Question

PYTHON

Given three unsorted lists, a name list (e.g. nameLst = [“Betty”, “Andrew”, “Zach”, “Cathy”, …]), an Exam1 score list (e.g. score1 = [50, 45, 48, 48, …]), and an Exam 2 score list (e.g. score2=[35, 48, 50, 37, …])

Write the following functions:

1. def makeLst (aNameLst, e1ScoreLst, e2ScoreLst) :

#the function will return a list of tuples as in the following form

#[(“Betty”, (50,35)), (“Andrew”, (45,48)), (“Zach”, (48,50)), (“Cathy”, (48,37)), …]

2. def personalAverage(aList) :

#takes a class list in the form of [(“Betty”, (50,35)), (“Andrew”, (45,48)), …]

#returns another class list in the form of [(“Betty”, 42.5), (“Andrew”, 46.5), …]

3. def sortByName (bList) :

#takes a class list in the form of [(“Betty”, 42.5), (“Andrew”, 46.5), …]

#returns another class list in the form of [(“Andrew”, 46.5), (“Betty”, 42.5), …]

#i.e. sort by name in ascending order

#note: you may call predefined sort() or sorted() in function body

4. def sortByAveScore (bList) :

#takes a class list in the form of [(“Betty”, 42.5), (“Andrew”, 46.5), …]

#returns another class list in the form of [(“Zach”, 48.0), (“Andrew”, 46.5), …]

#i.e. sort by average score in descending order

5. def classMean(aList) :

#takes a class list in the form of [(“Betty”, (50,35)), (“Andrew”, (45,48)), …]

#returns exam1MeanScore and exam2MeanScore for the class

6. def classMedian (aList) :

#similar to (5) but returns the exam1MedianScore and exam2MedianScore

7. def classStdDev (aList) :

#similar to (5) but returns the standard deviation for Exam1 and Exam2 respectively

8. Write a main function that performs the following tasks and then call the main function to run.

Tasks in main function:

1. Read in or assigned three lists; if lengths of the three lists not equal, reject it (either terminate the program or ask user to re-enter if data sets are read in.)
2. Test all of the above functions using the following data sets:
3. nameLst = [“Betty”, “Andrew”, “Zach”, “Cathy”, “Jay”, “Kevin”, “Cassie”, “Matt”, “Lux”, “Xavier”, “Peter”, “John”, “Jamie”, “Joe”, “Ellen”, “Nancy”, “Ray”, “Bryce”, “Gordon”, “Gee”]

score1 = [50, 45, 48, 48, 46, 49, 35, 38, 42, 50, 44, 37, 35, 32, 48, 32, 44, 45, 41, 46]

score2 = [35, 48, 50, 37, 45, 41, 47, 50, 47, 41, 47, 48, 46, 49, 41, 38, 46, 42, 33, 39]

4. Test all of the above functions using the following data sets:

nameLst = [], score1=[], score2=[]

5. Perform one more test using the following data sets:

nameLst = [‘A’, ‘D’, ‘W’, ‘K’, ‘Z’]

score1 = [30, 20, 50]

score2 = [50, 30, 25, 49, 23, 42]

Special notes:

(1)        you may call predefined sort() or sorted() in the body of any of the above functions

(2)        you cannot use the statistics module

Answer 1

Python code to perform the given tasks

import math

def makeLst(aNameLst,e1ScoreLst,e2ScoreLst):#task1
if len(aNameLst) == 0 :
alist = []
else:
a = list(zip(e1ScoreLst,e2ScoreLst))
alist = list(zip(aNameLst,a))
return alist

def personalAverage(alist):#task2
if len(alist) == 0:
blist = []
else:
k = list(zip(*alist))
names = list(k[0])
b = list(k[1])
avg = []
for x in b:
avg.append((x[0]+x[1])/2)
blist = list(zip(names,avg))
return blist

def sortByName(blist):#task3
if len(blist) == 0:
h = []
else:
h = sorted(blist)
return h

