blocks|key|1610685|text|如果标准化的意思是删除重复的字符，这应该是可行的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1610686|re.sub(r'(\w)\1%2B',+r'\1',+'yeeeesssss')++//+yes|code-block|syntax|javascript|1610687|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

IF by normalizing, you mean remove repeated characters, this should work:

<pre><code>re.sub(r'(\w)\1+', r'\1', 'yeeeesssss') // yes
</code></pre>

blocks|key|1610868|text|没有更多的信息是不可能回答你的问题的。正如您所说的，您希望从iterable中删除重复项。您可以使用itertools.groupby来做到这一点|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1610869|>>>+"".join(c+for+c,+_+in+groupby("yeeessssss"))
'yes'|code-block|syntax|javascript|1610870|当然，这将删除所有重复的：|1610871|>>>+dedupe+=+lambda+s:+"".join(c+for+c,+_+in+groupby(s))
>>>+dedupe("hello")
'helo'
>>>+dedupe("Mississippi")
'Misisipi'|1610872|我认为你的问题可能要困难得多；也就是说，如何将可能有重复字母的单词规范化为实际的英语单词。这基本上是不可能做到的--+beeeeeee或feeeed会变成什么？--但是，通过大量的努力，您可能可以通过各种启发式方法来近似它。|1610873|一个简单的方法是查看单词是否在字典中，如果没有，则一次删除一个重复的字母，直到它存在为止。这将是非常低效的，但可能会有效。|1610874|另一种方法是使用自然语言库将单词转换为某种“范式”。这可能是通过它的发音，它如何拼写，或其他东西。然后，您可以找到最接近该正常形式的单词，并使用它来给出您的去重单词。|1610875|另一种方法是定义字符串之间的某种“修改距离”，其中您为“删除字符”、“插入字符”和“修改字符”的每个操作分配固定的成本。然后，您可以在此度量下计算与输入最接近的单词。这是一个经过充分研究的问题，因为它用于生物信息学，并且有一种优雅的动态编程方法。不幸的是，解决这个问题也真的很有挑战性(一个相关的问题是我在本科时的一个为期几周的课程作业)。|1610876|1610877|;tl，dr|1610878|只需删除重复项就可以了。作为一个英语单词，找到最佳近似值是非常困难的。|1610879|entityMap^0|1E|H|0|0|0|0|1N|8|1W|6|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1E|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1F|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1G|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]|$1|X|3|Y|5|6|7|1I|8|@]|D|@]|E|$]]|$1|Z|3|10|5|6|7|1J|8|@]|D|@]|E|$]]|$1|11|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|12|$]]

It's impossible to answer your question without more information. As you've stated it, you want to remove duplicates from an iterable. You can do that with <code>itertools.groupby</code>:

<pre><code>&gt;&gt;&gt; "".join(c for c, _ in groupby("yeeessssss"))
'yes'
</code></pre>

Of course, that will remove all duplicates:

<pre><code>&gt;&gt;&gt; dedupe = lambda s: "".join(c for c, _ in groupby(s))
&gt;&gt;&gt; dedupe("hello")
'helo'
&gt;&gt;&gt; dedupe("Mississippi")
'Misisipi'
</code></pre>

I think your question is probably much more difficult; namely, how to normalise words which might have duplicate letters into actual English words. This is basically impossible to do precisely -- what would <code>beeeeeee</code> or <code>feeeed</code> become? -- but, with a lot of effort, you could probably approximate it by any of various heuristics.

One simple one would be to see if the word is in a dictionary, and if not, remove duplicate letters one at a time until it is. This will be very inefficient, but might work.

Another way would be to use a natural-language library to convert the word to some "normal form". This might be by how it sounds, how it is spelled, or something else. You could then find the closest word to that normal form and use it to give your de-duplicated word.

Yet another way would be to define some sort of "modification distance" between strings, where you assign a fixed cost to each of the operations "delete a character", "insert a character", and "modify a character". You could then compute the closest word to the input under this metric. This is a well-studied problem because it is used in bioinformatics, and there is an elegant dynamic programming approach to it. Unfortunately, it's also really quite challenging to work out (a related question was a several-week coursework project in my undergraduate degree).

