Making a python iterator go backwards?

An iterator is by definition an object with the next() method — no mention of prev() whatsoever. Thus, you either have to cache your results so you can revisit them or reimplement your iterator so it returns results in the sequence you want them to be.

Am I missing something or couldn’t you use the technique described in the Iterator section in the Python tutorial?

>>> class reverse_iterator:
...     def __init__(self, collection):
...         self.data = collection
...         self.index = len(self.data)
...     def __iter__(self):
...         return self
...     def next(self):
...         if self.index == 0:
...             raise StopIteration
...         self.index = self.index - 1
...         return self.data[self.index]
...     
>>> for each in reverse_iterator(['a', 1, 2, 3, 'c', 17]):
...     print each
... 
17
c
3
2
1
a

I know that this doesn’t walk the iterator backwards, but I’m pretty sure that there is no way to do that in general. Instead, write an iterator that walks a discrete collection in reverse order.

Edit you can also use the reversed() function to get a reversed iterator for any collection so that you don’t have to write your own:

>>> it = reversed(['a', 1, 2, 3, 'c', 17])
>>> type(it)
<type 'listreverseiterator'>
>>> for each in it:
...  print each
... 
17
c
3
2
1
a

Based on your question, it sounds like you want something like this:

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

i think thi will help you to solve your problem

    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))

You can wrap your iterator in an iterator helper to enable it to go backward. It will store the iterated values in a collection and reuse them when going backwards.

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]

The usage can be similar to this one:

my_iter = iter(my_iterable_source)
memory_iterator = MemoryIterator(my_iter)
try:
    if forward:
        print(next(memory_iterator))
    else:
        print(memory_iterator.prev())
except StopIteration:
    pass

ls = [' a', 5, ' d', 7, 'bc',9, ' c', 17,  '43', 55, 'ab',22, 'ac']
direct = -1
l = ls[::direct]
for el in l:
    print el

Where direct is -1 for reverse or 1 for ordinary.

Python you can use a list and indexing to simulate an iterator:

a = [1,2,3]

current = 1

def get_next(a):
  current = a[a.index(current)+1%len(a)]
  return current
def get_last(a):
  current = a[a.index(current)-1]
  return current # a[-1] >>> 3 (negative safe)

if your list contains duplicates then you would have to track your index separately:

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)

please see this function made by Morten Piibeleht. It yields a (previous, current, next) tuple for every element of an iterable.

https://gist.github.com/mortenpi/9604377

You can enable an iterator to move backwards by following code.

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]

Usage:

>>> 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

An iterator that visits the elements of a list in reverse order:

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

I came here looking for a bi-directional iterator. Not sure if this is what the OP was looking for but it is one way to make a bi-directional iterator—by giving it an attribute to indicate which direction to go in next:

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

Demo:

my_counter = BidirectionalCounter(10)
print(next(my_counter))
print(next(my_counter))
my_counter.reverse()
print(next(my_counter))
print(next(my_counter))

Output:

11
12
11
10

I edited the python code from dilshad (thank you) and used the following Python 3 based code to step between list item’s back and forth or let say bidirectional:

# bidirectional class
class bidirectional_iterator:
def __init__(self):
    self.data = ["MyData", "is", "here", "done"]
    self.index = -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 == -1:
        raise StopIteration
    return self.data[self.index]

Example:

>>> r = bidirectional_iterator()
>>> itr=iter(r)
>>> print (next(itr))
MyData
>>> print (next(itr))
is
>>> print (next(itr))
here
>>> print (reversed(itr))
is
>>> print (reversed(itr))
MyData
>>> print (next(itr))
is

This is a common situation when we need to make an iterator go back one step. Because we should get the item and then check if we should break the loop. When breaking the loop, the last item may be requied in later usage.

Except of implementing an iteration class, here is a handy way make use of builtin itertools.chain :

from itertools import chain

>>> iterator = iter(range(10))
>>> for i in iterator:
...     if i <= 5:
...         print(i)
...     else:
...         iterator = chain([i], iterator)  # push last value back
...         break
...         
0
1
2
3
4
5
>>> for i in iterator:
...     print(i)
...     
6
7
8
9