让python迭代器倒退?
有没有办法让python列表迭代器倒退 基本上我有这个让python迭代器倒退?,python,list,iterator,Python,List,Iterator,有没有办法让python列表迭代器倒退 基本上我有这个 class IterTest(object): def __init__(self, data): self.data = data self.__iter = None def all(self): self.__iter = iter(self.data) for each in self.__iter: mtd = getattr
class IterTest(object):
def __init__(self, data):
self.data = data
self.__iter = None
def all(self):
self.__iter = iter(self.data)
for each in self.__iter:
mtd = getattr(self, type(each).__name__)
mtd(each)
def str(self, item):
print item
next = self.__iter.next()
while isinstance(next, int):
print next
next = self.__iter.next()
def int(self, item):
print "Crap i skipped C"
if __name__ == '__main__':
test = IterTest(['a', 1, 2,3,'c', 17])
test.all()
运行此代码将导致输出:
a
1
2
3
Crap i skipped C
我知道为什么它会给我输出,但是有没有办法让str()方法后退一步
编辑
好吧,也许让我更清楚一点。我不想做完全相反的事情,基本上我想知道的是,是否有一种简单的方法可以实现python中的双向迭代器?不,一般来说,不能让python迭代器倒退。但是,如果您只想后退一步,可以尝试以下方法:
def str(self, item):
print item
prev, current = None, self.__iter.next()
while isinstance(current, int):
print current
prev, current = current, self.__iter.next()
class buffered:
def __init__(self,it):
self.it = iter(it)
self.buf = []
def __iter__(self): return self
def __next__(self):
if self.buf:
return self.buf.pop()
return next(self.it)
def push(self,item): self.buf.append(item)
if __name__=="__main__":
b = buffered([0,1,2,3,4,5,6,7])
print(next(b)) # 0
print(next(b)) # 1
b.push(42)
print(next(b)) # 42
print(next(b)) # 2
然后,您可以随时在prev
中访问上一个元素
如果您确实需要双向迭代器,您可以自己实现一个,但它可能会带来比上述解决方案更大的开销:
class bidirectional_iterator(object):
def __init__(self, collection):
self.collection = collection
self.index = 0
def next(self):
try:
result = self.collection[self.index]
self.index += 1
except IndexError:
raise StopIteration
return result
def prev(self):
self.index -= 1
if self.index < 0:
raise StopIteration
return self.collection[self.index]
def __iter__(self):
return self
类双向迭代器(对象):
定义初始化(自我,集合):
self.collection=collection
self.index=0
def next(自我):
尝试:
结果=自收集[自索引]
自索引+=1
除索引器外:
提出停止迭代
返回结果
def prev(自我):
self.index-=1
如果自指数<0:
提出停止迭代
返回self.collection[self.index]
定义(自我):
回归自我
根据定义,迭代器是一个带有
next()
方法的对象——没有提到prev()
。因此,您必须缓存结果,以便重新访问它们,或者重新实现迭代器,以便它按您希望的顺序返回结果。我是否遗漏了一些内容,或者您不能使用
我知道这不会使迭代器倒退,但我很确定,一般来说,没有办法做到这一点。相反,编写一个迭代器,以相反的顺序遍历离散集合
编辑您还可以使用该函数为任何集合获取反向迭代器,这样您就不必编写自己的迭代器:
>>> it = reversed(['a', 1, 2, 3, 'c', 17])
>>> type(it)
<type 'listreverseiterator'>
>>> for each in it:
... print each
...
17
c
3
2
1
a
>it=reversed(['a',1,2,3',c',17])
>>>类型(it)
>>>对于其中的每一项:
... 打印每个
...
17
C
3.
2.
1.
A.
根据你的问题,听起来你想要这样的东西:
def str(self, item):
print item
prev, current = None, self.__iter.next()
while isinstance(current, int):
print current
prev, current = current, self.__iter.next()
class buffered:
def __init__(self,it):
self.it = iter(it)
self.buf = []
def __iter__(self): return self
def __next__(self):
if self.buf:
return self.buf.pop()
return next(self.it)
def push(self,item): self.buf.append(item)
if __name__=="__main__":
b = buffered([0,1,2,3,4,5,6,7])
print(next(b)) # 0
print(next(b)) # 1
b.push(42)
print(next(b)) # 42
print(next(b)) # 2
我想这会帮助你解决你的问题
class TestIterator():
def __init__(self):`
self.data = ["MyData", "is", "here","done"]
self.index = -1
#self.index=len(self.data)-1
def __iter__(self):
return self
def next(self):
self.index += 1
if self.index >= len(self.data):
raise StopIteration
return self.data[self.index]
def __reversed__(self):
self.index = -1
if self.index >= len(self.data):
raise StopIteration
return self.data[self.index]
r = TestIterator()
itr=iter(r)
print (next(itr))
print (reversed(itr))
您可以将迭代器包装在迭代器帮助器中,使其能够向后运行。它将迭代的值存储在一个集合中,并在返回时重用它们
class MemoryIterator:
def __init__(self, iterator : Iterator):
self._iterator : Iterator = iterator
self._array = []
self._isComplete = False
self._pointer = 0
def __next__(self):