<hr>

;tl,dr

Just removing duplicates is easy. Finding the best approximation as an English word is Very Hard.

blocks|key|2840118|text|这似乎类似于您需要使用拼写检查器所做的事情。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2840119|一种常用的解决方案是使用Soundex函数将单词简化为“听起来像什么”，然后将其与已知的有效单词字典进行比较。我不认为它是万无一失的，但这是一个可以让你朝着正确的方向开始的想法。|2840120|http://en.wikipedia.org/wiki/Soundex|offset|length|2840121|Soundex并不是唯一的选择。还有Metaphone和其他几个类似的算法可能会起作用。|2840122|这里有一个关于Soundex和Python的问题：Soundex+algorithm+in+Python+(homework+help+request)|2840123|最难的部分可能是找到一本好字典，但我在这个搜索中幸运地找到了：http://www.bing.com/search?q=download%2Bword%2Blist&qs=n&form=QBRE&pq=download%2Bword%2Blist&sc=8-18&sp=-1&sk=|2840124|无论你做什么，都不会是完美的。正如一些评论所指出的那样，在英语中有许多复杂的东西需要处理(就这一点而言，任何语言都是如此)。例如，区分"too“和"to”取决于上下文。微软和其他公司已经将经过多年开发的开发团队投入到拼写检查器中，拼写检查器仍然不能100%25正确地进行拼写检查，仍然需要人工干预。我想你在单词规范化方面也会面临同样的问题。|2840125|entityMap|0|LINK|mutability|MUTABLE|url|1|https://stackoverflow.com/questions/1626217/soundex-algorithm-in-python-homework-help-request|2|http://www.bing.com/search?q=download%2Bword%2Blist&qs=n&form=QBRE&pq=download%2Bword%2Blist&sc=8-18&sp=-1&sk=^0|0|0|0|10|0|0|0|P|1F|1|0|V|2U|2|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|12|8|@]|9|@$F|13|G|14|1|15]]|A|$]]|$1|H|3|I|5|6|7|16|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|17|8|@]|9|@$F|18|G|19|1|1A]]|A|$]]|$1|L|3|M|5|6|7|1B|8|@]|9|@$F|1C|G|1D|1|1E]]|A|$]]|$1|N|3|O|5|6|7|1F|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|E]]|W|$5|S|T|U|A|$V|X]]|Y|$5|S|T|U|A|$V|Z]]]]

This seems similar to what you'd need to do using a spell checker. 

One often used solution is to use Soundex functions to reduce the word to "what it sounds like" and then compare it against a known valid-word dictionary. I don't think it would be fool-proof, but it's an idea that may start you off in the right direction. 

<a href="http://en.wikipedia.org/wiki/Soundex" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Soundex</a>

Soundex isn't the only option. There are also Metaphone and several other similar algorithms that might work. 

There's a previous question about Soundex with Python here: <a href="https://stackoverflow.com/questions/1626217/soundex-algorithm-in-python-homework-help-request">Soundex algorithm in Python (homework help request)</a>

The hardest part is probably finding a good dictionary, but I've had luck with this search: <a href="http://www.bing.com/search?q=download+word+list&amp;qs=n&amp;form=QBRE&amp;pq=download+word+list&amp;sc=8-18&amp;sp=-1&amp;sk=" rel="nofollow noreferrer">http://www.bing.com/search?q=download+word+list&amp;qs=n&amp;form=QBRE&amp;pq=download+word+list&amp;sc=8-18&amp;sp=-1&amp;sk=</a>

No matter what you do, it's not going to be perfect. As pointed out by some of the comments, there are a lot of complexites to deal with in the English language (and any language, for that matter). Differentiating between "too" and "to", for example depend on the context. Microsoft and others have put teams of developers through years of development into spell-checkers, and spell-checkers are still not able to do it correctly 100% of the time, and still require human intervention. I think you'll face the same issue with word normalization.

blocks|key|1115524|text|使用enchant模块检查返回的单词是否为英文单词：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1115525|import+enchant,itertools
d_us=+enchant.Dict("en_US")
d_uk=+enchant.Dict("en_UK")
words=[]
teks=teks='yeeeessssssss'
for+x+in+itertools.permutations(set(teks)):
++++if+d_us.check(''.join(x))+or+d_uk.check(''.join(x)):
++++++words.append(''.join(x))|code-block|syntax|javascript|1115526|entityMap^0|2|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

use the <code>enchant</code> module to check if the returned word is a english word or not :

<pre><code>import enchant,itertools
d_us= enchant.Dict("en_US")
d_uk= enchant.Dict("en_UK")
words=[]
teks=teks='yeeeessssssss'
for x in itertools.permutations(set(teks)):
 if d_us.check(''.join(x)) or d_uk.check(''.join(x)):
 words.append(''.join(x))
</code></pre>

How can I 'normalise'

<pre><code>word = 'yeeeessssssss'
</code></pre>

to

<pre><code>word = 'yes'
</code></pre>

How can I remove duplicate letters in strings?

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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我如何才能“正常化”？word = 'yeeeessssssss'至word = 'yes'

问如何删除字符串中的重复字母？
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问如何删除字符串中的重复字母？EN