# EECS 1015: Final programming exam (Lists, tuples, dictionaries, sets, assertions, classes, multi-file) solved

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EECS 1015:
TASK 1 – Word frequency in a song lyric [15 pnts]
Topics covered: variables, expressions, control structures, string processing, functions, and lists.
In the utilities.py file, there is a variable named lyrics that is bound to a long string that has some of
the words to the song “Happy.” Note that the string includes newline characters, commas, and periods.
In the main() function, you should pass the lyrics variable as an argument to the function task1(). Your
task is to extract the words from the lyrics string and count the number of times each word appears in the
string. Follow the procedure below. You are welcome to introduce additional functions. If you write
additional functions, write them in the final.py file (not in utilities.py).
Perform the following steps:
String pre-processing
(1) Convert the string passed to your task1() function to lower case
(2) Remove all commas and periods from the string.
Counting the words and printing out the information
(3) Split the string to find all the words.
(4) For each word, determine the number of times it appears.
(5) Print out each word, its count, and a string of * representing the count.
For example, if the string “the” appears three (3) times and the string “you” appears four (4) times, print:
the [3] ***
you [4] ****
Hint: When printing out the word from the lyrics, you can use “%10s” in the string formatting. This notation
will result in the string always taking ten (10) characters.
what [3] ***
along [4] ****
i’m [6] ******
she’s [1] *
the [3] ***
take [1] *
roof [1] *
….
you [7] *******
Note: There are many ways to perform this task. You may find using the “set” data type discussed in lecture
10 useful. Set can be used to find the unique items in a list. For example:
x = set([10,20,10,20,30]) -> x = {10,20,30}.
Also, remember that you can always convert a set to a list as follows:
x = list({10,20,30}) -> x = [10,20,30]
Page 3/10
Task 2 – “Invert” a dictionary’s keys and values [25 pnts]
Topics covered: variables, expressions, control structures, functions, lists and dictionaries.
In the utilities.py file, there is a variable named nameDict that is bound to a dictionary object. The
dictionary object’s keys are strings (representing a person’s name), and the value associated with each key is
the person’s age (e.g., “Abdullah”:18, “Su”:21, …, “Priya”:18).
In the main() function, you should pass the nameDict variable as an argument to the function Task2().
Task2 function should create a new dictionary that “inverts” the keys and values in nameDict. In particular,
your new dictionary will have keys that are the ages; the associated value for each key will be a list of names.
Once you have “inverted” the dictionary, you should do the following.
(1) Print out the new dictionary variable (this way, we can see you constructed it correctly)
(2) For each key in the new dictionary, print out the age and then each person’s name in the list as shown
below.
You are welcome to introduce additional functions. If you write additional functions, write them in the
final.py file (not in utilities.py).
{17: [‘Dirk’, ‘Sarah’, ‘Bosko’, ‘Cameron’], 18: [‘Mahesh’, ‘Abdullah’, ‘Franck’,
‘Xavier’, ‘Maxim’, ‘Priya’], 19: [‘Kai’, ‘Bailey’, ‘Ollie’, ‘Parsa’, ‘Ming’,
‘Javier’], 20: [‘Jol’, ‘Seonjoo’, ‘Beatrice’], 21: [‘Su’, ‘Abbo’, ‘Olivia’], 22:
[‘Pravel’, ‘Urzula’], 23: [‘Katja’], 24: [‘Svetlana’]}
Age 17
Dirk
Sarah
Bosko
Cameron
Age 18
Mahesh
Abdullah
Franck
Xavier
Maxim
Priya

