julius-datajunkie

import unittest

def SumMultiplesOfThreeOrFive(number):
    return sum(i for i in range(1,number) if (i % 3 == 0 or i % 5 == 0))
    #for i in range(1,number):
    #    if (i % 3 == 0 or i % 5 == 0):
    #       sum += i
    #return sum

class TestSumMultiplesOfThreeOrFive(unittest.TestCase):

julius-datajunkie

import unittest

def EvenFibonacciNumbers(limit):
    #Find all even fibonacci numbers
    fibonacciSeries = FibonacciGenerator(limit)
    return sum(i for i in fibonacciSeries if i % 2 == 0)
    
def FibonacciGenerator(limit):
    a = 0
    b = 1

julius-datajunkie

import unittest
from math import sqrt
def isPrime(i):
    if (i == 2 or i == 3):
        return True
    else:
        for j in range(3, int(sqrt(i)+1)):
            if ( i  % j == 0):
                return False
    return True

julius-datajunkie

import unittest

def LargestIntDividesEleven(numberOfDigits):
    largestInt = 10**(numberOfDigits)-1
    return largestInt - (largestInt % 11)

def reverse(number): 
    ''' 
    This method does not speed things up, but it does provide a rather 
    interesting way of reversing any integers without converting it to 

julius-datajunkie

import unittest
def LongestZeroSequence(n):
    '''
    Input: n is a string of a binary sequence
    Output: the length of the longest consecutive zero sequence
    when encounter 1, stop counting and save the max sequence so far.
    output the max sequence
    ''' 
    maxSequence = 0       
    seq = 0

julius-datajunkie

# Sieve of Eratosthenes
# Code by David Eppstein, UC Irvine, 28 Feb 2002
# http://code.activestate.com/recipes/117119/

def PrimesGenerator():
    """ 
    Generate an infinite sequence of prime numbers.
    """
    # Maps composites to primes witnessing their compositeness.
    # This is memory efficient, as the sieve is not "run forward"

julius-datajunkie

import numpy as np
import matplotlib.pyplot as plt

def drawdown2(series):
	MDD = 0
	peak = -99999
	length = len(series)
	DD = np.zeros(length)
	for i in range(length):
		if (series[i] > peak): # peak will be the maximum value seen so far (0 to i)