-
diff --git a/doc/pub/DimRed/html/._DimRed-bs018.html b/doc/pub/DimRed/html/._DimRed-bs018.html index 1dc23b946..26757403b 100644 --- a/doc/pub/DimRed/html/._DimRed-bs018.html +++ b/doc/pub/DimRed/html/._DimRed-bs018.html @@ -233,12 +233,7 @@ n = 100 mean = (-1, 2) cov = [[4, 2], [2, 2]] X = np.random.multivariate_normal(mean, cov, n) -# Print the X-matrix -print(X) -
-Try to add to this code your own calculation of the covariance matrix. -
Now we are going to implement the PCA algorithm. We will break it down into various substeps. @@ -258,31 +253,31 @@ $$ When you are done with these steps, print out \( \mu_n \) to verify it is close to \( \mu \) and plot your mean centered data to verify it is centered at the origin! Compare your code with the functionality from Scikit-Learn discussed above. -The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simply through the mean() function. +The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simple through the mean() function.
df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
-display(df)
-
# we center it ourselves
X_centered = X - X.mean(axis=0)
-# test that we get the same as Pandas
-print(X_centered-df)
-Alternatively, you could also have used the functions we discussed earlier for scaling the data set. -That is, we could have used the StandardScaler function in Scikit-Learn, a function which -ensures that for each feature/predictor we study the mean value is -zero and the variance is one (every column in the design/feature -matrix). You would then not get the same results, since we divide by the variance. For diagonal covariance matrix elements will then be one, while the non-diagonal ones will be divided by \( 2\sqrt{2} \) for our specific case. +Alternatively, we could use the functions we discussed +earlier for scaling the data set. That is, we could have used the +StandardScaler function in Scikit-Learn, a function which ensures +that for each feature/predictor we study the mean value is zero and +the variance is one (every column in the design/feature matrix). You +would then not get the same results, since we divide by the +variance. The diagonal covariance matrix elements will then be one, +while the non-diagonal ones need to be divided by \( 2\sqrt{2} \) for our +specific case.
-Now we are going to use the mean centered data to compute the sample covariance of the data. Recall it is given by: +Now we are going to use the mean centered data to compute the sample covariance of the data. $$ \begin{equation*} \Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}_i^T \bar{x}_i = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n) @@ -294,11 +289,29 @@ We can write our own code or simply use either the functionaly of numpy o
-
print(np.cov(X.T))
-print(df.cov())
+print(df.cov())
print(np.cov(X_centered.T))
+Note that the way we define the covariance matrix here has a factor \( n-1 \) instead of \( n \).
+Our own code here is not very elegant and asks for improvements.
+
+
+
+
# extract the relevant columns from the centered design matrix of dim n x 2 x = X_centered[:,[0]]
+y = X_centered[:,[1]]
+Cov = np.zeros((2,2))
+Cov[0,1] = np.sum(x.T@y)/(n-1.0)
+Cov[0,0] = np.sum(x.T@x)/(n-1.0)
+Cov[1,1] = np.sum(y.T@y)/(n-1.0)
+Cov[1,0]= Cov[0,1]
+print("Centered covariance using own code")
+print(Cov)
+plt.plot(x, y, 'x')
+plt.axis('equal')
+plt.show()
+
+
Depending on the number of points \( n \), we will get results that are close to the covariance values defined above.