if self._isComplete or self._pointer < len(self._array):
if self._isComplete and self._pointer >= len(self._array):
raise StopIteration
value = self._array[self._pointer]
self._pointer = self._pointer + 1
return value
try:
value = next(self._iterator)
self._pointer = self._pointer + 1
self._array.append(value)
return value
except StopIteration:
self._isComplete = True
def prev(self):
if self._pointer - 2 < 0:
raise StopIteration
self._pointer = self._pointer - 1
return self._array[self._pointer - 1]
其中direct是反向的
-1
,或普通的1
。Python您可以使用列表和索引来模拟迭代器:
a=[1,2,3]
电流=1
def get_下一步(a):
电流=a[a.指数(电流)+1%长度(a)]
回流
def get_last(a):
电流=a[a.指数(电流)-1]
返回电流#a[-1]>>>3(负安全)
如果列表中包含重复项,则必须单独跟踪索引:
a =[1,2,3]
index = 0
def get_next(a):
index = index+1 % len(a)
current = a[index]
return current
def get_last(a):
index = index-1 % len(a)
current = a[index-1]
return current # a[-1] >>> 3 (negative safe)
请参阅Morten Piibeleht制作的此函数。它为iterable的每个元素生成一个(上一个、当前个、下一个)元组
您可以通过以下代码使迭代器向后移动
class EnableBackwardIterator:
def __init__(self, iterator):
self.iterator = iterator
self.history = [None, ]
self.i = 0
def next(self):
self.i += 1
if self.i < len(self.history):
return self.history[self.i]
else:
elem = next(self.iterator)
self.history.append(elem)
return elem
def prev(self):
self.i -= 1
if self.i == 0:
raise StopIteration
else:
return self.history[self.i]
class EnableBackardificator:
定义初始化(self,迭代器):
self.iterator=迭代器
self.history=[None,]
self.i=0
def next(自我):
self.i+=1
如果self.i
用法:
>>> prev = lambda obj: obj.prev() # A syntactic sugar.
>>>
>>> a = EnableBackwardIterator(iter([1,2,3,4,5,6]))
>>>
>>> next(a)
1
>>> next(a)
2
>>> a.next() # The same as `next(a)`.
3
>>> prev(a)
2
>>> a.prev() # The same as `prev(a)`.
1
>>> next(a)
2
>>> next(a)
3
>>> next(a)
4
>>> next(a)
5
>>> next(a)
6
>>> prev(a)
5
>>> prev(a)
4
>>> next(a)
5
>>> next(a)
6
>>> next(a)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
prev=lambda obj:obj.prev()#语法糖。
>>>
>>>a=使能反向计数器(iter([1,2,3,4,5,6]))
>>>
>>>下一(a)
1.
>>>下一(a)
2.
>>>与“next(a)”相同。
3.
>>>上(a)
2.
>>>a.prev()#与“prev(a)”相同。
1.
>>>下一(a)
2.
>>>下一(a)
3.
>>>下一(a)
4.
>>>下一(a)
5.
>>>下一(a)
6.
>>>上(a)
5.
>>>上(a)
4.
>>>下一(a)
5.
>>>下一(a)
6.
>>>下一(a)
回溯(最近一次呼叫最后一次):
文件“”,第1行,在
停止迭代
按相反顺序访问列表元素的迭代器:
class ReverseIterator:
def __init__(self,ls):
self.ls=ls
self.index=len(ls)-1
def __iter__(self):
return self
def __next__(self):
if self.index<0:
raise StopIteration
result = self.ls[self.index]
self.index -= 1
return result
类反转器:
定义初始化(自我,ls):
self.ls=ls
自索引=len(ls)-1
定义(自我):
回归自我
定义下一个(自我):
如果self.index我来这里寻找一个双向迭代器。不确定这是否是OP所寻找的,但这是一种生成双向迭代器的方法,方法是为其提供一个属性,以指示下一步的方向:
class BidirectionalCounter:
"""An iterator that can count in two directions (up
and down).
"""
def __init__(self, start):
self.forward = True
# Code to initialize the sequence
self.x = start
def __iter__(self):
return self
def __next__(self):
if self.forward:
return self.next()
else:
return self.prev()
def reverse(self):
self.forward = not self.forward
def next(self):
"""Compute and return next value in sequence.
"""
# Code to go forward
self.x += 1
return self.x
def prev(self):
"""Compute and return previous value in sequence.
"""
# Code to go backward
self.x -= 1
return self.x
演示:
输出:
11
12
11
10
是的,我试图避免这种情况,但是,因为这会增加相当多的恼人的开销:/在上面添加了一个双向迭代器
示例,因为我已经看到您更新了您的问题,但是这可能会比我的第一个解决方案引入更多的开销。请注意,此类不会生成适当的迭代器。您可以在其实例上手动调用.next()和.prev(),但不能利用迭代器的功能,比如在for
循环或列表理解中传递它。这将引发一个TypeError:iter()返回了类型为“双向迭代器”的非迭代器。
。奇怪的是,它适合我(Python 2.7.1,Mac OS X)。在Python中,迭代器只需要一个next()
方法和一个返回自身的\uuuuuuuuuuuuuuuuuuuuuuuuuuuu
。@etuardu:您使用的是Python 3吗?iircnext
已重命名为\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu?你明确地想要能够倒退,还是仅仅想要结果倒退?哦。。。。是否要停止中间迭代并进行备份?您可能需要缓存已遍历的列表部分,并使用reversed()遍历它
my_counter = BidirectionalCounter(10)
print(next(my_counter))
print(next(my_counter))
my_counter.reverse()
print(next(my_counter))
print(next(my_counter))
11
12
11
10