有没有一种方法可以替代硒元素，或者说是美素4？

Question

有趣的问题。我用硒刮掉一个投注点，然后用bs4处理。问题是，网站加载赔率信息的方式与加载团队名称的方式不同。例如：

London v Tokyo            2/1   4/1
Amsterdam v Helsinki      5/1   3/1

New York v California     7/1   10/1

当我把它拉出来，迭代它，结果是这样的：

Names = [London, Tokyo, Amsterdam, Helsinki]
Odds = [2/1, 5/1, 4/1, 3/1, 7/1, 10/1]

概率是从上到下，从左到右，在不同的长度块加载。也就是说，当我试着把名字和赔率拼在一起时，它们就不匹配了。你知道吗

我的问题是，我怎样才能避开这个问题？我希望最终把信息公布出来，这样球队的名字后面就有了胜算：

Games = [London, 2/1, Tokyo, 4/1, Amsterdam, 5/1, Helsinki, 3/1, New York, 7/1, California, 10/1]

**更新** 网站是：https://www.bet365.com/#/AC/B151/C1/D50/E2/F163/ 如果你得到一个登录页，那么只需点击通过。然后在左侧面板上显示“电子竞技”，然后在中间页显示“所有比赛”。你知道吗

代码：

from selenium import webdriver
from bs4 import BeautifulSoup

url = "https://www.bet365.com/#/AC/B151/C1/D50/E2/F163/"
driver = webdriver.Chrome()
driver.get(url)

# Then i'm navigating to the "All Matches" page

soup = BeautifulSoup(driver.page_source, 'html.parser')
teams = driver.find_elements_by_class_name("sl-CouponParticipantWithBookCloses_Name")
odds_raw = driver.find_elements_by_class_name("gl-ParticipantOddsOnly_Odds")

odds = []
teams_text = []
new_teams = []
new_odds = []

for name in teams:
teams_text.append(name.text)

球队像街区一样进入，例如：“伦敦对东京”。 所以为了将团队名称分开，我迭代并拆分它们

for name in teams_text:
first, second = name.split(" v ")
new_teams.append(first)
new_teams.append(second)

然后我把收到的赔率分成小部分，然后把它们变成小数：

for odd in odds_raw:
odds.append(odd.text)
for odd in odds:
first, second = odd.split("/")
new_odd = (int(first) / int(second)) + 1
new_odds.append(round(new_odd, 2))

所以现在我有一个所有团队名称的列表，还有一个十进制奇数值的列表。这就是我的问题所在。bet365产生比赛赔率的方式是在每个游戏分区不同长度的垂直块中。你知道吗

所以如果几率是这样的：

Division 1
London v Tokyo        1   2
Amsterdam v Helsinki  3   4
Division 2
New York v California 5   6
Division 3
Sydney v Brisbane     7   8
Bali v Singapore      9   10
Berlin v Paris        11  12

当我把它们拉出来的时候，几率会是这样的：

[1, 3, 2, 4, 5, 6, 7, 9, 11, 8, 10, 12]

在划分长度不同的地方，我很难弄清楚如何接近它。你知道吗

Answer 1

您可以使用如下for循环实现所需的输出：

Names = ["London", "Tokyo", "Amsterdam", "Helsinki","New York","California"]
Odds = [2/1, 5/1, 4/1, 3/1, 7/1, 10/1]
start_nmb = 1 

for nmb, odd in enumerate(Odds):
    Names.insert(start_nmb, odd)
    start_nmb += 2

输出：

['London', 2.0, 'Tokyo', 5.0, 'Amsterdam', 4.0, 'Helsinki', 3.0, 'New York', 7.0, 'California', 10.0]

希望这有帮助！你知道吗

Answer 2

这是一个冗长的方法来尝试。赔率的奇数行（由循环确定）进入第1组（第1组与第2组的左侧）。即使排成2队。列表列表被展平。然后@user942640按照答案here所示将列表合并为备用成员。你知道吗

注意：这依赖于等长列表来返回准确的结果。你知道吗

import itertools
from bs4 import BeautifulSoup as bs
#your existing code to get to page and wait for presence of all elements
soup = bs(driver.page_source, 'lxml')
teams = [item.text.split(' v ') for item in soup.select('.sl-CouponParticipantWithBookCloses_NameContainer')]

i = 0
team1 = []
team2 = []

for item in soup.select('.sl-MarketCouponValuesExplicit2'):
    if i % 2 == 0:
        team1.append([i.text for i in item.select('div:not(.gl-MarketColumnHeader )')])
    else:
        team2.append([i.text for i in item.select('div:not(.gl-MarketColumnHeader )')])
    i+=1

team1 =  [item for sublist in team1 for item in sublist]
team2 =  [item for sublist in team2 for item in sublist]
teams = [item for sublist in teams for item in sublist]
team_odds =  [x for x in itertools.chain.from_iterable(itertools.zip_longest(team1,team2)) if x]
final = [x for x in itertools.chain.from_iterable(itertools.zip_longest(teams, team_odds)) if x]
print(final)

---

比如（注意概率不断更新）：

from selenium import webdriver
import itertools
from bs4 import BeautifulSoup as bs
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC

driver = webdriver.Chrome()
driver.get('https://www.bet365.com/#/HO/')
driver.get('https://www.bet365.com/#/AC/B151/C1/D50/E2/F163/')
WebDriverWait(driver,10).until(EC.presence_of_all_elements_located((By.CSS_SELECTOR, ".sl-MarketCouponValuesExplicit2")))
soup = bs(driver.page_source, 'lxml')
teams = [item.text.split(' v ') for item in soup.select('.sl-CouponParticipantWithBookCloses_NameContainer')]

i = 0
team1 = []
team2 = []

for item in soup.select('.sl-MarketCouponValuesExplicit2'):
    if i % 2 == 0:
        team1.append([i.text for i in item.select('div:not(.gl-MarketColumnHeader )')])
    else:
        team2.append([i.text for i in item.select('div:not(.gl-MarketColumnHeader )')])
    i+=1

team1 =  [item for sublist in team1 for item in sublist]
team2 =  [item for sublist in team2 for item in sublist]
teams = [item for sublist in teams for item in sublist]

team_odds =  [x for x in itertools.chain.from_iterable(itertools.zip_longest(team1,team2)) if x]
final = [x for x in itertools.chain.from_iterable(itertools.zip_longest(teams, team_odds)) if x]
print(final)

Answer 3

可以使用regex来捕获元素。你知道吗

import re
s = '''London v Tokyo 2/1 4/1 Amsterdam v Helsinki 5/1 3/1 New York v California 7/1 10/1'''
re.findall(r'(\w+)\s+v\s+(\w+)\s+(\d+/\d+)\s+(\d+/\d+)', s)

[('London', 'Tokyo', '2/1', '4/1'),
 ('Amsterdam', 'Helsinki', '5/1', '3/1'),
 ('York', 'California', '7/1', '10/1')]