PyDictionary | Dictionary Module for Python | Dictionary library
kandi X-RAY | PyDictionary Summary
kandi X-RAY | PyDictionary Summary
Current Version: 2.0.1 Created By Pradipta Bora 2020.
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Top functions reviewed by kandi - BETA
- Prints the meanings
- Return the word meanings of a word
- Get a dictionary of the meanings
- Return a BeautifulSoup object
- Prints antonyms
- Get a list of antonyms
- Return a list of antonyms
- Print the synonyms
- Get the synonyms for a given term
- Get synonyms
- Return a list of translations to given language
- Translate a single word
PyDictionary Key Features
PyDictionary Examples and Code Snippets
Lines of Code : 24License : Strong Copyleft (CC BY-SA 4.0)
Copyclass ListItem(OneLineListItem):
# Here define all the necessary attrs., methods apart from the defaults (if you need any).
RecycleView:
id: scroll
#size_word = dic.meaning('book')
print(list(word.values())[0])
def handle_input(x, y, letter):
global word #add this line
word += letter
print(word)
coords = make_coords(x, y)
for list in grid:
for button in list:
button.update(disabled=True)
for coord in coords:
xfrom PyDictionary import PyDictonary
dictionary = PyDictionary()
valid_word = bool(dictionary.meaning("hdaoij")) # False
valid_word = bool(dictionary.meaning("hello")) # True
valid_word = bool(dictionary.meaning(in>>> 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', 'HPS C:\> py -0
Installed Pythons found by C:\windows\py.exe Launcher for Windows
-3.9-64 *
-3.7-32
PS C:\> py -3.7 -m pip list
Package Version
------------------ ---------
beautifulsoup4 4.8.1from itertools import groupby
data = [
{'date': '2020-02-02', 'id': '1', 'dept': '20020', 'CNT': '1', 'rep_level': 'form1'},
{'date': '2020-02-02', 'id': '1', 'dept': '20020', 'CNT': '0', 'rep_level': 'form2'},
{'date': '2020-flat_list = [i for sublist in synonyms if sublist is not None for i in sublist]
flat_list
>>>
['syn1', 'syn2', 'syn3', 'syn4', 'syn5', 'syn6', 'syn7', 'syn8']
import itertools
list(itertools.chain(*[x for x in synonyms if x]))
import string
# Python Dictionary
# I manually created these word relationship - primary_word:synonyms
word_relationship = {"father": ['dad', 'daddy', 'old man', 'pa', 'pappy', 'papa', 'pop'],
"mother": ["mamma", "momma", "mama"Community Discussions
Trending Discussions on PyDictionary
QUESTION
I'm making a kivy app to find the rhyming words for a word entered by the user. It displays all the rhyming words as OneLineListItems in an MDList which is inside a kivy RecycleView. On clicking on one of these OneLineListItems it displays the definition of the word on the right-hand side of the screen. However, when I click on a OneLineListItem its definition takes very long to appear and sometimes it lags so badly that the app closes. Am I doing something wrong or is it just my computer? Code below:
ANSWER
Answered 2022-Jan-25 at 08:32First create a custom class for the data-class like following:
QUESTION
What I am trying to achieve is the ability to choose one random meaning of a word with PyDictionary, using this code:
...ANSWER
Answered 2021-Oct-18 at 03:18dic is returning a dict object, not a list - so you can't use indexes to get the first item.
You can do this instead
QUESTION
This seems like an incredibly silly problem, but it's doing my nut in. I just started working with Python a few days ago, after spending a while in other languages, and I've run into a very weird problem. I'm attempting to set a top-level string variable called 'word' to the concatenation of itself and a letter. Python is able to find the variable just fine, as demonstrated below.
However, any time I try to concatenate, I suddenly get an error telling me Python is unable to find the variable.
The yellow squiggle is my IDE telling me that the variable can't be found. I've tried literally directly copy+pasting other solutions from other stack-exchange posts. It's a little embarrassing to be having this much trouble with concatenation, but I've tried most of the stuff I can think of.
Edit:Here's my code. As you can see, "word" is initialized at the top of the file:
...ANSWER
Answered 2021-Aug-11 at 22:08Since your word is defined outside of the function, you need to redeclare it as a global variable:
QUESTION
I thought about using
...ANSWER
Answered 2021-Jun-15 at 11:27I saw this post a while ago: How to check if a word is an English word with Python?
QUESTION
Basically, I downloaded the library using pip install PyDictionary and I made sure it exists by writing the same thing again and this appeared:
ANSWER
Answered 2021-May-07 at 20:18What you're having issues with is that your pip command, is not connected to your installation of python3.
If you're using Windows, then you should have a "python launcher" that you can use to double-check your installed versions of Python like this:
QUESTION
Using nltk to generate a list of synonyms to based on an input list of keywords.
I'm getting "TypeError: 'NoneType' object is not iterable" when nltk doesn't have a synonym (i.e., when "None" is returned).
...ANSWER
Answered 2021-Jan-20 at 18:28You can remove the None values and use itertools -
QUESTION
From today, I started getting error while installing modules from requirements.txt, I tried to find the error module and remove it but I couldn't find.
ANSWER
Answered 2021-Jan-17 at 12:41Create a list of all the dependencies and run the following code.
QUESTION
I am trying to write a GUI that emulates the Hangman game using Tkinter. So far I have gotten the GUI to create a label that updates based on the letters the user has guessed correctly, but the terminal still gives an error : "TypeError: argument of type 'StringVar' is not iterable". I have looked at other solutions to this error, but haven't been able to figure out how to solve the problem.
It's not done yet - but here is the code so far:
...ANSWER
Answered 2020-Oct-08 at 23:09This line is causing the error:
QUESTION
I am looking for an English Dictionary dump for some text analysis in Python. This would include a word and some of its attributes (noun/verb, its forms, tenses, and probably origin too!). So, I envision these as columns of a data frame. I have gone through numerous threads where folks have suggested some sources but I believe none of those fulfill the above requirements (some are just word lists, others are words with just meanings). Moreover, they kind of look non-exhaustive (very small corpus whereas I am targeting to have ~500000 words). Is there a dump available from authoritative sources like Oxford or Merriam Webster? Also, there is a PyDictionary module. Is it possible to fetch such a dump from this module?
...ANSWER
Answered 2020-Sep-16 at 16:33WordNet is a corpus of words, their synonyms, hyponyms, and meronyms, grouped by synsets and made available for free give that you follow their license. https://wordnet.princeton.edu/. Since this is a popular choice, you can find this corpus in almost any data format with a little searching. Database contains 155,327 words.
BabelNet is another corpus that have aggregated WordNet, Wikipedia, and many other sources into a database of 91,218,220 glossary definitions covering many languages. https://babelnet.org/
If you want to use the Oxford dictionary and Merriam Webster, they are commercial products who dont throw around their database with unlimited access. Both have API interfaces you can gain access to with a registered API key.
QUESTION
@commands.command(aliases=['lookup'])
async def define(self, message, *, arg):
dictionary=PyDictionary()
Define = dictionary.meaning(arg)
length = len(arg.split())
if length == 1:
embed = discord.Embed(description="**Noun:** " + Define["Noun"][0] + "\n\n**Verb:** " + Define["Verb"][0], color=0x00ff00)
embed.set_author(name = ('Defenition of the word: ' + arg),
icon_url=message.author.avatar_url)
await message.send(embed=embed)
else:
CommandError = discord.Embed(description= "A Term must be only a single word" , color=0xfc0328)
await message.channel.send(embed=CommandError)
ANSWER
Answered 2020-Aug-20 at 05:16You can test if key exists with the following code:
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