english-words | text file containing 479k English words | Natural Language Processing library

by dwyl Python Version: Current License: Unlicense

X-Ray Key Features Code Snippets(2)Community Discussions(8)Vulnerabilities Install Support

kandi X-RAY | english-words Summary

english-words is a Python library typically used in Artificial Intelligence, Natural Language Processing applications. english-words has no bugs, it has no vulnerabilities, it has a Permissive License and it has medium support. However english-words build file is not available. You can download it from GitHub.

List Of English Words.

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Quality

Security

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Reuse

Support

english-words has a medium active ecosystem.

It has 9014 star(s) with 1683 fork(s). There are 198 watchers for this library.

It had no major release in the last 6 months.

There are 74 open issues and 33 have been closed. On average issues are closed in 164 days. There are 21 open pull requests and 0 closed requests.

It has a neutral sentiment in the developer community.

The latest version of english-words is current.

Quality

english-words has no bugs reported.

Security

english-words has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

License

english-words is licensed under the Unlicense License. This license is Permissive.

Permissive licenses have the least restrictions, and you can use them in most projects.

Reuse

english-words releases are not available. You will need to build from source code and install.

english-words has no build file. You will be need to create the build yourself to build the component from source.

Top functions reviewed by kandi - BETA

kandi has reviewed english-words and discovered the below as its top functions. This is intended to give you an instant insight into english-words implemented functionality, and help decide if they suit your requirements.

Load all valid words from a file .

Get all kandi verified functions for this library.

english-words Key Features

No Key Features are available at this moment for english-words.

english-words Examples and Code Snippets

Calculates the English count of English words .

python

Lines of Code : 14

License : Permissive (MIT License)

Copy

def get_english_count(message: str) -> float:
    message = message.upper()
    message = remove_non_letters(message)
    possible_words = message.split()

    if possible_words == []:
        return 0.0

    matches = 0
    for word in possible_w

Loads English words .

python

Lines of Code : 7

License : Permissive (MIT License)

Copy

def load_dictionary() -> dict[str, None]:
    path = os.path.split(os.path.realpath(__file__))
    english_words: dict[str, None] = {}
    with open(path[0] + "/dictionary.txt") as dictionary_file:
        for word in dictionary_file.read().split(

Community Discussions

Trending Discussions on english-words

How do I compare elements in a list without nested loops?

Can't display const text in Flutter

How to speed up a "append to file" kind of program?

Importing each line from text file from Resources in Unity to list in C#

How can you filter out related words in large list quickly

is there a way to check if a word in a generated list, is inside a text file?

What is the Time Complexity of LeetCode 273. Integer to English Words?

How to use/import a large dictionary without taking a massive performance hit?

QUESTION

How do I compare elements in a list without nested loops?

Asked 2022-Feb-22 at 09:08

So I have a ginormous list of 466,550 words and I need to compare the words with each other to check the similarity between them. Similarity between two words is defined as the number of common characters between them in my case. But if I try :

...

ANSWER

Answered 2022-Feb-22 at 09:08

I think an important part of this problem is how you interpret the clause "highest similarity with other words in dictionary" in the problem description. If you take that to mean the highest sum of similarities with all the other words, you can compute that sum in a more efficient way than your nested loops.

Rather than directly comparing all words, you can instead count how many different words contain each letter. Then in a second pass, you can add up the number of other words that share each of a word's letters, which is almost the sum of similarities we want. We just need to subtract out the word's similarity with itself (which is the number of unique letters it contains), since we're only supposed to compare to other words.

Here's a function that computes a list of similarities in the same order the words are in:

Source https://stackoverflow.com/questions/71216420

QUESTION

Can't display const text in Flutter

Asked 2021-Dec-11 at 12:56

I am using english-words package and I try to get random word and display it in Text but I got error. What I am doing wrong?

...

ANSWER

Answered 2021-Dec-11 at 12:56

Remove the const keyword before the Center widget. you cant use a const widget if that widget has a none-constant data/variable at the build time.

