Minor corrections.
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% Author: FOSSEE
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% Title page
\title[Matrices \& Curve Fitting]{Python for Science and Engg: Matrices \& Least Square Fit}
\author[FOSSEE] {FOSSEE}
\institute[IIT Bombay] {Department of Aerospace Engineering\\IIT Bombay}
\date[] {22 April, 2010\\Day 1, Session 4}
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%% the beginning of each subsection:
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\begin{frame}<beamer>
\frametitle{Outline}
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\begin{document}
\begin{frame}
\titlepage
\end{frame}
\begin{frame}
\frametitle{Outline}
\tableofcontents
% \pausesections
\end{frame}
\section{Matrices}
\begin{frame}
\frametitle{Matrices: Introduction}
\alert{All matrix operations are done using \kwrd{arrays}}
\end{frame}
\begin{frame}[fragile]
\frametitle{Matrices: Initializing}
\begin{lstlisting}
In []: c = array([[11,12,13],
[21,22,23],
[31,32,33]])
In []: c
Out[]:
array([[11, 12, 13],
[21, 22, 23],
[31, 32, 33]])
\end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{Initializing some special matrices}
\begin{small}
\begin{lstlisting}
In []: ones((3,5))
Out[]:
array([[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.]])
In []: ones_like([1, 2, 3, 4])
Out[]: array([1, 1, 1, 1])
In []: identity(2)
Out[]:
array([[ 1., 0.],
[ 0., 1.]])
\end{lstlisting}
Also available \alert{\typ{zeros, zeros_like}}
\end{small}
\end{frame}
\begin{frame}[fragile]
\frametitle{Accessing elements}
\begin{small}
\begin{lstlisting}
In []: c
Out[]:
array([[11, 12, 13],
[21, 22, 23],
[31, 32, 33]])
In []: c[1][2]
Out[]: 23
In []: c[1,2]
Out[]: 23
In []: c[1]
Out[]: array([21, 22, 23])
\end{lstlisting}
\end{small}
\end{frame}
\begin{frame}[fragile]
\frametitle{Changing elements}
\begin{small}
\begin{lstlisting}
In []: c[1,1] = -22
In []: c
Out[]:
array([[ 11, 12, 13],
[ 21, -22, 23],
[ 31, 32, 33]])
In []: c[1] = 0
In []: c
Out[]:
array([[11, 12, 13],
[ 0, 0, 0],
[31, 32, 33]])
\end{lstlisting}
\end{small}
How to access one \alert{column}?
\end{frame}
\begin{frame}[fragile]
\frametitle{Slicing}
\begin{small}
\begin{lstlisting}
In []: c[:,1]
Out[]: array([12, 0, 32])
In []: c[1,:]
Out[]: array([0, 0, 0])
In []: c[0:2,:]
Out[]:
array([[11, 12, 13],
[ 0, 0, 0]])
In []: c[1:3,:]
Out[]:
array([[ 0, 0, 0],
[31, 32, 33]])
\end{lstlisting}
\end{small}
\end{frame}
\begin{frame}[fragile]
\frametitle{Slicing \ldots}
\begin{small}
\begin{lstlisting}
In []: c[:2,:]
Out[]:
array([[11, 12, 13],
[ 0, 0, 0]])
In []: c[1:,:]
Out[]:
array([[ 0, 0, 0],
[31, 32, 33]])
In []: c[1:,:2]
Out[]:
array([[ 0, 0],
[31, 32]])
\end{lstlisting}
\end{small}
\end{frame}
\begin{frame}[fragile]
\frametitle{Striding}
\begin{small}
\begin{lstlisting}
In []: c[::2,:]
Out[]:
array([[11, 12, 13],
[31, 32, 33]])
In []: c[:,::2]
Out[]:
array([[11, 13],
[ 0, 0],
[31, 33]])
In []: c[::2,::2]
Out[]:
array([[11, 13],
[31, 33]])
\end{lstlisting}
\end{small}
\end{frame}
\begin{frame}[fragile]
\frametitle{Shape of a matrix}
\begin{lstlisting}
In []: c
Out[]:
array([[11, 12, 13],
[ 0, 0, 0],
[31, 32, 33]])
In []: c.shape
Out[]: (3, 3)
\end{lstlisting}
\emphbar{Shape specifies shape or dimensions of a matrix}
\end{frame}
\begin{frame}[fragile]
\frametitle{Elementary image processing}
\begin{small}
\begin{lstlisting}
In []: a = imread('lena.png')
In []: imshow(a)
Out[]: <matplotlib.image.AxesImage object at 0xa0384cc>
\end{lstlisting}
\end{small}
\typ{imread} returns an array of shape (512, 512, 4) which represents an image of 512x512 pixels and 4 shades.\\
\typ{imshow} renders the array as an image.
