pycld3 | Python3 bindings for the Compact Language Detector v3 | Natural Language Processing library

by bsolomon1124 C++ Version: 0.22 License: Apache-2.0

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

kandi X-RAY | pycld3 Summary

pycld3 is a C++ library typically used in Artificial Intelligence, Natural Language Processing applications. pycld3 has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

This package contains Python bindings (via Cython) to Google's CLD3 library. The library outputs BCP-47-style language codes. For some languages, output is differentiated by script. Language and script names from Unicode CLDR. It supports over 100 languages/scripts. See full list of supported languages/scripts in Google's CLD3 documentation.

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Support

pycld3 has a low active ecosystem.

It has 121 star(s) with 5 fork(s). There are 3 watchers for this library.

It had no major release in the last 12 months.

There are 8 open issues and 13 have been closed. On average issues are closed in 76 days. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of pycld3 is 0.22

Quality

pycld3 has 0 bugs and 1 code smells.

Security

pycld3 has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

pycld3 code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

pycld3 is licensed under the Apache-2.0 License. This license is Permissive.

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

Reuse

pycld3 releases are not available. You will need to build from source code and install.

Installation instructions are not available. Examples and code snippets are available.

It has 180 lines of code, 4 functions and 3 files.

It has medium code complexity. Code complexity directly impacts maintainability of the code.

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pycld3 Key Features

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pycld3 Examples and Code Snippets

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Community Discussions

Trending Discussions on pycld3

Create new column that checks equality with another column in Python

Find the most popular element from a row in Python

QUESTION

Create new column that checks equality with another column in Python

Asked 2020-Aug-31 at 21:32

df
   pycld3   spacy   seqtolang   langid  langdetect  text_language
0   lt     unknown      ro        en         pl         unknown
1   bg     unknown      fi        en         tl         unknown
3   no        id        in        de         no         no
4   en        en        zh        en         en         en
5   en        en        en        en         en         en

...

ANSWER

Answered 2020-Aug-31 at 20:44

df
Out[6]: 
  one  two three
0  10  1.2   4.2
1  15   70  0.03
2   8    5     0
df['new'] = df['one']==df['two']
df
Out[8]: 
  one  two three    new
0  10  1.2   4.2  False
1  15   70  0.03  False
2   8    5     0  False

df['new'] = df['new'].astype(int)
df
Out[10]: 
  one  two three  new
0  10  1.2   4.2    0
1  15   70  0.03    0
2   8    5     0    0

Source https://stackoverflow.com/questions/63677875

QUESTION

Find the most popular element from a row in Python

Asked 2020-Aug-31 at 17:52

I want to create a new column that is populated by the most popular element in the four rows mentioned (I have more than 4 rows in the dataset).

...

ANSWER

Answered 2020-Aug-31 at 17:52

Try with:

Source https://stackoverflow.com/questions/63675720

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