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For a list or other collection, use Python’s built-in max() and min() functions. For example, max(numbers) returns the largest item and min(numbers) returns the smallest. If you must process a one-pass iterator, track both values in a single loop instead.
Find the largest and smallest values with built-ins
Pass the collection as the single argument to each function:
numbers = [12, -4, 7, 0]
largest = max(numbers)
smallest = min(numbers)
print(largest) # 12
print(smallest) # -4max() and min() also accept two or more separate positional arguments, such as max(12, -4, 7, 0). For a collection, use the iterable form shown above. See the Python 3.13.16 built-in functions documentation for the function signatures and behavior.
Handle empty collections and tied values
Empty input
Calling max() or min() on an empty iterable without a default raises ValueError. If an empty collection is possible, either check it first or supply a meaningful default:
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numbers = []
if numbers:
largest = max(numbers)
smallest = min(numbers)
else:
largest = smallest = NoneChoose a default that clearly represents your application’s “no value” case. An arbitrary number such as 0 can be mistaken for an actual minimum or maximum.
Ties
If multiple items have the same extreme value, each function returns the first matching item encountered in the iterable.
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Choose an item by a field with key=
When a collection contains records rather than plain numbers, use key to compare a field. The function returns the original selected item, not the key value:
people = [
{"name": "Mina", "score": 82},
{"name": "Jules", "score": 94},
{"name": "Ravi", "score": 76},
]
highest_scorer = max(people, key=lambda person: person["score"])
lowest_scorer = min(people, key=lambda person: person["score"])Here, the lambda supplies the score used for comparison, while highest_scorer and lowest_scorer remain the selected dictionaries.
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Find both extrema in one pass
Lists can be traversed again, but iterators and streams may be consumed as they are read. Calling max(iterator) and then min(iterator) on the same one-pass iterator does not calculate both results from the full input: the first call advances it. The Python Functional Programming HOWTO explains that functions such as min() consume iterators.
For a stream, initialize both extrema from the first item and update them as each later item arrives:
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def extrema(values):
iterator = iter(values)
try:
first = next(iterator)
except StopIteration:
raise ValueError("extrema() requires at least one value")
smallest = largest = first
for value in iterator:
if value < smallest:
smallest = value
if value > largest:
largest = value
return smallest, largest
smallest, largest = extrema(iter([12, -4, 7, 0]))Initializing from actual input handles negative-only data without guessing numeric bounds. The explicit empty-input check prevents trying to use a nonexistent first value.
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Which approach should you use?
- List or re-iterable collection: Use
min()andmax()for straightforward, readable code. - One-pass iterator or stream: Update both extrema in the same loop, or recreate the iterator if it can be safely recreated.
- Exercise that requires comparisons: Use a loop initialized from the first item rather than assuming values are positive.
- Records compared by a field: Pass a
keyfunction to choose the record while preserving the original item.
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