Fix - NameError: name 'reduce' is not defined in Python

Fix - NameError: name 'reduce' is not defined in Python #

The Python "NameError: name 'reduce' is not defined" occurs when we use the reduce() function without importing it first. To solve the error, import the function from the functools module before using it - from functools import reduce.

Here is an example of how the error occurs.

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def do_math(accumulator, current):
    return accumulator + current

# ⛔️ NameError: name 'reduce' is not defined
print(reduce(do_math, [1, 2, 3], 0))  # 👉️ 6

To solve the error, we have to import the reduce function from the functools module.

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from functools import reduce


def do_math(accumulator, current):
    return accumulator + current


print(reduce(do_math, [1, 2, 3], 0))  # 👉️ 6

The reduce function takes the following 3 parameters:

You will often see the reduce() function getting passed a lambda:

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from functools import reduce


# 👇️ 6
print(
    reduce(
        lambda accumulator, current: accumulator + current,
        [1, 2, 3],
        0
    )
)

The lambda function in the example takes the accumulated value and the current value as parameters and returns the sum of the two.

If we provide a value for the initializer argument, it is placed before the items of the iterable in the calculation.

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from functools import reduce


def do_math(accumulator, current):
    print(accumulator) # 👉️ is 0 on first iteration
    return accumulator + current


# 👇️ 6
print(
    reduce(
        do_math, [1, 2, 3],
        0  # 👈️ initializer set to 0
    )
)

In the example, we passed 0 for the initializer argument, so the value of the accumulator will be 0 on the first iteration.

The value of the accumulator would get set to the first element in the iterable if we didn't pass a value for the initializer.

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def do_math(accumulator, current):
    print(accumulator)  # 👉️ is 1 on first iteration
    return accumulator + current


# 👇️ 6
print(
    reduce(
        do_math, [1, 2, 3],
    )
)

If the iterable is empty and the initializer is provided, the initializer is returned.

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from functools import reduce


def do_math(accumulator, current):
    return accumulator + current


# 👇️ 100
print(
    reduce(
        do_math,
        [],
        100  # 👈️ initializer set to 100
    )
)

If the initializer is not provided and the iterable contains only 1 item, the first item is returned.

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from functools import reduce


def do_math(accumulator, current):
    return accumulator + current


# 👇️ 123
print(
    reduce(
        do_math,
        [123],
    )
)

The list in the example only contains a single element and we didn't provide a value for the initializer, so the reduce() function returned the list element.