olooney

'''A simple wrapper around BeautifulSoup to quickly find the URLs of all resources referenced by an HTML page.'''

import urllib2
import urlparse
import BeautifulSoup

def make_soup(url):
    page = urllib2.urlopen(url)
    contents = page.read()
    soup = BeautifulSoup.BeautifulSoup(contents)

olooney

import poplib, email, getpass

M = poplib.POP3_SSL('pop.gmail.com', 995)
username = raw_input("gmail username: ")
M.user(username)
M.pass_(getpass.getpass('%s@gmail.com password: ' % username))
status, emails, octets = M.list()

if status.startswith('+OK'):
    print 'new emails:', len(emails)

olooney

"""Solves the 8-queens problem (and its N-queens generalization.)"""

# It is clear that since no two queens can occupy the same column, and there are
# as many columns as queens to place, then each column must contain one and only one
# queen.  The only question is, on which row to place each queen?

# A "left-board" is a way to represent a partial solution to the n-queens problem.
# It is simply a list of integers, each indicating the row occupied by the queen
# in each column. Because it is a partial solution, it can have fewer than 8 elements,
# or even be empty. For example, [0, 2] indicates a that the queen in the left-most

olooney

def collapse_intervals(intervals):
    '''intervals is an iterable of 2-tuples representing the start and end of
    closed intervals.  returns a list of 2-tuples after collapsing any
    overlapping intervals. For example:

        collapse_intervals([(1,5), (9,10), (4,7)]) -> [(1,7), (9,10)]

    The algorithm doesn't care if intervals are out of order, or if two points
    of the same interval are out of order. It also doesn't require the points

olooney

 cat ~/.ssh/id_rsa.pub | ssh $USER@$SERVER "umask 077; test -d ~/.ssh || mkdir ~/.ssh ; cat >> ~/.ssh/authorized_keys"

olooney

# http://en.wikipedia.org/wiki/RSA_(algorithm)

p = 61
q = 53
n = p*q
totient = (p-1)*(q-1)

# here, e is 2^4 + 1; in real world applications, e is often 2^16 + 1
e = 17

olooney

"""
Example using Python's itertools and generators to implement the classic FizzBuzz problem using pythonic iterators.


Output looks like this:

	$ python fizz_buzz_with_iterator.py
	1
	2
	Fizz

olooney

#! /usr/bin/env python
"""
Simple wrapper around Github's gist HTTP API.

"""
import argparse
import logging
import requests
import json
from getpass import getuser, getpass

olooney

logs/

olooney

arima.simulation <- function(linear=1, exponential=0, seasonal=0, season=25, random=1, ahead=10) {
  sim.data <- (1:100 * linear/100) + exponential*exp( (1:100/100))/exp(1) + seasonal*sin(1:100 * 2 * pi / season ) + random*rnorm(100);
  sims <- ts(sim.data, frequency=25)
  sim.model <- auto.arima(sims);
  sim.past.future <- c(as.vector(fitted(sim.model)),as.vector(forecast(sim.model, h=ahead)$mean));
  ylimits = c( min(sim.past.future), max(sim.past.future))
  
  plot(sim.data, pch=20, col=rgb(0,0,1, alpha=0.5), ylim=ylimits, xlim=c(0,length(sim.past.future)));
  lines(seq_along(sim.past.future), sim.past.future, col=rgb(0,1,0, alpha=0.75));
  return(sim.model);