def followers(self, level=-1):
"""Find all the followers of this guy
"""
self_depth = self.get('boss_path').count('/') - 1
ww = Wiseguy()
ww.set('boss_path', "%s%s/" % (self.get('boss_path'), '[0-9]+'))
followers = Mafia.find(ww, find_all=True)
if level > 0:
followers = filter(
lambda item: (item.get('boss_path').count('/') - 1) - self_depth <= level,
followers
)
return followers
def heir(self):
"""Find the heir of this guy. Calculate according to this formula:
- Firstly, oldest wiseguy at the same level.
- Second candidate is immediate and oldest subordinate.
- Finally, return None if none match (only if no subordinates or if just
one).
"""
path = self.get('boss_path')
depth = path.count('/') - 1
ww = Wiseguy()
# Simple regex to find the boss paths of same size. Which also implies
# they are all at the same level in the tree.
ww.set('boss_path', "^\/([0-9]+\/){%d}$" % depth)
# Run the search and viola
heir_candidates = Mafia.find(ww, find_all=True)
if len(heir_candidates) == 0:
heir_candidates = self.followers()
if len(heir_candidates) == 0:
return None
date_sorted_wiseguys = []
for wiseguy in heir_candidates:
if wiseguy.get('id') != self.get('id'):
if wiseguy.get('active') == "1":
date_sorted_wiseguys.append((wiseguy, wiseguy.get('date_of_initiation')))
date_sorted_wiseguys = sorted(
date_sorted_wiseguys,
key=operator.itemgetter(1)
)
return date_sorted_wiseguys[0][0]
def find_one(cls, *args, **kwargs):
"""Find a wiseguy
"""
# Since rows are dicts anyway, I will just use this shortcut
return Mafia.find(Wiseguy.from_dict(kwargs), find_all=False)
def save(self):
Mafia.update(self)
