Lecture "Brute-force algorithms", exercise 4

[essepuntato]

Write in Python the function def my_range(stop_number) which behave like the built-in function range() introduced in Section "Insertion sort" and returns a proper list, and accompany the function with the related test case. It is not possible to use the built-in function range() in the implementation.

[arcangelo7]

We didn't talk about the reverse() method in the classroom (different from reversed(), because it returns a list and not an iterator), so there's probably a smarter way to do this exercise.

def my_range(stop_number):
    output_range_object = list()
    current_number = stop_number

    while current_number >= 1:
        output_range_object.append(current_number - 1)
        current_number -= 1

    output_range_object.reverse()
    return output_range_object

def test_my_range(stop_number, expected):
    result = my_range(stop_number)

    if result == expected:
        return True
    else:
        return False

print(test_my_range(5, [0, 1, 2, 3, 4]))
# It prints True

[sntcristian]

[arcangelo7]

Replying to https://github.com/comp-think/2019-2020/issues/19#issuecomment-549853632 by @sntcristian:

I didn't understand what the if len(output_range) != 0 control is for. I think it works even without it

[sntcristian]

Right! I haven't thought it was that simple.

[NoonShin]

def my_range(stop_number):
    output = list()
    i = 0
    while i < stop_number:
        output.append(i)
        i += 1
    return output

def test_my_range(input, expected):
    return my_range(input) == expected

print(test_my_range(5, range(5)))

[FrancescoFernicola]

[essepuntato]

Hi all,

please find attached my personal solution – also available online:

# Test case for the function
def test_my_range(stop_number, expected):
    result = my_range(stop_number)
    if expected == result:
        return True
    else:
        return False

# Code of the function
def my_range(stop_number):
    l = list()
    while stop_number > 0:
        stop_number = stop_number - 1
        l.insert(0, stop_number)
    return l

# Tests
print(test_my_range(0, []))
print(test_my_range(1, [0]))
print(test_my_range(4, [0, 1, 2, 3]))

Please be aware that, formally speaking, range(5) is not a list. Thus, checking if the object returned by the function range is a list will return False anyway.

[NoonShin]

Thanks for the heads-up about the range function @essepuntato. I mistakenly used a Python 2.7 interpreter for testing my solution to this one and it worked fine on that (since in Python 2 range used to create lists instead of iterators).

[aschimmenti]

def my_range(num): output_list = [] '''n = 0

start from 0 which is the first element

'''while n < num:

keep appending the n to the list until n is equal to num-1

'''''output_list.append(n) '''''''''n = n+1 ''''''return output_list print(my_range(10) == list(range(10)))

it prints True