python - 从图书馆目录中抓取信息

Question

我正在开展一个项目，从特定图书馆抓取图书的目录信息。到目前为止，我的脚本可以从表格中删除所有单元格。但是，我对如何只返回新不列颠图书馆的特定单元格感到困惑。

import requests
from bs4 import BeautifulSoup

mypage = 'http://lci-mt.iii.com/iii/encore/record/C__Rb1872125__S%28*%29%20f%3Aa%20c%3A47__P0%2C3__Orightresult__U__X6?lang=eng&suite=cobalt'
response = requests.get(mypage)

soup = BeautifulSoup(response.text, 'html.parser')

data = []
table = soup.find('table', attrs={'class':'itemTable'})


rows = table.find_all('tr')
for row in rows:
    cols = row.find_all('td')
    cols = [ele.text.strip() for ele in cols]
    data.append([ele for ele in cols if ele]) # Get rid of empty values

for index, libraryinfo in enumerate(data):
    print(index, libraryinfo)

这是脚本中新不列颠图书馆的示例输出:

["New Britain, Main Library - Children's Department", 'J FIC PALACIO', 'Check Shelf']

与其归还所有单元格，我如何只归还与新不列颠图书馆有关的单元格？我只想要图书馆名称和结账状态。

期望的输出是:

["New Britain, Main Library - Children's Department", 'Check Shelf']

可以有多个单元格，因为一本书在同一个图书馆可以有多本。

Answer 1

为了根据特定字段(示例中的第一个字段)简单地过滤掉数据，您可以建立一个理解:

[element for element in data if 'New Britain' in element[0]]

您提供的示例消除了使数据元素具有不同大小的空值。这使得更难知道哪个字段对应于每个数据组件。使用字典，我们可以使数据更易于理解和处理。

一些字段内部似乎有空 block (只有类似空格的字符 ['\n', '\r', '\t', ' ']).所以 strip 不会删除那些。将它与简单的正则表达式结合可以帮助改善这一点。我写了一个简单的函数来做到这一点:

def squish(s):
    return re.sub(r'\s+', ' ', s)

总结一下，我相信这会对你有所帮助:

import re

import requests
from bs4 import BeautifulSoup


def squish(s):
    return re.sub(r'\s+', ' ', s)


def filter_by_location(data, location_name):
    return [x for x in data if location_name.lower() in x['Location'].lower()]


mypage = 'http://lci-mt.iii.com/iii/encore/record/C__Rb1872125__S%28*%29%20f%3Aa%20c%3A47__P0%2C3__Orightresult__U__X6?lang=eng&suite=cobalt'
response = requests.get(mypage)

soup = BeautifulSoup(response.text, 'html.parser')

data = []
table = soup.find('table', attrs={'class':'itemTable'})

headers = [squish(element.text.strip()) for element in table.find('tr').find_all('th')]

for row in table.find_all('tr')[1:]:
    cols = [squish(element.text.strip()) for element in row.find_all('td')]
    data.append({k:v for k, v in zip(headers, cols)})

filtered_data = filter_by_location(data, 'New Britain')
for x in filtered_data:
    print('Location: {}'.format(x['Location']))
    print('Status: {}'.format(x['Status']))
    print()

运行后得到如下结果:

Location: New Britain, Jefferson Branch - Children's Department
Status: Check Shelf

Location: New Britain, Main Library - Children's Department
Status: Check Shelf

Location: New Britain, Main Library - Children's Department
Status: Check Shelf