Python:为什么这段代码要花很长时间(无限循环?)
我正在谷歌应用程序引擎中开发一个应用程序。我的方法之一是采用永不完成,这让我觉得它陷入了一个无限循环。我已经盯着它看了,但是想不出来 免责声明:我正在使用运行我的测试。也许它的行为很奇怪 这就是有问题的函数:Python:为什么这段代码要花很长时间(无限循环?),python,google-app-engine,recursion,infinite-loop,Python,Google App Engine,Recursion,Infinite Loop,我正在谷歌应用程序引擎中开发一个应用程序。我的方法之一是采用永不完成,这让我觉得它陷入了一个无限循环。我已经盯着它看了,但是想不出来 免责声明:我正在使用运行我的测试。也许它的行为很奇怪 这就是有问题的函数: def _traverseForwards(course, c_levels): ''' Looks forwards in the dependency graph ''' result = {'nodes': [], 'arcs': []} if c_leve
def _traverseForwards(course, c_levels):
''' Looks forwards in the dependency graph '''
result = {'nodes': [], 'arcs': []}
if c_levels == 0:
return result
model_arc_tails_with_course = set(_getListArcTailsWithCourse(course))
q_arc_heads = DependencyArcHead.all()
for model_arc_head in q_arc_heads:
for model_arc_tail in model_arc_tails_with_course:
if model_arc_tail.key() in model_arc_head.tails:
result['nodes'].append(model_arc_head.sink)
result['arcs'].append(_makeArc(course, model_arc_head.sink))
# rec_result = _traverseForwards(model_arc_head.sink, c_levels - 1)
# _extendResult(result, rec_result)
return result
起初,我认为这可能是一个递归错误,但我注释掉了递归,问题仍然存在。如果使用c_levels=0
调用此函数,它将正常运行
它引用的模型:
class Course(db.Model):
dept_code = db.StringProperty()
number = db.IntegerProperty()
title = db.StringProperty()
raw_pre_reqs = db.StringProperty(multiline=True)
original_description = db.StringProperty()
def getPreReqs(self):
return pickle.loads(str(self.raw_pre_reqs))
def __repr__(self):
return "%s %s: %s" % (self.dept_code, self.number, self.title)
class DependencyArcTail(db.Model):
''' A list of courses that is a pre-req for something else '''
courses = db.ListProperty(db.Key)
def equals(self, arcTail):
for this_course in self.courses:
if not (this_course in arcTail.courses):
return False
for other_course in arcTail.courses:
if not (other_course in self.courses):
return False
return True
class DependencyArcHead(db.Model):
''' Maintains a course, and a list of tails with that course as their sink '''
sink = db.ReferenceProperty()
tails = db.ListProperty(db.Key)
def _makeArc(source, sink):
return {'source': source, 'sink': sink}
def _getListArcTailsWithCourse(course):
''' returns a LIST, not SET
there may be duplicate entries
'''
q_arc_heads = DependencyArcHead.all()
result = []
for arc_head in q_arc_heads:
for key_arc_tail in arc_head.tails:
model_arc_tail = db.get(key_arc_tail)
if course.key() in model_arc_tail.courses:
result.append(model_arc_tail)
return result
它引用的实用程序函数:
class Course(db.Model):
dept_code = db.StringProperty()
number = db.IntegerProperty()
title = db.StringProperty()
raw_pre_reqs = db.StringProperty(multiline=True)
original_description = db.StringProperty()
def getPreReqs(self):
return pickle.loads(str(self.raw_pre_reqs))
def __repr__(self):
return "%s %s: %s" % (self.dept_code, self.number, self.title)
class DependencyArcTail(db.Model):
''' A list of courses that is a pre-req for something else '''
courses = db.ListProperty(db.Key)
def equals(self, arcTail):
for this_course in self.courses:
if not (this_course in arcTail.courses):
return False
for other_course in arcTail.courses:
if not (other_course in self.courses):
return False
return True
class DependencyArcHead(db.Model):
''' Maintains a course, and a list of tails with that course as their sink '''
sink = db.ReferenceProperty()
tails = db.ListProperty(db.Key)
def _makeArc(source, sink):
return {'source': source, 'sink': sink}
def _getListArcTailsWithCourse(course):
''' returns a LIST, not SET
there may be duplicate entries
'''
q_arc_heads = DependencyArcHead.all()
result = []
for arc_head in q_arc_heads:
for key_arc_tail in arc_head.tails:
model_arc_tail = db.get(key_arc_tail)
if course.key() in model_arc_tail.courses:
result.append(model_arc_tail)
return result
我是不是漏掉了一些很明显的东西,还是盖恩特做错了
另外,使此运行变慢的测试在数据存储中的任何类型的模型都不超过5个。我知道这可能很慢,但我的应用程序只执行一次,然后将其缓存。忽略注释掉的递归,我不认为这应该是一个无限循环-您只是在有限结果集上执行一些for循环 然而,这看起来确实很慢。在整个表上循环,然后在每个嵌套循环中执行更多的数据存储查询。这类请求似乎不太可能在GAE上及时完成,除非您的表非常非常小
一些粗略的数字: 如果
H
=dependencyard
中的实体数和T
=每个dependencyard
中尾部的平均数,则:
正在处理\u getListArcTailsWithCourse
查询(估计不足)。在“最坏”的情况下,此函数返回的H*T
结果将包含
元素H*T
\u遍历所有这些结果
次,从而执行另一个H*(H*T)查询H
- 即使
和H
仅为10秒左右,您也可能要进行数千次查询。如果他们更大,那么。。。(这将忽略在取消递归调用注释时所做的任何其他查询)T