\end{frame}
\begin{frame}[fragile]
\frametitle{Slicing \& Striding Exercises}
\begin{itemize}
\item Crop the image to get the top-left quarter
\item Crop the image to get only the face
\item Resize image to half by dropping alternate pixels
\end{itemize}
\end{frame}
\begin{frame}[fragile]
\frametitle{Solutions}
\begin{small}
\begin{lstlisting}
In []: imshow(a[:256,:256])
Out[]: <matplotlib.image.AxesImage object at 0xb6f658c>
In []: imshow(a[200:400,200:400])
Out[]: <matplotlib.image.AxesImage object at 0xb757c2c>
In []: imshow(a[::2,::2])
Out[]: <matplotlib.image.AxesImage object at 0xb765c8c>
\end{lstlisting}
\end{small}
\end{frame}
\begin{frame}[fragile]
\frametitle{Transpose of a Matrix}
\begin{lstlisting}
In []: a = array([[ 1, 1, 2, -1],
...: [ 2, 5, -1, -9],
...: [ 2, 1, -1, 3],
...: [ 1, -3, 2, 7]])
In []: a.T
Out[]:
array([[ 1, 2, 2, 1],
[ 1, 5, 1, -3],
[ 2, -1, -1, 2],
[-1, -9, 3, 7]])
\end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{Matrix Addition}
\begin{lstlisting}
In []: b = array([[3,2,-1,5],
[2,-2,4,9],
[-1,0.5,-1,-7],
[9,-5,7,3]])
In []: a + b
Out[]:
array([[ 4. , 3. , 1. , 4. ],
[ 4. , 3. , 3. , 0. ],
[ 1. , 1.5, -2. , -4. ],
[ 10. , -8. , 9. , 10. ]])
\end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{Elementwise Multiplication}
\begin{lstlisting}
In []: a*b
Out[]:
array([[ 3. , 2. , -2. , -5. ],
[ 4. , -10. , -4. , -81. ],
[ -2. , 0.5, 1. , -21. ],
[ 9. , 15. , 14. , 21. ]])
\end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{Matrix Multiplication}
\begin{lstlisting}
In []: dot(a, b)
Out[]:
array([[ -6. , 6. , -6. , -3. ],
[-64. , 38.5, -44. , 35. ],
[ 36. , -13.5, 24. , 35. ],
[ 58. , -26. , 34. , -15. ]])
\end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{Inverse of a Matrix}
\begin{lstlisting}
\end{lstlisting}
\begin{small}
\begin{lstlisting}
In []: inv(a)
Out[]:
array([[-0.5 , 0.55, -0.15, 0.7 ],
[ 0.75, -0.5 , 0.5 , -0.75],
[ 0.5 , -0.15, -0.05, -0.1 ],
[ 0.25, -0.25, 0.25, -0.25]])
\end{lstlisting}
\end{small}
\emphbar{Try this: \typ{I = dot(a, inv(a))}}
\end{frame}
\begin{frame}[fragile]
\frametitle{Determinant and sum of all elements}
\begin{lstlisting}
In []: det(a)
Out[]: 80.0
\end{lstlisting}
\begin{lstlisting}
In []: sum(a)
Out[]: 12
\end{lstlisting}
\end{frame}
%%use S=array(X,Y)
\begin{frame}[fragile]
\frametitle{Eigenvalues and Eigen Vectors}
\begin{small}
\begin{lstlisting}
In []: e = array([[3,2,4],[2,0,2],[4,2,3]])
In []: eig(e)
Out[]:
(array([-1., 8., -1.]),
array([[-0.74535599, 0.66666667, -0.1931126 ],
[ 0.2981424 , 0.33333333, -0.78664085],
[ 0.59628479, 0.66666667, 0.58643303]]))
In []: eigvals(e)
Out[]: array([-1., 8., -1.])