def sortByAveScore(blist):#task4
if len(blist)==0:
k = []
else:
k = blist
k.sort(key = lambda x: x[1],reverse = True)
#print(k)
return k

def classMean(alist):#task5
if len(alist) == 0:
e1_mean = 'Nan'
e2_mean = 'Nan'
else:
k = list(zip(*alist))
names = list(k[0])
b = list(k[1])
e = list(zip(*b))
e1 = list(e[0])
e2 = list(e[1])
e1_mean = sum(e1)/len(e1)
e2_mean = sum(e2)/len(e2)
return e1_mean,e2_mean

def classMedian(alist):#task6
if len(alist)==0:
median1 = 'Nan'
median2 = 'Nan'
else:
k = list(zip(*alist))
names = list(k[0])
b = list(k[1])
e = list(zip(*b))
e1 = list(e[0])
e2 = list(e[1])
e1.sort()
e2.sort()
n1 = len(e1)
if n1 % 2 == 0:
m1 = e1[n1//2]
m2 = e1[n1//2 - 1]
median1 = (m1+m2)/2
else:
median1 = e1[n1//2]
n2 = len(e2)
if n2 % 2 == 0:
h1 = e2[n2//2]
h2 = e2[n2//2 - 1]
median2 = (h1+h2)/2
else:
median2 = e2[n2//2]
return median1,median2

def classStdDev(alist):3task7
if len(alist)==0:
std1 = 'Nan'
std2 = 'Nan'
else:
k = list(zip(*alist))
names = list(k[0])
b = list(k[1])
e = list(zip(*b))
e1 = list(e[0])
e2 = list(e[1])
e1_mean = sum(e1)/len(e1)
e2_mean = sum(e2)/len(e2)
v1 = 0
for x in e1:
v1 += (x-e1_mean)**2
v1 /= len(e1)
std1 = math.sqrt(v1)
v2 = 0
for x in e2:
v2 += (x-e2_mean)**2
v2 /= len(e2)
std2 = math.sqrt(v2)
return std1,std2

def main():
def call(nameLst,score1,score2):
if len(nameLst) == len(score1) == len(score2):
mlist = makeLst(nameLst,score1,score2)
avglist = personalAverage(mlist)
slist = sortByName(avglist)
salist = sortByAveScore(avglist)
print(classMean(mlist))
print(classMedian(mlist))
print(classStdDev(mlist))
else:
print("Please re-enter the data..")
#TestCase1
nameLst = ["Betty", "Andrew", "Zach", "Cathy", "Jay", "Kevin", "Cassie", "Matt", "Lux", "Xavier", "Peter", "John", "Jamie", "Joe", "Ellen", "Nancy", "Ray", "Bryce", "Gordon", "Gee"]
score1 = [50, 45, 48, 48, 46, 49, 35, 38, 42, 50, 44, 37, 35, 32, 48, 32, 44, 45, 41, 46]
score2 = [35, 48, 50, 37, 45, 41, 47, 50, 47, 41, 47, 48, 46, 49, 41, 38, 46, 42, 33, 39]
print("TEST CASE 1")
call(nameLst,score1,score2)

#TestCase2
nameLst = []
score1 = []
score2 = []
print("TEST CASE 2")
call(nameLst,score1,score2)

#TestCase3
nameLst = ['A', 'D', 'W', 'K', 'Z']
score1 = [30, 20, 50]
score2 = [50, 30, 25, 49, 23, 42]
print("TEST CASE 3")
call(nameLst,score1,score2)

if __name__ == '__main__':
main()
Screenshots for the above code:

Output of the program:

Note - Values nan for mean ,median and standard deviation are returnd when there is no data,that means when scores and names lists are empty.

Can check the accuracy of the mean,median and standard deviation for the data in a list using numpy built in functions as follows:

import numpy as np

a = [50, 45, 48, 48, 46, 49, 35, 38, 42, 50, 44, 37, 35, 32, 48, 32, 44, 45, 41, 46]

np.mean(a)#mean

np.median(a)#median

np.std(a)#standard deviation