functools.reduce

Learn accumulation with functools.reduce.

functools.reduce

functools.reduce folds a sequence into a single value. It applies a two-argument function repeatedly: at each step it combines the accumulated value with the next item.

The syntax is reduce(fn, seq, initial). fn(accumulator, item) is called for each item.

from functools import reduce

nums = [1, 2, 3, 4]
total = reduce(lambda a, b: a + b, nums, 0)
print(total)  # 10

Without an initial value, the first item of the sequence is used. Providing an initial value is safer for empty sequences.

Product and chaining with reduce

reduce is not just for sums. Use lambda a, b: a * b for products; the initial value must be 1 (since 1 is the multiplicative identity).

The initial value matters: 0 for sums, 1 for products, "" for string concatenation. A wrong initial value corrupts the result.

Without an initial value, reduce on an empty sequence raises TypeError. Safe code always provides one.

Exercise Compute a product with reduce

Compute the product of nums = [2, 3, 4] with reduce. Use lambda a, b: a * b as the combiner and 1 as the initial value. Print the result.

Show example solution

from functools import reduce

nums = [2, 3, 4]
result = reduce(lambda a, b: a * b, nums, 1)
print(result)

Test yourself

Complete the reduce call

from functools import reduce

nums = [1, 2, 3, 4]
total = ____(lambda a, b: a + b, nums, 0)
print(total)

Show answer

from functools import reduce

nums = [1, 2, 3, 4]
total = reduce(lambda a, b: a + b, nums, 0)
print(total)

Match the reduce steps

fn(accumulator, item)

Per-step call
initial=0

Starting accumulator
lambda a, b: a+b

Summing combiner
reduce

Fold operation
What should the reduce initial value be for a product?
- 1
- 0
- Empty string ""
- None
Show answer

1 — 1 is the multiplicative identity; starting at 1 keeps the product correct.

Next lesson: Summary - Advanced decorators

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