Python 对next()和iter()的确切工作方式感到困惑
在尝试实现迭代器时,我对next()感到困惑。我制作了一个简单的测试脚本,其中迭代器的工作方式与我预期的一样:Python 对next()和iter()的确切工作方式感到困惑,python,iterator,next,Python,Iterator,Next,在尝试实现迭代器时,我对next()感到困惑。我制作了一个简单的测试脚本,其中迭代器的工作方式与我预期的一样: class Object: def __init__(self, name): self.name = name def prin(self): print self.name class Some: def __init__(self): self.data = list() def __iter__(
class Object:
def __init__(self, name):
self.name = name
def prin(self):
print self.name
class Some:
def __init__(self):
self.data = list()
def __iter__(self):
return self.SomeIter(self, len(self.data))
def fill(self, obj):
self.data.append(obj)
def printMe(self):
for entry in self:
print entry.name
class SomeIter:
def __init__(self, some, end):
self.index = 0
self.end = end
self.name = some.data[self.index].name
self.data = some.data
def next(self):
if self.index == self.end:
raise StopIteration
else:
self.name = self.data[self.index].name
self.index += 1
return self
########################################################################
someX = Some()
obj1 = Object("A")
obj2 = Object("B")
someX.fill(obj1)
someX.fill(obj2)
someX.fill(obj2)
for obj in someX:
print obj.name
我得到“abb”作为输出。那很好。但是我还有一个树类的迭代器。next()-方法的工作原理基本相同。我首先更新实例,然后返回它。但是对于树迭代器,第一个元素被跳过。这对我来说很有意义,因为我只有在更新实例后才返回self。但是,在上面的实现中,我为什么会得到不同的行为,在这种情况下,实例会被更新,然后也会被返回
########################################################################
# RIGHT-HAND-CORNER-BOTTOM-UP-POST-ORDER-TRAVERSAL-ITERATOR
########################################################################
class RBPIter:
"""!
@brief Right hand corner initialised iterator, traverses tree bottom
up, right to left
"""
def __init__(self, tree):
self.current = tree.get_leaves(tree.root)[-1] # last leaf is right corner
self.csi = len(self.current.sucs)-1 # right most index of sucs
self.visited = list() # visisted nodes
self.tree = tree
self.label = self.current.label
########################################################################
def __iter__(self):
return self
########################################################################
def begin(self):
return self.tree.get_leaves(self.tree.root)[-1]
########################################################################
def end(self):
return self.tree.root
########################################################################
def find_unvisited(self, node):
"""!
@brief finds rightmost unvisited node transitively dominated by node
"""
leaves = self.tree.get_leaves(self.tree.root)
# loop through leaves from right to left, as leaves are listed
# in order, thus rightmost list elememt is rightmost leaf
for i in range(len(leaves)-1, -1, -1):
# return the first leaf, that has not been visited yet
if leaves[i] not in self.visited:
self.label = leaves[i].label
return leaves[i]
# return None if all leaves have been visited
return None
########################################################################
def go_up(self, node):
"""!
@brief sets self.current to pred of self.current,
appends current node to visited nodes, reassignes csi
"""
self.visited.append(self.current)
self.current = self.current.pred
if self.current.sucs[0] not in self.visited:
self.current = self.find_unvisited(self.current)
self.label = self.current.label
self.csi = len(self.current.sucs)-1
self.visited.append(self.current)
########################################################################
def next(self):
"""!
