itertools 库中的 tee() 函数

Question

这是一个从列表中获取最小值、最大值和平均值的简单示例。下面的两个函数有相同的结果。我想知道这两个函数之间的区别。为什么要使用itertools.tee()？它提供了什么优势？from statistics import medianfrom itertools import teepurchases = [1, 2, 3, 4, 5]def process_purchases(purchases):    min_, max_, avg = tee(purchases, 3)    return min(min_), max(max_), median(avg)def _process_purchases(purchases):    return min(purchases), max(purchases), median(purchases)def main():    stats = process_purchases(purchases=purchases)    print("Result:", stats)    stats = _process_purchases(purchases=purchases)    print("Result:", stats)if __name__ == '__main__':    main()

Answer 1

迭代器在 python 中只能迭代一次。之后，它们“耗尽”并且不再返回更多值。

map()您可以在、和许多其他函数zip()中看到这一点：filter()

purchases = [1, 2, 3, 4, 5]

double = map(lambda n: n*2, purchases)

print(list(double))

# [2, 4, 6, 8, 10]

print(list(double))

# [] <-- can't use it twice

如果将迭代器传递给两个函数，例如 的返回值，您可以看到它们之间的区别map()。在这种情况下_process_purchases()失败，因为用尽了迭代器并且没有为和min()留下任何值。max()median()

tee()接受一个迭代器并给你两个或更多，允许你多次使用传递给函数的迭代器：

from itertools import tee

from statistics import median

purchases = [1, 2, 3, 4, 5]

def process_purchases(purchases):

    min_, max_, avg = tee(purchases, 3)

    return min(min_), max(max_), median(avg)

def _process_purchases(purchases):

    return min(purchases), max(purchases), median(purchases)

double = map(lambda n: n*2, purchases)

_process_purchases(double)

# ValueError: max() arg is an empty sequence

double = map(lambda n: n*2, purchases)

process_purchases(double)

# (2, 10, 6)