我有很多巴士站。 每个列表代表公共汽车在一次旅程中停靠。 行程路径是相同的，但公共汽车可能会停在任何一站。
我想在ORDER中创建所有停靠点的完整列表。
下面是三个巴士旅程样本。
bus_1 = ["5171","1337","1341","1350","1352","1357","320","278","10","15","215","218","1623","1624","7347"]
bus_2 = ["5171","2976","2979","2981","2991","2992","1326","1327","1329","1331","1336","1337","1339","1340","1342","1345","1350","1354","1357","1359","320","278","12","15","17","21","205","215","216","218","1624","1626","1627","7347"]
bus_3 = ["5171","2977" "2978","2991","1325","1326","1327","1329","1330","1331","1332","1333","1337","1340","1341","1342","1344","1345","1347","1348","1352","1353","1354","1355","1357","1359","320","278","10","12","14","15","17","19","21","85","204","215","218","219","220","1622","1623","1624","1625","1626","1627","7347"]
这可能吗？
答案 0 :(得分：2)
嗯，这是我的解决方案 - 请检查它是否满足您的需求。我已经用字典替换了子路由的单独变量，希望它不是问题。
routes = {
'bus_1': ["5171","1337","1341","1350","1352","1357","320","278","10","15","215","218","1623","1624","7347"],
'bus_2': ["5171","2976","2979","2981","2991","2992","1326","1327","1329","1331","1336","1337","1339","1340","1342","1345","1350","1354","1357","1359","320","278","12","15","17","21","205","215","216","218","1624","1626","1627","7347"],
'bus_3': ["5171","2977", "2978","2991","1325","1326","1327","1329","1330","1331","1332","1333","1337","1340","1341","1342","1344","1345","1347","1348","1352","1353","1354","1355","1357","1359","320","278","10","12","14","15","17","19","21","85","204","215","218","219","220","1622","1623","1624","1625","1626","1627","7347"]
}
stops = set()
stop_pairs = set()
for route in routes.values():
stops.update(route)
for i in range(1,len(route)):
# each pair is correctly ordered
stop_pairs.add(tuple(route[i-1:i+1]))
# at this point, 'stops' is the set of all stops
# and 'stop_pairs' is a set of tuples representing sequence of stops
ordered_stops = []
while stops:
# let's look for the minimal elements in stops
minimals = set()
for stop in stops:
if not any((s, stop) in stop_pairs for s in stops):
# there is no such s that s precedes stop
minimals.add(stop)
ordered_stops.append(minimals)
stops.difference_update(minimals)
print(ordered_stops)
我的结果是：
[{'5171'}，{'2977'，'2976'}，{'2979'，'2978'}，{'2981'}，{'2991'}，{'1325'，'2992' }，{'1326'}，{'1327'}，{'1329'}，{'1330'}，{'1331'}，{'1336'，'1332'}，{'1333'}，{' 1337'}，{'1339'}，{'1340'}，{'1341'}，{'1342'}，{'1344'}，{'1345'}，{'1347'，'1350'}， {'1348'}，{'1352'}，{'1353'}，{'1354'}，{'1355'}，{'1357'}，{'1359'}，{'320'}，{' 278'}，{'10'}，{'12'}，{'14'}，{'15'}，{'17'}，{'19'}，{'21'}，{'205' ，''85'}，{'204'}，{'215'}，{'216'}，{'218'}，{'219'}，{'220'}，{'1622'}，{' 1623'}，{'1624'}，{'1625'}，{'1626'}，{'1627'}，{'7347'}]
更新：我刚刚注意到我的程序可以轻松推广以生成变体解决方案，并相应地进行编辑。现在，结果列表由具有相同优先级的停靠点组成（即，每个集合中的停靠点可以重新洗牌）。显然，有六个这样的集合：{'2977'，'2976'}，{'2979'，'2978'}，{'1325'，'2992'}，{'1336'，'1332'}，{' 1347'，'1350'}，{'205'，'85'}。
答案 1 :(得分：0)
我能够接近但我认为你需要更多的路线来获得完整的路线。它无法添加的某些值没有足够相似的周围元素，但这可能会让您走上正确的道路。
bus_1 = ["5171","1337","1341","1350","1352","1357","320","278","10","15","215","218","1623","1624","7347"]
bus_2 = ["5171","2976","2979","2981","2991","2992","1326","1327","1329","1331","1336","1337","1339","1340","1342","1345","1350","1354","1357","1359","320","278","12","15","17","21","205","215","216","218","1624","1626","1627","7347"]
bus_3 = ["5171","2977", "2978","2991","1325","1326","1327","1329","1330","1331","1332","1333","1337","1340","1341","1342","1344","1345","1347","1348","1352","1353","1354","1355","1357","1359","320","278","10","12","14","15","17","19","21","85","204","215","218","219","220","1622","1623","1624","1625","1626","1627","7347"]
buses=[bus_1,bus_2,bus_3]
added=set()
final=bus_1[:]
temp=bus_1+bus_2+bus_3
i=0
x=0
while set(final) != set(temp):
# get 2 elements which we'll see if they're in final route
if x<len(buses):
t=buses[x][i:i+2]
else:
t=final[i:i+2]
try:
a=final.index(t[0])
b=final.index(t[1])
except:
a=b=-1
# check if in final and not next to each other, and not added yet
for q,bus in enumerate(buses):
