blocks|key|792074|text|我认为这应该可以回答你的问题，我知道我的改变字母的列表并不完全相同，但我认为从这里开始你应该能够实现它。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|792075|它把每个s变成b，b变成n，n变成s。|792076|lettersToChange+=+['s','b','n']

randomLetters+=+['k','l','m','n','s','s','b','n','b','n','s','b','n','b','s']
for+i+in+range(len(randomLetters)):
++++if+randomLetters[i]+in+lettersToChange:
++++++++pos+=+lettersToChange.index(randomLetters[i])
++++++++randomLetters[i]+=+lettersToChange[(pos%2B1)%253]|code-block|syntax|javascript|792077|所以在randomLetters列表之前和之后：|792078|+['k',+'l',+'m',+'n',+'s',+'s',+'b',+'n',+'b',+'n',+'s',+'b',+'n',+'b',+'s']
+['k',+'l',+'m',+'s',+'b',+'b',+'n',+'s',+'n',+'s',+'b',+'n',+'s',+'n',+'b']|792079|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I think this should answer your question, I know my list that changes the letters isn't exactly the same, but I think from here you should be able to implement it. 

It changes each s to b, b to n and n to s.

<pre><code>lettersToChange = ['s','b','n']

randomLetters = ['k','l','m','n','s','s','b','n','b','n','s','b','n','b','s']
for i in range(len(randomLetters)):
 if randomLetters[i] in lettersToChange:
 pos = lettersToChange.index(randomLetters[i])
 randomLetters[i] = lettersToChange[(pos+1)%3]
</code></pre>

So before and after of the randomLetters list:

<pre><code> ['k', 'l', 'm', 'n', 's', 's', 'b', 'n', 'b', 'n', 's', 'b', 'n', 'b', 's']
 ['k', 'l', 'm', 's', 'b', 'b', 'n', 's', 'n', 's', 'b', 'n', 's', 'n', 'b']
</code></pre>

blocks|key|748915|text|据我所知，您有一个特定的状态序列:S+->+B+->+S+->+N+->+B+->+N+->+have|type|unstyled|depth|inlineStyleRanges|entityRanges|data|748916|这可以用生成器来解决，比如：|748917|import+itertools
state+=+itertools.cycle(['S',+'B',+'S',+'N',+'B',+'N'])|code-block|syntax|javascript|748918|对next(state)的后续调用将按顺序生成列表，当列表到达末尾时从开始开始。示例：|748919|>>>+next(state)
'S'
>>>+next(state)
'B'
>>>+next(state)
'S'
>>>+next(state)
'N'
>>>+next(state)
'B'
>>>+next(state)
'N'
>>>+next(state)
'S'
>>>+next(state)
'B'
>>>+next(state)
'S'
>>>+next(state)
'N'
>>>+next(state)
'B'|748920|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

As I understand, you have a specific sequence of states:
S -> B -> S -> N -> B -> N -> [begin again]

This could be solved with a generator such as:

<pre><code>import itertools
state = itertools.cycle(['S', 'B', 'S', 'N', 'B', 'N'])
</code></pre>

Subsequent calls of next(state) will yield the list in sequence, starting at the beginning when it reaches the end. Example:

<pre><code>&gt;&gt;&gt; next(state)
'S'
&gt;&gt;&gt; next(state)
'B'
&gt;&gt;&gt; next(state)
'S'
&gt;&gt;&gt; next(state)
'N'
&gt;&gt;&gt; next(state)
'B'
&gt;&gt;&gt; next(state)
'N'
&gt;&gt;&gt; next(state)
'S'
&gt;&gt;&gt; next(state)
'B'
&gt;&gt;&gt; next(state)
'S'
&gt;&gt;&gt; next(state)
'N'
&gt;&gt;&gt; next(state)
'B'
</code></pre>

blocks|key|748922|text|如果我理解正确的话，你必须有一个索引，因为你必须区分，哪一个S、B或N应该前进到它的下一个状态。大概是这样的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|748923|TRAVERSE_ORDER+=+"SBSNBN"
EXAMPLE_INPUT+=+["S",+"B",+"N"]

def+traverse_single_item(state,+state_index=0):
++++next_state+=+TRAVERSE_ORDER.find(state,+state_index)+%2B+1
++++if+next_state+==+len(TRAVERSE_ORDER):
++++++++next_state+=+0
++++return+(TRAVERSE_ORDER[next_state],+next_state)

new_state+=+[traverse_single_item(*elem)+for+elem+in+EXAMPLE_INPUT]
print(new_state)

newer_state+=+[traverse_single_item(*elem)+for+elem+in+new_state]
print(newer_state)

newest_state+=+[traverse_single_item(*elem)+for+elem+in+newer_state]
print(newest_state)|code-block|syntax|javascript|748924|如果在没有索引的情况下调用函数，则假定它是遍历顺序中的第一个匹配项。请注意星号：*elem。它为函数调用解包元组。|748925|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If i understand correctly, you must have an index, because you have to tell apart, which S or B or N should advance to it's next state. Something like that:

<pre><code>TRAVERSE_ORDER = "SBSNBN"
EXAMPLE_INPUT = ["S", "B", "N"]

def traverse_single_item(state, state_index=0):
 next_state = TRAVERSE_ORDER.find(state, state_index) + 1
 if next_state == len(TRAVERSE_ORDER):
 next_state = 0
 return (TRAVERSE_ORDER[next_state], next_state)

new_state = [traverse_single_item(*elem) for elem in EXAMPLE_INPUT]
print(new_state)

newer_state = [traverse_single_item(*elem) for elem in new_state]
print(newer_state)

newest_state = [traverse_single_item(*elem) for elem in newer_state]
print(newest_state)
</code></pre>

If you call the function without an index, it assumes to be the first occurence in the traverse order.
Please note the asterisk: *elem. It unpacks the tuple for the function call.

i know this will be a very novice question to ask here but i kind of blank out and need this to get resolved immediately.

I want to change values from S -> B and so on as mention in below code. For that i can create array or list "A = [S, B, N]" or whatever required. But need to traverse in this manner.

<pre><code>S --&gt; B 
B --&gt; S
S --&gt; N
N --&gt; B
B --&gt; N
N --&gt; S
</code></pre>

I tried creating 2 lists A and B with same values [S, B, N] and traversing through both of them with 

<pre><code>i=0;i&lt;length.A;i++
 j=0;j&lt;length.B;j++
</code></pre>

and 

<pre><code>i=0;i&lt;length.A;i++
 j=i;j&lt;length.B;j++
</code></pre>

But in scenario 1: S changed to B, then it tries changes from S -> N. Which is not possible.
In Scenario 2: We hit 4 out of 6 but not all 6 required outputs.

I could've researched more but due to lack of time, i couldn't.

Also I've written the code in pseudo manner.

Basic loop concept with array/list in python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我知道这将是一个非常新手的问题，但我有点空白，需要立即解决这个问题。我想更改S、->、B等值，就像下面代码中提到的那样。为此，我可以创建数组或列表"A = S，B，N“或任何需要的东西。但是需要以这种方式遍历。S --> B B --> SS --> NN --> BB --> NN --> S我尝试使用相同的值S、B...

问python中数组/列表的基本循环概念
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