OPS435 Python Lab 4

OBJECTIVES

This lab will provide you will additional scripting tools to help us write even more effective Python scripts to be applied to practical application involving VM management and deployment in future labs.

The first investigation in this lab will focus on Data Structures. In Wikipedia (http://searchsqlserver.techtarget.com/definition/data-structure)
"A data structure is defined as a specialized format for organizing and storing data. Any data structure is designed to organize data to suit a specific purpose so that it can be accessed and worked with in appropriate ways."

Each data structure has its own advantages and limitations. This lab will emphasize the most important differences as they relate to Python scripting.

The second investigation will focus closely on strings. You have been using and storing strings since our first class, however in this lab we will dive into the more complex nature of string manipulation. Finally, this lab will cover how to use a variety of different regular expression functions, for searching and input validation.

PYTHON REFERENCE

As you develop your Python scripting skills, you may start to be "overwhelmed" with the volume of information that you have absorbed over these labs. One way to help, is to write what you have learned in your labs into your lab logbook. Also, in programming, it is important to use online references in order to obtain information regarding Python scripting techniques and tools.

Below is a table with links to useful online Python reference sites (by category). You may find these references useful when performing assignments, etc.

Data Structures	Lists & List Comprehension	Strings	Regular Expressions	Miscellaneous
Tuples Sets	Lists More on Lists List Comprehensions	Strings String Comparisons	Regular Expression Operations Regular Expressions (HOWTO)	Dictionaries

INVESTIGATION 1: DATA STRUCTURES

In this investigation, you will learn several tools when using data structures in Python scripting.

These tools include tuples, sets, dictionaries, and more advanced list functions.

PART 1 - Tuples

Many often confuse a tuple with a list (which you learned about in a previous lab). A tuple is a type of list whose values cannot be changes. In fact, the structure of a tuple cannot be changed (like adding, removing list elements).

There are many advantages to using tuples when creating Python scripts:

• Data protection (eg. values are are NOT allowed to change like income tax rate, social insurance number, etc)
• The data structure in a tuple cannot be changed (eg. structure cannot be corrupted)
• Tuples can be used as keys in data dictionaries (which are NOT allowed to change)
• Tuples allow for faster access than lists

Term to indicate that a data structure cannot be changed is called immutable (as opposed to "mutable" which means the data structure can be changed).

Perform the Following Steps:

1. Launch your ipython3 shell:

   ipython3

   Let's create two tuples, so we can learn how to use them and learn how they differ from lists.
   
   Note: tuples are defined by using parenthesis ( ) as opposed to lists are defined by using square brackets [ ]
   
   
2. Issue the following:

   t1 = ('Prime', 'Ix', 'Secundus', 'Caladan')
   t2 = (1, 2, 3, 4, 5, 6)

3. Values from a tuple can be retrieved in the same way as a list. For example, issue the following:

   t1[0]
   t2[2:4]

4. You can also check to see whether a value exists inside a tuple or not. To demonstrate, issue the following:

   'Ix' in t1
   'Geidi' in t1

   Let's now see how a tuple differs from a list. We will now create a list and note the difference between them.
   
   
5. Issue the following to create a list:

   list2 = [ 'uli101', 'ops235', 'ops335', 'ops435', 'ops535', 'ops635' ]

6. See if you can change the value of your list by issuing the following:

   list2[0]= 'ica100'
   list2[0]
   print(list2)

   .You should have been successful in changing the value of your list.
   
   
7. Now, try changing the value of your previously-created tuple by issuing:

   t2[1] = 10

   Did it work? Once created the tuple values will not be able to change.
   
   If you would like a tuple with different values than the tuple you currently have, then you must create a new one.
   
   
8. To create a new tuple, issue the following:

   t3 = t2[2:3]

9. You can use most of the basic operations with tuples as you did with lists.
   
   
10. To demonstrate, issue the following:

    len(t1)     # list the length of the tuple
    t1 * 3      # repetition
    t1 + t2     # concatenation, remember this is creating a new tuple, not modifying

11. Also, as with lists, you can use loops with tuples. Issue the following to demonstrate:

    for item in t1:
        print('item: ' + item)

PART 2 - Sets

So far, you have been exposed to two structures that are used to contain data: lists and tuples. You can modify the values within a list as well as modify the structure of a list (i.e. add and remove elements), whereby you cannot with a tuple.

