我得到了一个结果列表，希望保留比时间线更新和比书签旧的元素。有没有比迭代整个列表并删除符合条件的元素更方便的方法？你能给我介绍一下如何具体吗？我获取数据然后进行排序的方式
results = A.all().search(self.request.get( q )).filter("published =", True)
results = sorted(results, key=lambda x: x.modified, reverse=True)
然后，我想保留比bookmark和timeline更早的元素，其中这些变量由HTTPGET定义，或者如果为空，则定义为
bookmark = datetime.now()
timeline = datetime.now () - timedelta (days = 50)
我希望你能理解我想做的事情（就像分页一样），并提前感谢你的任何想法。
如果您正在使用SearchableModel并进行搜索查询（从片段中可以看出您是这样的），则无法在不需要分解索引的情况下应用排序顺序或不等式过滤器，正如您在上一个主题问题中所述。因此，您不能将这些过滤器作为查询的一部分来应用，因此手动过滤结果是最好的（也是唯一的）选择。
这
bookmark = datetime.now()
timeline = datetime.now () - timedelta (days = 50)
results = A.all().search(self.request.get( q )).filter("published =", True)
results = results.filter( modified__gte=bookmark ).filter( modified__lte=timeline )
results = sorted(results, key=lambda x: x.modified, reverse=True)
还是这个？
bookmark = datetime.now()
timeline = datetime.now () - timedelta (days = 50)
results = A.all().search(self.request.get( q )).filter("published =", True)
results = [ r for r in results if r.modified >= bookmark and r.modified <= timeLine ]
results = sorted(results, key=lambda x: x.modified, reverse=True)
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