disambigufile 0.3.2

Class with file-like interface to a file found in provided search path

Class with file-like interface to a file found in provided search path

Features

• Search a path for a file that matches a pattern

• Search a path for a file inside directories that match a pattern

• Basic file-like interfaces:

  • DisFile(…).open()

  • with DisFile(…) as f: …

• Get path from config file (if optini module installed)

Examples

Simple usage:

from disambigufile import DisFile
path = '/bin:/usr/bin:/usr/local/bin'
try:
    print(DisFile('^ls', path=path))
except Exception as e:
    print(f"unable to disambiguate file; exception: {e}")

Using a with statement to open the disambiguated file:

from disambigufile import DisFile
path = 'path1:path2'
try:
    with DisFile(r'^asdf', path=path) as f:
        print(f.read())
except Exception as e:
    print(f"unable to disambiguate file; exception: {e}")

With more specific exception handling:

from disambigufile import DisFile
import disambigufile
path = '/bin:/usr/bin:/usr/local/bin'
try:
    print(DisFile('^ls', path=path))
except disambigufile.Error as e:
    # will only catch module-specific exceptions
    print(f"unable to disambiguate file; exception: {e}")

Match a file inside of a matched directory:

from disambigufile import DisFile
# search for unique file matching ~/Datasets/*2019-08-19*/data*
path='~/Datasets'
try:
    hit = DisFile(
        pattern='2019-08-19',
        path=path,
        subpattern='^data',
    )
    print(hit)
except disambigufile.Error as e:
    print(f"unable to disambiguate file; exception: {e}")

License

• Free software: MIT license

Documentation

• https://disambigufile.readthedocs.io/

Credits

This package was created with Cookiecutter and the audreyr/cookiecutter-pypackage project template.