@brief advances iterator
"""
# if current node is a leaf, go to its predecessor
if self.current.suc_count == 0 or self.current in self.visited:
self.go_up(self.current)
# if current node is not a leaf, find the next unvisited
else:
self.current = self.find_unvisited(self.current)
if self.current == self.end():
raise StopIteration
return self
编辑1:
我比较了两个迭代器的输出,它们不同,Someter将第一个元素输出了两次:
next: A
A
next: A
B
next: B
B
next: B
另一个迭代器不:
next: a
s
next: s
i
next: i
r
next: r
t
next: t
t
next: t
s
next: s
e
next: e
t
next: t
否则,“下一步:a”将出现2次
编辑2:
这对我来说真的毫无意义
查看这些调用和输出:
someXIter = iter(someX)
print someXIter.next().name
print someXIter.next().name
print someXIter.next().name
输出:
next: A
A
next: A
B
next: B
B
使用此代码:
class SomeIter:
def __init__(self, some, end):
self.index = 0
self.end = end
self.name = some.data[self.index].name
self.data = some.data
def next(self):
print "next: ", self.name
if self.index == self.end:
raise StopIteration
else:
self.name = self.data[self.index].name
self.index += 1
return self
为什么这对我来说毫无意义?因为,在第一次调用next()时,它会打印“next:A”,然后更新实例,并打印函数调用的返回值,这也是“A”。哇?为什么不是“B”,因为返回值应该是更新的实例。Python2.7
要成为迭代器,必须实现迭代器协议:
- 定义对象和下一个对象
必须返回obj.\uuuuu iter\uuuuuu
self
- 一旦引发了
,对StopIteration
(obj.next()
)的后续调用必须引发next(obj)
StopIteration
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
,必须返回一个实现迭代器协议的对象。如果类中的项包含在类似列表的内置类型中,\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
可以简单地返回iter(list)
我猜整个概念中隐含的是迭代器必须跟踪它在迭代中的位置
如果希望对象是具有不同迭代序列的迭代器,可以定义不同的方法来跟踪迭代并前进到下一项,然后在obj.next()
中使用这些方法
简单的例子:
class Thing(object):
def __init__(self, name):
self.name = name
def __str__(self):
return self.name
def __repr__(self):
return 'Thing({})'.format(self.name)
class Some(object):
def __init__(self):
self.data = None
# need something to keep track of the iteration sequence
self.__index = None
# type of iteration do you want to do
self.direction = self.forward
def __iter__(self):
# reset the iteration
self.__index = None
return self
def next(self):
try:
return self.direction()
except IndexError:
raise StopIteration
def forward(self):
if self.__index is None:
self.__index = -1
self.__index += 1
return self.data[self.__index]
def reverse(self):
if self.__index is None:
self.__index = 0
self.__index -= 1
return self.data[self.__index]
用法
因此,也许只需保持next
简单,并将树遍历逻辑放在不同的方法中。这可能会使测试这种逻辑变得更容易。您可以随时添加不同的行为:
def odd_then_even(self):
if self.__index is None:
self.__index = -1
self.__odd = True
self.__index += 2
try:
return self.data[self.__index]
except IndexError:
if not self.__odd:
raise IndexError
self.__odd = False
self.__index = 0
return self.data[self.__index]
>>> some.direction = some.odd_then_even
>>> for thing in some:
print thing,
B D A C
>>>
我很难理解您的内部类解决方案是如何工作的,但有一件事情看起来不太对劲,那就是迭代器对象的next
方法正在返回self,而next
似乎应该返回序列/集合中的下一项。当您对一组内容进行迭代时,迭代应该提供单个内容,而不是迭代器对象的修改实例。您是否可以将再现问题的代码量降至最低?但是第一段代码已经是一个最小化的例子了?它也包含执行代码。输出也在我的帖子里。我不明白你想让我怎样把它最小化。你为什么把SomeIter
作为一个类来实现呢?一是因为我认为迭代器必须是对象,二是因为我在实际程序中需要几个迭代器。你对next
或你的实现感到困惑吗<代码>下一步
应返回下一项,并在耗尽时引发停止迭代
。Python的哪个版本?你为什么不使用新型的类呢?
def odd_then_even(self):
if self.__index is None:
self.__index = -1
self.__odd = True
self.__index += 2
try:
return self.data[self.__index]
except IndexError:
if not self.__odd:
raise IndexError
self.__odd = False
self.__index = 0
return self.data[self.__index]
>>> some.direction = some.odd_then_even
>>> for thing in some:
print thing,
B D A C
>>>