# HG changeset patch # User Puneeth Chaganti # Date 1255013535 -19800 # Node ID 1c1d6aaa2be33ac2f2609cfb3753e43056a734e4 # Parent d0679d2f9339c75645e7fe9b543cfd673d9787be Added first cut of Day2 handout; Minor edits to sessions 2 and 3. diff -r d0679d2f9339 -r 1c1d6aaa2be3 day2/handout.tex --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/day2/handout.tex Thu Oct 08 20:22:15 2009 +0530 @@ -0,0 +1,423 @@ +\documentclass[12pt]{article} +\usepackage{amsmath} +\title{Python Workshop\\Problems and Exercises} +\author{Asokan Pichai\\Prabhu Ramachandran} +\begin{document} +\maketitle + +\section{Matrices and Arrays \& 2D Plotting} +\subsection{Matrices and Arrays} +\begin{verbatim} +# Simple array math example +>>> import numpy as np +>>> a = np.array([1,2,3,4]) +>>> b = np.arange(2,6) +>>> b +array([2,3,4,5]) +>>> a*2 + b + 1 # Basic math! +array([5, 8, 11, 14]) + +# Pi and e are defined. +>>> x = np.linspace(0.0, 10.0, 1000) +>>> x *= 2*np.pi/10 # inplace. +# apply functions to array. +>>> y = np.sin(x) +>>> z = np.exp(y) + +>>> x = np.array([1., 2, 3, 4]) +>>> np.size(x) +4 +>>> x.dtype # What is a.dtype? +dtype('float64') +>>> x.shape +(4,) +>>> print np.rank(x), x.itemsize +1 8 +>>> x[0] = 10 +>>> print x[0], x[-1] +10.0 4.0 + +>>> a = np.array([[ 0, 1, 2, 3], +... [10,11,12,13]]) +>>> a.shape # (rows, columns) +(2, 4) +# Accessing and setting values +>>> a[1,3] +13 +>>> a[1,3] = -1 +>>> a[1] # The second row +array([10,11,12,-1]) + +>>> b=np.array([[0,2,4,2],[1,2,3,4]]) +>>> np.add(a,b,a) +>>> np.sum(x,axis=1) + +>>> np.greater(a,4) +>>> np.sqrt(a) + +>>> np.array([2,3,4]) +array([2, 3, 4]) + +>>> np.linspace(0, 2, 4) +array([0.,0.6666667,1.3333333,2.]) + +>>>np.ones([2,2]) +array([[ 1., 1.], + [ 1., 1.]]) + +>>>a = np.array([[1,2,3],[4,5,6]]) +>>>np.ones_like(a) +array([[1, 1, 1], + [1, 1, 1]]) + +>>> a = np.array([[1,2,3], [4,5,6], + [7,8,9]]) +>>> a[0,1:3] +array([2, 3]) +>>> a[1:,1:] +array([[5, 6], + [8, 9]]) +>>> a[:,2] +array([3, 6, 9]) +>>> a[...,2] +array([3, 6, 9]) + +>>> a[0::2,0::2] +array([[1, 3], + [7, 9]]) +# Slices are references to the +# same memory! + +>>> np.random.rand(3,2) +array([[ 0.96276665, 0.77174861], + [ 0.35138557, 0.61462271], + [ 0.16789255, 0.43848811]]) +>>> np.random.randint(1,100) +42 +\end{verbatim} + +\subsection{Problem Set} +\begin{verbatim} + >>> from scipy import misc + >>> A=misc.imread(name) + >>> misc.imshow(A) +\end{verbatim} +\begin{enumerate} + \item Convert an RGB image to Grayscale. $ Y = 0.5R + 0.25G + 0.25B $. + \item Scale the image to 50\%. + \item Introduce some random noise. + \item Smooth the image using a mean filter. + \\\small{Each element in the array is replaced by mean of all the neighbouring elements} + \\\small{How fast does your code run?