In this section, you will learn about sets. A set has similar characteristics as a list, but there are two major differing characteristics:

• Sets are un-ordered
• Sets cannot contain duplicate values

These characteristics make sets very useful for performing tasks such as comparisons: finding similarities or differences in multiple sets. In fact, you are not permitted to create duplicate entries within sets, since the new duplicate entries will be automatically removed. Also, sets are considered to be fast!

Perform the Following Steps:

1. Within your ipython3 shell, create a couple sets to work with by issuing the following:

   s1 = {'Prime', 'Ix', 'Secundus', 'Caladan'}
   s2 = {1, 2, 3, 4, 5}
   s3 = {4, 5, 6, 7, 8}

   Note: Sets are defined by using braces { } as opposed to tuples that use parenthesis ( ), or lists that use square brackets [ ]
   
   
2. Try to issue the following to access a set through the index.

   s1[0]

   This should have created an error, this is not how to access data inside a set because they are un-ordered. Instead, you should use the method (used in the previous section) to check to see if a value is contained within the set.
   
   
3. To demonstrate, issue the following:

   'Ix' in s1
   'Geidi' in s1

   
   Sets can be combined, but it is important to note that any duplicate values (shared among sets) will be deleted.
   
   
4. Issue the following, and note the items (and values) that are common to the following sets:

   s2
   s3

5. Now, issue the following to return a set containing only UNIQUE values (no duplicates) from both sets:

   s2 | s3         # returns a set containing all values from both sets
   s2.union(s3)    # same as s2 | s3

   Notice that both methods above provides the same result, but the first method requires less keystrokes.
   
   Instead of combining sets, we can display values that are common to both sets. This is known in mathematical terms as an intersection between the lists.
   
   
6. To demonstrate intersection between sets s2 and s3, issue the following:

   s2 & s3             # returns a set containing all values that s2 and s3 share
   s2.intersection(s3) # same as s2 & s3

7. Sets can also have their values compared against other sets. First find out what items are in s2 but not in s3. This is also called a difference. But notice that it only shows values that s2 contains, specifically values that s3 doesn't have. So this isn't really the true difference between the sets.

   s2
   s3
   s2 - s3             # returns a set containing all values in s2 that are not found s3
   s2.difference(s3)   # same as s2 - s3

8. In order to see every difference between both sets, you need to find the symmetric difference. This will return a set that shows all numbers that both sets do not share together.
   
   
9. To demonstrate, issue the following:

   s2 ^ s3                     # returns a set containing all values that both sets DO NOT share
   s2.symmetric_difference(s3) # same as s2 ^ s3

   These powerful features can be useful and efficient. Unfortunately, lists cannot perform these operations, unless we have to convert the lists into sets. In order to that, you should first perform a comparison, then convert the list to a set.

There are two problems with performing the above-mentioned technique:

• Sets are un-ordered so if the list order is important this will cause problems and remove order
• Sets cannot contain duplicate values, if the list contains any duplicate values they will be deleted.

However, if the list does not have any of the above requirements this is a great solution to some problems.

1. To demonstrate, issue the following:

   l2 = [1, 2, 3, 4, 5]
   l3 = [4, 5, 6, 7, 8]
   new_list = list(set(l2).intersection(set(l3)))  # set() can make lists into sets. list() can make sets into lists
   new_list

Create a Python Script Demonstrating Comparing Sets

Perform the Following Instructions

1. Create the ~/ops435/lab4/lab4a.py script. The purpose of this script will be to demonstrate the different way of comparing sets. There will be three functions, each returning a different set comparison.
2. Use this template to get started:

