CSV 读入字典答案

【问题标题】：CSV reading into a dictionaryCSV 读入字典
【发布时间】：2015-05-05 15:02:26
【问题描述】：

我正在尝试让程序读取 CSV 文件（“airports.csv”）并将其保存到字典中，在那里我可以输入密钥，这将调用与该密钥关联的机场对象。

当我尝试打印整个列表以确保它正常工作时，它会打印密钥但会打印{'GKA: <__main__.Airport object as 0x02E92570>}'

由于问题是基于使用字典对象而不是列表，您能否解释一下我哪里出错了。我可能混合了其中的一些并且走错了路，因为我总是用列表而不是对象来制作字典。

我已附上 CSV 文件和教程。

如果有人可以告诉我如何正确调用字典中的对象。谢谢

    import csv
    class Airport:

        def __init__(self, idx=-1, airportname='', cityname='', countryname='', code3='',
        code4='',lat=0,long=0, altitude=0,timezone='',DST='',Tz=''):

            self.idx=idx
            self.airportname=airportname
            self.cityname=cityname
            self.countryname=countryname
            self.code3=code3
            self.code4=code4
            self.lat=lat
            self.long=long
            self.lat=lat
            self.altitude=altitude
            self.timezone=timezone
            self.DST=DST
            self.Tz=Tz

        def dictairportChosen(self,filename):

            self.__airportDict={}

            f=open(filename, encoding="utf8")
            csvreader = csv.reader(f)

            for idx, airportname, cityname, countryname, code3, code4, lat, long, altitude, timezone, DST, TZ,  in csvreader:
                airport=Airport(idx, airportname, cityname, countryname, code3, code4, lat, long, altitude, timezone, DST, TZ)
                self.__airportDict[code3]=airport
                print(self.__airportDict)


    aAirportChosen=Airport()
    aAirportChosen.dictairportChosen("airports.csv")

这是 CSV 文件的 sn-p：

    1,"Goroka","Goroka","Papua New Guinea","GKA","AYGA",-6.081689,145.391881,5282,10,"U","Pacific/Port_Moresby"
    2,"Madang","Madang","Papua New Guinea","MAG","AYMD",-5.207083,145.7887,20,10,"U","Pacific/Port_Moresby"
    3,"Mount Hagen","Mount Hagen","Papua New Guinea","HGU","AYMH",-5.826789,144.295861,5388,10,"U","Pacific/Port_Moresby"
    4,"Nadzab","Nadzab","Papua New Guinea","LAE","AYNZ",-6.569828,146.726242,239,10,"U","Pacific/Port_Moresby"
    5,"Port Moresby Jacksons Intl","Port Moresby","Papua New Guinea","POM","AYPY",-9.443383,147.22005,146,10,"U","Pacific/Port_Moresby"
    6,"Wewak Intl","Wewak","Papua New Guinea","WWK","AYWK",-3.583828,143.669186,19,10,"U","Pacific/Port_Moresby"
    7,"Narsarsuaq","Narssarssuaq","Greenland","UAK","BGBW",61.160517,-45.425978,112,-3,"E","America/Godthab"
    8,"Nuuk","Godthaab","Greenland","GOH","BGGH",64.190922,-51.678064,283,-3,"E","America/Godthab"

实际教程的问题是： 2. 写代码至读从这 CSV 和创造一个字典的飞机场对象代替这列表在这字典和飞机场对象。你应该结尾向上和一个字典那有这飞机场代码作为一个钥匙和一个飞机场目的作为这抬头价值允许这代码以下至运行：

airportLookupDict={‘DUB’, airport(….),‘LHR’, airport(….)}
myairport=airportLookupDict.get(‘DUB’)
print(myairport.name)
outputs:
DUBLIN

【问题讨论】：

为什么你认为有问题？您有一本字典，其中键是机场代码，值是 Airport 实例 - 您期待什么？
那是从我试图将字典放在它自己的类中时开始的。在那里改变了它并且它起作用了。谢谢

标签： python class csv dictionary

【解决方案1】：

未经测试，但您可以尝试

self.__airportDict[code3]=airport.__dict__

来自Python dictionary from an object's fields

【讨论】：

【解决方案2】：

您的代码没有问题。它按问题中的预期工作。如果您想访问列表的各个元素，只需将实体名称添加到列表中，如下所示：

    for keys, values in self.__airportDict.items():
        print keys
        print values.airportname

【讨论】：