Workshop materials: comparison day1/session4.tex

equal deleted inserted replaced

-:2299700a8b97
+:682b6f66fe11
 In []: identity(2)
 Out[]:
 array([[ 1.,  0.],
 [ 0.,  1.]])
 \end{lstlisting}
-Also available \alert{\typ{zeros, zeros_like, empty, empty_like}}
+Also available \alert{\typ{zeros, zeros_like}}
 \end{small}
 \end{frame}
 \begin{frame}[fragile]
 In []: c[1][2]
 Out[]: 23
 In []: c[1,2]
 Out[]: 23
+\end{lstlisting}
+\end{frame}
+\begin{frame}[fragile]
+\frametitle{Changing elements}
+\begin{small}
+\begin{lstlisting}
 In []: c[1]
 Out[]: array([21, 22, 23])
-\end{lstlisting}
-\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],
 array([[11, 12, 13],
 [ 0,  0,  0],
 [31, 32, 33]])
 \end{lstlisting}
 \end{small}
-How to change one \alert{column}?
+How to access one \alert{column}?
 \end{frame}
 \begin{frame}[fragile]
 \frametitle{Slicing}
 \begin{small}
 \end{lstlisting}
 \emphbar{Shape specifies shape or dimensions of a matrix}
 \end{frame}
 \begin{frame}[fragile]
-\frametitle{Slicing \& Striding Exercises}
+\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{lstlisting}
+\end{small}
-\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}
+\end{frame}
 \begin{frame}[fragile]
 \frametitle{Solutions}
 \begin{small}
 \begin{lstlisting}
 In []: imshow(a[:256,:256])
 [-1, -9,  3,  7]])
 \end{lstlisting}
 \end{frame}
 \begin{frame}[fragile]
-\frametitle{Sum of all elements}
-\begin{lstlisting}
-In []: sum(a)
-Out[]: 12
-\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],
 \end{lstlisting}
 \end{small}
 \end{frame}
 \begin{frame}[fragile]
-\frametitle{Determinant}
+\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}
 %% \end{frame}
 \section{Least Squares Fit}
 \begin{frame}[fragile]
 \frametitle{$L$ vs. $T^2$ - Scatter}
-\vspace{-0.15in}
+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}
-\vspace{-0.15in}
+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. $T^2$ }
 \frametitle{$L$ vs. $T^2$ - Least Square Fit}
-\vspace{-0.15in}
+This is what our intention is.
+\vspace{-0.1in}
 \begin{figure}
 \includegraphics[width=4in]{data/least-sq-fit}
 \end{figure}
 \end{frame}
-\begin{frame}
+\begin{frame}[fragile]
-\frametitle{Least Square Fit Curve}
+\frametitle{Matrix Formulation}
-\begin{center}
-\begin{itemize}
-\item $L \alpha T^2$
-\item Best Fit Curve $\rightarrow$ Linear
-\begin{itemize}
-\item Least Square Fit
-\end{itemize}
-\item \typ{lstsq()}
-\end{itemize}
-\end{center}
-\end{frame}
-\begin{frame}[fragile]
-\frametitle{\typ{lstsq}}
 \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 \\
 \begin{frame}[fragile]
 \frametitle{Getting $L$ and $T^2$}
 %If you \alert{closed} IPython after session 2
 \begin{lstlisting}
-In []: l = []
+In []: L = []
 In []: t = []
 In []: for line in open('pendulum.txt'):
 ....     point = line.split()
-....     l.append(float(point[0]))
+....     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 []: 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 = 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}
 \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 = 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.
+\item Now plot \typ{Tline} vs. \typ{L}, to get the Least squares fit line.
 \end{itemize}
 \begin{lstlisting}
-In []: plot(l, Tline)
+In []: plot(L, Tline)
 \end{lstlisting}
 \end{frame}
 \begin{frame}[fragile]
 \frametitle{Least Squares Fit}

changeset 379	682b6f66fe11
parent 378	2299700a8b97
child 382	41c34770d63a