Diagonalize the sample covariance matrix to obtain the principal components
diff --git a/doc/pub/DimRed/html/DimRed-bs.html b/doc/pub/DimRed/html/DimRed-bs.html
index 59e9e22ca..2fdf5d407 100644
--- a/doc/pub/DimRed/html/DimRed-bs.html
+++ b/doc/pub/DimRed/html/DimRed-bs.html
@@ -224,7 +224,7 @@ MathJax.Hub.Config({
[2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University
-
Jan 1, 2020
Jan 2, 2020
diff --git a/doc/pub/DimRed/html/DimRed-reveal.html b/doc/pub/DimRed/html/DimRed-reveal.html
index d40e41dbd..5a9a90a8f 100644
--- a/doc/pub/DimRed/html/DimRed-reveal.html
+++ b/doc/pub/DimRed/html/DimRed-reveal.html
@@ -148,7 +148,7 @@ MathJax.Hub.Config({
[2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University
-
Jan 1, 2020
Jan 2, 2020
@@ -1039,12 +1039,7 @@ n = 100
mean = (-1, 2)
cov = [[4, 2], [2, 2]]
X = np.random.multivariate_normal(mean, cov, n)
-# Print the X-matrix
-print(X)
-Try to add to this code your own calculation of the covariance matrix. -
Now we are going to implement the PCA algorithm. We will break it down into various substeps. @@ -1068,31 +1063,31 @@ $$ When you are done with these steps, print out \( \mu_n \) to verify it is close to \( \mu \) and plot your mean centered data to verify it is centered at the origin! Compare your code with the functionality from Scikit-Learn discussed above. -The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simply through the mean() function. +The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simple through the mean() function.
df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
-display(df)
-
# we center it ourselves
X_centered = X - X.mean(axis=0)
-# test that we get the same as Pandas
-print(X_centered-df)
-Alternatively, you could also have used the functions we discussed earlier for scaling the data set. -That is, we could have used the StandardScaler function in Scikit-Learn, a function which -ensures that for each feature/predictor we study the mean value is -zero and the variance is one (every column in the design/feature -matrix). You would then not get the same results, since we divide by the variance. For diagonal covariance matrix elements will then be one, while the non-diagonal ones will be divided by \( 2\sqrt{2} \) for our specific case. +Alternatively, we could use the functions we discussed +earlier for scaling the data set. That is, we could have used the +StandardScaler function in Scikit-Learn, a function which ensures +that for each feature/predictor we study the mean value is zero and +the variance is one (every column in the design/feature matrix). You +would then not get the same results, since we divide by the +variance. The diagonal covariance matrix elements will then be one, +while the non-diagonal ones need to be divided by \( 2\sqrt{2} \) for our +specific case.
-Now we are going to use the mean centered data to compute the sample covariance of the data. Recall it is given by: +Now we are going to use the mean centered data to compute the sample covariance of the data.
$$
\begin{equation*}
@@ -1106,11 +1101,29 @@ We can write our own code or simply use either the functionaly of numpy o
-
print(np.cov(X.T))
-print(df.cov())
+print(df.cov())
print(np.cov(X_centered.T))
+Note that the way we define the covariance matrix here has a factor \( n-1 \) instead of \( n \).
+Our own code here is not very elegant and asks for improvements.
+
+
+
+
# extract the relevant columns from the centered design matrix of dim n x 2 x = X_centered[:,[0]]
+y = X_centered[:,[1]]
+Cov = np.zeros((2,2))
+Cov[0,1] = np.sum(x.T@y)/(n-1.0)
+Cov[0,0] = np.sum(x.T@x)/(n-1.0)
+Cov[1,1] = np.sum(y.T@y)/(n-1.0)
+Cov[1,0]= Cov[0,1]
+print("Centered covariance using own code")
+print(Cov)
+plt.plot(x, y, 'x')
+plt.axis('equal')
+plt.show()
+
+
Depending on the number of points \( n \), we will get results that are close to the covariance values defined above.
Diagonalize the sample covariance matrix to obtain the principal components
diff --git a/doc/pub/DimRed/html/DimRed-solarized.html b/doc/pub/DimRed/html/DimRed-solarized.html
index a15038ff7..0c1ae2da3 100644
--- a/doc/pub/DimRed/html/DimRed-solarized.html
+++ b/doc/pub/DimRed/html/DimRed-solarized.html
@@ -179,7 +179,7 @@ MathJax.Hub.Config({
[2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University
-
Jan 1, 2020
Jan 2, 2020
@@ -1019,12 +1019,7 @@ n = 100
mean = (-1, 2)
cov = [[4, 2], [2, 2]]
X = np.random.multivariate_normal(mean, cov, n)
-# Print the X-matrix
-print(X)
-Try to add to this code your own calculation of the covariance matrix. -
Now we are going to implement the PCA algorithm. We will break it down into various substeps. @@ -1044,31 +1039,31 @@ $$ When you are done with these steps, print out \( \mu_n \) to verify it is close to \( \mu \) and plot your mean centered data to verify it is centered at the origin! Compare your code with the functionality from Scikit-Learn discussed above. -The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simply through the mean() function. +The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simple through the mean() function.
