Circular list iterator in Python

Question

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.

Is there a neat way to do it in Python?

Answer 1

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'

Answer 2

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'
#....

Answer 3

Or you can do 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

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

Answer 4

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]

Answer 5

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']

Answer 6

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

Answer 7

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

Answer 8

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))
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print(next(a))
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print(next(a))
print(next(a))