Updated for day1 of GRD workshop.

1 files changed, 36 insertions, 43 deletions

diff --git a/day1/session4.tex b/day1/session4.tex
index 99ca05b..42227f1 100644
--- a/day1/session4.tex
+++ b/day1/session4.tex
@@ -164,7 +164,7 @@ 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}
 
@@ -182,8 +182,6 @@ In []: c[1][2]
 Out[]: 23
 In []: c[1,2]
 Out[]: 23
-In []: c[1]
-Out[]: array([21, 22, 23])
   \end{lstlisting}
 \end{frame}
 
@@ -191,6 +189,9 @@ Out[]: array([21, 22, 23])
   \frametitle{Changing elements}
   \begin{small}
   \begin{lstlisting}
+In []: c[1]
+Out[]: array([21, 22, 23])
+
 In []: c[1,1] = -22
 In []: c
 Out[]: 
@@ -206,7 +207,7 @@ array([[11, 12, 13],
        [31, 32, 33]])
   \end{lstlisting}
   \end{small}
-How to change one \alert{column}?
+How to access one \alert{column}?
 \end{frame}
 
 \begin{frame}[fragile]
@@ -294,23 +295,28 @@ Out[]: (3, 3)
 \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{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}
@@ -345,14 +351,6 @@ array([[ 1,  2,  2,  1],
 \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],
@@ -410,11 +408,16 @@ array([[-0.5 ,  0.55, -0.15,  0.7 ],
 \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)
@@ -467,7 +470,8 @@ Out[]: array([-1.,  8., -1.])
 \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}
@@ -475,37 +479,24 @@ Out[]: array([-1.,  8., -1.])
 
 \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}
-\frametitle{Least Square Fit Curve}
-\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}}
+\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
@@ -535,11 +526,11 @@ Out[]: array([-1.,  8., -1.])
 \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]))
   ....
   ....
@@ -549,7 +540,7 @@ In []: for line in open('pendulum.txt'):
 \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}}
@@ -558,7 +549,7 @@ In []: t = array(t)
 \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}
@@ -586,13 +577,15 @@ In []: coef = result[0]
 \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}