How do I find an item in a list in Python?

Three questions, three tools. First: is it there at all?

fruits = ["apple", "banana", "cherry"]

print("banana" in fruits)
print("mango" in fruits)

True
False

Where is it?

index() returns the position of the first match, and raises ValueError if there isn't one. Check first or catch it:

fruits = ["apple", "banana", "cherry", "banana"]

print(fruits.index("banana"))

try:
    fruits.index("mango")
except ValueError:
    print("no mango")

1
no mango

Every position, not just the first:

fruits = ["apple", "banana", "cherry", "banana"]
print([i for i, f in enumerate(fruits) if f == "banana"])

[1, 3]

The first one that matches a condition

Usually you're not looking for an exact value but for "the first order over 100" or "the user called Hedy". next() with a generator stops at the first match, and the second argument is what you get when nothing matches:

orders = [
    {"id": "A-1", "total": 40},
    {"id": "A-2", "total": 180},
    {"id": "A-3", "total": 250},
]

big = next((o for o in orders if o["total"] > 100), None)
print(big)

huge = next((o for o in orders if o["total"] > 1000), None)
print(huge)

{'id': 'A-2', 'total': 180}
None

Leave out the None and a search with no match raises StopIteration, which is a confusing error to meet in the middle of a program.

Looking things up again and again?

x in some_list checks items one by one. Fine for a few hundred, slow for a big list searched in a loop. If you look up by a key, build a dictionary once; if you only ask "is it there?", use a set. The set membership tip has the measured difference.

orders = [{"id": "A-1", "total": 40}, {"id": "A-2", "total": 180}]

by_id = {o["id"]: o for o in orders}      # build once
print(by_id["A-2"]["total"])              # then every lookup is instant

180