if x!=q and t[0] in bus and t[1] in bus and a+1==b and (t[0],t[1]) not in added:
u=bus.index(t[0])
v=bus.index(t[1])
final[a:b]=bus[u:v]
added.add((t[0],t[1]))
i=0
x=0
if x<len(buses):
for q,bus in enumerate(buses):
if q==x and i+2<len(bus):
i+=1
elif q==x and i+2==len(bus):
x+=1
i=0
elif x==len(buses)+1 and i+2<len(final):
i+=1
else:
x+=1
i=0
if x>len(buses)+1:
break
print("final route(incomplete)")
print(final)
print('unique stops')
print(set(temp))
#this is my result
#mine=['5171', '2976', '2979', '2981', '2991', '2992', '1326', '1327', '1329', '1330', '1331', '1336', '1337', '1339', '1340', '1341', '1350', '1352', '1353', '1354', '1355', '1357', '1359', '320', '278', '10', '12', '14', '15', '17', '19', '21', '205', '215', '216', '218', '219', '220', '1622', '1623', '1624', '1625', '1626', '1627', '7347']
#this is what you probably want
#goal=['5171', '2976', '2977', '2978', '2979', '2981', '2991', '2992', '1325', '1326', '1327', '1329', '1330', '1331', '1332', '1333', '1336', '1337', '1339', '1340', '1341', '1342', '1344', '1345', '1347', '1348', '1350', '1352', '1353', '1354', '1355', '1357', '1359', '320', '278', '10', '12', '14', '15', '17', '19', '21', '85','204', 205', '215', '216', '218', '219', '220', '1622', '1623', '1624', '1625', '1626', '1627', '7347']
#unable to add
#['204', '85', '1348', '1347', '1342', '1344', '1345', '1333', '1332', '1325', '2978']
答案 2 :(得分：0)
# All three buses in 1 2d list.
route = ["5171","1337","1341","1350","1352","1357","320","278","10","15","215","218","1623","1624","7347"],["5171","2976","2979","2981","2991","2992","1326","1327","1329","1331","1336","1337","1339","1340","1342","1345","1350","1354","1357","1359","320","278","12","15","17","21","205","215","216","218","1624","1626","1627","7347"],["5171","2977","2978","2991","1325","1326","1327","1329","1330","1331","1332","1333","1337","1340","1341","1342","1344","1345","1347","1348","1352","1353","1354","1355","1357","1359","320","278","10","12","14","15","17","19","21","85","204","215","218","219","220","1622","1623","1624","1625","1626","1627","7347"]
temp = set()
out = {}
# remove dupes by adding each stop to a set.
for journey in route:
for stop in journey:
temp.add(stop)
# create a dictionary where the key is a unique stop
# the value is a list of two empty sets.
for i in temp:
out[i] = [set(),set()]
for i in temp:
for journey in route:
if i in journey:
# add stops which occur before this stop to one dictionary
for before in journey[0:journey.index(i)]:
out[i][0].add(before)
# and stops which occur after this stop to the other dictionary
for after in journey[journey.index(i):-1]:
out[i][1].add(after)
# create a template final list of unique stops from the set.
final_list = list(temp)
# iterate ove the dictionary
for key, value in out.items():
# get list of what stops should be ordered before the current stop
before_set = value[0]
# and a list of those that should be after.
after_set = value[1]
for b in before_set:
# if the current value is not where is should be move it.
if final_list.index(b) > final_list.index(key):
temp = final_list.index(key)
final_list.insert(final_list.index(b), key)
final_list.pop(temp)
for a in after_set:
# if the current value is not where is should be move it.
if final_list.index(a) < final_list.index(key):
temp = final_list.index(key)
final_list.insert(final_list.index(a), key)
final_list.pop(temp+1)
# and run again in the opposite direction.
final_list = final_list[::-1]
for key, value in out.items():
before_set = value[0]
after_set = value[1]
for b in before_set:
if final_list.index(b) > final_list.index(key):
temp = final_list.index(key)
final_list.insert(final_list.index(b), key)
final_list.pop(temp)
for a in after_set:
if final_list.index(a) < final_list.index(key):
temp = final_list.index(key)
final_list.insert(final_list.index(a), key)
final_list.pop(temp+1)
print(final_list)