該方法旨在分析排名靠前的所有商店。如果它們在任何其他排名列表中的位置不低于第一位，則它們屬于該前級并被添加到“級別”列表中。接下來，將它們從領先者中刪除，并調整所有列表，以便有新的領先者。重復這個過程，直到沒有商店離開。
def rank_stores(rankings):
"""
Rank stores with rankings by volume sales with over lap between lists.
:param rankings: list of rankings of stores also in lists.
:return: Ordered list with sets of items at same rankings.
"""
rank_global = []
# Evaluate all stores in the number one postion, if they are not below
# number one somewhere else, then they belong at this level.
# Then remove them from the front of the list, and repeat.
while sum([len(x) for x in rankings]) > 0:
tops = []
# Find out which of the number one stores are not in a lower position
# somewhere else.
for rank in rankings:
if not rank:
continue
else:
top = rank[0]
add = True
for rank_test in rankings:
if not rank_test:
continue
elif not rank_test[1:]:
continue
elif top in rank_test[1:]:
add = False
break
else:
continue
if add:
tops.append(top)
# Now add tops to total rankings list,
# then go through the rankings and pop the top if in tops.
rank_global.append(set(tops))
# Remove the stores that just made it to the top.
for rank in rankings:
if not rank:
continue
elif rank[0] in tops:
rank.pop(0)
else:
continue
return rank_global
對于提供的排名：
ranking_1 = ['J','A','Z','B','C']
ranking_2 = ['A','H','K','B']
ranking_3 = ['Q','O','A','N','K']
rankings = [ranking_1, ranking_2, ranking_3]
然后調用函數：
rank_stores(rankings)
結果是：
[{'J', 'Q'}, {'O'}, {'A'}, {'H', 'N', 'Z'}, {'K'}, {'B'}, {'C'}]
在某些情況下，可能沒有足夠的信息來確定明確的排名。試試這個順序。
['Z', 'A', 'B', 'J', 'K', 'F', 'L', 'E', 'W', 'X', 'Y', 'R', 'C']
我們可以得出以下排名：
a = ['Z', 'A', 'B', 'F', 'E', 'Y']
b = ['Z', 'J', 'K', 'L', 'X', 'R']
c = ['F', 'E', 'W', 'Y', 'C']
d = ['J', 'K', 'E', 'W', 'X']
e = ['K', 'F', 'W', 'R', 'C']
f = ['X', 'Y', 'R', 'C']
g = ['Z', 'F', 'W', 'X', 'Y', 'R', 'C']
h = ['Z', 'A', 'E', 'W', 'C']
i = ['L', 'E', 'Y', 'R', 'C']
j = ['L', 'E', 'W', 'R']
k = ['Z', 'B', 'K', 'L', 'W', 'Y', 'R']
rankings = [a, b, c, d, e, f, g, h, i, j, k]
調用函數：
rank_stores(rankings)
結果是：
[{'Z'},
{'A', 'J'},
{'B'},
{'K'},
{'F', 'L'},
{'E'},
{'W'},
{'X'},
{'Y'},
{'R'},
{'C'}]
在這種情況下，沒有足夠的信息來確定“J”相對于“A”和“B”的位置。只是它在“Z”和“K”之間的范圍內。
當在數百個排名和商店中相乘時，某些商店將無法按絕對數量正確排名。