df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
-display(df)
-
# we center it ourselves
X_centered = X - X.mean(axis=0)
-# test that we get the same as Pandas
-print(X_centered-df)
-Alternatively, you could also have used the functions we discussed earlier for scaling the data set. -That is, we could have used the StandardScaler function in Scikit-Learn, a function which -ensures that for each feature/predictor we study the mean value is -zero and the variance is one (every column in the design/feature -matrix). You would then not get the same results, since we divide by the variance. For diagonal covariance matrix elements will then be one, while the non-diagonal ones will be divided by \( 2\sqrt{2} \) for our specific case. +Alternatively, we could use the functions we discussed +earlier for scaling the data set. That is, we could have used the +StandardScaler function in Scikit-Learn, a function which ensures +that for each feature/predictor we study the mean value is zero and +the variance is one (every column in the design/feature matrix). You +would then not get the same results, since we divide by the +variance. The diagonal covariance matrix elements will then be one, +while the non-diagonal ones need to be divided by \( 2\sqrt{2} \) for our +specific case.
-Now we are going to use the mean centered data to compute the sample covariance of the data. Recall it is given by: +Now we are going to use the mean centered data to compute the sample covariance of the data. $$ \begin{equation*} \Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}_i^T \bar{x}_i = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n) @@ -1080,11 +1075,29 @@ We can write our own code or simply use either the functionaly of numpy o
-
print(np.cov(X.T))
-print(df.cov())
+print(df.cov())
print(np.cov(X_centered.T))
+Note that the way we define the covariance matrix here has a factor \( n-1 \) instead of \( n \).
+Our own code here is not very elegant and asks for improvements.
+
+
+
+
# extract the relevant columns from the centered design matrix of dim n x 2 x = X_centered[:,[0]]
+y = X_centered[:,[1]]
+Cov = np.zeros((2,2))
+Cov[0,1] = np.sum(x.T@y)/(n-1.0)
+Cov[0,0] = np.sum(x.T@x)/(n-1.0)
+Cov[1,1] = np.sum(y.T@y)/(n-1.0)
+Cov[1,0]= Cov[0,1]
+print("Centered covariance using own code")
+print(Cov)
+plt.plot(x, y, 'x')
+plt.axis('equal')
+plt.show()
+
+
Depending on the number of points \( n \), we will get results that are close to the covariance values defined above.
Diagonalize the sample covariance matrix to obtain the principal components
diff --git a/doc/pub/DimRed/html/DimRed.html b/doc/pub/DimRed/html/DimRed.html
index 424c95f51..479e67b4f 100644
--- a/doc/pub/DimRed/html/DimRed.html
+++ b/doc/pub/DimRed/html/DimRed.html
@@ -184,7 +184,7 @@ MathJax.Hub.Config({
[2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University
-
Jan 1, 2020
Jan 2, 2020
@@ -1024,12 +1024,7 @@ n = 100
mean = (-1, 2)
cov = [[4, 2], [2, 2]]
X = np.random.multivariate_normal(mean, cov, n)
-# Print the X-matrix
-print(X)
-Try to add to this code your own calculation of the covariance matrix. -
Now we are going to implement the PCA algorithm. We will break it down into various substeps. @@ -1049,31 +1044,31 @@ $$ When you are done with these steps, print out \( \mu_n \) to verify it is close to \( \mu \) and plot your mean centered data to verify it is centered at the origin! Compare your code with the functionality from Scikit-Learn discussed above. -The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simply through the mean() function. +The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simple through the mean() function.
