def test_nonascii_chars(self):
passwordmetrics.configure()
# In 8-bit strings, ö is a part of Latin-1
entropy, unkown = passwordmetrics._character_entropy(b'abcdefgh\xf6')
self.assertEqual(unkown, set())
# But this is not!
entropy, unkown = passwordmetrics._character_entropy('abcdefgh\N{LATIN CAPITAL LETTER H WITH STROKE}')
self.assertEqual(unkown, set('\N{LATIN CAPITAL LETTER H WITH STROKE}'))
# If you want these to work, you have to make a custom configuration.
if 'unicode' not in locals():
unicode = str
groups = {'lowercase': set(unicode(string.ascii_lowercase)),
'uppercase': set(unicode(string.ascii_uppercase)),
'digits': set(unicode(string.digits)),
'punctuation': set(unicode(string.punctuation)),
'whitespace': set(unicode(string.whitespace)),
'non-printable': set(unicode((i)) for i in range(128) if chr(i) not in string.printable),
'extras': set('åäöö\N{LATIN CAPITAL LETTER H WITH STROKE}'),
}
passwordmetrics.configure(groups=groups)
# This will still raise an error,
entropy, unkown = passwordmetrics._character_entropy('abcdefghö\N{LATIN CAPITAL LETTER H WITH STROKE}')
self.assertEqual(unkown, set())
def test_character_entropy(self):
entropy, unkown = passwordmetrics._character_entropy('abcdefgh')
self.assertAlmostEqual(entropy, 37.6035177451287)
entropy, unkown = passwordmetrics._character_entropy('12345678')
self.assertAlmostEqual(entropy, 26.5754247590989)
# Repeated characters count as one:
entropy, unkown = passwordmetrics._character_entropy('aAaAaAaA')
self.assertAlmostEqual(entropy, 11.4008794362821)
# As good as it gets for an 8 char password
entropy, unkown = passwordmetrics._character_entropy('xyFg98%!')
self.assertAlmostEqual(entropy, 52.4367108134211)
def test_substitutions(self):
# This long password gets a reasonable score on characters alone
self.assertAlmostEqual(passwordmetrics._character_entropy('Tr0ub4dor&3')[0], 65.54588851677637)
# But much worse when considering it uses a word, even though it's misspelled and only
# appears once in the whole corpus (because I put it there).
self.assertAlmostEqual(passwordmetrics.metrics('Tr0ub4dor&3')['entropy'], 42.11714046349914)
# Although still better than if there was no substitutions in the word
self.assertAlmostEqual(passwordmetrics.metrics('Troubador&3')['entropy'], 31.332505617941614)
words, rest = passwordmetrics._find_words('batterybattery', self.words)
self.assertEquals(rest, '')
words, rest = passwordmetrics._find_words('b4tteryb4ttery', self.words)
self.assertEquals(rest, '4')
words, rest = passwordmetrics._find_words('b4ttery8attery', self.words)
self.assertEquals(rest, '48')
