Loops (while and for)

Education objectives

• blocks while and for

• keywords in, break and continue

• built-in functions range and enumerate

Loops with the keyword while

i = 0
while i < 4:
    i += 1
print("i =", i)

i = 4

i = 0
while i < 4:
    i += 1
    print("i =", i)

i = 1
i = 2
i = 3
i = 4

Exercise 9

• Edit a script with Spyder that calculates the average of a set of numbers. For example numbers = [67, 12, 2, 9, 23, 5]

  • using the functions sum and len

  • manually (without sum), using the keyword while

  • check that the 2 methods give the same results with

  assert avg0 == avg1

• Run the script

  • in Spyder,

  • in a IPython session opened from another terminal,

  • with the command python.

Solution to Exercise 9

numbers = [67, 12, 2, 9, 23, 5]

avg0 = sum(numbers) / len(numbers)

tmp = 0
i = 0
while i < len(numbers):
    tmp += numbers[i]
    i = i + 1
avg1 = tmp / len(numbers)

assert avg0 == avg1

Simulating do while stop_condition construction

while True:
    if stop_condition:
        break
    # content of the do-while loop

Loops with the keyword for

values = range(5)
for i in values:
    print("i =", i)

i = 0
i = 1
i = 2
i = 3
i = 4

The build-in function range() is very useful for loops. It creates a range object, which is an iterable, immutable sequence of numbers.

# syntax is range(start, stop, step)
range(0, 4, 1)

range(0, 4)

(default start is 0, default step is 1)

range(4)

range(0, 4)

list(range(1, 8, 2))

[1, 3, 5, 7]

range() is memory efficient because it does not store all the numbers in memory.

range(1_000_000_000_000_000)

range(0, 1000000000000000)

Warning

Do not try list(range(1_000_000_000_000_000)) at home! Can you estimate the order of magnitude of memory it would use?

It is common to use range() directly in for loops:

for idx in range(4):
    print(idx, end=", ")

0, 1, 2, 3,

for loops are of course not limited to integers:

groceries = ["Carrots", "Cabbage", "Milk", "Onions", "Pepper"]

print("Groceries list")
for grocery in groceries:
    print("-", grocery)

Groceries list
- Carrots
- Cabbage
- Milk
- Onions
- Pepper

The built-in function enumerate is very useful to access indices

print("My top 5 groceries:")
for index, grocery in enumerate(groceries):
    print(f"{index}. {grocery}")

My top 5 groceries:
0. Carrots
1. Cabbage
2. Milk
3. Onions
4. Pepper

Loops: keywords continue and break

• continue: passes the block in the loop and continues the loop.

for x in range(1, 8):
    if x == 5:
        continue
    print(x, end=", ")

1, 2, 3, 4, 6, 7,

• break: stop the loop.

for x in range(1, 8):
    if x == 5:
        break
    print(x, end=", ")

1, 2, 3, 4,

Exercise 10

• Extend your script with another method (using a for loop) to compute the average.

• In IPython, try to understand how the function enumerate works. Use it in your script.

Solution to Exercise 10

l = [67, 12, 2, 9, 23, 5]
# simple implementation with sum and len
avg0 = sum(l) / len(l)

# now with for and without sum
avg2 = 0
for e in l:
    avg2 += e
avg2 /= len(l)

# now with for and enumerate, but without sum and len
avg3 = 0
for i, e in enumerate(l):
    avg3 += e
avg3 /= i + 1

# and now let's check:
assert avg2 == avg0

Exercise 11

We build a list:

from random import randint, shuffle

n = 20
i_removed = randint(0, n - 1)
print("integer remove from the list:", i_removed)
numbers = list(range(n))
numbers.remove(i_removed)
shuffle(numbers)
print(f"shuffled list:\n  {numbers}")

integer remove from the list: 7
shuffled list:
  [15, 4, 5, 18, 6, 8, 3, 1, 0, 12, 16, 11, 14, 9, 13, 17, 19, 10, 2]

One element has been removed:

• Find this element (given that you can change the ordering of numbers).

• Find this element (given that you cannot change the ordering of numbers).

Solution to Exercise 11

# we can change ordering, let's sort
print(numbers)
l_sorted = sorted(numbers)
print(l_sorted)
missing = None
for idx, elem in enumerate(l_sorted):
    if elem != idx:
        missing = idx
        break
if missing is None:
    missing = len(numbers)

print(f"{missing = }")
assert missing == i_removed

[15, 4, 5, 18, 6, 8, 3, 1, 0, 12, 16, 11, 14, 9, 13, 17, 19, 10, 2]
[0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
missing = 7

# we cannot sort -> higher complexity
for elem in range(len(numbers) + 1):
    if elem not in numbers:
        break
missing = elem

print(f"{missing = }")
assert missing == i_removed

missing = 7

# another solution
actual_sum = sum(numbers)
len_numbers = len(numbers)
original_sum = (len_numbers + 1) * (len_numbers) // 2
missing = original_sum - actual_sum

print(f"{missing = }")
assert missing == i_removed

missing = 7

list: list comprehension

They are iterable so they are often used to make loops. We have already seen how to use the keyword for. For example to build a new list (side note: x**2 computes x^2):

l0 = [1, 4, 10]
l1 = []
for number in l0:
    l1.append(number**2)

print(l1)

[1, 16, 100]

There is a more readable (and slightly more efficient) method to do such things, the list comprehension:

l1 = [number**2 for number in l0]
print(l1)

[1, 16, 100]

# list comprehension with a condition
[s for s in ["a", "bbb", "e"] if len(s) == 1]

['a', 'e']

# lists comprehensions can be cascaded
[(x, y) for x in [1, 2] for y in ["a", "b"]]

[(1, 'a'), (1, 'b'), (2, 'a'), (2, 'b')]

Exercise 12 (advanced)

• Write a function extract_patterns(text, n=3) extracting the list of patterns of size n=3 from a long string (e.g. if text = "basically", patterns would be the list ['bas', 'asi', 'sic', ..., 'lly']). Use list comprehension, range, slicing. Use a sliding window.

• You can apply your function to a long “ipsum lorem” string (ask to your favorite web search engine).

Solution to Exercise 12 (advanced)

text = "basically"


def extract_patterns(text, n=3):
    pat = [text[i : i + n] for i in range(len(text) - n + 1)]
    return pat


print("patterns=", extract_patterns(text))
print("patterns=", extract_patterns(text, n=5))

patterns= ['bas', 'asi', 'sic', 'ica', 'cal', 'all', 'lly']
patterns= ['basic', 'asica', 'sical', 'icall', 'cally']