df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
-display(df)
-
# we center it ourselves
X_centered = X - X.mean(axis=0)
-# test that we get the same as Pandas
-print(X_centered-df)
-Alternatively, you could also have used the functions we discussed earlier for scaling the data set. -That is, we could have used the StandardScaler function in Scikit-Learn, a function which -ensures that for each feature/predictor we study the mean value is -zero and the variance is one (every column in the design/feature -matrix). You would then not get the same results, since we divide by the variance. For diagonal covariance matrix elements will then be one, while the non-diagonal ones will be divided by \( 2\sqrt{2} \) for our specific case. +Alternatively, we could use the functions we discussed +earlier for scaling the data set. That is, we could have used the +StandardScaler function in Scikit-Learn, a function which ensures +that for each feature/predictor we study the mean value is zero and +the variance is one (every column in the design/feature matrix). You +would then not get the same results, since we divide by the +variance. The diagonal covariance matrix elements will then be one, +while the non-diagonal ones need to be divided by \( 2\sqrt{2} \) for our +specific case.
-Now we are going to use the mean centered data to compute the sample covariance of the data. Recall it is given by: +Now we are going to use the mean centered data to compute the sample covariance of the data. $$ \begin{equation*} \Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}_i^T \bar{x}_i = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n) @@ -1085,11 +1080,29 @@ We can write our own code or simply use either the functionaly of numpy o
-
print(np.cov(X.T))
-print(df.cov())
+print(df.cov())
print(np.cov(X_centered.T))
+Note that the way we define the covariance matrix here has a factor \( n-1 \) instead of \( n \).
+Our own code here is not very elegant and asks for improvements.
+
+
+
+
# extract the relevant columns from the centered design matrix of dim n x 2 x = X_centered[:,[0]]
+y = X_centered[:,[1]]
+Cov = np.zeros((2,2))
+Cov[0,1] = np.sum(x.T@y)/(n-1.0)
+Cov[0,0] = np.sum(x.T@x)/(n-1.0)
+Cov[1,1] = np.sum(y.T@y)/(n-1.0)
+Cov[1,0]= Cov[0,1]
+print("Centered covariance using own code")
+print(Cov)
+plt.plot(x, y, 'x')
+plt.axis('equal')
+plt.show()
+
+
Depending on the number of points \( n \), we will get results that are close to the covariance values defined above.
Diagonalize the sample covariance matrix to obtain the principal components
diff --git a/doc/pub/DimRed/ipynb/DimRed.ipynb b/doc/pub/DimRed/ipynb/DimRed.ipynb
index 42b7a9f71..bded76346 100644
--- a/doc/pub/DimRed/ipynb/DimRed.ipynb
+++ b/doc/pub/DimRed/ipynb/DimRed.ipynb
@@ -10,7 +10,7 @@
" \n",
"**Morten Hjorth-Jensen**, Department of Physics, University of Oslo and Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University\n",
"\n",
- "Date: **Jan 1, 2020**\n",
+ "Date: **Jan 2, 2020**\n",
"\n",
"Copyright 1999-2020, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license\n",
"\n",
@@ -1111,17 +1111,13 @@
"n = 100\n",
"mean = (-1, 2)\n",
"cov = [[4, 2], [2, 2]]\n",
- "X = np.random.multivariate_normal(mean, cov, n)\n",
- "# Print the X-matrix\n",
- "print(X)"
+ "X = np.random.multivariate_normal(mean, cov, n)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
- "Try to add to this code your own calculation of the covariance matrix.\n",
- "\n",
"Now we are going to implement the PCA algorithm. We will break it down into various substeps.\n",
"\n",
"### Compute the sample mean and center the data\n",
@@ -1161,7 +1157,7 @@
"When you are done with these steps, print out $\\mu_n$ to verify it is\n",
"close to $\\mu$ and plot your mean centered data to verify it is\n",
"centered at the origin! Compare your code with the functionality from **Scikit-Learn** discussed above.\n",
- "The following code elements perform these operations using **pandas** or using our own functionality for doing so. The latter, using **numpy** is rather simply through the **mean()** function."
