbasis.
According to the Rayleigh quotient theorem, the maximum is achieved by $\lambda_1$ when $\bv = \bv_1$.

Therefore, $\bv_1$ is called the first **principal component** of the data,
and one may continue to take $\bv_2,\bv_3,\ldots$ as the next principal components.

##### Algorithm (principal component analysis)

Input: a data represented by an $N\times d$ matrix $X$ and the desired number $k$ of principal components
Output: the principal components represented by the column vectors of a matrix $P$

1. Let $J$ be the $N\times N$ all-ones matrix.
Define $X_0 = (I - \frac{1}{n}J)X$.
2. Calculate the covariance matrix $C= \frac{1}{N-1}X_0\trans X_0$.
3. Find a diagonal matrix $D$, whose diagonal entries are $\lambda_1 \geq \cdots \geq \lambda_d$, and an orthogonal matrix $Q$ such that $C = QDQ\trans$.
3. Let $P$ be the $d\times k$ matrix obtained by the first $k$ columns of $Q$.

A few points to emphasize:

- The $Q$ matrix is the same as the $V$ in the algorithm in 606.
- The eigenvalues $\lambda_1 \geq \cdots \geq \lambda_d$ are the variances when the data are projected onto the columns of $Q$.

## Side stories

- explained variance ratio
- `scipy.linalg.eigh`

## Experiments

##### Exercise 1

執行以下程式碼。

<!-- eng start -->
Run the code below.
<!-- eng end -->

### code
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

np.random.seed(0)
print_ans = False

mu = np.random.randint(-3,4, (2,))
cov = np.array([[1,1.9],
                [1.9,4]])
X = np.random.multivariate_normal(mu, cov, (20,))
mu = X.mean(axis=0)
X0 = X - mu
# u,s,vh = np.linalg.svd(X0)
C = (1 / (19)) * X0.T.dot(X0)
vals,vecs = np.linalg.eigh(C)
vals = vals[::-1]
vecs = vecs[:,::-1]
P = vecs[:,:1]

plt.axis('equal')
plt.scatter(*X.T)
plt.scatter(*mu, c="red")
plt.arrow(*mu, *(vecs[:,0]), head_width=0.3, color="red")

xs = X0.dot(P)
ys = np.zeros_like(xs)
plt.scatter(xs, ys)

pretty_print(LatexExpr("Q ="))
print(vecs)

if print_ans:
    print("red point: center")
    print("red arrow: first principal component")
    print("orange points: projection of blue points onto the red arrow")
    print("red arrow = first column of Q")

##### Exercise 1(a)

若藍點為給定的資料。
改變不同的 `seed`，說明紅點、紅箭頭、橘點的意思。

<!-- eng start -->
Let the blue points be the dataset. Try different `seed` and explore the meaning of the red point, the red arrow, and the orange point.
<!-- eng end -->

##### Exercise 1(b)

令 $X_0$ 的列向量為將藍點資料置中過後的資料點、
而 $C = \frac{1}{N-1}X_0\trans X_0$ 為其共變異數矩陣。
若 $C = QDQ\trans$ 為 $C$ 的譜分解，其中 $D$ 的對角線項由大到小排列。
說明紅箭頭與 $Q$ 的關係。

<!-- eng start -->
Let $X_0$ be the matrix whose rows record the data from the blue points, centered at their mean. Let $C = \frac{1}{N-1}X_0\trans X_0$ be the covariance matrix. Suppose $C = QDQ\trans$ is the spectral decomposition of $C$ such that the diagonal entries of $D$ is arranged from the largest to the smallest. Explain the relation between the red arrow and $Q$.
<!-- eng end -->

:::info
What do the experiments try to tell you? (open answer)

...
:::

## Exercises

##### Exercise 2

令 $X$ 是已置中的 $N\times d$ 資料矩陣、
而 $C = \frac{1}{N-1}X_0\trans X_0$ 為其共變異數矩陣。
令 $\bv$ 為一 $\mathbb{R}^d$ 中的單位向量。
說明 $\bu = X\bv$ 為將 $X$ 的所有樣本點投影在 $\bv$ 方向的值、
並說明 $\bu$ 的變異數為 $\bv\trans C\bv$。

