循环遍历列表的排列并将函数应用于它们

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()

路线列表包含排列，我希望它循环遍历每个排列中每个机场对的 distance(airport1,airport2) 函数。

每当我运行它时，它只会计算相同的两个机场之间的距离 24 次，而我想要它做的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？然后创建一个列表来保存所有这些计算？

Answer 1

我认为从for i in AirportCombinations 开始，您的循环中的事情有些过于复杂了。

我已经简化了该循环（并为清晰起见重命名了您的变量），但保留了您的方法，即为每条路线转换为列表。

你还有一个错字：totaldistance += distance 应该是 totaldistance += legcost，我想。

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，留给读者看Zen of Python 的哪一部分更重要 - 这更扁平（比嵌套更好），但这是否胜过“可读性计数”...

zip 使用documented behaviour:

当最短的输入迭代用完时，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

我认为你的问题在这里：

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)

当你应该附加 i[value] 时，你正在附加 value 到 route

或者更好的是，直接遍历i：

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)

附： - 在嵌套循环的两个级别中使用 i 只是令人困惑......