形成网络刮擦数据BS4

Question

输出FulfordRoad所需的输出是这样的：

WaterLane
york
york
YO104PA
YO306PQ

line1城 postCodeFulfordRoadyorkYO104PAWaterLaneyorkYO306PQ代码

import requests
from bs4 import BeautifulSoup
import pandas as pd



list1 = []



response3 = requests.get("https://stores.aldi.co.uk/yorkshire-amp-humber/york")
soup3 = BeautifulSoup(response3.text, "html.parser")
    
    
try:
    for a1 in soup3.find_all('span', attrs={'class':'Address-field Address-line1'}):
        line1 = a1.get_text()
        print(line1)

    for a2 in soup3.find_all('span', attrs={'class':'Address-field Address-city'}):
        line2 = a2.get_text()
        print(line2)

    for a3 in soup3.find_all('span', attrs={'class':'Address-field Address-postalCode'}):
        line3 = a3.get_text()
        print(line3)

  
except:
        pass
    
data = pd.DataFrame(list1) 

我真的很感谢您能给我解决这个问题的任何支持。 thanks， s

跟着您的方法，我会这样做（

单一的for-lop

）：

list1 = []
for addr in soup3.find_all("div", class_="Address"):
    line1 = addr.find("span", class_="Address-line1").get_text()
    city = addr.find("span", class_="Address-city").get_text()
    postcode = addr.find("span", class_="Address-postalCode").get_text()
    list1.append([line1, city, postcode])

df = pd.DataFrame(list1, columns=["line1", "city", "postcode"])

另一个变体：

from collections import defaultdict

data = defaultdict(list)

for addr in soup3.find_all("div", class_="Address"):
    data["line1"].append(addr.find("span", class_="Address-line1").get_text())
    data["city"].append(addr.find("span", class_="Address-city").get_text())
    data["postcode"].append(addr.find("span", class_="Address-postalCode").get_text())

df = pd.DataFrame(data)

Answer 1

输出：

print(df)

          line1  city  postcode
0  Fulford Road  York  YO10 4PA
1    Water Lane  York  YO30 6PQ