Looping patterns

Numbering with enumerate()

To print a numbered list, you could keep a counter yourself, as in Module 3. enumerate()enumerate()A built-in that gives each item of a list together with its position, as (position, item) pairs. start= sets the first number. does it for you: each round it gives a pair, the position and the item, ready to unpack (lesson 4.3).

runners = ["Tigist", "Abel", "Meron"]
for position, name in enumerate(runners, start=1):
    print(f"{position}. {name}")

Without start=1 it counts from 0, like indexes. Remove it and see. If you forget enumerate altogether, Python tries to unpack each name:

Two lists together with zip()

When two lists line up, item by item, zip() walks them together. Each round gives one item from each:

items = ["teff", "coffee", "sugar"]
prices = [1200, 450, 90]
for item, price in zip(items, prices):
    print(f"{item}: {price} birr")

What if one list is shorter? Predict.

names = ["Hana", "Abel", "Selam"]
scores = [88, 92]
for name, score in zip(names, scores):
    print(name, score)

Hana 88
Abel 92

That silence can hide a bug: if a score is missing, nothing tells you. When the lists should match, check len() of both first.

Another order: reversed() and sorted()

reversed() goes through a list backwards, and sorted() (lesson 4.2) goes through it in order. Neither changes the list:

for n in reversed([1, 2, 3]):
    print(n)
names = ["Selam", "Abel", "Hana"]
for name in sorted(names):
    print(name)
print(names)

in instead of a flag

To ask "is Selam registered?" you could loop with a flag: start it False, and set it True when you find her.

registered = ["Hana", "Selam", "Abel"]
found = False
for name in registered:
    if name == "Selam":
        found = True
print(found)
print("Selam" in registered)

The last line asks the same question with in. When all you want to know is "is it there?", use in. Keep the flag loop for when you need to do more along the way.

Don't remove while you loop

You want to remove every 2 from a list. This looks right. Predict what it prints.

nums = [1, 2, 2, 3]
for n in nums:
    if n == 2:
        nums.remove(n)
print(nums)

[1, 2, 3]

Step through it and watch n. After the first 2 is removed, the loop's next value is 3: the second 2 was never looked at.

[1, 2, 3]

Two fixes. Collect the items you want to keep into a new list (the collect plan from lesson 4.2), or loop over a copy, nums[:], and remove from the original:

nums = [1, 2, 2, 3]
kept = []
for n in nums:
    if n != 2:
        kept.append(n)
print(kept)

nums = [1, 2, 2, 3]
for n in nums[:]:
    if n == 2:
        nums.remove(n)
print(nums)

A slice makes a new list, so the loop walks the copy while the original changes.