} +\end{enumerate} + +\subsection{2D Plotting} +\begin{verbatim} +$ ipython -pylab +>>> x = linspace(0, 2*pi, 1000) +>>> plot(x, sin(x)) +>>> plot(x, sin(x), 'ro') +>>> xlabel(r'$\chi$', color='g') +# LaTeX markup! +>>> ylabel(r'sin($\chi$)', color='r') +>>> title('Simple figure', fontsize=20) +>>> savefig('/tmp/test.eps') + +# Set properties of objects: +>>> l, = plot(x, sin(x)) +# Why "l,"? +>>> setp(l, linewidth=2.0, color='r') +>>> l.set_linewidth(2.0) +>>> draw() # Redraw. +>>> setp(l) # Print properties. +>>> clf() # Clear figure. +>>> close() # Close figure. + +>>> w = arange(-2,2,.1) +>>> plot(w,exp(-(w*w))*cos) +>>> ylabel('$f(\omega)$') +>>> xlabel('$\omega$') +>>> title(r"$f(\omega)=e^{-\omega^2} + cos({\omega^2})$") +>>> annotate('maxima',xy=(0, 1), + xytext=(1, 0.8), + arrowprops=dict( + facecolor='black', + shrink=0.05)) + +>>> x = linspace(0, 2*np.pi, 1000) +>>> plot(x, cos(5*x), 'r--', + label='cosine') +>>> plot(x, sin(5*x), 'g--', + label='sine') +>>> legend() +# Or use: +>>> legend(['cosine', 'sine']) + +>>> figure(1) +>>> plot(x, sin(x)) +>>> figure(2) +>>> plot(x, tanh(x)) +>>> figure(1) +>>> title('Easy as 1,2,3') + +\end{verbatim} + +\subsection{Problem Set} + \begin{enumerate} + \item Write a function that plots any regular n-gon given n. + \item Consider the logistic map, $f(x) = kx(1-x)$, plot it for + $k=2.5, 3.5$ and $4$ in the same plot. + + \item Consider the iteration $x_{n+1} = f(x_n)$ where $f(x) = + kx(1-x)$. Plot the successive iterates of this process. + \item Plot this using a cobweb plot as follows: + \begin{enumerate} + \item Start at $(x_0, 0)$ + \item Draw line to $(x_i, f(x_i))$; + \item Set $x_{i+1} = f(x_i)$ + \item Draw line to $(x_i, x_i)$ + \item Repeat from 2 for as long as you want + \end{enumerate} +\end{enumerate} + +\section{Advanced Numpy} + +\subsection{Broadcasting} +\begin{verbatim} +>>> a = np.arange(4) +>>> b = np.arange(5) +>>> a+b +>>> a+3 +>>> c=np.array([3]) +>>> a+c +>>> b+c + +>>> a = np.arange(4) +>>> a+3 +array([3, 4, 5, 6]) + +>>> x = np.ones((3, 5)) +>>> y = np.ones(8) +>>> (x[..., None] + y).shape +(3, 5, 8) + +\end{verbatim} + +\subsection{Copies \& Views} +\begin{verbatim} +>>> a = np.array([[1,2,3],[4,5,6]]) +>>> b = a +>>> b is a +>>> b[0,0]=0; print a +>>> c = a.view() +>>> c is a +>>> c.base is a +>>> c.flags.owndata +>>> d = a.copy() +>>> d.base is a +>>> d.flags.owndata + +>>> a = np.arange(1,9) +>>> a.shape=3,3 +>>> b = a[0,1:3] +>>> c = a[0::2,0::2] +>>> a.flags.owndata +>>> b.flags.owndata +>>> b.base +>>> c.base is a + +>>> b = a[np.array([0,1,2])] +array([[1, 2, 3], + [4, 5, 6], + [7, 8, 9]]) +>>> b.flags.owndata +>>> abool=np.greater(a,2) +>>> c = a[abool] +>>> c.flags.owndata + +\end{verbatim} + +\section{Scipy} +\subsection{Linear Algebra} +\begin{verbatim} +>>> import scipy as sp +>>> from scipy import linalg +>>> A=sp.mat(np.arange(1,10)) +>>> A.shape=3,3 +>>> linalg.inv(A) +>>> linalg.det(A) +>>> linalg.norm(A) +>>> linalg.expm(A) #logm +>>> linalg.sinm(A) #cosm, tanm, ... + +>>> A = sp.mat(np.arange(1,10)) +>>> A.shape=3,3 +>>> linalg.lu(A) +>>> linalg.eig(A) +>>> linalg.eigvals(A) +\end{verbatim} + +\subsection{Solving Linear Equations} + +\begin{align*} + 3x + 2y - z & = 1 \\ + 2x - 