blocks|key|458566|text|嗯，http://docs.python.org/library/itertools.html#itertools.cycle没有这样的start元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|458567|也许你只是开始了这个循环，去掉了你不喜欢的第一个元素。|458568|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/itertools.html#itertools.cycle^0|2|1P|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]]|C|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]]]

Hm, <a href="http://docs.python.org/library/itertools.html#itertools.cycle" rel="nofollow">http://docs.python.org/library/itertools.html#itertools.cycle</a> doesn't have such a start element.

Maybe you just start the cycle anyway and drop the first elements that you don't like.

blocks|key|458623|text|使用算术mod运算符。假设你从位置k开始，那么k应该像这样更新：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|458624|k+=+(k+%2B+1)+%25+len(l)|code-block|syntax|javascript|458625|如果你想从某个元素开始，而不是从索引开始，你总是可以像k+=+l.index(x)一样查找它，其中x是所需的项。|458626|entityMap^0|4|3|H|1|N|1|0|0|R|E|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

Use the arithmetic <code>mod</code> operator. Suppose you're starting from position <code>k</code>, then <code>k</code> should be updated like this:

<pre><code>k = (k + 1) % len(l)
</code></pre>

If you want to start from a certain element, not index, you can always look it up like <code>k = l.index(x)</code> where x is the desired item.

blocks|key|672879|text|import+itertools+as+it
l+=+[1,+2,+3,+4]
list(it.islice(it.dropwhile(lambda+x:+x+!=+4,+it.cycle(l)),++10))
#+returns:+[4,+1,+2,+3,+4,+1,+2,+3,+4,+1]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|672880|所以你想要的迭代器是：|unstyled|672881|it.dropwhile(lambda+x:+x+!=+4,+it.cycle(l))|672882|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>import itertools as it
l = [1, 2, 3, 4]
list(it.islice(it.dropwhile(lambda x: x != 4, it.cycle(l)), 10))
# returns: [4, 1, 2, 3, 4, 1, 2, 3, 4, 1]
</code></pre>

so the iterator you want is:

<pre><code>it.dropwhile(lambda x: x != 4, it.cycle(l))
</code></pre>

blocks|key|458689|text|看看itertools模块。它提供了所有必要的功能。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|458690|from+itertools+import+cycle,+islice,+dropwhile

L+=+[1,+2,+3,+4]

cycled+=+cycle(L)++#+cycle+thorugh+the+list+'L'
skipped+=+dropwhile(lambda+x:+x+!=+4,+cycled)++#+drop+the+values+until+x==4
sliced+=+islice(skipped,+None,+10)++#+take+the+first+10+values

result+=+list(sliced)++#+create+a+list+from+iterator
print(result)|code-block|syntax|javascript|458691|输出：|458692|[4,+1,+2,+3,+4,+1,+2,+3,+4,+1]|458693|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/itertools.html^0|2|9|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

Look at <a href="http://docs.python.org/library/itertools.html" rel="noreferrer">itertools</a> module. It provides all the necessary functionality.

<pre><code>from itertools import cycle, islice, dropwhile

L = [1, 2, 3, 4]

cycled = cycle(L) # cycle thorugh the list 'L'
skipped = dropwhile(lambda x: x != 4, cycled) # drop the values until x==4
sliced = islice(skipped, None, 10) # take the first 10 values

result = list(sliced) # create a list from iterator
print(result)
</code></pre>

Output:

<pre><code>[4, 1, 2, 3, 4, 1, 2, 3, 4, 1]
</code></pre>

blocks|key|672975|text|我不是一个很喜欢导入模块的人，因为你只需要几行代码就可以完成你自己的事情。以下是我没有导入的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|672976|def+cycle(my_list,+start_at=None):
++++start_at+=+0+if+start_at+is+None+else+my_list.index(start_at)
++++while+True:
++++++++yield+my_list[start_at]
++++++++start_at+=+(start_at+%2B+1)+%25+len(my_list)|code-block|syntax|javascript|672977|这将返回一个循环列表的(无限)迭代器。要获取循环中的下一个元素，必须使用next语句：|offset|length|style|CODE|672978|>>>+it1+=+cycle([101,102,103,104])
>>>+next(it1),+next(it1),+next(it1),+next(it1),+next(it1)
(101,+102,+103,+104,+101)+#+and+so+on+...
>>>+it1+=+cycle([101,102,103,104],+start_at=103)
>>>+next(it1),+next(it1),+next(it1),+next(it1),+next(it1)
(103,+104,+101,+102,+103)+#+and+so+on+...|672979|entityMap^0|0|0|10|4|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I'm not such a big fan of importing modules when you can do things by your own in a couple of lines. Here's my solution without imports:

<pre><code>def cycle(my_list, start_at=None):
 start_at = 0 if start_at is None else my_list.index(start_at)
 while True:
 yield my_list[start_at]
 start_at = (start_at + 1) % len(my_list)
</code></pre>

This will return an (infinite) iterator looping your list. To get the next element in the cycle you must use the <code>next</code> statement:

<pre><code>&gt;&gt;&gt; it1 = cycle([101,102,103,104])
&gt;&gt;&gt; next(it1), next(it1), next(it1), next(it1), next(it1)
(101, 102, 103, 104, 101) # and so on ...
&gt;&gt;&gt; it1 = cycle([101,102,103,104], start_at=103)
&gt;&gt;&gt; next(it1), next(it1), next(it1), next(it1), next(it1)
(103, 104, 101, 102, 103) # and so on ...
</code></pre>

blocks|key|458752|text|可以使用下面这样的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|458753|def+my_cycle(data,+start=None):
++k+=+0+if+not+start+else+start
++while+True:
++++yield+data[k]
++++k+=+(k+%2B+1)+%25+len(data)|code-block|syntax|javascript|458754|然后运行：|458755|for+val+in+my_cycle([0,1,2,3],+2):
++print(val)|458756|基本上与前面的一个答案相同。是我的错。|458757|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Can use something like this:

<pre><code>def my_cycle(data, start=None):
 k = 0 if not start else start
 while True:
 yield data[k]
 k = (k + 1) % len(data)
</code></pre>

Then run:

<pre><code>for val in my_cycle([0,1,2,3], 2):
 print(val)
</code></pre>

Essentially the same as one of the previous answers. My bad.

Say I have a list:

<pre><code>l = [1, 2, 3, 4]
</code></pre>

And I want to cycle through it. Normally, it would do something like this,

<pre><code>1, 2, 3, 4, 1, 2, 3, 4, 1, 2...
</code></pre>

I want to be able to start at a certain point in the cycle, not necessarily an index, but perhaps matching an element. Say I wanted to start at whatever element in the list <code>==4</code>, then the output would be,

<pre><code>4, 1, 2, 3, 4, 1, 2, 3, 4, 1...
</code></pre>

How can I accomplish this?

Cycle through list starting at a certain element

假设我有一个列表：l = [1, 2, 3, 4]我想通过它循环。通常，它会做这样的事情，1, 2, 3, 4, 1, 2, 3, 4, 1, 2...我希望能够从循环中的某个点开始，不一定是索引，但可能会匹配一个元素。假设我想从列表==4中的任何元素开始，那么输出将是，4, 1, 2, 3, 4, 1, 2, 3, ...

问循环从某个元素开始的列表
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