+ "The following code elements perform these operations using **pandas** or using our own functionality for doing so. The latter, using **numpy** is rather simple through the **mean()** function."
]
},
{
@@ -1175,27 +1171,27 @@
"df = pd.DataFrame(X)\n",
"# Pandas does the centering for us\n",
"df = df -df.mean()\n",
- "display(df)\n",
- "\n",
"# we center it ourselves\n",
- "X_centered = X - X.mean(axis=0)\n",
- "# test that we get the same as Pandas\n",
- "print(X_centered-df)"
+ "X_centered = X - X.mean(axis=0)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
- "Alternatively, you could also have used the functions we discussed earlier for scaling the data set. \n",
- "That is, we could have used the **StandardScaler** function in **Scikit-Learn**, a function which \n",
- "ensures that for each feature/predictor we study the mean value is\n",
- "zero and the variance is one (every column in the design/feature\n",
- "matrix). You would then not get the same results, since we divide by the variance. For diagonal covariance matrix elements will then be one, while the non-diagonal ones will be divided by $2\\sqrt{2}$ for our specific case.\n",
+ "Alternatively, we could use the functions we discussed\n",
+ "earlier for scaling the data set. That is, we could have used the\n",
+ "**StandardScaler** function in **Scikit-Learn**, a function which ensures\n",
+ "that for each feature/predictor we study the mean value is zero and\n",
+ "the variance is one (every column in the design/feature matrix). You\n",
+ "would then not get the same results, since we divide by the\n",
+ "variance. The diagonal covariance matrix elements will then be one,\n",
+ "while the non-diagonal ones need to be divided by $2\\sqrt{2}$ for our\n",
+ "specific case.\n",
"\n",
"### Compute the sample covariance\n",
"\n",
- "Now we are going to use the mean centered data to compute the sample covariance of the data. Recall it is given by:"
+ "Now we are going to use the mean centered data to compute the sample covariance of the data."
]
},
{
@@ -1223,11 +1219,40 @@
},
"outputs": [],
"source": [
- "print(np.cov(X.T))\n",
"print(df.cov())\n",
"print(np.cov(X_centered.T))"
]
},
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Note that the way we define the covariance matrix here has a factor $n-1$ instead of $n$.\n",
+ "Our own code here is not very elegant and asks for improvements."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# extract the relevant columns from the centered design matrix of dim n x 2 x = X_centered[:,[0]]\n",
+ "y = X_centered[:,[1]]\n",
+ "Cov = np.zeros((2,2))\n",
+ "Cov[0,1] = np.sum(x.T@y)/(n-1.0)\n",
+ "Cov[0,0] = np.sum(x.T@x)/(n-1.0)\n",
+ "Cov[1,1] = np.sum(y.T@y)/(n-1.0)\n",
+ "Cov[1,0]= Cov[0,1]\n",
+ "print(\"Centered covariance using own code\")\n",
+ "print(Cov)\n",
+ "plt.plot(x, y, 'x')\n",
+ "plt.axis('equal')\n",
+ "plt.show()"
+ ]
+ },
{
"cell_type": "markdown",
"metadata": {},
@@ -1275,7 +1300,7 @@
},
{
"cell_type": "code",
- "execution_count": 14,
+ "execution_count": 15,
"metadata": {
"collapsed": false
},
@@ -1586,7 +1611,7 @@
},
{
"cell_type": "code",
- "execution_count": 15,
+ "execution_count": 16,
"metadata": {
"collapsed": false
},
@@ -1633,7 +1658,7 @@
},
{
"cell_type": "code",
- "execution_count": 16,
+ "execution_count": 17,
"metadata": {
"collapsed": false
},
@@ -1657,7 +1682,7 @@
},
{
"cell_type": "code",
- "execution_count": 17,
+ "execution_count": 18,
"metadata": {
"collapsed": false
},
@@ -1681,7 +1706,7 @@
},
{
"cell_type": "code",
- "execution_count": 18,
+ "execution_count": 19,
"metadata": {