\end{lstlisting}
\end{small}
\end{frame}
%% \begin{frame}[fragile]
%% \frametitle{Computing Norms}
%% \begin{lstlisting}
%% In []: norm(e)
%% Out[]: 8.1240384046359608
%% \end{lstlisting}
%% \end{frame}
%% \begin{frame}[fragile]
%% \frametitle{Singular Value Decomposition}
%% \begin{small}
%% \begin{lstlisting}
%% In []: svd(e)
%% Out[]:
%% (array(
%% [[ -6.66666667e-01, -1.23702565e-16, 7.45355992e-01],
%% [ -3.33333333e-01, -8.94427191e-01, -2.98142397e-01],
%% [ -6.66666667e-01, 4.47213595e-01, -5.96284794e-01]]),
%% array([ 8., 1., 1.]),
%% array([[-0.66666667, -0.33333333, -0.66666667],
%% [-0. , 0.89442719, -0.4472136 ],
%% [-0.74535599, 0.2981424 , 0.59628479]]))
%% \end{lstlisting}
%% \end{small}
%% \inctime{15}
%% \end{frame}
\section{Least Squares Fit}
\begin{frame}[fragile]
\frametitle{$L$ vs. $T^2$ - Scatter}
Linear trend visible.
\vspace{-0.1in}
\begin{figure}
\includegraphics[width=4in]{data/L-Tsq-points}
\end{figure}
\end{frame}
\begin{frame}[fragile]
\frametitle{$L$ vs. $T^2$ - Line}
This line does not make any mathematical sense.
\vspace{-0.1in}
\begin{figure}
\includegraphics[width=4in]{data/L-Tsq-Line}
\end{figure}
\end{frame}
\begin{frame}[fragile]
\frametitle{$L$ vs. $T^2$ - Least Square Fit}
This is what our intention is.
\vspace{-0.1in}
\begin{figure}
\includegraphics[width=4in]{data/least-sq-fit}
\end{figure}
\end{frame}
\begin{frame}[fragile]
\frametitle{Matrix Formulation}
\begin{itemize}
\item We need to fit a line through points for the equation $T^2 = m \cdot L+c$
\item In matrix form, the equation can be represented as $T_{sq} = A \cdot p$, where $T_{sq}$ is
$\begin{bmatrix}
T^2_1 \\
T^2_2 \\
\vdots\\
T^2_N \\
\end{bmatrix}$
, A is
$\begin{bmatrix}
L_1 & 1 \\
L_2 & 1 \\
\vdots & \vdots\\
L_N & 1 \\
\end{bmatrix}$
and p is
$\begin{bmatrix}
m\\
c\\
\end{bmatrix}$
\item We need to find $p$ to plot the line
\end{itemize}
\end{frame}
\begin{frame}[fragile]
\frametitle{Getting $L$ and $T^2$}
%If you \alert{closed} IPython after session 2
\begin{lstlisting}
In []: L = []
In []: t = []
In []: for line in open('pendulum.txt'):
.... point = line.split()
.... L.append(float(point[0]))
.... t.append(float(point[1]))
....
....
\end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{Getting $L$ and $T^2$ \dots}
\begin{lstlisting}
In []: L = array(L)
In []: t = array(t)
\end{lstlisting}
\alert{\typ{In []: tsq = t*t}}
\end{frame}
\begin{frame}[fragile]
\frametitle{Generating $A$}
\begin{lstlisting}
In []: A = array([L, ones_like(L)])
In []: A = A.T
\end{lstlisting}
%% \begin{itemize}
%% \item A is also called a Van der Monde matrix
%% \item It can also be generated using \typ{vander}
%% \end{itemize}
%% \begin{lstlisting}
%% In []: A = vander(L, 2)
%% \end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{\typ{lstsq} \ldots}
\begin{itemize}
\item Now use the \typ{lstsq} function
\item Along with a lot of things, it returns the least squares solution
\end{itemize}
\begin{lstlisting}
In []: result = lstsq(A,tsq)
In []: coef = result[0]
\end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{Least Square Fit Line \ldots}
We get the points of the line from \typ{coef}
\begin{lstlisting}
In []: Tline = coef[0]*L + coef[1]
In []: Tline.shape
\end{lstlisting}
\begin{itemize}
\item Now plot \typ{Tline} vs. \typ{L}, to get the Least squares fit line.
