搜集Wikipedia信息(表) [英] Scraping Wikipedia information (table)

问题描述

我需要在Wikipedia上按区域刮取有关Elenco dei comuni的信息.我想创建一个数组，该数组可以让我将每个乡镇与相应区域相关联，即:

I would need to scrape information regarding Elenco dei comuni per regione on Wikipedia. I would like to create an array that can allow me to associate each comune to the corresponding region, i.e. something like this:

'Abbateggio': 'Pescara' -> Abruzzo

我试图使用 BeautifulSoup 和 requests 来获取信息，如下所示:

I tried to get information using BeautifulSoup and requests as follows:

from bs4 import BeautifulSoup as bs
import requests

     with requests.Session() as s: # use session object for efficiency of tcp re-use
        s.headers = {'User-Agent': 'Mozilla/5.0'}
        r = s.get('https://it.wikipedia.org/wiki/Comuni_d%27Italia')
        soup = bs(r.text, 'html.parser')
        for ele in soup.find_all('h3')[:6]:
            tx = bs(str(ele),'html.parser').find('span', attrs={'class': "mw-headline"})
            if tx is not None:
                print(tx['id'])

但是它不起作用(它向我返回了一个空列表).我使用Google Chrome浏览器查看的信息如下:

however it does not work (it returns me an empty list). The information that I have looked at using Inspect of Google Chrome are the following:

<span class="mw-headline" id="Elenco_dei_comuni_per_regione">Elenco dei comuni per regione</span> (table)

<a href="/wiki/Comuni_dell%27Abruzzo" title="Comuni dell'Abruzzo">Comuni dell'Abruzzo</a> 

(此字段应随每个区域而变化)

(this field should change for each region)

然后< table class ="wikitable可排序查询表排序">

请问您如何获得这种结果的建议?任何帮助和建议，将不胜感激.

Could you please give me advice on how to get such results? Any help and suggestion will be appreciated.

示例:

我有一个字: comunediabbateggio .这个词包括 Abbateggio .我想知道哪个地区可以与该城市相关联(如果存在).来自Wikipedia的信息需要创建一个数据集，该数据集可以让我检查该字段并将其关联到一个地区的城市/城市.我应该期望的是:

I have a word: comunediabbateggio. This word includes Abbateggio. I would like to know which region can be associated with that city, if it exists. Information from Wikipedia needs to create a dataset that can allow me to check the field and associate to comuni/cities a region. What I should expect is:

WORD                         REGION/STATE
comunediabbateggio           Pescara

我希望这可以为您提供帮助.很抱歉，如果不清楚.另一个可能会更好地理解英语的英语示例如下:

I hope this can help you. Sorry if it was not clear. Another example for English speaker that might be slightly better for understanding is the following:

除了上面的意大利语链接，您还可以考虑以下内容: https://en.wikipedia.org/wiki/List_of_comuni_of_Italy .对于每个地区(伦巴第大区，威尼托大区，西西里岛……)，我需要收集有关各省公社列表的信息.如果您点击 ...的社区列表的链接，则会有一个列出该社区的表，例如 https://en.wikipedia.org/wiki/List_of_communes_of_the_Province_of_Agrigento .

Instead of the Italian link above, you can also consider the following: https://en.wikipedia.org/wiki/List_of_comuni_of_Italy . For each region (Lombardia, Veneto, Sicily, ... ) I would need to collect information about the list of communes of the Provinces. if you click in a link of List of Communes of ... , there is a table that list the comune, e.g. https://en.wikipedia.org/wiki/List_of_communes_of_the_Province_of_Agrigento.

推荐答案

import re
import requests
from bs4 import BeautifulSoup
import pandas as pd
from tqdm import tqdm



target = "https://en.wikipedia.org/wiki/List_of_comuni_of_Italy"


def main(url):
    with requests.Session() as req:
        r = req.get(url)
        soup = BeautifulSoup(r.content, 'html.parser')

        provinces = [item.find_next("span").text for item in soup.findAll(
            "span", class_="tocnumber", text=re.compile(r"\d[.]\d"))]

        search = [item.replace(
            " ", "_") if " " in item else item for item in provinces]

        nested = []
        for item in search:
            for a in soup.findAll("span", id=item):
                goes = [b.text.split("of ")[-1]
                        for b in a.find_next("ul").findAll("a")]
                nested.append(goes)

        dictionary = dict(zip(provinces, nested))

        urls = [f'{url[:24]}{b.get("href")}' for item in search for a in soup.findAll(
            "span", id=item) for b in a.find_next("ul").findAll("a")]
    return urls, dictionary


def parser():
    links, dics = main(target)
    com = []
    for link in tqdm(links):
        try:
            df = pd.read_html(link)[0]
            com.append(df[df.columns[1]].to_list()[:-1])
        except ValueError:
            com.append(["N/A"])
    com = iter(com)
    for x in dics:
        b = dics[x]
        dics[x] = dict(zip(b, com))
    print(dics)


parser()

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