   #!/usr/bin/env python3
   
   def join_sets(set1, set2):
       # join_sets will return a set that has every value from both set1 and set2 inside it
   
   def match_sets(set1, set2):
       # match_sets will return a set that contains all values found in both set1 and set2
   
   def diff_sets(set1, set2):
       # diff_sets will return a set that contains all different values which are not shared between the sets
   
   if __name__ == '__main__':
       set1 = set(range(1,10))
       set2 = set(range(5,15))
       print('set1: ', set1)
       print('set2: ', set2)
       print('join: ', join_sets(set1, set2))
       print('match: ', match_sets(set1, set2))
       print('diff: ', diff_sets(set1, set2))

• The match_sets() function should return a set that contains all values found in both sets
• The diff_sets() function should return a set that contains all values which are not shared between both sets
• The join_sets() function should return a set that contains all values from both sets
• All three functions should accept two arguments both are sets
• The script should show the exact output as the samples
• The script should contain no errors

Sample Run 1:

run lab4a.py
set1:  {1, 2, 3, 4, 5, 6, 7, 8, 9}
set2:  {5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
join:  {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
match:  {8, 9, 5, 6, 7}
diff:  {1, 2, 3, 4, 10, 11, 12, 13, 14}

Sample Run 2(with import):

import lab4a
set1 = {1,2,3,4,5}
set2 = {2,1,0,-1,-2}
lab4a.join_sets(set1,set2)
{-2, -1, 0, 1, 2, 3, 4, 5}
lab4a.match_sets(set1,set2)
{1, 2}
lab4a.diff_sets(set1,set2)
{-2, -1, 0, 3, 4, 5}

3. Exit the ipython3 shell, download the checking script and check your work. Enter the following commands from the bash shell.

cd ~/ops435/lab4/
pwd #confirm that you are in the right directory
ls CheckLab4.py || wget matrix.senecac.on.ca/~acoatley-willis/CheckLab4.py
python3 ./CheckLab4.py -f -v lab4a

4. Before proceeding, make certain that you identify any and all errors in lab4a.py. When the checking script tells you everything is OK before proceeding to the next step.

Create a Python Script Demonstrating Comparing Lists

Perform the Following Instructions

1. Create the ~/ops435/lab4/lab4b.py script. The purpose of this script will be to improve the previous script to perform the same joins, matches, and diffs, but this time on lists.
2. Use the following as a template:

   #!/usr/bin/env python3
   
   def join_lists(list1, list2):
       # join_lists will return a list that contains every value from both list1 and list2 inside it
   
   def match_lists(list1, list2):
       # match_lists will return a list that contains all values found in both list1 and list2
   
   def diff_lists(list1, list2):
       # diff_lists will return a list that contains all different values, which are not shared between the lists
   
   if __name__ == '__main__':
       list1 = list(range(1,10))
       list2 = list(range(5,15))
       print('list1: ', list1)
       print('list2: ', list2)
       print('join: ', join_lists(list1, list2))
       print('match: ', match_lists(list1, list2))
       print('diff: ', diff_lists(list1, list2))

• The match_lists() function should return a list that contains all values found in both lists
• The diff_lists() function should return a list that contains all values which are not shared between both lists
• The join_lists() function should return a list that contains all values from both sets
• All three functions should accept two arguments both are lists
• The script should show the exact output as the samples
• The script should contain no errors

Sample Run 1:

run lab4b.py
list1:  [1, 2, 3, 4, 5, 6, 7, 8, 9]
list2:  [5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
join:  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
match:  [8, 9, 5, 6, 7]
diff:  [1, 2, 3, 4, 10, 11, 12, 13, 14]

Sample Run 2(with import):

import lab4b
list1 = [1,2,3,4,5]
list2 = [2,1,0,-1,-2]
join_lists(list1,list2)
[0, 1, 2, 3, 4, 5, -2, -1]
match_lists(list1,list2)                                                                                                                  
[8, 9, 5, 6, 7]
diff_lists(list1,list2)                                                                                                                   
[1, 2, 3, 4, 10, 11, 12, 13, 14]

3. Exit the ipython3 shell, download the checking script and check your work. Enter the following commands from the bash shell.

cd ~/ops435/lab4/
pwd #confirm that you are in the right directory
ls CheckLab4.py || wget matrix.senecac.on.ca/~acoatley-willis/CheckLab4.py
python3 ./CheckLab4.py -f -v lab4b

4. Before proceeding, make certain that you identify any and all errors in lab4b.py. When the checking script tells you everything is OK before proceeding to the next step.