Age 24
Svetlana
Page 4/10
Task 3 – Poor person’s graphics — draw a circle in a 20×20 raster (2D lists) [30 points]
Topics covered: variables, expressions, control structures, functions, and lists of lists, testing (assert)
In the utilities.py file, there is a variable named raster that is bound to a list of lists. The raster
variable has 20 lists, each list with 20 empty strings. This “list of lists” is effectively a 2D list that logically
makes up a 20×20 grid. In computer graphics, such grids are called a raster. Our simple drawing program will
place either a “*” or a ” ” (space) in each location of the 20×20 grid. This can be done by access the
element as follows: raster[5][10] = “*” would place a “*” at the 6th row and 11th column of the raster;
raster[3][3] = ” ” would place a ” ” at the 4th row and 4th column of the raster. Task 3 goal is to
draw a circle in the 20×20 raster based on a user-supplied circle radius between 1 and 9.
In the main() function, you should pass the raster variable as an argument to the function Task3().
Your function should work as follows:
(1) Ask the user for a radius from [1 to 9]. If the user inputs a value that is not 1-9, then raise an assertion
error with the following statement: “Radius must be between 1-9″.
(2) Loop through the 20×20 raster to draw the circle using the algorithm sketched out below:
loop through i -> 0 to 19
loop through j -> 0 to 19
at raster location i,j
# convert i,j to an x,y coordinate by
# subtracting 10 from the i,j
# (subtracting 10, centers the circle at 10,10)
x = i – 10
y = j – 10
if sqrt(x2 + y2) <= radius
place a ‘*’ in the i,j location on the raster
otherwise
place a ‘ ‘ in the i,j location on the raster
The sqrt() function from the math module has been imported for you in the final.py file.
For example:
Consider the yellow raster location (i=4 and j=6) with in the figure above. The radius of the circle to draw is 8.
Let’s determine if the raster location i=4 and j=6 should be a ” ” or a “*”?
x = 4-10 = -6
y = 6-10 = -4
sqrt(-6*-6 + -4*-4) = 7.22
7.22 is less than or equal to 8, so raster[4][6]=”*”
(3) After you draw the circle, ask the user if they would like to do it again?
See sample output on the next page.
Page 5/10
Example output for Task 3 (also see accompanying video)

*
*******
*********
*********
*********
***********
*********
*********
*********
*******
*

