如何只取列表元素的某一部分？

时间：2021-07-01 08:47:58

标签： python python-3.x list

这是网站的链接：

website

这是我的脚本：

import requests
from requests import get
from bs4 import BeautifulSoup
import pandas as pd
import numpy as np
import time
from selenium import webdriver
from selenium.webdriver.support.select import Select
from selenium.webdriver.support.ui import WebDriverWait     
from selenium.webdriver.common.by import By     
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.keys import Keys


PATH = "driver\chromedriver.exe"

options = webdriver.ChromeOptions() 
options.add_argument("--disable-gpu")
options.add_argument("--window-size=1200,900")
options.add_argument('enable-logging')

driver = webdriver.Chrome(options=options, executable_path=PATH)

wait = WebDriverWait(driver, 20)

driver.get('https://fr.hotels.com/search.do?destination-id=10398359&q-check-in=2021-06-26&q-check-out=2021-06-27&q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER')
driver.maximize_window()
time.sleep(2)


webdriver.ActionChains(driver).send_keys(Keys.ESCAPE).perform()
time.sleep(2)

wait.until(EC.visibility_of_element_located((By.CSS_SELECTOR, 'button[class="uolsaJ"]'))).click()

links = []

def is_element_visible(xpath):
    wait1 = WebDriverWait(driver, 2)
    try:
        wait1.until(EC.visibility_of_element_located((By.XPATH, xpath)))
        return True
    except Exception:
        return False

while not is_element_visible("//div[@id='20']"):
    my_elems = driver.find_elements_by_xpath('//a[@class="_61P-R0"]')

    links = [my_elem.get_attribute("href") for my_elem in my_elems]


    driver.execute_script("window.scrollBy(0, 1000)")
    time.sleep(5)


print(links)

这是输出：

['https://fr.hotels.com/ho716157152/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho397103/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho1098309152/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho449686/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho315896/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho1574324896/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho288352/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho748227104/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho225263/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho225250/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho405210/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho547798/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho252584/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho351562/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho714011808/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho424335/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho442661/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho437481/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3']

那些是酒店的网址，我想知道如何有一个特定的部分。

我希望在每个 URL 中都包含这些 ID：

'https://fr.hotels.com/ho437481/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3'

-> 437481

有点像重新创建列表，但只是使用这些数字而不是 URL。

类似的东西：

['716157152', '397103', '1098309152' ... , '437481']

6 个答案:

答案 0 :(得分：2)

您可以使用正则表达式，但如果结构始终为 https://fr.hotels.com/ho[your_id]/[...]，则 split 就足够了：

hotel_ids = [link.split('/')[3][2:] for link in links]

split 将字符串转换为类似 ['https:', '', 'fr.hotels.com', 'ho[your_id]'] 的列表，因此 id 将始终位于第 4 个位置（索引 = 3），而 [2:] 去掉了开头的 'ho '.

答案 1 :(得分：1)

您可以在获得 links

后执行此操作 
links = [s.split('/')[3][2:] for s in links]

# Output

['716157152', '397103', '1098309152', '449686', '315896', '1574324896', '288352', '748227104', '225263', '225250', '405210', '547798', '252584', '351562', '714011808', '424335', '442661', '437481']

答案 2 :(得分：0)

我更喜欢其他答案，但正则表达式也是一个可行的选择。

import re

in_arr = ['https://fr.hotels.com/ho716157152/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho397103/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho1098309152/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho449686/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho315896/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho1574324896/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho288352/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho748227104/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho225263/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho225250/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho405210/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho547798/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho252584/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho351562/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho714011808/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho424335/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho442661/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho437481/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3']

regex = "(?<=\.com\/ho)[\w]+"
out = map(lambda x: re.findall(regex, x)[0], in_arr)

print(list(out))

输出：

['716157152', '397103', '1098309152', '449686', '315896', '1574324896', '288352', '748227104', '225263', '225250', '405210', '547798', '252584', '351562', '714011808', '424335', '442661', '437481']

答案 3 :(得分：0)

对字符串使用正则表达式：

>>> s = 'https://fr.hotels.com/ho437481/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3'
>>> m = re.match('https://fr.hotels.com/ho(\d+)/',s)
>>> m.group(1)
'437481'

您可以将它放在一个函数中，然后对 URL 列表使用 map，或者使用 for 循环，甚至是列表解析。

答案 4 :(得分：0)

#!/usr/bin/env python3

urls = [
    "https://fr.hotels.com/ho716157152/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho397103/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho1098309152/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho449686/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho315896/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho1574324896/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho288352/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho748227104/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho225263/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho225250/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho405210/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho547798/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho252584/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho351562/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho714011808/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho424335/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho442661/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3",
    "https://fr.hotels.com/ho437481/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3"
]

def get_id(url):
    return url.split('/')[3][2:]

ids = [get_id(url) for url in urls]
print(ids)

答案 5 :(得分：0)

这可以使用正则表达式解决：

import re
urls = ['https://fr.hotels.com/ho716157152/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3', 'https://fr.hotels.com/ho397103/?q-rooms=1&q-room-0-adults=2&q-room-0-children=0&sort-order=BEST_SELLER&ZSX=0&SYE=3']
print([re.findall(r'\d+',url)[0] for url in urls])
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