156

I need to iterate over a circular list, possibly many times, each time starting with the last visited item.

The use case is a connection pool. A client asks for connection, an iterator checks if pointed-to connection is available and returns it, otherwise loops until it finds one that is available.

How can I do this neatly in Python?

If you instead need an immediately created list of the results up to a certain length, rather than iterating on demand: see Repeat list to max number of elements for general techniques, and How to replicate array to specific length array for Numpy-specific techniques.

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11 Answers 11

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266

Use itertools.cycle, that's its exact purpose:

from itertools import cycle

lst = ['a', 'b', 'c']

pool = cycle(lst)

for item in pool:
    print(item)

Output:

a b c a b c ...

(Loops forever, obviously)

In order to manually advance the iterator and pull values from it one by one, simply call next(pool):

>>> next(pool)
'a'
>>> next(pool)
'b'
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6 Comments

You are printing items in a loop. What I want to leave the loop and come back later? (I want to start where I left off).
@user443854 use pool.next() to get the single next item from the cycle
@user443854 FWIW this is a much better answer than mine. No reason to go around re-implementing library functions!
pool.next() didn't work for me, only next(pool). Probably because of Python 3?
@fjsj that is correct, on Python 3 you need to use next(iterator) (which BTW also works just fine on Python 2.x, and therefore is the canonical form that should be used). See Is generator.next() visible in python 3.0? for a more in-depth explanation. Updated my answer accordingly.
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76

The correct answer is to use itertools.cycle. But, let's assume that library function doesn't exist. How would you implement it?

Use a generator:

def circular():
    while True:
        for connection in ['a', 'b', 'c']:
            yield connection

Then, you can either use a for statement to iterate infinitely, or you can call next() to get the single next value from the generator iterator:

connections = circular()
next(connections) # 'a'
next(connections) # 'b'
next(connections) # 'c'
next(connections) # 'a'
next(connections) # 'b'
next(connections) # 'c'
next(connections) # 'a'
#....
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6 Comments

Nice! How does it know to start over when the list is exhausted?
@user443854 the while True means to repeat forever
@juanchopanza: Yep; itertools.cycle is a better answer. This shows how you could write the same functionality if itertools isn't available :)
Does the simple generator also save a copy of each element like itertools.cycle does? Or would the simple generator be a more memory-efficient design? Per the cycle docs: Note, this member of the toolkit may require significant auxiliary storage (depending on the length of the iterable).
@dthor this generator creates a list with three elements and literates over it, then destroys the list and creates a new one, in perpetuity. That documentation for cycle implies that the input iterable is converted to list before its generator starts, since iterable is only "good for one pass over the set of values".
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16

You can do it like this:

conn = ['a', 'b', 'c', 'd', 'e', 'f']
conn_len = len(conn)
index = 0
while True:
    print(conn[index])
    index = (index + 1) % conn_len

This prints a b c d e f a b c... forever

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Comments

5

you can accomplish this with append(pop()) loop:

l = ['a','b','c','d']
while True:
    print l[0]
    l.append(l.pop(0))

or for i in range() loop:

l = ['a','b','c','d']
ll = len(l)
while True:
    for i in range(ll):
       print l[i]

or simply:

l = ['a','b','c','d']

while True:
    for i in l:
       print i

all of which print:

>>>
a
b
c
d
a
b
c
d
...etc.

of the three I'd be prone to the append(pop()) approach as a function

servers = ['a','b','c','d']

def rotate_servers(servers):
    servers.append(servers.pop(0))
    return servers

while True:
    servers = rotate_servers(servers)
    print servers[0]
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2 Comments

Upvoting this because it helped me with a completely different use case where I simply want to iterate over a list a number of times, each time with the start element advancing one step. My use case is to iterate over the players in a game of poker, advancing the dealer puck one player forward for each round.
Removing an item from the front of a Python list is slow since all elements of the list have to be shifted. The official documentation warns against this (see link). Use a deque instead!
4

If you wish to cycle n times, implement the ncycles itertools recipe:

from itertools import chain, repeat


def ncycles(iterable, n):
    "Returns the sequence elements n times"
    return chain.from_iterable(repeat(tuple(iterable), n))


list(ncycles(["a", "b", "c"], 3))
# ['a', 'b', 'c', 'a', 'b', 'c', 'a', 'b', 'c']
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3

You need a custom iterator -- I'll adapt the iterator from this answer.

from itertools import cycle

class ConnectionPool():
    def __init__(self, ...):
        # whatever is appropriate here to initilize
        # your data
        self.pool = cycle([blah, blah, etc])
    def __iter__(self):
        return self
    def __next__(self):
        for connection in self.pool:
            if connection.is_available:  # or however you spell it
                return connection
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3

In order to avoid infinite loop, I have used length of array to iterate only until size of list is double.You can implement your own pre condition .Idea is to avoid infinite loop.

#Implement Circular Linked List
from itertools import cycle
list=[1,2,3,4,5]
lstlength=len(list)*2
print(lstlength)
pool=cycle(list)
i=0
#To avoid infinite loop break when you have iterated twice size of the list
for items in pool:
    print(items)
    if i >lstlength:
        break
    i += 1
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0 
class A(object):
    def __init__(self, l):
        self.strt = 0
        self.end = len(l)
        self.d = l

    def __iter__(self):
        return self

    def __next__(self):
        val = None
        if self.strt>=self.end:
            self.strt=0
        val = self.d[self.strt]
        self.strt += 1
        return val

a= A([8,9,7,66])
print(next(a))
print(next(a))
print(next(a))
print(next(a))
print(next(a))
print(next(a))
print(next(a))
print(next(a))
print(next(a))
print(next(a))
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0

For those, who may be interested. To loop forward or backward starting from given index:

def loop_fwd(arr, index):
  while True:
    arr_index = index % len(arr)
    yield arr_index, arr[arr_index]
    index += 1


def loop_bcw(arr, index):
  while True:
    arr_index = index % len(arr)
    yield arr_index, arr[arr_index]
    index -= 1


forward_it = loop_fwd([1,2,3,4,5], 3)
backward_it = loop_bcw([1,2,3,4,5], 3)

print('forward:')
for i in range(10):
  print(next(forward_it))


print('backward:')
for i in range(10):
  print(next(backward_it))
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0 
size = len(array)
for index in range(0, 2 * size):
  # here when it reaches size, the index becomes 0 again and continues
  value = array[index % size]
  print(value)
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1 Comment

Thank you for your interest in contributing to the Stack Overflow community. This question already has quite a few answers—including one that has been extensively validated by the community. Are you certain your approach hasn’t been given previously? If so, it would be useful to explain how your approach is different, under what circumstances your approach might be preferred, and/or why you think the previous answers aren’t sufficient. Can you kindly edit your answer to offer an explanation?
0

This is fairly neat. Doesn't require list size and the generator provides lazy evaluation, no itertools requirement also.

def cycle_through_list(lst):
    while True:
        yield from lst

Usage:

cycler = cycle_through_list([1, 2, 3, 4])
for _ in range(12):
    print(next(cycler), end="")

Output:

123412341234
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