I have the following python3 code
class Test:
pos = [0,0]
actions = []
def bar(self, target):
for i in target:
def _():
print(i,end="")
self.actions.append(_)
foo = Test()
foo.bar("abcd")
for i in foo.actions:
i()
Which is meant to output:
abcd
but instead it outputs:
dddd
I'm pretty sure the function is using the value of i when executing (the last value i had) and not i's value the function _ is declared, which is what I want.
The general solution to this is to store the value as a default parameter value, like this:
class Test:
pos = [0,0]
actions = []
def bar(self, target):
for i in target:
def _(i=i): # This is the only changed line
print(i,end="")
self.actions.append(_)
foo = Test()
foo.bar("abcd")
for i in foo.actions:
i()
>>> abcd
Let's output some things:
class Test:
pos = [0, 0]
actions = []
def bar(self, target):
for i in target:
print(f"i={i} id={id(i)}")
def _():
print(f"_i={i} _id={id(i)}")
print(i, end="")
self.actions.append(_)
Output:
i=a id=2590411675120
i=b id=2590411458416
i=c id=2590411377456
i=d id=2590411377200
_i=d _id=2590411377200
d_i=d _id=2590411377200
d_i=d _id=2590411377200
d_i=d _id=2590411377200
See, the i in def _ overrides every time for loop iterates and eventually last value is what you get.
How to solve this? Pass i as an argument:
from functools import partial
class Test:
pos = [0, 0]
actions = []
def bar(self, target):
for i in target:
print(f"i={i} id={id(i)}")
def _(i):
print(f"_i={i} _id={id(i)}")
print(i, end="")
self.actions.append(partial(_, i))
Output:
i=a id=2618064721392
i=b id=2618064504688
i=c id=2618064423728
i=d id=2618064423472
_i=a _id=2618064721392
a_i=b _id=2618064504688
b_i=c _id=2618064423728
Let's remove print statements now:
from functools import partial
class Test:
pos = [0, 0]
actions = []
def bar(self, target):
for i in target:
def _(i):
print(i, end="")
self.actions.append(partial(_, i))
foo = Test()
foo.bar("abcd")
for i in foo.actions:
i()
# Output: abcd
the reason why you are getting the last value of i is that the function print takes reference to i and not literal value of it and since it's in a loop you will get the last value i had. a workaround to this problem would be as Mandera said to set function default parameter and this actually works because the value of i is stored in the function's attribute __defaults__, which is responsible of storing default parameters values.
So the final code would be :
class Test:
pos = [0,0]
actions = []
def bar(self, target):
for i in target:
def foo(p=i):print(p,end="")
self.actions.append(foo)
if __name__=="__main__":
foo = Test()
foo.bar("abcd")
for i in foo.actions:
i()
Note
the function foo isn't overridden
I have a generator function in a class:
self.s = [['a',1],['b',2],['c',3]
def generator(self):
for r in self.s:
yield r
In another function I initalize it as a variable:
var = self.generator()
And it's yielded as necessary:
>>> next(var) # returns ['a',1]
>>> next(var) # returns ['b',2]
>>> next(var) # returns ['c',3]
Can defining the generator be done in one line, however? I've considered the below:
var = lambda: [(yield r) for r in self.s]
>>> next(var) # SyntaxError: 'yield' inside list comprehension
Here's a minimal code I'm working with:
class Foo():
def __init__(self):
self.s = {}
def generator(self):
for r in self.s:
yield r
def fetch(self):
if not self.s:
self.fetch_gen = self.generator()
self.s['a'] = 1
self.s['b'] = 2
self.s['c'] = 3
try:
var = self.s.get(next(self.fetch_gen))
except StopIteration:
return None
return var
BAR = Foo()
while True:
OUT = BAR.fetch()
if OUT is None:
break
print(OUT)
Output is below:
1
2
3
I just wanted to see if I could get rid of Foo.generator and instead declare the generator in one line.
You are returning a list comprehension. You can just do:
var = (r for r in self.s)
that will generate a generator with the values you want. You test it later with next(var) is in your code.
Is there any way to do something like this in python 2.7?
def scaleit(g, k):
for item in g:
yield item*k
promise = ??????
# defines a generator for reference but not use:
# other functions can make use of it,
# but it requires a call to promise.fulfill() to
# define what the generator is going to yield;
# promise raises an error if next() is called
# before the promise is fulfilled
f = scaleit(promise, 3)
promise.fulfill(range(10))
for item in f:
print item
Yes; generators don't run until they're actually iterated, so you can just defer iterating the fulfilled promise's value until requested:
class Promise(object):
def fulfill(self, result):
self.result = result
def __iter__(self):
return iter(self.result)
def scaleit(g, k):
for item in g:
yield item*k
promise = Promise()
f = scaleit(promise, 3)
promise.fulfill(range(10))
print list(f)
Is this what you want?
def scaleit(g, k):
for item in g:
yield item * k
class Promise(object):
def __init__(self):
self.g = None
def fulfill(self, g):
self.g = iter(g)
def __iter__(self):
return self
def next(self):
return next(self.g)
promise = Promise()
f = scaleit(promise, 3)
promise.fulfill(range(10))
for item in f:
print item
I think you want the send() method on generators:
def gen():
reply = yield None
if not reply: # reply will be None if send() wasn't called
raise ValueError("promise not fulfilled")
yield 5
g1 = gen()
next(g1) # advance to the first yield
g1.send(True)
print(next(g1)) # prints 5
g2 = gen()
next(g2)
# forget to send
print(next(g2)) # raises ValueError
I'm creating an interator like so:
some_list = [1,2,5,12,30,75,180,440]
iter(some_list)
I have a need to access the current value of an iterator again. Is there a current() method that allows me to stay on the same position?
You certainly can make a class which will allow you to do this:
from collections import deque
class RepeatableIter(object):
def __init__(self,iterable):
self.iter = iter(iterable)
self.deque = deque([])
def __iter__(self):
return self
#define `next` and `__next__` for forward/backward compatability
def next(self):
if self.deque:
return self.deque.popleft()
else:
return next(self.iter)
__next__ = next
def requeue(self,what):
self.deque.append(what)
x = RepeatableIter([1, 2, 3, 4, 5, 6])
count = 0
for i in x:
print i
if i == 4 and count == 0:
count += 1
x.requeue(i)
The question is really why would you want to?
You can use numpy.nditer to build your iterator, then you have many amazing options including the current value.
import numpy
rng = range(100)
itr = numpy.nditer([rng])
print itr.next() #0
print itr.next() #1
print itr.next() #2
print itr.value #2 : current iterator value
Adapting the third example from this answer:
class CurrentIterator():
_sentinal = object()
_current = _sentinal
#property
def current(self):
if self._current is self._sentinal:
raise ValueError('No current value')
return self._current
def __init__(self, iterable):
self.it = iter(iterable)
def __iter__(self):
return self
def __next__(self):
try:
self._current = current = next(self.it)
except StopIteration:
self._current = self._sentinal
raise
return current
next = __next__ # for python2.7 compatibility
Some interesting points:
use of _sentinal so an error can be raised if no current value exists
use of property so current looks like a simple attribute
use of __next__ and next = __next__ for Python 2&3 compatibility
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