使用Python xmltodict阅读第三标签
XML文件的部分:使用Python xmltodict阅读第三标签
<?xml version="1.0" encoding="UTF-8"?>
<Station>
<Code>HT</Code>
<Type>knooppuntIntercitystation</Type>
<Namen>
<Kort>Den Bosch</Kort>
<Middel>'s-Hertogenbosch</Middel>
<Lang>'s-Hertogenbosch</Lang>
</Namen>
<Land>NL</Land>
<Synoniemen>
<Synoniem>Hertogenbosch ('s)</Synoniem>
<Synoniem>Den Bosch</Synoniem>
</Synoniemen>
</Station>
<Station>
<Code>ALMO</Code>
<Type>stoptreinstation</Type>
<Namen>
<Kort>Oostvaard</Kort>
<Middel>Oostvaarders</Middel>
<Lang>Almere Oostvaarders</Lang>
</Namen>
<Land>NL</Land>
<Synoniemen></Synoniemen>
</Station>
<Station>
<Code>ATN</Code>
<Type>stoptreinstation</Type>
<Namen>
<Kort>Aalten</Kort>
<Middel>Aalten</Middel>
<Lang>Aalten</Lang>
</Namen>
<Land>NL</Land>
<Synoniemen></Synoniemen>
</Station>
<Station>
<Code>ASA</Code>
<Type>intercitystation</Type>
<Namen>
<Kort>Amstel</Kort>
<Middel>Amsterdam Amstel</Middel>
<Lang>Amsterdam Amstel</Lang>
</Namen>
<Land>NL</Land>
<Synoniemen></Synoniemen>
</Station>
我的Python代码阅读XMLFILE:
import xmltodict
def leesXML(filename):
with open(filename) as mijnXMLBestand:
inhoud = mijnXMLBestand.read()
xmldictionary = xmltodict.parse(inhoud)
return xmldictionary
stationsdictionary = leesXML('stations.xml')
stations = stationsdictionary['Stations']['Station']
def program():
for station in stations:
type = station['Type']
Code = station['Code']
print(Code + ' - ' + type)
print('Dit zijn de codes en types van de 4 stations:')
program()
用程序( )我会得到一个类型和列表代码标记出来的XML文件。
现在,我试图找出如何让郎标记出来,并做出similair列表像一个以上,但只有代码和郎标签
我做了一个新的变量:
thirdtag = stationsdictionary['Stations']['Station']['Namen']
当我打印这我得到一个错误
TypeError: list indices must be integers or slices, not str
谁能说什么我做错了吗?请使用简单易读的Python代码。我是一名学生,开始学习python。
感谢
的stations
对象是list
实例:
>>>stations.__class__
<class 'list'>
因此,你必须访问给予list
整数索引元素(或片,但并不适用于这里),像此:
>>>stations[0]
OrderedDict([('Code', 'HT'), ('Type', 'knooppuntIntercitystation'), ('Namen', OrderedDict([('Kort', 'Den Bosch'), ('Middel', "'s-Hertogenbosch"), ('Lang', "'s-Hertogenbosch")])), ('Land', 'NL'), ('Synoniemen', OrderedDict([('Synoniem', ["Hertogenbosch ('s)", 'Den Bosch'])]))])
这最后一个对象是一个Dict
实例(更具体地是OrderedDict
),所以你要找的是:
>>> stations[0]['Namen']
OrderedDict([('Kort', 'Den Bosch'), ('Middel', "'s-Hertogenbosch"), ('Lang', "'s-Hertogenbosch")])
编辑:为了回答在评论你的最后一个问题,你可以做到以下几点:
>>>station_names = []
>>>for station in stations:
>>> station_names.append(station['Namen'])
>>>#or by comprehension
>>>station_names = [station['Namen'] for station in stations]
在你将获得任何情况下:
>>> station_names
[OrderedDict([('Kort', 'Den Bosch'), ('Middel', "'s-Hertogenbosch"), ('Lang', "'s-Hertogenbosch")]), OrderedDict([('Kort', 'Oostvaard'), ('Middel', 'Oostvaarders'), ('Lang', 'Almere Oostvaarders')])]
很好的解释先生。 –
所以我现在提出: insideName = stations [0] ['Name'] >>> print(insideName ['Lang'] 我得到一个站的正确答案,但是如何打印出所有的字典Namen的值?所以不用把[0]放在那里,因为这是一个特定的站,我一次都需要它们。 –
请你分享一个完整的文件例子吗? –
好吧,我只是改变了它。 –