from numpy import * import pylab x = [0, 0]; A = [ [.5, 0], [0, .5] ]; b1 = [0, 0]; b2 = [.5, 0]; b3 = [.25, sqrt(3)/4]; pylab.ion() # animation on #Note the comma after line. This is placed here because plot returns a list of lines that are drawn. line, = pylab.plot(x[0],x[1],'m.',markersize=6) pylab.axis([0,1,0,1]) data1 = [] data2 = [] iter = 0 while True: r = fix(random.rand()*3) if r==0: x = dot(A,x)+b1 if r==1: x = dot(A,x)+b2 if r==2: x = dot(A,x)+b3 data1.append(x[0]) data2.append(x[1]) line.set_xdata(data1) # update the data line.set_ydata(data2) pylab.draw() # draw the points again iter += 1 print iterThis is the result:
Thursday, June 30, 2011
Animation with matplotlib, bringin the Sierpinski triangle to life
Few post ago we have seen how to plot the Sierpinski triangle. This post is an update that shows how to create an animation where at each step a new point of the fractal is plotted:
Tuesday, June 28, 2011
Searching for IP address using regular expression
This snippet finds an IP address in a string using regex:
import re ip = re.compile('(([2][5][0-5]\.)|([2][0-4][0-9]\.)|([0-1]?[0-9]?[0-9]\.)){3}' +'(([2][5][0-5])|([2][0-4][0-9])|([0-1]?[0-9]?[0-9]))') match = ip.search("Your ip address is 192.168.0.1, have fun!") if match: print 'IP address found:', print match.group(), # matching substring print 'at position',match.span() # indexes of the substring found else: print 'IP address not found'The output will be
IP address found: 192.168.0.1 at position (19, 30)
Thursday, June 9, 2011
Crawling the web with SGMLParser
In this example we will use SGMLParser in order to build a simple web crawler.
import urllib from random import choice from sgmllib import SGMLParser class LinkExplorer(SGMLParser): def reset(self): SGMLParser.reset(self) self.links = [] # list with the urls def start_a(self, attrs): """ fill the links with the links in the page """ for k in attrs: if k[0] == 'href' and k[1].startswith('http'): self.links.append(k[1]) def explore(parser,s_url,maxvisit=10,iter=0): """ pick a random link in the page s_url and follow its links recursively """ if iter < maxvisit: # it will stop after maxvisit iteration print '(',iter,') I am in',s_url usock = urllib.urlopen(s_url) # download the page parser.reset() # reset the list parser.feed(usock.read()) # parse the current page if len(parser.links) > 0: explore(parser,choice(parser.links),maxvisit,iter+1) else: # if the page has no links to follow print 'the page has no links' # test the crawler starting from the python's website parser = LinkExplorer() explore(parser,"http://www.python.org/")Let's go!
( 0 ) I am in http://www.python.org/ ( 1 ) I am in http://wiki.python.org/moin/NumericAndScientific ( 2 ) I am in http://numpy.scipy.org/ ( 3 ) I am in http://sphinx.pocoo.org/ ( 4 ) I am in http://www.bitbucket.org/birkenfeld/sphinx/issues/ ( 5 ) I am in http://blog.bitbucket.org ( 6 ) I am in http://haproxy.1wt.eu/ ( 7 ) I am in http://www.olivepeak.com/blog/posts/read/free-your-port-80-with-haproxy ( 8 ) I am in http://www.olivepeak.com/peaknotes/ ( 9 ) I am in http://notes.olivepeak.com/account/create
Tuesday, June 7, 2011
Sierpinski fractal
As described in the introduction of Numerical Computing in Matlab we can generate fractals using affine transformations. The following script uses it to generate the famous Sierpinski triangle:
from numpy import * import pylab x = [0, 0]; A = [ [.5, 0], [0, .5] ]; b1 = [0, 0]; b2 = [.5, 0]; b3 = [.25, sqrt(3)/4]; for i in range(3000): # 3000 points will be generated r = fix(random.rand()*3) if r==0: x = dot(A,x)+b1 if r==1: x = dot(A,x)+b2 if r==2: x = dot(A,x)+b3 pylab.plot(x[0],x[1],'m.',markersize=2) pylab.show()It will take a while, here's the result
Friday, June 3, 2011
The Collatz conjecture
The Collatz conjecture is a famous unsolved problem in number theory. Start with any
positive integer n. Repeat the following steps:
The unanswered question is, does the process always terminate?
Let to see a script that generate the sequence of number involved by the algorithm:
The Collatz conjecture is is also known as the 3n + 1 conjecture, the Ulam conjecture, Kakutani's problem, the Thwaites conjecture, Hasse's algorithm, or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers, or as wondrous numbers.
References
positive integer n. Repeat the following steps:
- If n = 1, stop.
- If n is even, replace it with n/2.
- If n is odd, replace it with 3n + 1
The unanswered question is, does the process always terminate?
Let to see a script that generate the sequence of number involved by the algorithm:
import sys from pylab import * # take n from the command line n = int(sys.argv[1]) i = 0 old_n = n while n > 1: if n%2 == 0: # if n is even n = n/2 else: n = 3*n+1 i += 1; plot([i-1, i],[old_n,n],'-ob') old_n = n title('hailstone sequence for '+sys.argv[1]+', length is '+str(i)) show()The script will plot the sequence:
$ python collatz.py 25
The Collatz conjecture is is also known as the 3n + 1 conjecture, the Ulam conjecture, Kakutani's problem, the Thwaites conjecture, Hasse's algorithm, or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers, or as wondrous numbers.
References
Thursday, June 2, 2011
Integration of a function using quad from scipy
Example of integration of a function using scipy:
from scipy.integrate import quad f = lambda x: 4-x*x # function to integrate p = lambda x: (-x*x*x)/3 + 4*x # the primitive of f # let's try to integrate f in the interval [0 2] r = quad(f,0,2) # integration using quad r_analytic = p(2.0)-p(0.0) # analytic integration print 'Results' print r[0] print r_analyticNumerical and analytic result compared:
Results 5.33333333333 5.33333333333
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