From 7b477df3d2285020a657193ed92ddccb6f02a193 Mon Sep 17 00:00:00 2001
From: Puneeth Chaganti
Date: Thu, 8 Oct 2009 20:22:15 +0530
Subject: Added first cut of Day2 handout; Minor edits to sessions 2 and 3.

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+\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_<TAB>
+>>> 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}
-- 
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