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()？它有什么优势？

from statistics import median
from itertools import tee

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)

def main():
    stats = process_purchases(purchases=purchases)
    print("Result:", stats)
    stats = _process_purchases(purchases=purchases)
    print("Result:", stats)

if __name__ == '__main__':
    main()

Answer 1

迭代器只能在 python 中迭代一次。之后它们是 "exhausted" 并且没有 return 更多值。

您可以在 map()、zip()、filter() 和许多其他函数中看到这一点：

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 map() 中的 return 值，您可以看到它们之间的区别。在这种情况下 _process_purchases() 失败，因为 min() 耗尽了迭代器并且没有为 max() 和 median() 留下任何值。

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)

itertools 库中的 tee() 函数

tee() function from itertools library

python

itertools

tee