blocks|key|921053|text|使用set来存储单词列表，因为查找它们会更快：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|921054|with+open("english_words.txt")+as+word_file:
++++english_words+=+set(word.strip().lower()+for+word+in+word_file)

def+is_english_word(word):
++++return+word.lower()+in+english_words

print+is_english_word("ham")++#+should+be+true+if+you+have+a+good+english_words.txt|code-block|syntax|javascript|921055|要回答问题的第二部分，复数应该已经在一个很好的单词列表中，但是如果您出于某种原因想要从列表中专门排除这些复数，您确实可以编写一个函数来处理它。但是英语的复数规则非常棘手，所以我一开始就把复数包含在单词列表中。|921056|至于在哪里可以找到英文词表，我只是通过谷歌搜索“英语词表”就找到了几个。这里有一个：http://www.sil.org/linguistics/wordlists/english/wordlist/wordsEn.txt，如果你特别想要其中一种方言，你可以在谷歌上搜索英国英语或美国英语。|offset|length|921057|entityMap|0|LINK|mutability|MUTABLE|url|http://www.sil.org/linguistics/wordlists/english/wordlist/wordsEn.txt^0|0|0|0|16|1X|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@$K|Y|L|Z|1|10]]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

Using a set to store the word list because looking them up will be faster:

<pre><code>with open("english_words.txt") as word_file:
 english_words = set(word.strip().lower() for word in word_file)

def is_english_word(word):
 return word.lower() in english_words

print is_english_word("ham") # should be true if you have a good english_words.txt
</code></pre>

To answer the second part of the question, the plurals would already be in a good word list, but if you wanted to specifically exclude those from the list for some reason, you could indeed write a function to handle it. But English pluralization rules are tricky enough that I'd just include the plurals in the word list to begin with.

As to where to find English word lists, I found several just by Googling "English word list". Here is one: <a href="http://www.sil.org/linguistics/wordlists/english/wordlist/wordsEn.txt" rel="noreferrer">http://www.sil.org/linguistics/wordlists/english/wordlist/wordsEn.txt</a> You could Google for British or American English if you want specifically one of those dialects.

blocks|key|1135680|text|为了(更)强大和灵活，可以使用专用的拼写检查库，比如PyEnchant。有一个tutorial，或者你可以直接进入：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1135681|>>>+import+enchant
>>>+d+=+enchant.Dict("en_US")
>>>+d.check("Hello")
True
>>>+d.check("Helo")
False
>>>+d.suggest("Helo")
['He+lo',+'He-lo',+'Hello',+'Helot',+'Help',+'Halo',+'Hell',+'Held',+'Helm',+'Hero',+"He'll"]
>>>|code-block|syntax|javascript|1135682|PyEnchant附带了几个字典(en_GB，en_US，de_DE，fr_FR)，但是如果你想要更多的语言，可以使用任何OpenOffice+ones。|1135683|似乎有一个名为inflect的多元化库，但我不知道它是否有用。|1135684|entityMap|0|LINK|mutability|MUTABLE|url|https://pypi.org/project/pyenchant/|1|https://pyenchant.github.io/pyenchant/tutorial.html|2|http://wiki.services.openoffice.org/wiki/Dictionaries|3|http://pypi.python.org/pypi/inflect^0|Q|9|Q|9|0|13|8|1|0|0|0|9|1P|F|2|0|7|7|7|7|3|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]|$9|18|A|19|1|1A]]|E|$]]|$1|F|3|G|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|M|3|N|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@$9|1L|A|1M|1|1N]]|E|$]]|$1|O|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]|Y|$5|R|S|T|E|$U|Z]]|10|$5|R|S|T|E|$U|11]]]]

For (much) more power and flexibility, use a dedicated spellchecking library like <a href="https://pypi.org/project/pyenchant/" rel="noreferrer"><code>PyEnchant</code></a>. There's a <a href="https://pyenchant.github.io/pyenchant/tutorial.html" rel="noreferrer">tutorial</a>, or you could just dive straight in:
<pre><code>&gt;&gt;&gt; import enchant
&gt;&gt;&gt; d = enchant.Dict(&quot;en_US&quot;)
&gt;&gt;&gt; d.check(&quot;Hello&quot;)
True
&gt;&gt;&gt; d.check(&quot;Helo&quot;)
False
&gt;&gt;&gt; d.suggest(&quot;Helo&quot;)
['He lo', 'He-lo', 'Hello', 'Helot', 'Help', 'Halo', 'Hell', 'Held', 'Helm', 'Hero', &quot;He'll&quot;]
&gt;&gt;&gt;
</code></pre>
<code>PyEnchant</code> comes with a few dictionaries (en_GB, en_US, de_DE, fr_FR), but can use any of the <a href="http://wiki.services.openoffice.org/wiki/Dictionaries" rel="noreferrer">OpenOffice ones</a> if you want more languages.
There appears to be a pluralisation library called <a href="http://pypi.python.org/pypi/inflect" rel="noreferrer"><code>inflect</code></a>, but I've no idea whether it's any good.