PART 3 - Dictionaries

In Python a Dictionary is a set of key-value pairs. Dictionaries are unordered, like sets, however any value can be retrieved from a dictionary if you know the key. This section will go over how to create, access, and change dictionaries, providing a new tool to store and manipulate data with.

Perform the Following Steps:

1. Start the ipython3 shell:

   ipython3

2. Start by creating a new dictionary to practice with:

   dict_york = {'Address': '70 The Pond Rd', 'City': 'Toronto', 'Postal Code': 'M3J3M6'}

3. The syntax here is to use {} to create a dictionary and placing key:value pairs inside separated by commas.
4. Take a close look at all the available functions available to dictionary objects

   dir(dict_york)
   help(dict_york)

5. All values can be viewed by using the dictionary.values() function. This function provides a list containing all values

   help(dict_york.values)
   dict_york.values()

6. All keys can be viewed by using the dictionary.keys() function. This function provides a list containing all keys

   help(dict_york.keys)
   dict_york.keys()

7. We can retrieve individual values from a dictionary by provide the key associated with the value

   dict_york['Address']
   dict_york['Postal Code']

8. Dictionary keys can be any immutable values, such as: strings, numbers, and tuples. Trying adding a couple new keys and values to the dictionary

   dict_york['Country'] = 'Canada'
   dict_york
   dict_york.values()
   dict_york.keys()

9. Study the output and add another key:value pair

   dict_york['Province'] = 'BC'
   dict_york
   dict_york.values()
   dict_york.keys()

10. Dictionary keys must be unique. Attempting to add a key that already exists in the dictionary will overwrite the existing value for that key

    dict_york['Province'] = 'ON'
    dict_york
    dict_york.values()
    dict_york.keys()

11. These lists that contain the values and keys of the dictionary are not real python lists, they are view of the dictionary
12. However we can change these from views into usable lists by using the list() function, the index can be used to access individual values

    list_of_keys = list(dict_york.keys())
    list_of_keys[0]

13. Lists can be changed into sets if we would like to perform comparisons with another set

    set_of_keys = set(dict_york.keys())
    set_of_values = set(dict_york.values())
    set_of_keys | set_of_values

14. The lists can be iterated over in a for loop

    list_of_keys = list(dict_york.keys())
    for key in list_of_keys:
        print(key)
    for value in dict_york.values()
        print(value)

15. The values and keys can be looped over using the index as well
16. The range() function provides a list of numbers in a range.
17. The len() provides a the number of items in a list. Used together len() and range() create a list of usable indexes for a specific list

    list_of_keys = list(dict_york.keys())
    list_of_values = list(dict_york.values())
    list_of_indexes = range(0, len(dict_york.keys()))
    list_of_indexes
    list_of_keys[0]
    list_of_values[0]

18. Using this this list of indexes we are able to pair the keys and values of two separate lists

    list_of_keys = list(dict_york.keys())
    list_of_values = list(dict_york.values())
    for index in range(0, len(list_of_keys)):
        print(list_of_keys[index] + '--->' + list_of_values[index])

19. Looping using indexes is not the best way to loop through a dictionary. A new dictionary could be created using this method, but this is not good

    list_of_keys = list(dict_york.keys())
    list_of_values = list(dict_york.values())
    new_dictionary = {}
    for index in range(0, len(list_of_keys)):
        new_dictionary[list_of_keys[index]] = list_of_values[index]

20. The above method uses a lot of memory and loops. The best method to create a dictionary from two lists is to use the zip() function

    list_of_keys = list(dict_york.keys())
    list_of_values = list(dict_york.values())
    new_dictionary = dict(zip(list_of_keys, list_of_values))

21. Loop through the keys in a dictionary also provides a easy way to get the value for each key at the same time

    for key in dict_york.keys():
        print(key + '--->' + dict_york[key])

