這是一個從列表中獲取最小值、最大值和平均值的簡單示例。下面的兩個函數有相同的結果。我想知道這兩個函數之間的區別。為什么要使用itertools.tee()？它提供了什么優勢？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()
1 回答
MYYA
TA貢獻1868條經驗 獲得超4個贊
迭代器在 python 中只能迭代一次。之后，它們“耗盡”并且不再返回更多值。
map()您可以在、和許多其他函數zip()中看到這一點：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
如果將迭代器傳遞給兩個函數，例如 的返回值，您可以看到它們之間的區別map()。在這種情況下_process_purchases()失敗，因為用盡了迭代器并且沒有為和min()留下任何值。max()median()
tee()接受一個迭代器并給你兩個或更多，允許你多次使用傳遞給函數的迭代器：
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)
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