如何在正则表达式中捕获子模式的所有重复

Question

我有一个格式化的字符串，它可以有任意长度的重复部分。例如，这是我想要解析的元数据示例。

File Name: chb03_34.edf
File Start Time: 01:51:23
File End Time: 2:51:23
Number of Seizures in File: 1
Seizure Start Time: 1982 seconds
Seizure End Time: 2029 seconds

File Name: chb23_07.edf
File Start Time: 11:03:16
File End Time: 11:45:56
Number of Seizures in File: 0

File Name: chb23_08.edf
File Start Time: 11:48:05
File End Time: 14:40:27
Number of Seizures in File: 2
Seizure 1 Start Time: 325 seconds
Seizure 1 End Time: 345 seconds
Seizure 2 Start Time: 5104 seconds
Seizure 2 End Time: 5151 seconds

File Name: chb23_09.edf
File Start Time: 14:40:47
File End Time: 18:41:13
Number of Seizures in File: 4
Seizure 1 Start Time: 2589 seconds
Seizure 1 End Time: 2660 seconds
Seizure 2 Start Time: 6885 seconds
Seizure 2 End Time: 6947 seconds
Seizure 3 Start Time: 8505 seconds
Seizure 3 End Time: 8532 seconds
Seizure 4 Start Time: 9580 seconds
Seizure 4 End Time: 9664 seconds

到目前为止，我已经创建了一个正则表达式，它可以捕获第一行，但只能在一个块中捕获最后一个癫痫发作（如果存在癫痫发作）。

import re

summary = "a formatted string read"


pattern = "File Name\: (.+)\nFile Start Time\: (.+)\nFile End Time\: (.+)\nNumber of Seizures in File\: (.+)(?:\n|\r|)(?:Seizure(?: | \d )Start Time\: (\d+) seconds\nSeizure(?: | \d )End Time\: (\d+) seconds(?:\n|\r|))*"
pattern = re.compile(pattern)

for p in pattern.finditer(summary):
    print(p.groups())

但是，例如最后一个块的这种模式的结果将仅捕获癫痫发作 4 的开始和结束时间。是否可以递归捕获重复的子模式？

编辑：使用regex 和模式The fourth bird 已在 cmets 中键入，我可以匹配字符串，但我在重复行中得到很多 None 值，也完全是 None 行。我怎样才能摆脱这些，或插入适当的值？

('chb23_06.edf', '08:57:57', '11:02:43', '1', '3962', '4075')
(None, None, None, None, None, None)
('chb23_07.edf', '11:03:16', '11:45:56', '0', None, None)
(None, None, None, None, None, None)
('chb23_08.edf', '11:48:05', '14:40:27', '2', '325', '345')
(None, None, None, None, '5104', '5151')
(None, None, None, None, None, None)
('chb23_09.edf', '14:40:47', '18:41:13', '4', '2589', '2660')
(None, None, None, None, '6885', '6947')
(None, None, None, None, '8505', '8532')
(None, None, None, None, '9580', '9664')
(None, None, None, None, None, None)
('chb23_10.edf', '18:41:40', '22:41:40', '0', None, None)
(None, None, None, None, None, None)
('chb23_16.edf', '13:46:32', '17:46:32', '0', None, None)
(None, None, None, None, None, None)
('chb23_17.edf', '17:46:42', '21:16:29', '0', None, None)
(None, None, None, None, None, None)
('chb23_19.edf', '02:28:28', '6:28:28', '0', None, None)
(None, None, None, None, None, None)
('chb23_20.edf', '06:28:36', '7:52:05', '0', None, None)
(None, None, None, None, None, None)

EDIT2：我完成了先前接受的答案的解决方案，但它有一些粗糙的边缘并且在某些文件中不起作用。我上传了一个有问题的文件。您可以在 here 中找到有问题的元数据示例的粘贴。

Answer 1

使用re，您可以捕获组中Seizure 字符串的可选迭代，然后从该组中捕获秒的数字值：

图案

File Name: (.+)\nFile Start Time: (.+)\nFile End Time: (.+)\nNumber of Seizures in File: (.+)((?:\nSeizure (?:\d )?Start Time: \d+ seconds\nSeizure (?:\d )?End Time: \d+ seconds)*)