blocks|key|921237|text|对于语义网方法，您可以运行sparql+query+against+WordNet+in+RDF+format。基本上只需使用urllib模块发出GET请求并返回JSON格式的结果，使用python+'json‘模块进行解析。如果它不是英文单词，你将得不到任何结果。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|921238|作为另一种想法，您可以查询Wiktionary's+API。|921239|entityMap|0|LINK|mutability|MUTABLE|url|http://wordnet.rkbexplorer.com/sparql/|1|https://stackoverflow.com/questions/2770547/is-there-an-api-available-to-retrieve-raw-wiktionary-data^0|D|16|0|0|D|G|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]|N|$5|I|J|K|C|$L|O]]]]

For a semantic web approach, you could run a <a href="http://wordnet.rkbexplorer.com/sparql/" rel="nofollow noreferrer">sparql query against WordNet in RDF format</a>. Basically just use urllib module to issue GET request and return results in JSON format, parse using python 'json' module. If it's not English word you'll get no results.

As another idea, you could query <a href="https://stackoverflow.com/questions/2770547/is-there-an-api-available-to-retrieve-raw-wiktionary-data">Wiktionary's API</a>.

blocks|key|921297|text|因为WordNet并不包含所有的英文单词，所以它在WordNet上不能很好地工作。基于没有附魔的NLTK的另一种可能性是NLTK的单词语料库|type|unstyled|depth|inlineStyleRanges|entityRanges|data|921298|>>>+from+nltk.corpus+import+words
>>>+"would"+in+words.words()
True
>>>+"could"+in+words.words()
True
>>>+"should"+in+words.words()
True
>>>+"I"+in+words.words()
True
>>>+"you"+in+words.words()
True|code-block|syntax|javascript|921299|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|$]]

It won't work well with WordNet, because WordNet does not contain all english words.
Another possibility based on NLTK without enchant is NLTK's words corpus

<pre><code>&gt;&gt;&gt; from nltk.corpus import words
&gt;&gt;&gt; "would" in words.words()
True
&gt;&gt;&gt; "could" in words.words()
True
&gt;&gt;&gt; "should" in words.words()
True
&gt;&gt;&gt; "I" in words.words()
True
&gt;&gt;&gt; "you" in words.words()
True
</code></pre>

blocks|key|921392|text|对于更快的基于NLTK的解决方案，您可以对单词集进行散列处理，以避免线性搜索。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|921393|from+nltk.corpus+import+words+as+nltk_words
def+is_english_word(word):
++++#+creation+of+this+dictionary+would+be+done+outside+of+
++++#+++++the+function+because+you+only+need+to+do+it+once.
++++dictionary+=+dict.fromkeys(nltk_words.words(),+None)
++++try:
++++++++x+=+dictionary[word]
++++++++return+True
++++except+KeyError:
++++++++return+False|code-block|syntax|javascript|921394|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|$]]

For a faster NLTK-based solution you could hash the set of words to avoid a linear search.

<pre><code>from nltk.corpus import words as nltk_words
def is_english_word(word):
 # creation of this dictionary would be done outside of 
 # the function because you only need to do it once.
 dictionary = dict.fromkeys(nltk_words.words(), None)
 try:
 x = dictionary[word]
 return True
 except KeyError:
 return False
</code></pre>

blocks|key|921427|text|使用pyEnchant.checker+SpellChecker：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|921428|from+enchant.checker+import+SpellChecker

def+is_in_english(quote):
++++d+=+SpellChecker("en_US")
++++d.set_text(quote)
++++errors+=+[err.word+for+err+in+d]
++++return+False+if+((len(errors)+>+4)+or+len(quote.split())+<+3)+else+True

print(is_in_english('“办理美国加州州立大学圣贝纳迪诺分校高仿成绩单Q/V2166384296加州州立大学圣贝纳迪诺分校学历学位认证'))
print(is_in_english('“Two+things+are+infinite:+the+universe+and+human+stupidity;+and+I\'m+not+sure+about+the+universe.”'))