22. Even better than the above, both key and value can be extracted in a single for loop using a special object

    for key, value in dict_york.items():
        print(key + ' | ' + value)

Create a Python Script for Managing Dictionaries

Perform the Following Instructions

1. Create the ~/ops435/lab4/lab4c.py script. The purpose of this script will be to create dictionaries, extract data from dictionaries, and to make comparisons between dictionaries.
2. Use the following as a template:

   #!/usr/bin/env python3
   
   # Dictionaries
   dict_york = {'Address': '70 The Pond Rd', 'City': 'Toronto', 'Country': 'Canada', 'Postal Code': 'M3J3M6', 'Province': 'ON'}
   dict_newnham = {'Address': '1750 Finch Ave E', 'City': 'Toronto', 'Country': 'Canada', 'Postal Code': 'M2J2X5', 'Province': 'ON'}
   # Lists
   list_keys = ['Address', 'City', 'Country', 'Postal Code', 'Province']
   list_values = ['70 The Pond Rd', 'Toronto', 'Canada', 'M3J3M6', 'ON']
   
   def create_dictionary(keys, values):
       # Place code here
   
   def split_dictionary(dictionary):
       # Place code here
          
   def shared_values(dict1, dict2):
       # Place code here
   
   if __name__ == '__main__':
       york = create_dictionary(list_keys, list_values)
       print('York: ', york)
       keys, values = split_dictionary(dict_newnham)
       print('Newnham Keys: ', keys)
       print('Newnham Values: ', values)
       keys, values = split_dictionary(york)
       print('York Keys: ', keys)
       print('York Values: ', values)
       common = shared_values(dict_york, dict_newnham)
       print('Shared Values', common)

• The script should contain three functions
• create_dictionary() accepts two lists as arguments keys and values, combines these lists together to create a dictionary
• create_dictionary() returns a dictionary that has the keys and associated values from the lists
• split_dictionary() accepts a single dictionary as a argument and splits the dictionary into two lists, keys and values
• split_dictionary() returns two lists: return keys, values
• shared_values() accepts two dictionaries as arguments finds all values that are shared between the two dictionaries
• shared_values() returns a set containing ONLY values found in BOTH dictionaries
• make sure the functions have the correct number of arguments required
• The script should show the exact output as the samples
• The script should contain no errors

Sample Run 1:

run lab4c.py
York:  {'Country': 'Canada', 'Postal Code': 'M3J3M6', 'Address': '70 The Pond Rd', 'Province': 'ON', 'City': 'Toronto'}
Newnham Keys:  ['Country', 'Postal Code', 'Address', 'Province', 'City']
Newnham Values:  ['Canada', 'M2J2X5', '1750 Finch Ave E', 'ON', 'Toronto']
York Keys:  ['Country', 'Postal Code', 'Address', 'Province', 'City']
York Values:  ['Canada', 'M3J3M6', '70 The Pond Rd', 'ON', 'Toronto']
Shared Values {'Canada', 'ON', 'Toronto'}

Sample Run 2(with import):

import lab4c
dict_york = {'Address': '70 The Pond Rd', 'City': 'Toronto', 'Country': 'Canada', 'Postal Code': 'M3J3M6', 'Province': 'ON'}
dict_newnham = {'Address': '1750 Finch Ave E', 'City': 'Toronto', 'Country': 'Canada', 'Postal Code': 'M2J2X5', 'Province': 'ON'}
list_keys = ['Address', 'City', 'Country', 'Postal Code', 'Province']
list_values = ['70 The Pond Rd', 'Toronto', 'Canada', 'M3J3M6', 'ON']

york = create_dictionary(list_keys, list_values)

york
{'Address': '70 The Pond Rd',
 'City': 'Toronto',
 'Country': 'Canada',
 'Postal Code': 'M3J3M6',
 'Province': 'ON'}

keys, values = split_dictionary(dict_newnham)

keys
['Country', 'Postal Code', 'Address', 'Province', 'City']

values
['Canada', 'M2J2X5', '1750 Finch Ave E', 'ON', 'Toronto']

keys, values = split_dictionary(york)

keys
['Country', 'Postal Code', 'Address', 'Province', 'City']

values
['Canada', 'M3J3M6', '70 The Pond Rd', 'ON', 'Toronto']

common = shared_values(dict_york, dict_newnham)

common
{'Canada', 'ON', 'Toronto'}

3. Exit the ipython3 shell, download the checking script and check your work. Enter the following commands from the bash shell.