2y + 4z & = -2 \\ + -x + \frac{1}{2}y -z & = 0 +\end{align*} +To Solve this, +\begin{verbatim} +>>> A = sp.mat([[3,2,-1],[2,-2,4] + ,[-1,1/2,-1]]) +>>> B = sp.mat([[1],[-2],[0]]) +>>> linalg.solve(A,B) +\end{verbatim} + +\subsection{Integrate} +\subsubsection{Quadrature} +Calculate the area under $(sin(x) + x^2)$ in the range $(0,1)$ +\begin{verbatim} +>>> def f(x): + return np.sin(x)+x**2 +>>> integrate.quad(f, 0, 1) +\end{verbatim} + +\subsubsection{ODE Integration} +Numerically solve ODEs\\ +\begin{align*} +\frac{dx}{dt} &=-e^{-t}x^2\\ + x(0) &=2 +\end{align*} +\begin{verbatim} +>>> def dx_dt(x,t): +... return -np.exp(-t)*x**2 +>>> x=integrate.odeint(dx_dt, 2, t) +>>> plt.plot(x,t) +\end{verbatim} + +\subsection{Interpolation} +\subsubsection{1D Interpolation} +\begin{verbatim} +>>> from scipy import interpolate +>>> interpolate.interp1d? +>>> x = np.arange(0,2*np.pi,np.pi/4) +>>> y = np.sin(x) +>>> fl = interpolate.interp1d( + x,y,kind='linear') +>>> fc = interpolate.interp1d( + x,y,kind='cubic') +>>> fl(np.pi/3) +>>> fc(np.pi/3) +\end{verbatim} + +\subsubsection{Splines} +Plot the Cubic Spline of $sin(x)$ +\begin{verbatim} +>>> x = np.arange(0,2*np.pi,np.pi/4) +>>> y = np.sin(x) +>>> tck = interpolate.splrep(x,y) +>>> X = np.arange(0,2*np.pi,np.pi/50) +>>> Y = interpolate.splev(X,tck,der=0) +>>> plt.plot(x,y,'o',x,y,X,Y) +>>> plt.show() +\end{verbatim} + +\subsection{Signal \& Image Processing} +Applying a simple median filter +\begin{verbatim} +>>> from scipy import signal, ndimage +>>> from scipy import lena +>>> A=lena().astype('float32') +>>> B=signal.medfilt2d(A) +>>> imshow(B) +\end{verbatim} +Zooming an array - uses spline interpolation +\begin{verbatim} +>>> b=ndimage.zoom(A,0.5) +>>> imshow(b) +\end{verbatim} + +\section{3D Data Visualization} +\subsection{Using mlab} +\begin{verbatim} +>>> from enthought.mayavi import mlab + +>>> mlab.test_ +>>> mlab.test_contour3d() +>>> mlab.test_contour3d?? +\end{verbatim} + +\subsubsection{Plotting Functions} +\begin{verbatim} +>>> from numpy import * +>>> t = linspace(0, 2*pi, 50) +>>> u = cos(t)*pi +>>> x, y, z = sin(u), cos(u), sin(t) + +>>> x = mgrid[-3:3:100j,-3:3:100j] +>>> z = sin(x*x + y*y) +>>> mlab.surf(x, y, z) +\end{verbatim} + +\subsubsection{Large 2D Data} +\begin{verbatim} +>>> mlab.mesh(x, y, z) + +>>> phi, theta = numpy.mgrid[0:pi:20j, +... 0:2*pi:20j] +>>> x = sin(phi)*cos(theta) +>>> y = sin(phi)*sin(theta) +>>> z = cos(phi) +>>> mlab.mesh(x, y, z, +... representation= +... 'wireframe') +\end{verbatim} + +\subsubsection{Large 3D Data} +\begin{verbatim} +>>> x, y, z = ogrid[-5:5:64j, +... -5:5:64j, +... -5:5:64j] +>>> mlab.contour3d(x*x*0.5 + y*y + + z*z*2) + +>>> mlab.test_quiver3d() +\end{verbatim} + +\subsection{Motivational Problem} +Atmospheric data of temperature over the surface of the earth. Let temperature ($T$) vary linearly with height ($z$)\\ +$T = 288.15 - 6.5z$ + +\begin{verbatim} +lat = linspace(-89, 89, 37) +lon = linspace(0, 360, 37) +z = linspace(0, 100, 11) +\end{verbatim} + +\begin{verbatim} +x, y, z = mgrid[0:360:37j,-89:89:37j, + 0:100:11j] +t = 288.15 - 6.5*z +mlab.contour3d(x, y, z, t) +mlab.outline() +mlab.colorbar() +\end{verbatim} + +\subsection{Lorenz equation} +\begin{eqnarray*} +\frac{d x}{dt} &=& s (y-x)\\ +\frac{d y}{d t} &=& rx -y -xz\\ +\frac{d z}{d t} &=& xy - bz\\ +\end{eqnarray*} + +Let $s=10,$ +$r=28,$ +$b=8./3.