模式匹配：

• File Name: (.+)\n 第 1 组，匹配文件名之后的所有内容：和换行符
• File Start Time: (.+)\n 第 2 组，匹配文件开始时间之后的所有内容：和换行符
• File End Time: (.+)\n 第 3 组，匹配文件结束时间之后的所有内容：和换行符
• Number of Seizures in File: (.+) 第 4 组，在 Number of Seizures in File 之后匹配所有内容：
• ( 第 5 组
  • (?:非捕获组作为一个整体匹配，然后可选地重复
    • \nSeizure (?:\d )?Start Time: \d+ seconds\n 匹配换行符并匹配 Seizure Start Time 和末尾的换行符
    • Seizure (?:\d )?End Time: \d+ seconds匹配扣押结束时间
  • )*关闭非捕获组并选择性地重复它
• )关闭第5组

Regex demo | Python demo

例如

pattern = re.compile(pattern)

for m in pattern.finditer(summary):
    print(m.group(1))
    print(m.group(2))
    print(m.group(3))
    print(m.group(4))
    print(re.findall(r"(\d+) seconds", m.group(5)))

每场比赛的输出如下所示：（或在没有 Seizure 值时为空列表，但您也可以对其进行测试）

chb23_08.edf
11:48:05
14:40:27
2
['325', '345', '5104', '5151']

Answer 2

如果您使用的是regex 模块，我建议您使用重复捕获。

为了清楚起见，我还添加了命名组：

import regex

pattern = regex.compile(
    r"File Name: (?P<name>.+)\n"
    r"File Start Time: (?P<start>.+)\n"
    r"File End Time: (?P<end>.+)\n"
    r"Number of Seizures in File: (?P<count>\d+)\n"
    r"(?:\n|(?:Seizure (?:\d )?Start Time: (?P<seizure_start>\d+) seconds\n"
    r"Seizure (?:\d )?End Time: (?P<seizure_end>\d+) seconds\n)*)"
)

summary = """File Name: chb03_34.edf
File Start Time: 01:51:23
File End Time: 2:51:23
Number of Seizures in File: 1
Seizure Start Time: 1982 seconds
Seizure End Time: 2029 seconds

File Name: chb23_07.edf
File Start Time: 11:03:16
File End Time: 11:45:56
Number of Seizures in File: 0

File Name: chb23_08.edf
File Start Time: 11:48:05
File End Time: 14:40:27
Number of Seizures in File: 2
Seizure 1 Start Time: 325 seconds
Seizure 1 End Time: 345 seconds
Seizure 2 Start Time: 5104 seconds
Seizure 2 End Time: 5151 seconds

File Name: chb23_09.edf
File Start Time: 14:40:47
File End Time: 18:41:13
Number of Seizures in File: 4
Seizure 1 Start Time: 2589 seconds
Seizure 1 End Time: 2660 seconds
Seizure 2 Start Time: 6885 seconds
Seizure 2 End Time: 6947 seconds
Seizure 3 Start Time: 8505 seconds
Seizure 3 End Time: 8532 seconds
Seizure 4 Start Time: 9580 seconds
Seizure 4 End Time: 9664 seconds
"""

for match in pattern.finditer(summary):
    print("Name:", match.group("name"))
    print("Seizure Count", match.group("count"))
    seizures = tuple(
        zip(match.captures("seizure_start"),match.captures("seizure_end")))
    for i, (start, end) in enumerate(seizures, start=1):
        print(f"Seizure #{i}: {start} -> {end}")

打印：

Name: chb03_34.edf
Seizure Count 1
Seizure #1: 1982 -> 2029
Name: chb23_07.edf
Seizure Count 0
Name: chb23_08.edf
Seizure Count 2
Seizure #1: 325 -> 345
Seizure #2: 5104 -> 5151
Name: chb23_09.edf
Seizure Count 4
Seizure #1: 2589 -> 2660
Seizure #2: 6885 -> 6947
Seizure #3: 8505 -> 8532
Seizure #4: 9580 -> 9664