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Mathematical Operations in Python

print("***** QUESTION 1 *****\n"); def factorial(n): """ Recursively calculat the factorial of n """ # Base case if n == 1: return 1 return n*factorial(n-1) def calnWk(n, k): return n**k def calnPk(n ,k): return factorial(n)/factorial(n-k) def calnCk(n, k): return calnPk(n, k)/factorial(k) print("\n***** QUESTION 2 *****\n"); # Use built-in functions to get the coefficient from sympy import * x, y = symbols('x y') exp = (2*x**2-4*y**3)**30 coeff_real = exp.expand().coeff(x**20).coeff(y**60) print("The coefficient calculated using built-in function for x^20*y^60 is " + str(coeff_real)) print("\n***** QUESTION 3 *****\n"); # Now get the coefficient using the functions from Part 1 # The coefficient of x^90 *y^10 can be calculated as a combination, selecting the element 90 from the 100 available elements # Coefficients of x and y: (ax^c + by^d)^n a = 2 b = -4 c = 2 d = 3 n = 30 # k = 20 # coeff_calc = calnCk(n,k)*a**(n-k) *b**k print("The coefficient calculated using the functions from Part 1 for for x^20*y^60 is: " + str(coeff_calc)) """ print("\n***** QUESTION 4 *****\n"); # Create a random 3x3 matrix import numpy as np A = np.random.uniform(low = 0, high = 1, size=(3,3)) # matrix of random integers from 1 to 10 b = np.matrix([[1],[0],[0]]) # Check if Ax = b has a solution print("Matrix A is:\n" + str(A)) print("\nVector b is:\n" + str(b)) x = np.linalg.solve(A,b) print("\nThe solution x for Ax = b is:\n" + str(x))

Text parsing in Python

Number of users = 500 Number of users with email id ending with .jp = 26 Number of users with email id ending with .uk = 20 Number of users with email id ending with .de = 8 Surnames that have duplicattes are ['Culbard', 'De', 'Schiell', 'Wisbey'] import csv users=0 first=True jp=0 uk=0 de=0 last_names=[] duplicates=[] print("Hello user.Welcome.") print("Analyzing mydata.csv file") with open('mydata.csv','r') as dataFile: reader=csv.reader(dataFile) for row in reader: if not first: users+=1 if row[1]: if row[1] in last_names: if row[1] not in duplicates: duplicates.append(row[1]) else: last_names.append(row[1]) if row[2]: if row[2][-3:]=='.jp': jp+=1 elifrow[2][-3:]=='.uk': uk+=1 elifrow[2][-3:]=='.de': de+=1 first=False dataFile.close() print("Completed analyzing") print("\nNumber of users = "+str(users)) print("Number of users with email id ending with .jp = "+str(jp)) print("Number of users with email id ending with .uk = "+str(uk)) print("Number of users with email id ending with .de = "+str(de)) print("Surnames that have duplicattes are "+str(duplicates)) f=open("Janet_results.txt","w") f.write("Number of users = "+str(users)) f.write("\nNumber of users with email id ending with .jp = "+str(jp)) f.write("\nNumber of users with email id ending with .uk = "+str(uk)) f.write("\nNumber of users with email id ending with .de = "+str(de)) f.write("\nSurnames that have duplicattes are "+str(duplicates)) f.close() print("\nThe results are saved in Janet_results.txt file.") print("Thank you for using the program.Good Bye.")