We Keep Coding

How to remove output 'None' when iterator is called?

Well, I wrote an iterator Divisors(n) that gets a natural number n as its argument and generates divisors of n (from the smallest to the largest). Here's code:
class Divisors:
def __init__(self, n):
self.n = n
self.i = 0
def __iter__(self):
return self
def __next__(self):
self.i += 1
if self.i == self.n + 1:
raise StopIteration
if self.n % self.i == 0:
return self.i
But, when I run it, I get:
>>> print([x for x in Divisors(12)])
[1, 2, 3, 4, None, 6, None, None, None, None, None, 12]
Question is: How to remove these None's from list to obtain:
[1, 2, 3, 4, 6, 12]
Yes, I know this can be done like this:
def divisors(n):
i = 1
while i <= n:
if n % i == 0:
yield i
i += 1
but I am interested in way I mentioned above.

You could keep incrementing self.i till it evenly divides the self.n:
class Divisors:
def __init__(self, n):
self.n = n
self.i = 0
def __iter__(self):
return self
def __next__(self):
self.i += 1
if self.i == self.n + 1:
raise StopIteration
# Keep incrementing
while self.n % self.i != 0:
self.i += 1
# Now it is guaranteed to divide
return self.i

I suspect this might work:
def __next__(self):
self.i += 1
if self.i == self.n + 1:
raise StopIteration
if self.n % self.i == 0:
return self.i
else
return self.__next__()
Some perfer the following style which should have a similar effect.
def __next__(self):
self.i += 1
if self.i == self.n + 1:
raise StopIteration
if self.n % self.i == 0:
return self.i
return self.__next__()

Doubly-linked list in Python

It turned out to be an error as a time limit was exceeded,
but I've already raised the StopIteration...
I think I did something wrong for my iteration part, but it's really hard to find the error. The test output keeps running and even printed out the None value. How does it happen?
class LinkedListIterator:
def __init__(self, head):
self.__current = head.get_next()
def __iter__(self):
return self
def __next__(self):
if self.__current == None:
raise StopIteration
else:
item = self.__current.get_data()
self.__current = self.__current.get_next()
return item
These were the inputs I used to run the program:
my_list = LinkedListDLL()
my_list.add_to_head(1)
print("Contents:", end=" ")
for node in my_list:
print(node, end=" ")
print()

This code is meant to stop iteration when it reaches the head of the list.
if self.__current == None:
raise StopIteration
However, you represent the head with a NodeDLL object which is different from None.
You could keep a reference to the head and check against that instead:
class LinkedListIterator:
def __init__(self, head):
self._head = head
self._current = head.get_next()
def __iter__(self):
return self
def __next__(self):
if self._current is self._head:
raise StopIteration
else:
item = self._current.get_data()
self._current = self._current.get_next()
return item

What you want to implement is the API of a MutableSequence with the implementation of a doubly-linked-list.
To do that in Python, you should rely on collections.abc which can guide you through the process of implementing all required methods.
By example, a linked-list is actually a class inheriting from MutableSequence.
from collections.abc import MutableSequence
class LinkedList(MutableSequence):
pass
ll = LinkedList()
On instantiation of a class which has some abstract methods not yet written, you will get a TypeError which will guide you through which methods need to be implemented.
TypeError: Can't instantiate abstract class LinkedList with abstract methods __delitem__, __getitem__, __len__, __setitem__, insert
In particular, note that a list or a linked-list is not an iterator, it is an iterable. What this means is the __iter__ method should not return self and rely on __next__, it should instead return a brand new iterator on the content of the linked-list.
In other words, you can iterate only once through an iterator and multiple times through and iterable.
Full implementation
It turns out I have a full implementation of a doubly-linked-list implemented that way. You can have a look.
from collections.abc import MutableSequence
class LinkedList(MutableSequence):
class _Node:
def __init__(self, value, _next=None, _last=None):
self.value, self._next, self._last = value, _next, _last
def __str__(self):
return f'Node({self.value})'
def __init__(self, iterable=()):
self.start = None
self.last = None
empty = object()
iterable = iter(iterable)
first = next(iterable, empty)
if first is empty:
return
current = self._Node(first)
self.start, self.last = current, current
for value in iterable:
new_node = self._Node(value, _last=self.last)
self.last._next = new_node
self.last = new_node
def __len__(self):
if self.start is None:
return 0
else:
return sum(1 for _ in self)
def __iter_nodes(self):
current = self.start
while current is not None:
yield current
current = current._next
def __reversed_iter_nodes(self):
current = self.last
while current is not None:
yield current
current = current._last
def __iter__(self):
for node in self.__iter_nodes():
yield node.value
def __reversed__(self):
for node in self.__reversed_iter_nodes():
yield node.value
def __get_node(self, index):
if index >= 0:
for item in self.__iter_nodes():
if index == 0:
return item
index -= 1
else:
for item in self.__reversed_iter_nodes():
if index == 0:
return item
index += 1
raise IndexError
def __getitem__(self, index):
if index >= 0:
for item in self:
if index == 0:
return item.value
index -= 1
else:
for item in reversed(self):
if index == 0:
return item.value
index += 1
raise IndexError
def __setitem__(self, key, value):
self[key].value = value
def __delitem__(self, key):
node = self[key]
if node._last:
node._last._next = node._next
if node._next:
node._next._last = node._last
def insert(self, index, value):
if index > len(self):
self.last = self._Node(value, _last=self.last)
else:
where = self.__get_node(index)
_last = where._last
new_node = self._Node(value, _next=where, _last=_last)
if _last:
_last._next = new_node
else:
self.start = new_node
where._last = new_node
Example
ll = LinkedList(range(1, 5))
print(*ll)
print(*reversed(ll))
ll.insert(2, 'foo')
print(*ll)
Output
1 2 3 4
4 3 2 1
1 2 foo 3 4