Edit: More on const

Source https://stackoverflow.com/questions/70315315

QUESTION

How to speed up a "append to file" kind of program?

Asked 2021-Aug-18 at 19:08

I was creating search algorithm and for that I needed data in huge amounts so I decided to create a file which will contain random words taken from the english-words module. For that I created this small code to be run from CMD:

...

ANSWER

Answered 2021-Aug-18 at 19:08

You can reduce the amount of work each iteration does (and reduce the number of iterations by a factor of 10) by building a complete line in each iteration.

Source https://stackoverflow.com/questions/68837412

QUESTION

Importing each line from text file from Resources in Unity to list in C#

Asked 2021-May-29 at 04:22

I have a text file in my Assets/Resources/Text, on Start() I want to import each line from the text file as a string in list. Using this code does it for me,

...

ANSWER

Answered 2021-May-29 at 04:22

To begin with Don't use Reources!

If against Unity's own recommendation you still want to do it for some reason then you have to load that resource using Resources.Load in your case as a TextAsset like e.g.

Source https://stackoverflow.com/questions/67744910

QUESTION

How can you filter out related words in large list quickly

Asked 2021-Mar-23 at 11:27

I am running this code on ~479k words and it is taking a very long time:

...

ANSWER

Answered 2021-Mar-23 at 11:07

this may work:

Source https://stackoverflow.com/questions/66760827

QUESTION

is there a way to check if a word in a generated list, is inside a text file?

Asked 2021-Jan-10 at 16:07

I want to check if a word in a generated list is inside a text file I made an anagram generator, and I want to see if the list created by that generator has a real word by checking if it's inside an English dictionary.

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ANSWER

Answered 2021-Jan-10 at 16:07

You are checking if the whole list is in the file. For example, if anagram = ['abc', 'acb'] There you are literally checking:

Source https://stackoverflow.com/questions/65655280

QUESTION

What is the Time Complexity of LeetCode 273. Integer to English Words?

Asked 2021-Jan-05 at 07:00

I am struggling to understand the time complexity of this solution. This question is about converting number to English words.

For Example, Input: num = 1234567891 Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One"

StringBuilder insert() has O(n) time Complexity.
I suspect the time Complexity is O(n^2). But I am not sure. Where n is the number of digits. Link to question: englishToWords

There is my code:

Code:

...

ANSWER

Answered 2021-Jan-01 at 18:55

It is O(log n), assuming n is num, since you get less than 2 words per digit of the numeric value of n, and the number of digits is ceil(log₁₀(n)).

Doubling the value will maybe add 2 more words, that's it.

Source https://stackoverflow.com/questions/65532583

QUESTION

How to use/import a large dictionary without taking a massive performance hit?

Asked 2020-Dec-16 at 19:49

I'm trying to use this word list I found for a somewhat ambitious hangman game for a discord bot, and it recommended to use the .json version of a file as a dictionary, if I were using Python, which I am. Only problem, it takes forever for it to go through(interpret?), presumably because it has 370102 lines, and considering this is going to be run on a raspberry pi, this probably isn't going to work out very well.

What would be the best way to go about this? I'm new to python and programming in general, so I'm not quite sure how to do so. Would it be faster if I were to use it in C? Maybe I could use an array somehow?

It doesn't have to be in a dictionary, it's just that the file was provided like that.

...

ANSWER

Answered 2020-Dec-16 at 19:40

For something as basic as dictionary lookup, put the words into a dbm file then read from that. This effectively works as an on-disk dictionary letting you look up keys and their values quickly without having to load the whole thing into memory.

Source https://stackoverflow.com/questions/65329900

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install english-words

You can download it from GitHub.
You can use english-words like any standard Python library. You will need to make sure that you have a development environment consisting of a Python distribution including header files, a compiler, pip, and git installed. Make sure that your pip, setuptools, and wheel are up to date. When using pip it is generally recommended to install packages in a virtual environment to avoid changes to the system.

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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