如何从列表中找到对象

问题描述

我使用下面的程序来创建从一个网站获得的城市列表。现在我想从我创建的列表中找到城市的名称（参数）。我怎么做？

换句话说，我如何从列表中找到一个对象？我试过：

检查 city 是否在 listOfCities $ b $ pre $ 如果listOfCities中的城市：
＃city在列表中


$ b

在列表中找到它的索引：

  i = listOfCities.index（city）
  

如果city不在$中，则会引发 IndexError c $ c> listOfCities 。

您可以使用HTMLParser来解析html而不是正则表达式。

完整的示例

 ＃！/ usr / bin / env python 
＃ -  *  -  coding： utf-8  -  *  -  
 from __future__ import unicode_literals 
 import cgi 
 
 try：
 from html.parser import HTMLParser 
除了ImportError：＃Python 2 
 from HTMLParser import HTMLParser 
 
 try：$ b $ from urllib.request import urlopen 
除了ImportError：＃Python 2 $ b $ from urllib2 import urlopen 
 
类CitiesParser（HTMLParser）：
从html中提取城市列表。
 def __init __（self，html）：
 HTMLParser .__ init __（self）
 self.cities = [] 
 self.incity = None 
 self.feed（html）
 
 def handle_starttag（self，tag，attrs）：
 self .incity = tag =='a'和（'c （self，tag）：
 self.incity = False 
 def handle_data（self，data）：$ b $如果self.incity ：
 self.cities.append（data.strip（））
 
＃下载并解析城市列表
 response = urlopen（http://weather.canoe.ca/ Weather / World.html）
 _，params = cgi.parse_header（response.headers.get（'Content-Type'，''））
 html = response.read（）。decode（params ['charset']）
 
＃找到城市
 cities = CitiesParser（html）.cities 
在['ArRiyāḍ'，'Riyadh']：$ b $如果城市是城市：
 print（％s is found％（city，））
 print（index is％d％（cities.index（city），））
 break 
 else：
 print（％r not found％（city，））
  

I used the following program to create a list of cities obtained from a website. Now I want to find the name of city (argument) from the list I created. How do I do that?

In other words, how do I find an object from a list? I tried: listOfCities.find (city), I got an error as the attribute find was not found.

def weatherNow (city):
  import urllib
  connection = urllib.urlopen("http://weather.canoe.ca/Weather/World.html")
  weather = connection.read()
  connection.close()
  cityLoc = weather.find('class="weatherred"')
  cityEnd = weather.find("</a>", cityLoc)
  if city != -1:
    listOfCities = []
    while cityLoc != -1:
      cityNames = weather[cityLoc+19:cityEnd-1]
      listOfCities.append(cityNames)
      cityLoc = weather.find('class="weatherred"', cityLoc+1)
      cityEnd = weather.find("</a>", cityLoc)

  print listOfCities

解决方案

To check whether city is in listOfCities:

if city in listOfCities:
   # city is in the list

To find its index in the list:

 i = listOfCities.index(city)

It raises IndexError if city is not in the listOfCities.

You could use HTMLParser to parse the html instead of regexes.

Complete example

#!/usr/bin/env python
# -*- coding: utf-8 -*-
from __future__ import unicode_literals
import cgi

try:
    from html.parser import HTMLParser
except ImportError: # Python 2
    from HTMLParser import HTMLParser

try:
    from urllib.request import urlopen
except ImportError: # Python 2
    from urllib2 import urlopen

class CitiesParser(HTMLParser):
    """Extract city list from html."""
    def __init__(self, html):
        HTMLParser.__init__(self)
        self.cities = []
        self.incity = None
        self.feed(html)

    def handle_starttag(self, tag, attrs):
        self.incity = tag == 'a' and ('class', 'weatherred') in attrs
    def handle_endtag(self, tag):
        self.incity = False
    def handle_data(self, data):
        if self.incity:
            self.cities.append(data.strip())

# download and parse city list
response = urlopen("http://weather.canoe.ca/Weather/World.html")
_, params = cgi.parse_header(response.headers.get('Content-Type', ''))
html = response.read().decode(params['charset'])

# find city
cities = CitiesParser(html).cities
for city in ['Ar Riyāḍ', 'Riyadh']:
    if city in cities:
        print("%s is found" % (city,))
        print("the index is %d" % (cities.index(city),))
        break
    else:
        print("%r is not found" % (city,))

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