Python Functions, Files, and Dictionaries - Week 4 - Assessment: More about Iteration - course_2_assessment_6

course_2_assessment_6.py

# Python Functions, Files, and Dictionaries - Week 4 - Assessment: More about Iteration - course_2_assessment_6



# 1) Write a function, sublist, that takes in a list of numbers as the parameter. 
# In the function, use a while loop to return a sublist of the input list. 
# The sublist should contain the same values of the original list up until it reaches the number 5 (
# it should not contain the number 5).

# Answer:

def sublist(lst):
    accum = 0
    sub_list = []
    while accum < len(lst):
        if lst[accum] == 5:
            break
        else:
            sub_list.append(lst[accum])
        accum = accum + 1
    return(sub_list)
      
original_list = [9, 8, 1, 2, 7, 3, 5, 9, 8, 6, 2]
sublist(original_list)
print(original_list)
print(sublist(original_list))


# 2) Write a function called check_nums that takes a list as its parameter, and contains a while loop that 
# only stops once the element of the list is the number 7. What is returned is a list of all of the numbers 
# up until it reaches 7.

# Answer:

def check_nums(lst):
    accum = 0
    sub_list = []
    while accum < len(lst):
        if lst[accum] == 7:
            break
        else:
            sub_list.append(lst[accum])
        accum = accum + 1
    return(sub_list)
      
original_list = [9, 8, 1, 2, 7, 3, 5, 9, 8, 6, 2]
check_nums(original_list)
print(original_list)
print(check_nums(original_list))



# 3) Write a function, sublist, that takes in a list of strings as the parameter. 
# In the function, use a while loop to return a sublist of the input list. 
# The sublist should contain the same values of the original list up until it reaches the string “STOP” 
# (it should not contain the string “STOP”).

# Answer:
 
def sublist(string_list):
    accum = 0
    sub_list = []
    while accum < len(string_list):
        if string_list[accum] == "STOP":
            break
        else:
            sub_list.append(string_list[accum])
        accum = accum + 1
    return sub_list

original_list = ['New', 'list', 'will', 'STOP', 'before', 'stop']
sublist(original_list)
print(original_list)
print(sublist(original_list))


# 4) Write a function called stop_at_z that iterates through a list of strings. 
# Using a while loop, append each string to a new list until the string that appears is “z”. 
# The function should return the new list.

# Answer:

def stop_at_z(lst):
    accum = 0
    new_list = []
    while accum < len(lst):
        if lst[accum] == 'z':
            break
        else:
            new_list.append(lst[accum])
        accum = accum + 1
    return new_list

original_list = ['k', 'j', 'f', 's', 'k', 'j', 'b', 'k', 'j', 'z', 'q', 'k','j']
stop_at_z(original_list)
print(original_list)
print(stop_at_z(original_list))



# 5) Below is a for loop that works. Underneath the for loop, rewrite the problem so that it does the same 
# thing, but using a while loop instead of a for loop. Assign the accumulated total in the while loop code 
# to the variable sum2. Once complete, sum2 should equal sum1.

# Answer:

sum1 = 0
sum2 = 0
i = 0 
lst = [65, 78, 21, 33]

for x in lst:
    sum1 = sum1 + x
    
while i < len(lst):
    sum2 = sum2 + lst[i]
    i = i + 1

print(sum1)
print(sum2)



# Challenge: Write a function called beginning that takes a list as input and contains a while loop that 
# only stops once the element of the list is the string ‘bye’. What is returned is a list that contains 
# up to the first 10 strings, regardless of where the loop stops. (i.e., if it stops on the 32nd element, 
# the first 10 are returned. If “bye” is the 5th element, the first 4 are returned.) 
# If you want to make this even more of a challenge, do this without slicing

# Answer:

def beginning(lst):
    accum = 0
    new_list = []
    while "bye" not in lst[accum] and accum < 10:
        new_list.append(lst[accum])
        accum = accum + 1
    return new_list
    
original_list = ['New', 'list', 'will', 'now', 'bye', 'stoped', 'before', 'accumulating', '10', 'items']
beginning(original_list)
print(original_list)
print(beginning(original_list))