Stop __next__ from returning None [duplicate]

Why is my iterator returning extra 'None' in the output. For the parameters/example below, I am getting [None,4,None] instead of the desired [4] Can anyone explain why I am getting the extra None and how I can fix it? The print out 'returning' only appears once so I am assuming only one item should be appended to the returning calling function.
code:
class Prizes(object):
def __init__(self,purchase,n,d):
self.purchase = purchase
self.length = len(purchase)
self.i = n-1
self.n = n
self.d = d
def __iter__(self):
return self
def __next__(self):
if self.i < self.length:
old = self.i
self.i += self.n
if (self.purchase[old])%(self.d) == 0:
print("returning")
return old+1
else:
raise StopIteration
def superPrize(purchases, n, d):
return list(Prizes(purchases, n, d))
purchases = [12, 43, 13, 465, 1, 13]
n = 2
d = 3
print(superPrize(purchases, n, d))
Output:
returning
[None, 4, None]

As people in the comments have pointed out, your line if (self.purchase[old])%(self.d) == 0: leads to the function returning without any return value. If there is no return value supplied None is implied. You need some way of continuing through your list to the next available value that passes this test before returning or raising StopIteration. One easy way of doing this is simply to add an extra else clause to call self.__next__() again if the test fails.
def __next__(self):
if self.i < self.length:
old = self.i
self.i += self.n
if (self.purchase[old])%(self.d) == 0:
print("returning")
return old+1
else:
return self.__next__()
else:
raise StopIteration

Functions return None if you don't have an explicit return statement. That's what happens in __next__ when if (self.purchase[old])%(self.d) == 0: isn't true. You want to stay in your __next__ until it has a value to return.
class Prizes(object):
def __init__(self,purchase,n,d):
self.purchase = purchase
self.length = len(purchase)
self.i = n-1
self.n = n
self.d = d
def __iter__(self):
return self
def __next__(self):
while self.i < self.length:
old = self.i
self.i += self.n
if (self.purchase[old])%(self.d) == 0:
return old+1
raise StopIteration
def superPrize(purchases, n, d):
return list(Prizes(purchases, n, d))
purchases = [12, 43, 13, 465, 1, 13]
n = 2
d = 3
print(superPrize(purchases, n, d))

Python iterator returning unwanted 'None'

Why is my iterator returning extra 'None' in the output. For the parameters/example below, I am getting [None,4,None] instead of the desired [4] Can anyone explain why I am getting the extra None and how I can fix it? The print out 'returning' only appears once so I am assuming only one item should be appended to the returning calling function.
code:
class Prizes(object):
def __init__(self,purchase,n,d):
self.purchase = purchase
self.length = len(purchase)
self.i = n-1
self.n = n
self.d = d
def __iter__(self):
return self
def __next__(self):
if self.i < self.length:
old = self.i
self.i += self.n
if (self.purchase[old])%(self.d) == 0:
print("returning")
return old+1
else:
raise StopIteration
def superPrize(purchases, n, d):
return list(Prizes(purchases, n, d))
purchases = [12, 43, 13, 465, 1, 13]
n = 2
d = 3
print(superPrize(purchases, n, d))
Output:
returning
[None, 4, None]

As people in the comments have pointed out, your line if (self.purchase[old])%(self.d) == 0: leads to the function returning without any return value. If there is no return value supplied None is implied. You need some way of continuing through your list to the next available value that passes this test before returning or raising StopIteration. One easy way of doing this is simply to add an extra else clause to call self.__next__() again if the test fails.
def __next__(self):
if self.i < self.length:
old = self.i
self.i += self.n
if (self.purchase[old])%(self.d) == 0:
print("returning")
return old+1
else:
return self.__next__()
else:
raise StopIteration

Functions return None if you don't have an explicit return statement. That's what happens in __next__ when if (self.purchase[old])%(self.d) == 0: isn't true. You want to stay in your __next__ until it has a value to return.
class Prizes(object):
def __init__(self,purchase,n,d):
self.purchase = purchase
self.length = len(purchase)
self.i = n-1
self.n = n
self.d = d
def __iter__(self):
return self
def __next__(self):
while self.i < self.length:
old = self.i
self.i += self.n
if (self.purchase[old])%(self.d) == 0:
return old+1
raise StopIteration
def superPrize(purchases, n, d):
return list(Prizes(purchases, n, d))
purchases = [12, 43, 13, 465, 1, 13]
n = 2
d = 3
print(superPrize(purchases, n, d))

Making a python iterator go backwards?

Is there anyway to make a python list iterator to go backwards?
Basically i have this
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()
Running this code results in the output:
a
1
2
3
Crap i skipped C
I know why it gives me the output, however is there a way i can step backwards in the str() method, by one step?
EDIT
Okay maybe to make this more clear. I don't want to do a full reverse, basically what i want to know if there is an easy way to do the equivalent of a bidirectional iterator in python?

No, in general you cannot make a Python iterator go backwards. However, if you only want to step back once, you can try something like this:
def str(self, item):
print item
prev, current = None, self.__iter.next()
while isinstance(current, int):
print current
prev, current = current, self.__iter.next()
You can then access the previous element any time in prev.
If you really need a bidirectional iterator, you can implement one yourself, but it's likely to introduce even more overhead than the solution above:
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

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

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.

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

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

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

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 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))

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)

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 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

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