"collapsed": false
},
@@ -1705,7 +1730,7 @@
},
{
"cell_type": "code",
- "execution_count": 19,
+ "execution_count": 20,
"metadata": {
"collapsed": false
},
@@ -1759,7 +1784,7 @@
},
{
"cell_type": "code",
- "execution_count": 20,
+ "execution_count": 21,
"metadata": {
"collapsed": false
},
@@ -1782,7 +1807,7 @@
},
{
"cell_type": "code",
- "execution_count": 21,
+ "execution_count": 22,
"metadata": {
"collapsed": false
},
@@ -1829,7 +1854,7 @@
},
{
"cell_type": "code",
- "execution_count": 22,
+ "execution_count": 23,
"metadata": {
"collapsed": false
},
diff --git a/doc/pub/DimRed/ipynb/ipynb-DimRed-src.tar.gz b/doc/pub/DimRed/ipynb/ipynb-DimRed-src.tar.gz
index 3dcbbe279..603419a5e 100644
Binary files a/doc/pub/DimRed/ipynb/ipynb-DimRed-src.tar.gz and b/doc/pub/DimRed/ipynb/ipynb-DimRed-src.tar.gz differ
diff --git a/doc/pub/DimRed/pdf/DimRed-minted.pdf b/doc/pub/DimRed/pdf/DimRed-minted.pdf
index d9bb19dfd..6c4542cb3 100644
Binary files a/doc/pub/DimRed/pdf/DimRed-minted.pdf and b/doc/pub/DimRed/pdf/DimRed-minted.pdf differ
diff --git a/doc/src/DimRed/DimRed.do.txt b/doc/src/DimRed/DimRed.do.txt
index 17ddc720e..0ad06f689 100644
--- a/doc/src/DimRed/DimRed.do.txt
+++ b/doc/src/DimRed/DimRed.do.txt
@@ -786,13 +786,8 @@ n = 100
mean = (-1, 2)
cov = [[4, 2], [2, 2]]
X = np.random.multivariate_normal(mean, cov, n)
-# Print the X-matrix
-print(X)
!ec
-
-Try to add to this code your own calculation of the covariance matrix.
-
Now we are going to implement the PCA algorithm. We will break it down into various substeps.
=== Compute the sample mean and center the data ===
@@ -812,27 +807,28 @@ and the mean-centered data $\bar{X} = \{ \bar{x}_1, \ldots, \bar{x}_n \}$ takes
When you are done with these steps, print out $\mu_n$ to verify it is
close to $\mu$ and plot your mean centered data to verify it is
centered at the origin! Compare your code with the functionality from _Scikit-Learn_ discussed above.
-The following code elements perform these operations using _pandas_ or using our own functionality for doing so. The latter, using _numpy_ is rather simply through the _mean()_ function.
+The following code elements perform these operations using _pandas_ or using our own functionality for doing so. The latter, using _numpy_ is rather simple through the _mean()_ function.
!bc pycod
df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
-display(df)
-
# we center it ourselves
X_centered = X - X.mean(axis=0)
-# test that we get the same as Pandas
-print(X_centered-df)
!ec
-Alternatively, you could also have used the functions we discussed earlier for scaling the data set.
-That is, we could have used the _StandardScaler_ function in _Scikit-Learn_, a function which
-ensures that for each feature/predictor we study the mean value is
-zero and the variance is one (every column in the design/feature
-matrix). You would then not get the same results, since we divide by the variance. For diagonal covariance matrix elements will then be one, while the non-diagonal ones will be divided by $2\sqrt{2}$ for our specific case.
+
+Alternatively, we could use the functions we discussed
+earlier for scaling the data set. That is, we could have used the
+_StandardScaler_ function in _Scikit-Learn_, a function which ensures
+that for each feature/predictor we study the mean value is zero and
+the variance is one (every column in the design/feature matrix). You
+would then not get the same results, since we divide by the
+variance. The diagonal covariance matrix elements will then be one,
+while the non-diagonal ones need to be divided by $2\sqrt{2}$ for our
+specific case.
