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

user_131427 0
Viewed 251431

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.

def is_english_word(word):
    pass # how to I implement is_english_word?

is_english_word(token.lower())

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?

wordnet nltk python
11 Answers

For All Linux/Unix Users

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 /usr/share/dict you have a words file. There is also a more specific american-english and british-english 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:

import re
file = open("/usr/share/dict/words", "r")
words = re.sub("[^\w]", " ",  file.read()).split()
file.close()
    
def is_word(word):
    return word.lower() in words
 
is_word("tarts") ## Returns true
is_word("jwiefjiojrfiorj") ## Returns False

Hope this helps!

user_125848730

I find that there are 3 package-based solutions to solve the problem. They are pyenchant, wordnet and corpus(self-defined or from ntlk). Pyenchant couldn't installed easily in win64 with py3. Wordnet doesn't work very well because it's corpus isn't complete. So for me, I choose the solution answered by @Sadik, and use 'set(words.words())' to speed up.

First:

pip3 install nltk
python3

import nltk
nltk.download('words')

Then:

from nltk.corpus import words
setofwords = set(words.words())

print("hello" in setofwords)
>>True
user_104821330

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.

user_91335260

Download this txt file https://raw.githubusercontent.com/dwyl/english-words/master/words_alpha.txt

then create a Set out of it using the following python code snippet that loads about 370k non-alphanumeric words in english

>>> with open("/PATH/TO/words_alpha.txt") as f:
>>>     words = set(f.read().split('\n'))
>>> len(words)
370106

From here onwards, you can check for existence in constant time using

>>> word_to_check = 'baboon'
>>> word_to_check in words
True

Note that this set might not be comprehensive but still gets the job done, user should do quality checks to make sure it works for their use-case as well.

user_29319460
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