cd ~/ops435/lab4/
pwd #confirm that you are in the right directory
ls CheckLab4.py || wget matrix.senecac.on.ca/~acoatley-willis/CheckLab4.py
python3 ./CheckLab4.py -f -v lab4c

4. Before proceeding, make certain that you identify any and all errors in lab4c.py. When the checking script tells you everything is OK before proceeding to the next step.

PART 4 - List Comprehension

We've already covered lists to a degree. Lets move into more advanced functions to use and generate lists. This is a very common practice in Python, understanding how to generate, manipulate, and apply functions to items inside a list can be incredibly useful. List comprehension is a way to build new lists from existing list and to do it faster than simply looping over lists.

Perform the Following Steps

1. Lets start with creating a list and applying some function to each item in the list. The below will print out the square of each item.

   l1 = [1, 2, 3, 4, 5]
   for item in l1:
       print(item ** 2)

2. To store these squares for later use, create a new list and append the squares to it. This will generate a new list that contains squared values in the same positions of the first list. This is using an existing list to create a new list.

   l1 = [1, 2, 3, 4, 5]
   l2 = []
   for item in l1:
       l2.append(item ** 2)
   l1
   l2

3. Move the squaring of numbers out into it's own separate function. While the squaring example is a simple function, this example could include a more complex function that does more processing on each item in the list.

   def square(number):
       return number ** 2
   
   l1 = [1, 2, 3, 4, 5]
   l2 = []
   for item in l1:
       l2.append(square(item))
   
   l1
   l2

4. The map function can be used to apply a function on each item in a list. This is exactly what happened above, however it gives much better syntax, removes the loop, including the variable that was created inside the loop. This will make the script a little more efficient while performing the same task.

   def square(number):
       return number ** 2
   
   l1 = [1,2,3,4,5]
   l2 = list(map(square, l1))
   
   l1
   l2

5. The above map function requires a function, and a list. This meant that before map() could be used a function needed to be defined earlier in the script. This entire process can be avoided through the use of anonymous functions. This is the ability to create a simple function without defining it, and pass it off for use. Below we will use lambda, which will return a function, and we can use that function immediately. The function takes 1 argument x, and it will perform a single operation on x, square it.

   square = lambda x: x ** 2
   l1 = [1,2,3,4,5]
   l2 = list(map(square, l1))
   
   l1
   l2

6. The above code is actually not particularly good, the whole purpose of using lambda here is we were avoiding the function definition and just quickly returning a function. However this does break down exactly what lambda does, it returns a function for use. Fix this by removing the square function and just use the return function from lambda. Now remember what map requires? map's first argument is a function, and map's second argument is a list. Here lambda will return a function and provide it as the first argument.

   l1 = [1,2,3,4,5]
   l2 = list(map(lambda x: x ** 2, l1))
   
   l1
   l2

7. Using the list comprehensions above our code will be faster and more efficient than using multiple variables and loops.

INVESTIGATION 2: STRINGS

Strings are in their most basic form a list of characters, or a bit of text. Strings store text so that we can use them later. In this section we will cover more than just displaying that text to the screen. Here, we will go over cutting strings into sub-strings, joining strings together, searching through strings, and matching strings against patterns.

LAB 4 SIGN-OFF (SHOW INSTRUCTOR)

Students should be prepared with all required commands (system information) displayed in a terminal (or multiple terminals) prior to calling the instructor for signoff.

Have Ready to Show Your Instructor:

✓ x

✓ x

✓ Lab4 logbook notes completed

Practice For Quizzes, Tests, Midterm & Final Exam

1. x
2. x
3. x