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我正在使用一个名为 OpenTripPlanner 的开源项目，它是我计划用来在给定时间模拟从一个点到另一个点的大量行程的工具。到目前为止，我已经设法找到包含有关行程的所有信息的 XML 文件所在的 URL。XML 是根据请求构建的，因此 URL 不是静态的。URL 看起来像这样：

http://localhost:8080/otp/routers/default/plan?fromPlace=48.40915,%20-71.04996&toPlace=48.41428,%20-71.06996&date=2017/12/04&time=8:00:00&mode=TRANSIT,WALK

（您需要运行 OpenTripPlanner 服务器才能打开它）

现在，我想读取这些 XML 文件并使用 python 3 进行一些数据分析，但我找不到读取文件的方法。我尝试使用 urllib.request 在本地下载文件，但是我从中获得的文件的格式很奇怪。它看起来像这样

{"requestParameters":{"date":"2017/12/04","mode":"TRANSIT,WALK","fromPlace":"48.40915, -71.04996","toPlace":"48.41428, -71.06996", "时间":"8:00:00"},"计划":{"日期":1512392400000,"从":{"名称":"原点","经度":-71.04996,"纬度":48.40915, "orig":"","vertexType":"NORMAL"},"to":{"name":"Destination","lon":-71.06996,"lat":48.41428,"orig":""," vertexType":"NORMAL"},"itineraries":[{"duration":1538,"startTime":1512392809000,"endTime":1512394347000,"walkTime":934,"transitTime":602,"waitingTime":2,"walkDistance":1189.6595112715966,"walkLimitExceeded":false,"elevationLost":0.0,"elevationGained":0.0,"transfers":0,"legs":[{"startTime":1512392809000,"endTime" :1512393537000,"departureDelay":0,"arrivalDelay":0,"realTime":false,"distance":926.553,"pathway":false,"mode":"WALK","route":"","agencyTimeZoneOffset ":-18000000,"interlineWithPreviousLeg":false,"from":{"name":"Origin","lon":-71.04996,"lat":48.40915,"departure":1512392809000,"orig":"", "vertexType":"NORMAL"},"to":{"name":"Roitelets / Martinets","stopId":"1:370","stopCode":"370","lon":-71.047688,"lat":48.401531,"arrival":1512393537000,"departure":1512393538000,"stopIndex":15,"stopSequence ":16,"vertexType":"TRANSIT"},"legGeometry":{"points":"s{mfHb{spL|ExBp@sDl@V@@lB|@j@FL?j@GbCk@|A] vEsA^KBA|C{@pCeACS~CuA`@Q","length":19},"rentedBike":false,"transitLeg":false,"duration":728.0,"steps":[{"distance": 131.991,"relativeDirection":"DEPART","streetName":"Rue D.-V.-Morrier","absoluteDirection":"SOUTH","stayOn":false,"area":false,"bogusName":false “隆”：-71.04961760502248，“纬度”：48。4090671692228,"elevation":[]},{"distance":72.319,"relativeDirection":"LEFT","streetName":"Rue Lorenzo-Genest","absoluteDirection":"EAST","stayOn":false, "面积":false,"bogusName":false,"lon":-71.0502299,"lat":48.4079519,"elevation":[]}

当我尝试在浏览器中打开文件时，我收到一条错误消息

XML Parsing Error: not well-formed
Location: http://localhost:63342/XML_reader/file.xml?_ijt=e1d6h53s4mh1ak94sqortejf9v
Line Number 1, Column 1: ...

我使用的脚本很简单，看起来像这样

import urllib.request

testfile = urllib.request.URLopener()
file_name = 'http://localhost:8080/otp/routers/default/plan?fromPlace=48.40915,%20-71.04996&toPlace=48.41428,%20-71.06996&date=2017/12/04&time=8:00:00&mode=TRANSIT,WALK'
testfile.retrieve(file_name, "file.xml")

如何使输出的 XML 文件格式正确？除了 urllib.request 之外，还有其他我可能想尝试的方法吗？

非常感谢

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1 回答 1

1

要将此文件作为 JSON 数据（不是 XML）导入，您需要 JSON 库

import urllib.request
import json
from pprint import pprint

testfile = urllib.request.URLopener()
file_name = 'http://localhost:8080/otp/routers/default/plan?fromPlace=48.40915,%20-71.04996&toPlace=48.41428,%20-71.06996&date=2017/12/04&time=8:00:00&mode=TRANSIT,WALK'
testfile.retrieve(file_name, "file.json")

data = json.load(open('file.json'))
pprint(data)
于 2018-01-23T21:04:59.157 回答