Searched refs:_diff_each_passwd_by_uid (Results 1 - 1 of 1) sorted by relevance

/sssd-io/src/tests/intg/
H A Dent.py258 def _diff_each_passwd_by_uid(pattern_dict): function
283 return _diff_each_passwd_by_uid(dict((p["uid"], p) for p in pattern_seq))
300 d = _diff_each_passwd_by_uid(pattern_dict)

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