def resource()

in ShortHand/lambda/resolve.py [0:0]

• 5 lines of code
• 3 McCabe index (conditional complexity)

def resource(name):
    """
    Returns resource types that match `name`, working right-to-left
    E.g. S3::Bucket will match AWS::S3::Bucket
    """

    return [
        key for key in SPEC["ResourceTypes"].keys()
        if key.endswith(name)
    ]