aboutcode-org / license-expression

Utility library to parse, normalize and compare License expressions for Python using a boolean logic engine. For expressions using SPDX or any other license id scheme.
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boolean-expression license-expression licensing python spdx spdx-license

================== license-expression

license-expression is a comprehensive utility library to parse, compare, simplify and normalize license expressions (such as SPDX license expressions) using boolean logic.

Software project licenses are often a combination of several free and open source software licenses. License expressions -- as specified by SPDX -- provide a concise and human readable way to express these licenses without having to read long license texts, while still being machine-readable.

License expressions are used by key FOSS projects such as Linux; several packages ecosystem use them to document package licensing metadata such as npm and Rubygems; they are important when exchanging software data (such as with SPDX and SBOM in general) as a way to express licensing precisely.

license-expression is a comprehensive utility library to parse, compare, simplify and normalize these license expressions (such as SPDX license expressions) using boolean logic like in: GPL-2.0-or-later WITH Classpath-exception-2.0 AND MIT.

It includes the license keys from SPDX https://spdx.org/licenses/ (version 3.25) and ScanCode license DB (version 32.3.0, last published on 2024-10-24). See https://scancode-licensedb.aboutcode.org/ to get started quickly.

license-expression is both powerful and simple to use and is a used as the license expression engine in several projects and products such as:

See also for details:

license-expression is also packaged for most Linux distributions. See below.

Alternative:

There is no known alternative library for Python, but there are several similar libraries in other languages (but not as powerful of course!):

Build and tests status

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Source code and download

Also available in several Linux distros:

Support

Description

This module defines a mini language to parse, validate, simplify, normalize and compare license expressions using a boolean logic engine.

This supports SPDX license expressions and also accepts other license naming conventions and license identifiers aliases to resolve and normalize any license expressions.

Using boolean logic, license expressions can be tested for equality, containment, equivalence and can be normalized or simplified.

It also bundles the SPDX License list (3.20 as of now) and the ScanCode license DB (based on latest ScanCode) to easily parse and validate expressions using the license symbols.

Usage examples

The main entry point is the Licensing object that you can use to parse, validate, compare, simplify and normalize license expressions.

Create an SPDX Licensing and parse expressions::

>>> from license_expression import get_spdx_licensing
>>> licensing = get_spdx_licensing()
>>> expression = ' GPL-2.0 or LGPL-2.1 and mit '
>>> parsed = licensing.parse(expression)
>>> print(parsed.pretty())
OR(
  LicenseSymbol('GPL-2.0-only'),
  AND(
    LicenseSymbol('LGPL-2.1-only'),
    LicenseSymbol('MIT')
  )
)

>>> str(parsed)
'GPL-2.0-only OR (LGPL-2.1-only AND MIT)'

>>> licensing.parse('unknwon with foo', validate=True, strict=True)
license_expression.ExpressionParseError: A plain license symbol cannot be used
as an exception in a "WITH symbol" statement. for token: "foo" at position: 13

>>> licensing.parse('unknwon with foo', validate=True)
license_expression.ExpressionError: Unknown license key(s): unknwon, foo

>>> licensing.validate('foo and MIT and GPL-2.0+')
ExpressionInfo(
    original_expression='foo and MIT and GPL-2.0+',
    normalized_expression=None,
    errors=['Unknown license key(s): foo'],
    invalid_symbols=['foo']
)

Create a simple Licensing and parse expressions::

>>> from license_expression import Licensing, LicenseSymbol
>>> licensing = Licensing()
>>> expression = ' GPL-2.0 or LGPL-2.1 and mit '
>>> parsed = licensing.parse(expression)
>>> expression = ' GPL-2.0 or LGPL-2.1 and mit '
>>> expected = 'GPL-2.0-only OR (LGPL-2.1-only AND mit)'
>>> assert str(parsed) == expected
>>> assert parsed.render('{symbol.key}') == expected

Create a Licensing with your own license symbols::

>>> expected = [
...   LicenseSymbol('GPL-2.0'),
...   LicenseSymbol('LGPL-2.1'),
...   LicenseSymbol('mit')
... ]
>>> assert licensing.license_symbols(expression) == expected
>>> assert licensing.license_symbols(parsed) == expected

>>> symbols = ['GPL-2.0+', 'Classpath', 'BSD']
>>> licensing = Licensing(symbols)
>>> expression = 'GPL-2.0+ with Classpath or (bsd)'
>>> parsed = licensing.parse(expression)
>>> expected = 'GPL-2.0+ WITH Classpath OR BSD'
>>> assert parsed.render('{symbol.key}') == expected

>>> expected = [
...   LicenseSymbol('GPL-2.0+'),
...   LicenseSymbol('Classpath'),
...   LicenseSymbol('BSD')
... ]
>>> assert licensing.license_symbols(parsed) == expected
>>> assert licensing.license_symbols(expression) == expected

And expression can be deduplicated, to remove duplicate license subexpressions without changing the order and without consider license choices as simplifiable::

>>> expression2 = ' GPL-2.0 or (mit and LGPL 2.1) or bsd Or GPL-2.0  or (mit and LGPL 2.1)'
>>> parsed2 = licensing.parse(expression2)
>>> str(parsed2)
'GPL-2.0 OR (mit AND LGPL 2.1) OR BSD OR GPL-2.0 OR (mit AND LGPL 2.1)'
>>> assert str(parsed2.simplify()) == 'BSD OR GPL-2.0 OR (LGPL 2.1 AND mit)'

Expression can be simplified, treating them as boolean expressions::

>>> expression2 = ' GPL-2.0 or (mit and LGPL 2.1) or bsd Or GPL-2.0  or (mit and LGPL 2.1)'
>>> parsed2 = licensing.parse(expression2)
>>> str(parsed2)
'GPL-2.0 OR (mit AND LGPL 2.1) OR BSD OR GPL-2.0 OR (mit AND LGPL 2.1)'
>>> assert str(parsed2.simplify()) == 'BSD OR GPL-2.0 OR (LGPL 2.1 AND mit)'

Two expressions can be compared for equivalence and containment:

>>> expr1 = licensing.parse(' GPL-2.0 or (LGPL 2.1 and mit) ')
>>> expr2 = licensing.parse(' (mit and LGPL 2.1)  or GPL-2.0 ')
>>> licensing.is_equivalent(expr1, expr2)
True
>>> licensing.is_equivalent(' GPL-2.0 or (LGPL 2.1 and mit) ',
...                         ' (mit and LGPL 2.1)  or GPL-2.0 ')
True
>>> expr1.simplify() == expr2.simplify()
True
>>> expr3 = licensing.parse(' GPL-2.0 or mit or LGPL 2.1')
>>> licensing.is_equivalent(expr2, expr3)
False
>>> expr4 = licensing.parse('mit and LGPL 2.1')
>>> expr4.simplify() in expr2.simplify()
True
>>> licensing.contains(expr2, expr4)
True

Development

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