使用渐近线完成程序

发布于 2024-12-05 13:00:46 字数 328 浏览 0 评论 0原文

这个问题可以用与语言无关的方式来回答，但我使用的是 python （仅供参考）。

我正在运行一个无限循环，当在最后一分钟发现<10个新对象时需要终止。

例如：


while True:
    newobjs = dig_more_objects(obj)
    if less than 10 newobjs have been discovered over the last minute
        break

编辑：问题是：我如何实现这一行：
如果最后一分钟发现的新对象少于 10 个

原文

This question could be answered in a language-agnostic fashion, but I am using python (fyi).

I am running an infinite loop which needs to terminate when during the last minute <10 new objects are discovered.

Eg:


while True:
    newobjs = dig_more_objects(obj)
    if less than 10 newobjs have been discovered over the last minute
        break

EDIT: The question is this: How do I implement this line:

if less than 10 newobjs have been discovered over the last minute

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永言不败 2024-12-12 13:00:46

这是一个粗略的尝试 - 根据 dig_more_objects 的性质，您可能需要调整条件：

import time
results = []
while True:
    mark = time.time()
    newobjs = dig_more_objects(obj)
    elapsed = time.time() - mark
    results.append((newobjs, elapsed))
    count = 0
    threshhold = 0
    for objs, elapsed in results[::-1]:
        count += len(objs)  # or +1 of dig_more_objects only returns one at a time
        threshhold += elapsed
        if threshhold > 60.0 and count < 10:
            break

Here's a rough stab at it -- depending on the nature of dig_more_objects you may need to adjust the condition:

import time
results = []
while True:
    mark = time.time()
    newobjs = dig_more_objects(obj)
    elapsed = time.time() - mark
    results.append((newobjs, elapsed))
    count = 0
    threshhold = 0
    for objs, elapsed in results[::-1]:
        count += len(objs)  # or +1 of dig_more_objects only returns one at a time
        threshhold += elapsed
        if threshhold > 60.0 and count < 10:
            break

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