问题:Python中的循环列表迭代器
我需要遍历一个循环列表,每次从最后一次访问的项目开始,可能要多次。
用例是一个连接池。客户端请求连接,迭代器检查指向的连接是否可用并返回,否则循环直到找到可用的连接。
有没有一种精巧的方法可以在Python中做到这一点?
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?
回答 0
使用itertools.cycle
,这是其确切目的:
from itertools import cycle
lst = ['a', 'b', 'c']
pool = cycle(lst)
for item in pool:
print item,
输出:
a b c a b c ...
(显然,永远循环)
为了手动推进迭代器并从中一一提取值,只需调用next(pool)
:
>>> next(pool)
'a'
>>> next(pool)
'b'
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'
回答 1
正确的答案是使用itertools.cycle。但是,让我们假设库函数不存在。您将如何实施?
使用生成器:
def circular():
while True:
for connection in ['a', 'b', 'c']:
yield connection
然后,您可以使用for
语句进行无限迭代,也可以调用next()
从生成器迭代器获取单个下一个值:
connections = circular()
next(connections) # 'a'
next(connections) # 'b'
next(connections) # 'c'
next(connections) # 'a'
next(connections) # 'b'
next(connections) # 'c'
next(connections) # 'a'
#....
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'
#....
回答 2
或者您可以这样做:
conn = ['a', 'b', 'c', 'd', 'e', 'f']
conn_len = len(conn)
index = 0
while True:
print(conn[index])
index = (index + 1) % conn_len
永远打印abcdefab c …
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
回答 3
您可以使用append(pop())
循环完成此操作:
l = ['a','b','c','d']
while 1:
print l[0]
l.append(l.pop(0))
或for i in range()
循环:
l = ['a','b','c','d']
ll = len(l)
while 1:
for i in range(ll):
print l[i]
或者简单地:
l = ['a','b','c','d']
while 1:
for i in l:
print i
所有这些打印:
>>>
a
b
c
d
a
b
c
d
...etc.
在这三个函数中,我倾向于使用append(pop())方法作为函数
servers = ['a','b','c','d']
def rotate_servers(servers):
servers.append(servers.pop(0))
return servers
while 1:
servers = rotate_servers(servers)
print servers[0]
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]
回答 4
您需要一个自定义迭代器-我将根据此答案改编迭代器。
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
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
回答 5
如果您希望循环n
时间,请实施ncycles
itertools配方:
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']
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']