>+False
>+True|code-block|syntax|javascript|921429|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|$]]

With pyEnchant.checker SpellChecker:

<pre><code>from enchant.checker import SpellChecker

def is_in_english(quote):
 d = SpellChecker("en_US")
 d.set_text(quote)
 errors = [err.word for err in d]
 return False if ((len(errors) &gt; 4) or len(quote.split()) &lt; 3) else True

print(is_in_english('“办理美国加州州立大学圣贝纳迪诺分校高仿成绩单Q/V2166384296加州州立大学圣贝纳迪诺分校学历学位认证'))
print(is_in_english('“Two things are infinite: the universe and human stupidity; and I\'m not sure about the universe.”'))

&gt; False
&gt; True
</code></pre>

blocks|key|921502|text|适用于所有Linux/Unix用户|type|unstyled|depth|inlineStyleRanges|entityRanges|data|921503|如果您的操作系统使用Linux内核，那么有一种简单的方法可以从英语/美语词典中获取所有单词。在目录/usr/share/dict中有一个words文件。还有更具体的american-english和british-english文件。它们包含该特定语言中的所有单词。你可以在每种编程语言中访问它，这就是为什么我想你可能想知道这一点。|offset|length|style|CODE|921504|现在，对于特定于python的用户，下面的python代码应该分配列表单词，使其具有每个单词的值：|921505|import+re
file+=+open("/usr/share/dict/words",+"r")
words+=+re.sub("[%5E\w]",+"+",++file.read()).split()

def+is_word(word):
++++return+word.lower()+in+words

is_word("tarts")+##+Returns+true
is_word("jwiefjiojrfiorj")+##+Returns+False|code-block|syntax|javascript|921506|希望这能有所帮助！|921507|entityMap^0|0|1D|F|1W|5|2A|G|2R|F|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]|$D|Y|E|Z|F|G]|$D|10|E|11|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|12|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|13|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|R|$]]

<h2>For All Linux/Unix Users</h2>

If your OS uses the Linux kernel, there is a simple way to get all the words from the English/American dictionary. In the directory <code>/usr/share/dict</code> you have a <code>words</code> file. There is also a more specific <code>american-english</code> and <code>british-english</code> files. These contain all of the words in that specific language. You can access this throughout every programming language which is why I thought you might want to know about this.

Now, for python specific users, the python code below should assign the list words to have the value of every single word:

<pre><code>import re
file = open("/usr/share/dict/words", "r")
words = re.sub("[^\w]", " ", file.read()).split()

def is_word(word):
 return word.lower() in words

is_word("tarts") ## Returns true
is_word("jwiefjiojrfiorj") ## Returns False
</code></pre>

Hope this helps!!!

blocks|key|921542|text|使用nltk.corpus而不是enchant。Enchant给出模棱两可的结果。例如:对于基准测试和基准测试，enchant返回true。它应该为benchmark返回false。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|921543|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

use nltk.corpus instead of enchant. Enchant gives ambiguous results. For example :
for benchmark and bench-mark enchant is returning true. It should suppose to return false for benchmark.

blocks|key|921578|text|你可以看到这个页面：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|921579|How+to+determine+language|offset|length|921580|我推荐langid|style|CODE|921581|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/39142778/^0|0|0|P|0|0|3|6|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|6|7|W|8|@$D|X|E|Y|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

you can see this page :
<a href="https://stackoverflow.com/questions/39142778/">How to determine language</a>
I recommend the <code>langid</code>

I want to check in a Python program if a word is in the English dictionary.

I believe nltk wordnet interface might be the way to go but I have no clue how to use it for such a simple task.

<pre><code>def is_english_word(word):
 pass # how to I implement is_english_word?

is_english_word(token.lower())
</code></pre>

In the future, I might want to check if the singular form of a word is in the dictionary (e.g., properties -> property -> english word). How would I achieve that?

How to check if a word is an English word with Python?

Python

Linux

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Unix

我想在Python程序中检查一个单词是否在英语词典中。我相信nltk wordnet接口可能是可行的，但我不知道如何使用它来完成如此简单的任务。def is_english_word(word):    pass # how to I implement is_english_word?is_english_word(...

问如何用Python检查一个单词是否为英文单词？
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