REGEX:我如何尽可能清晰地获得两个活动的团队

Question

我已经尝试了一段时间以成功地通过博彩公司网站解析并检索市场/赔率。

我来了，我可以获取硒网元元素的.TEXT属性，所以我有类似的东西：

编辑以展示更多示例

BPZ vs Griffin - League of Legends - Challenger Korea
Sat 2/25   1511 BPZ 1.645         
10:30PM   1512 Griffin   2.250
Team Battle Comics vs RisingStar Gaming - League of Legends - Challenger Korea
Sat 2/25   1513 Team Battle Comics 5.800         
11:59PM   1514 RisingStar Gaming   1.133
Going In vs Hala Ares - Dota 2 - Prodota Cup
Sat 2/25   1529 Going In 1.667         
1:30PM   1530 Hala Ares   2.200
Unicorns of Love vs G2 Esports - League of Legends - Intel Extreme Masters
Sat 2/25   1545 Unicorns of Love 2.750         
11:15AM   1546 G2 Esports   1.444

实际上几个小时的谷歌正则搜索和读取语法后，我无法做到的是在我需要的情况下提取该字符串的一部分。在上面的字符串中，如果我可以使用Regex将其过滤成看起来像这样的字典：

{'event':'BPZ vs Griffin - League of Legends',
 'outcome1':'BPZ',
 'outcome2':'Griffin',
 'outcome1odds':1.645,
 'outcome2odds':2.25,
 'date':'Sat 2/25',
 'time':'10:30PM'}

那我会非常高兴。我相当确定这是可能的，但是我遇到了太多的困难，将我的头围绕着正则努力以实现这一目标。非常感谢任何帮助和/或资源。

Answer 1

此模式应执行技巧：

(?P<event>(?P<outcome1>[^-]+?) vs (?P<outcome2>[^-]+) -.*?) -[^b]*?(?P<date>(?:Mon|Tue|Wed|Thu|Fri|Sat|Sun) d+/d+(?:/d+)?)[^.]*(?P<outcome1odds>d+.d+)s+(?P<time>d+:d+[AP]M)[^.]*(?P<outcome2odds>d+.d+)

这很长，但是作为交换，您可以使用.groupdict()函数直接获得所需的结果：

print(re.match(pattern, text).groupdict())

---

分解：

(?P<event>          # in a named capture group, match...
    (?P<outcome1>   # outcome1, which is...
        [^-]+?      # all text up to...
    ) 
     vs             # a literal " vs " 
    (?P<outcome2>   # outcome2 is...
        [^-]+       # all text up to...
    )
     -              # the next literal " -"
    .*?             # still inside the "event" group, match until...
)
 -                  # a literal " -"
[^b]*?             # skip forward to...
(?P<date>           # the date, which is...
    (?:Mon|Tue|Wed|Thu|Fri|Sat|Sun) # a weekday
     d+/d+(?:/d+)?   # followed by digits separated with /
)
[^.]*               # skip worward to...
(?P<outcome1odds> 
    d+.d+        # a floating point number
)
s+
(?P<time>           # match the time, which is...
    d+:d+         # digits separated with :
    [AP]M           # followed by AM or PM
)
[^.]*               # skip to...
(?P<outcome2odds> 
    d+.d+        # another floating point number
)

Answer 2

使用re.match（）和match.groupdict（）的解决方案（获取匹配的所有命名子组）方法：

s = '''
BPZ vs Griffin - League of Legends - Challenger Korea
Sat 2/25   1511 BPZ 1.645
10:30PM   1512 Griffin   2.250
'''
p = r'^(?P<event>[w ]+-[w ]+)s[ws-]+' 
    r'(?P<date>[A-Z]w+ d+/d{2})s+d+s(?P<outcome1>[w ]+)' 
    r'(?P<outcome1odds>d+.d+)s+(?P<time>d+:d+(AM|PM))s+d+s' 
    r'(?P<outcome2>[w ]+)(?P<outcome2odds>d+.d+)'
matches = re.match(p, s.strip(), re.M)
result = {k:v.strip() for k,v in matches.groupdict().items()}
print(result)

输出：

{'time': '10:30PM', 'event': 'BPZ vs Griffin - League of Legends', 'outcome2odds': '2.250', 'date': 'Sat 2/25', 'outcome2': 'Griffin', 'outcome1': 'BPZ', 'outcome1odds': '1.645'}

Answer 3

使用此长正则表达式，您可以在8组中找到数据：

(.*-.*)s-s.*n(w{3})s*(d+/d+)s*d+s*(w+)s*(d+.?d*)s*n(d+:d+ww)s*d+s*(w+)s*(d+.?d*)
Full match  0-104   `BPZ vs Griffin - League of Legends - Challenger Korea Sat 2/25 1511 BPZ 1.645
10:30PM 1512 Griffin 2.250`
Group 1.    n/a `BPZ vs Griffin - League of Legends`
Group 2.    n/a `Sat`
Group 3.    n/a `2/25`
Group 4.    n/a `BPZ`
Group 5.    n/a `1.645`
Group 6.    n/a `10:30PM`
Group 7.    n/a `Griffin`
Group 8.    n/a `2.250`
{'event':'$1',
 'outcome1':'$4',
 'outcome2':'$7',
 'outcome1odds':$5,
 'outcome2odds':$8,
 'date':'$2 $3',
 'time':'$6'}

https://regex101.com/r/2idfnb/2