通过列表的排列循环并将功能应用于它们

Question

我正在尝试获取5个机场输入的排列列表，以通过距离计算并对每个排列进行计算。

from airports import *
from math import *
import itertools
dicts=TravelLookUp()
dicts.dictAirport('airport.csv')
dicts.dictCurrency('countrycurrency.csv')
dicts.dictCurrencyRates('currencyrates.csv')
#print(airportdict.airportDict)
#airportdict.airportDict[] IS DICTIONARY
#print(test)
#print(AirportDict)
test= dicts.airportDict['DUB'].getLat()
print(test)
airportList=[]
airportHome=[]
airport1=input('Choose the first airport: ').upper()
airport2=input('Choose the second airport: ').upper()
airport3=input('Choose the third airport: ').upper()
airport4=input('Choose the fourth airport: ').upper()
airport5=input('Choose the fifth airport: ').upper()
airportHome=[airport1]
airportList=[airport2,airport3,airport4,airport5]
AirportCombinations = itertools.permutations(airportList,r=4)
def distance(airport1,airport2):
    distacnces=[]
    lat1=dicts.airportDict[airport1].getLat()
    long1=dicts.airportDict[airport1].getLong()
    lat2=dicts.airportDict[airport2].getLat()
    long2=dicts.airportDict[airport2].getLong()
    lat1R= (lat1*pi/180)
    long1R=(long1*pi/180)
    lat2R=(lat2*pi/180)
    long2R=(long2*pi/180)
    distance= acos(sin(lat1R)*sin(lat2R)*cos(long1R-long2R))*6373
    distances.append(distance)
    return distances
for i in AirportCombinations:
    i=list(i)
    route=[airportHome]
    for value in range(len(i)):
            route.append(value)
            totaldistance=0
            for i in range(len(route)-1):
                legcost=distance(route[i],route[i+1])
                totaldistance+=distance
                print(totaldistance)
distance()

路由列表保留排列，我希望它在每个排列中的每个机场对中循环浏览distance(airport1,airport2)功能。

每当我运行此操作时，它只会计算24次相同两个机场之间的距离，而我希望它要做的就是计算每个组合的距离。

如何通过distance(airport1,airport2)函数计算运行所有排列？然后创建一个列表以保存所有这些计算？

Answer 1

我认为您在循环中有些复杂的事情开始for i in AirportCombinations。

我已经简化了该循环（并为清晰重命名您的变量），但保留了您的方法，即转换为每条路线的列表。

您也有一个错别字：我认为totaldistance += distance应该是totaldistance += legcost。

for sub_route in AirportCombinations:
    route = [airportHome] + list(sub_route)
    totaldistance = 0
    for i in range(len(route)-1):
        legcost = distance(route[i], route[i+1])
        totaldistance += legcost
    
    #I've un-indented this so it prints for each permutation
    print('route ' + str(route) + ' has distance ' + str(totaldistance)) 
    # Add the distance to a list or dictionary here if you want them available later...

---

另一种方法

您可以完全摆脱内部循环，剩下的是读者，Python的禅宗的哪一部分更重要 - 这是扁平的（比嵌套更好），但是Trump"可读性计数"是否...

邮政编码利用已记录的行为：

当最短输入峰值耗尽时，迭代器停止

for sub_route in AirportCombinations:
    route = [airportHome] + list(sub_route)
    totaldistance = sum(distance(a[0], a[1]) for a in zip(route, route[1:]))        
    print('route ' + str(route) + ' has distance ' + str(totaldistance)) 
    # Add the distance to a list or dictionary here if you want them available later...

Answer 2

我认为您的问题在这里：

for i in AirportCombinations:
    i=list(i)
    route=[airportHome]        
    for value in range(len(i)): <<<
            route.append(value) <<<
            totaldistance=0
            for i in range(len(route)-1):
                legcost=distance(route[i],route[i+1])
                totaldistance+=distance
                print(totaldistance)

当您应该附加i[value]

时，您将value附加到route

或更好的是，直接通过i迭代：

for i in AirportCombinations:
    i=list(i)
    route=[airportHome]
    for value in i:
            route.append(value)
            totaldistance=0
            for i in range(len(route)-1):
                legcost=distance(route[i],route[i+1])
                totaldistance+=distance
                print(totaldistance)

P.S。 - 在两个嵌套环的两个级别中使用i只是令人困惑...