=== Compute the sample covariance ===
-Now we are going to use the mean centered data to compute the sample covariance of the data. Recall it is given by:
+Now we are going to use the mean centered data to compute the sample covariance of the data.
!bt
\begin{equation*}
\Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}_i^T \bar{x}_i = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n)
@@ -841,10 +837,26 @@ Now we are going to use the mean centered data to compute the sample covariance
where the data points $x_i \in \mathbb{R}^p$ (here in this example $p = 2$) are column vectors and $x^T$ is the transpose of $x$.
We can write our own code or simply use either the functionaly of _numpy_ or that of _pandas_, as follows
!bc pycod
-print(np.cov(X.T))
print(df.cov())
print(np.cov(X_centered.T))
!ec
+Note that the way we define the covariance matrix here has a factor $n-1$ instead of $n$.
+Our own code here is not very elegant and asks for improvements.
+!bc pycod
+# extract the relevant columns from the centered design matrix of dim n x 2 x = X_centered[:,[0]]
+y = X_centered[:,[1]]
+Cov = np.zeros((2,2))
+Cov[0,1] = np.sum(x.T@y)/(n-1.0)
+Cov[0,0] = np.sum(x.T@x)/(n-1.0)
+Cov[1,1] = np.sum(y.T@y)/(n-1.0)
+Cov[1,0]= Cov[0,1]
+print("Centered covariance using own code")
+print(Cov)
+plt.plot(x, y, 'x')
+plt.axis('equal')
+plt.show()
+!ec
+
Depending on the number of points $n$, we will get results that are close to the covariance values defined above.
diff --git a/doc/src/DimRed/Programs/PCAsimple.py b/doc/src/DimRed/Programs/PCAsimple.py
index bfbccb78e..9e7716348 100644
--- a/doc/src/DimRed/Programs/PCAsimple.py
+++ b/doc/src/DimRed/Programs/PCAsimple.py
@@ -2,28 +2,47 @@ import numpy as np
import pandas as pd
from IPython.display import display
import matplotlib.pyplot as plt
-from sklearn.model_selection import train_test_split
-from sklearn.preprocessing import MinMaxScaler, StandardScaler
-
-n = 10000
+n = 1000
mean = (-1, 2)
cov = [[4, 2], [2, 2]]
-print(cov)
X = np.random.multivariate_normal(mean, cov, n)
-print(np.cov(X.T))
df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
-correlation_matrix = df.cov()
-print(correlation_matrix)
+print("Centered covariance with Pandas")
+covarianceX = df.cov()
+
+print(covarianceX)
# we center it ourselves
X_centered = X - X.mean(axis=0)
+print("Centered covariance using numpy")
print(np.cov(X_centered.T))
-#print("test that we get the same as Pandas")
-#print(X_centered-X_train_scaled)
+# extract the relevant columns from the centered design matrix
+x = X_centered[:,[0]]
+y = X_centered[:,[1]]
+Cov = np.zeros((2,2))
+cov_xy = np.sum(x.T@y)/(n-1.0)
+cov_xx = np.sum(x.T@x)/(n-1.0)
+cov_yy = np.sum(y.T@y)/(n-1.0)
+
+Cov[0,0]= cov_xx
+Cov[1,1]= cov_yy
+Cov[0,1]= cov_xy
+Cov[1,0]= Cov[0,1]
+print("Centered covariance using own code")
+print(Cov)
+
+plt.plot(x, y, 'x')
+plt.axis('equal')
+plt.show()
+
+
+
+
+"""
#Now we do an SVD
U, s, V = np.linalg.svd(X_centered)
c1 = V.T[:, 0]
@@ -38,6 +57,6 @@ print("Check that we get the same")
print(X2D-X2Dsl)
print(pca.components_.T[:, 0])
-
+"""