Python为何无可迭代对象乘法辅助函数？技术疑问

Great question—this gets to some core design choices in Python, so let's unpack it.

First, let's clarify what "multiplication" means for iterables. When you do something like [1,2] * 3 or "abc" * 2, you're using sequence repetition—a behavior built into sequence types (lists, tuples, strings, etc.) via their __mul__ magic method. This isn't a universal "iterable multiplication" because not all iterables support it (think generators or file objects, which are one-time use).

Why no dedicated times(iterable) function? #

Python follows the philosophy of "There should be one—and preferably only one—obvious way to do it." For sequences, that obvious way is already the * operator. Adding a separate times function would duplicate functionality, which goes against Python's goal of keeping the language clean and uncluttered.

What about operator.mul? #

You're right that mul isn't a built-in function—it lives in the operator module for a reason: to keep the global namespace from being flooded with operator-specific functions. But here's a fun fact: operator.mul actually does work with sequences that support __mul__! Try this:

import operator
operator.mul([1,2,3], 2)  # Returns [1,2,3,1,2,3]

It's just a generic function that delegates to the object's __mul__ method, so it works for numbers and sequences. It doesn't get top billing in math docs because the math module focuses on numerical math, not sequence operations.

What about non-sequence iterables? #

If you want to repeat a non-sequence iterable (like a generator) multiple times, Python doesn't have a single built-in function for it, but you can easily build it with itertools.chain:

from itertools import chain

def times(iterable, n):
    return chain.from_iterable(iterable for _ in range(n))

# Example: Repeat a generator 3 times
gen = (x for x in [1,2])
list(times(gen, 3))  # Returns [1,2,1,2,1,2]

This isn't built-in because the use case is less common than sequence repetition, and Python prefers to give you flexible building blocks rather than specialized functions for every edge case.

内容的提问来源于stack exchange，提问作者Torxed