Try again: Y/N? y

*
*********
***********
*************
***************
*****************
*****************
*****************
*****************
*******************
*****************
*****************
*****************
*****************
***************
*************
***********
*********
*
Try again: Y/N?
Page 6/10
Task 4 – Robust statistics for a corrupted list (Class/Objects) [30 pnts]
Topics covered: variables, expressions, control structures, functions, lists, classes/objects
In the utilities.py file, there is a variable named numList that is bound to a list of numbers. For Task4,
you will write code for a class named listStats that is used to compute the mean (also called the average),
trimmed mean, and median of a list. The trimmed mean and median are known as “robust statistics”
because they can give good answers even when some of the data in the list has been “corrupted”. Task4 will
examine how the mean, trimmed mean, and median perform on corrupted input.
Your listStat class should have the following methods that provide the described functionality:
__init__( parameters: numList, corruption_level )
This method is the object’s constructor or initializer. The constructor is passed a numList and a
“corruption level”. The corruption level is a value between 0 and 100.
Create an instance variable that stores a copy of the numList.
Next “corrupt” the numList as follows:
For each item in the numList, generate a random number between 1 and 100.
If the random number is <= the corruption_level parameter, then corrupt the item as follows:
Generate another random number between 1 and 100.
If the number is > 50, set the item in the list to -1; otherwise set the item to 100.
computeMean()
This method computes the mean (i.e., standard average) of the corrupted list.
The mean is the sum of all items in the list divided by the number of items.
Print out the result as “Standard Mean XXX” where XXX is formatted to have 2 decimal places.
computeTrimmedMean()
A “trimmed mean” is computed as follows.
(1) sort the list
(2) Remove the first 10 items and last 10 items from the list
The removal of these items is where the term “trim” comes from, which means to “cut” or “remove.”
(3) Now, compute the average of the new list.
– Note: sum only the items in the trimmed list and divide by the number of items in the trimmed list.
Print out the result as “Trimmed Mean XXX” where XXX is formatted to have 2 decimal places.
computeMedian()
The median is computed by sorting the list and taking the middle value.
The middle value is the length of the list divided by 2 using integer division (//).
Take care to make sure this method sorts the full corrupted list and not the trimmed list.
Print out the result as “Median XXX”. Note that the median will be an integer.
printStatistics()
This method prints out the corruption level and the corrupted list (see output example on next page).
This method then calls the computeMean(), computeTrimmedMean(), and computeMedian() methods.
Recall that you call a method within another method using the self keyword (see Lecture 9
slides 46-47 for an example in the “putting it all together” example). The mean/trimmed mean/median
methods will print out their results as described above.
Page 7/10
IMPORTANT: write your class in the utilities.py file. In task4() in the main file (final.py), declare
three listStats objects with the following corruption levels (0, 15, and 30). For each of these three objects,
call the printStatistics() method as follows:
obj1 = listStats(numList, 0) # no corruption
obj2 = listStats(numList, 15) # 15% chance of corruption
obj3 = listStats(numList,30) # 30% chance of corruption
obj1.printStatistics() # print stats
obj2.printStatistics() # print stats
obj3.printStatistics() # print stats
Example output for Task 4 (also see accompanying video)
Corruption level 0%
[21, 20, 29, 16, 16, 19, 26, 19, 26, 23, 19, 25, 16, 18, 18, 18, 21, 15, 18, 14, 22, 17,
32, 30, 22, 28, 26, 17, 19, 16, 17, 17, 19, 21, 20, 17, 18, 34, 23, 18, 23, 22, 17, 22,
16, 11, 21, 26, 20, 27]
Standard Mean 20.70
Trimmed Mean 19.90
Median 20
Corruption level 15%
[21, 20, 29, 100, 16, 19, 26, 19, 26, 23, -1, 25, 16, 18, 18, 18, 100, 15, 18, 14, 22,
17, 32, 30, 22, 28, 100, -1, 19, 16, 17, -1, 19, 21, 20, 17, 18, 34, 23, 18, 23, -1, 100,
-1, -1, 11, -1, 26, 20, 27]
Standard Mean 24.28
Trimmed Mean 20.03
Median 19
Corruption level 30%
[100, 20, 29, 16, -1, 19, 26, 19, 100, 23, 19, 100, 16, 18, 18, 100, 21, 15, 18, 14, 22,
100, 32, 100, 22, -1, 26, 17, -1, 16, 17, 17, -1, 21, 100, -1, 18, 34, 23, 18, 23, 100,
17, 22, 16, 11, 21, -1, 20, 27]
Standard Mean 30.50
Trimmed Mean 20.30
Median 20
Notice how the trimmed mean and median give similar results even when the list is corrupted.
way.
See pages below on how to submit your final exam’s code.
MAKE SURE TO SELECT final with websubmit.
Note how items are corrupted as either
-1 or 100.
Page 8/10
4. SUBMISSIONS (EECS web-submit)
You will submit your final using the EECS web submit.
Click on the following URL: https://webapp.eecs.yorku.ca/submit
STEP 1 — If you don’t have an EECS account,
has a passport York account).
If you do have an EECS account, enter here
and go to STEP 3.
STEP 1 — If you don’t have an EECS account,
has a passport York account).
If you do have an EECS account, enter here
and go to STEP 3.
STEP 1 — If you don’t have an EECS account,
has a passport York account).
If you do have an EECS account, enter here
and go to STEP 3.
STEP 2 – Enter your passport York
Page 9/10
STEP 3 – Select the correct menu option
as follows. Term “F”, Course “1015”,
Assignment “Final”.
STEP 3 cont’ – Select your file. Make sure
you attached both files final.py and
utilities.py
STEP 3 cont’ – once you have entered
everything above, click “Submit Files”.
STEP 4 – Confirm that you have entered
everything in correctly. If you make a
mistake here and submit to the wrong
course, or wrong lab, we won’t be able to
tell and will mark your lab as not submitted.
Please double check before clicking OK.
Final
utilities.py
final.py
Page 10/10
For more details on websubmit, see EECS department instructions:
https://wiki.eecs.yorku.ca/dept/tdb/services:submit:websubmit
STEP 5 – After you submit, your webpage
will refresh and show that you have
submitted the files and the time.
I recommend you logout.
You can resubmit the file if you make