\end{itemize}
\begin{lstlisting}
In []: plot(L, Tline)
\end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{Least Squares Fit}
\vspace{-0.15in}
\begin{figure}
\includegraphics[width=4in]{data/least-sq-fit}
\end{figure}
\end{frame}
\section{Summary}
\begin{frame}
\frametitle{What did we learn?}
\begin{itemize}
\item Matrices
\begin{itemize}
\item Initializing
\item Accessing elements
\item Slicing and Striding
\item Transpose
\item Addition
\item Multiplication
\item Inverse of a matrix
\item Determinant
\item Eigenvalues and Eigen vector
%% \item Norms
%% \item Singular Value Decomposition
\end{itemize}
\item Least Square Curve fitting
\end{itemize}
\end{frame}
\end{document}
%% Questions for Quiz %%
%% ------------------ %%
\begin{frame}[fragile]
\frametitle{\incqno }
\begin{lstlisting}
In []: a = array([[1, 2],
[3, 4]])
In []: a[1,0] = 0
\end{lstlisting}
What is the resulting array?
\end{frame}
\begin{frame}[fragile]
\frametitle{\incqno }
\begin{lstlisting}
In []: x = array(([1,2,3,4],
[2,3,4,5]))
In []: x[-2][-3] = 4
In []: print x
\end{lstlisting}
What will be printed?
\end{frame}
%% \begin{frame}[fragile]
%% \frametitle{\incqno }
%% \begin{lstlisting}
%% In []: x = array([[1,2,3,4],
%% [3,4,2,5]])
%% \end{lstlisting}
%% What is the \lstinline+shape+ of this array?
%% \end{frame}
\begin{frame}[fragile]
\frametitle{\incqno }
\begin{lstlisting}
In []: x = array([[1,2,3,4]])
\end{lstlisting}
How to change \lstinline+x+ to \lstinline+array([[1,2,0,4]])+?
\end{frame}
\begin{frame}[fragile]
\frametitle{\incqno }
\begin{lstlisting}
In []: x = array([[1,2,3,4],
[3,4,2,5]])
\end{lstlisting}
How do you get the following slice of \lstinline+x+?
\begin{lstlisting}
array([[2,3],
[4,2]])
\end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{\incqno }
\begin{lstlisting}
In []: x = array([[9,18,27],
[30,60,90],
[14,7,1]])
\end{lstlisting}
What is the output of \lstinline+x[::3,::3]+
\end{frame}
\begin{frame}[fragile]
\frametitle{\incqno }
\begin{lstlisting}
In []: a = array([[1, 2],
[3, 4]])
\end{lstlisting}
How do you get the transpose of this array?
\end{frame}
\begin{frame}[fragile]
\frametitle{\incqno }
\begin{lstlisting}
In []: a = array([[1, 2],
[3, 4]])
In []: b = array([[1, 1],
[2, 2]])
In []: a*b
\end{lstlisting}
What does this produce?
\end{frame}
\begin{frame}
\frametitle{\incqno }
What command do you use to find the inverse of a matrix and its
eigenvalues?
\end{frame}
%% \begin{frame}
%% \frametitle{\incqno }
%% The file \lstinline+datafile.txt+ contains 3 columns of data. What
%% command will you use to read the entire data file into an array?
%% \end{frame}
%% \begin{frame}
%% \frametitle{\incqno }
%% If the contents of the file \lstinline+datafile.txt+ is read into an
%% $N\times3$ array called \lstinline+data+, how would you obtain the third
%% column of this data?
%% \end{frame}