<!-- eng start -->
Let $X$ be a $N\times d$ data matrix whose rows are centered at their mean. Let $C = \frac{1}{N-1}X_0\trans X_0$ be the covariance matrix. Fix a unit vector $\bv$ in $\mathbb{R}^d$. Show that $\bu = X\bv$ records the length of data points in $X$ projected on $\bv$ and show that the variance of $\bu$ is $\bv\trans C\bv$.
<!-- eng end -->

##### Exercise 3

執行以下程式碼。

<!-- eng start -->
Run the code below.
<!-- eng end -->

import numpy as np

X = np.random.randn(3,4)
u,s,vh = np.linalg.svd(X) 
vals,vecs = np.linalg.eigh(X.T.dot(X))
vals = vals[::-1]
vecs = vecs[:,::-1]
print("V =")
print(vh.T)
print("Q =")
print(vecs)
print("singular values:", s)
print("eigenvalues:", vals)

##### Exercise 3(a)

說明為什麼印出來的兩個矩陣會一樣。
（除了有的行會有正負號的差別。）

<!-- eng start -->
Explain why the two output matrices are the same (except that some of the columns are negated).
<!-- eng end -->

##### Exercise 3(b)

說明印出來的奇異值和特徵值有什麼關係。

<!-- eng start -->
Explain the relation between the singular values and the eigenvalues in the output.
<!-- eng end -->

##### Exercise 3(c)

解釋

vals = vals[::-1]
vecs = vecs[:,::-1]

的用處。

<!-- eng start -->
Explain the meaning of

vals = vals[::-1]
vecs = vecs[:,::-1]

<!-- eng end -->

##### Exercise 4

有時候我們只須要最大的幾個特徵值。
所以如果把所有的特徵值都算出來是不必要的浪費時間。
查找資源如何利用 SciPy 中的 `linalg.eigh` 找最大的幾個特徵值。

<!-- eng start -->
Sometimes we only need the first few largest eigenvalues, and it is a waste of time to compute all eigenvalues. Search online and find out how to find the largest eigenvalues through `linalg.eigh` in SciPy.
<!-- eng end -->

##### Exercise 5

給定 $k$ 時，

$$
p = \frac{\sum_{i=1}^k \lambda_i}{\sum_{i=1}^d \lambda_i}
$$

稱作**變異數解釋率（explained variance ratio）**。
當解釋率夠高的時候，表示投影後的資料足以代表原資料的重要特徵。

觀察下圖，其橫軸為 $k$，縱軸為解釋率。
判斷 $k$ 應該要取多少，以及為什麼。

<!-- eng start -->
For a given $k$,

$$
p = \frac{\sum_{i=1}^k \lambda_i}{\sum_{i=1}^d \lambda_i}
$$

is known as the **explained variance ratio** . When this rate is high enough, it means the projected data is representative enough for the original data.

Observe the figure below, where the $x$-axis represents $k$ and the $y$-axis represents the explained variance ratio. Determine which $k$ is preferred and give your reasons.
<!-- eng end -->

import numpy as np
import matplotlib.pyplot as plt

X = np.genfromtxt('hidden_text.csv', delimiter=',')
print("shape of X =", X.shape)
X = X - X.mean(axis=0)
u,s,vh = np.linalg.svd(X)
vals = s**2

plt.plot(np.arange(1,101), vals.cumsum() / vals.sum())
plt.gca().set_xlim(0,10)

##### Exercise 6

找一個共變異數矩陣 `cov` 使得 `np.random.multivariate_normal` 產出來的資料大約是一個楕圓形，且：

- 長軸指向 $45^\circ$ 角、短軸指向 $135^\circ$ 角；
- 長軸是大約是短軸的 $2$ 倍。

<!-- eng start -->
Find a covariance matrix `cov` such that the data generated by `np.random.multivariate_normal` is roughly an ellipse such that:

- the major axis is on $45^\circ$, while the minor axis is on $135^\circ$;
- the major axis is about twice the length of the minor axis.
<!-- eng end -->

import numpy as np
import matplotlib.pyplot as plt

mu = np.array([0,0])
cov = np.array([
    [1,0],
    [0,1]
])
X = np.random.multivariate_normal(mu, cov, (1000,))
plt.axis('equal')
plt.scatter(*X.T)