$ + +\begin{verbatim} +x, y, z = mgrid[-50:50:20j,-50:50:20j, + -10:60:20j] + +\end{verbatim} + +\end{document} diff -r d0679d2f9339 -r 1c1d6aaa2be3 day2/session2.tex --- a/day2/session2.tex Thu Oct 08 19:10:53 2009 +0530 +++ b/day2/session2.tex Thu Oct 08 20:22:15 2009 +0530 @@ -324,7 +324,7 @@ \item Integrating Functions given fixed samples \item Numerical integrators of ODE systems \end{itemize} - Calculate $\int^1_0(sin(x) + x^2)dx$ + Calculate the area under $(sin(x) + x^2)$ in the range $(0,1)$ \begin{lstlisting} >>> def f(x): return np.sin(x)+x**2 @@ -340,11 +340,11 @@ x(0)&=2 \end{align*} \begin{lstlisting} - def dx_dt(x,t): +>>> def dx_dt(x,t): return -np.exp(-t)*x**2 - x=integrate.odeint(dx_dt, 2, t) - plt.plot(x,t) +>>> x=integrate.odeint(dx_dt, 2, t) +>>> plt.plot(x,t) \end{lstlisting} \inctime{10} \end{frame} @@ -369,15 +369,15 @@ \begin{frame}[fragile] \frametitle{Interpolation - Splines} - Cubic Spline of $sin(x)$ + Plot the Cubic Spline of $sin(x)$ \begin{lstlisting} - x = np.arange(0,2*np.pi,np.pi/4) - y = np.sin(x) - tck = interpolate.splrep(x,y) - X = np.arange(0,2*np.pi,np.pi/50) - Y = interpolate.splev(X,tck,der=0) - plt.plot(x,y,'o',x,y,X,Y) - plt.show() +>>> x = np.arange(0,2*np.pi,np.pi/4) +>>> y = np.sin(x) +>>> tck = interpolate.splrep(x,y) +>>> X = np.arange(0,2*np.pi,np.pi/50) +>>> Y = interpolate.splev(X,tck,der=0) +>>> plt.plot(x,y,'o',x,y,X,Y) +>>> plt.show() \end{lstlisting} \inctime{10} \end{frame} @@ -403,16 +403,16 @@ \frametitle{Signal \& Image Processing} Applying a simple median filter \begin{lstlisting} - from scipy import signal, ndimage - from scipy import lena - A=lena().astype('float32') - B=signal.medfilt2d(A) - imshow(B) +>>> from scipy import signal, ndimage +>>> from scipy import lena +>>> A=lena().astype('float32') +>>> B=signal.medfilt2d(A) +>>> imshow(B) \end{lstlisting} Zooming an array - uses spline interpolation \begin{lstlisting} - b=ndimage.zoom(A,0.5) - imshow(b) +>>> b=ndimage.zoom(A,0.5) +>>> imshow(b) \end{lstlisting} \inctime{5} \end{frame} diff -r d0679d2f9339 -r 1c1d6aaa2be3 day2/session3.tex --- a/day2/session3.tex Thu Oct 08 19:10:53 2009 +0530 +++ b/day2/session3.tex Thu Oct 08 20:22:15 2009 +0530 @@ -192,7 +192,7 @@ \inctime{10} \end{frame} -\section{Tools at your disposal:} +\section{Tools at your disposal} \subsection{Mayavi2.0} @@ -280,7 +280,7 @@ \end{frame} \begin{frame}[fragile] - \frametitle{Using mlab:} + \frametitle{Using mlab} \begin{lstlisting} >>> from enthought.mayavi import mlab @@ -288,7 +288,7 @@ \vspace*{0.5in} - \myemph{\Large Try these:} + \myemph{\Large Try these} \vspace*{0.25in}