python中的多行模式匹配

Question

定期计算机生成的消息（简化）：

Hello user123,

- (604)7080900
- 152
- minutes

Regards

使用 python，如何在之间提取“(604)7080900”、“152”、“分钟”（即遵循前导 "- " 模式的任何文本）两个空行（空行是“Hello user123”之后的 \n\n 和“Regards”之前的 \n\n）。如果结果字符串列表存储在数组中就更好了。谢谢！

编辑：两个空白行之间的行数不固定。

第二次编辑：

例如

hello

- x1
- x2
- x3

- x4

- x6
morning
- x7

world

x1 x2 x3 很好，因为所有行都被 2 个空行包围，出于同样的原因，x4 也很好。 x6 不好，因为它后面没有空白行，x7 不好，因为它前面没有空白。 x2 很好（不像 x6、x7），因为前面的线是好线，后面的线也很好。

当我发布问题时，这个条件可能不清楚：

a continuous of good lines between 2 empty lines

good line must have leading "- "
good line must follow an empty line or follow another good line
good line must be followed by an empty line or followed by another good line

谢谢

Answer 1

>>> import re
>>>
>>> x="""Hello user123,
...
... - (604)7080900
... - 152
... - minutes
...
... Regards
... """
>>>
>>> re.findall("\n+\n-\s*(.*)\n-\s*(.*)\n-\s*(minutes)\s*\n\n+",x)
[('(604)7080900', '152', 'minutes')]
>>>

Answer 2

最简单的方法是遍历这些行（假设您有一个行列表或一个文件，或者将字符串拆分为一个行列表），直到看到一行 '\n' ，然后检查每一行是否以 '- ' 开头（使用 startswith 字符串方法）并将其切片，存储结果，直到找到另一个空行。例如：

# if you have a single string, split it into lines.
L = s.splitlines()
# if you (now) have a list of lines, grab an iterator so we can continue
# iteration where it left off.
it = iter(L)
# Alternatively, if you have a file, just use that directly.
it = open(....)

# Find the first empty line:
for line in it:
    # Treat lines of just whitespace as empty lines too. If you don't want
    # that, do 'if line == ""'.
    if not line.strip():
        break
# Now starts data.
for line in it:
    if not line.rstrip():
        # End of data.
        break
    if line.startswith('- '):
        data.append(line[:2].rstrip())
    else:
        # misformed data?
        raise ValueError, "misformed line %r" % (line,)

编辑：由于您详细说明了您想要执行的操作，因此这里是循环的更新版本。它不再循环两次，而是收集数据直到遇到“坏”行，并在遇到块分隔符时保存或丢弃收集的行。它不需要显式迭代器，因为它不会重新启动迭代，因此您只需向它传递一个行列表（或任何可迭代的）即可：

def getblocks(L):
    # The list of good blocks (as lists of lines.) You can also make this
    # a flat list if you prefer.
    data = []
    # The list of good lines encountered in the current block
    # (but the block may still become bad.)
    block = []
    # Whether the current block is bad.
    bad = 1
    for line in L:
        # Not in a 'good' block, and encountering the block separator.
        if bad and not line.rstrip():
            bad = 0
            block = []
            continue
        # In a 'good' block and encountering the block separator.
        if not bad and not line.rstrip():
            # Save 'good' data. Or, if you want a flat list of lines,
            # use 'extend' instead of 'append' (also below.)
            data.append(block)
            block = []
            continue
        if not bad and line.startswith('- '):
            # A good line in a 'good' (not 'bad' yet) block; save the line,
            # minus
            # '- ' prefix and trailing whitespace.
            block.append(line[2:].rstrip())
            continue
        else:
            # A 'bad' line, invalidating the current block.
            bad = 1
    # Don't forget to handle the last block, if it's good
    # (and if you want to handle the last block.)
    if not bad and block:
        data.append(block)
    return data

这里它正在运行：

>>> L = """hello
...
... - x1
... - x2
... - x3
...
... - x4
...
... - x6
... morning
... - x7
...
... world""".splitlines()
>>> print getblocks(L)
[['x1', 'x2', 'x3'], ['x4']]

Answer 3

>>> s = """Hello user123,

- (604)7080900
- 152
- minutes

Regards
"""
>>> import re
>>> re.findall(r'^- (.*)', s, re.M)
['(604)7080900', '152', 'minutes']

Answer 4

l = """Hello user123,

- (604)7080900
- 152
- minutes

Regards  

Hello user124,

- (604)8576576
- 345
- minutes
- seconds
- bla

Regards"""

这样做：

result = []
for data in s.split('Regards'): 
    result.append([v.strip() for v in data.split('-')[1:]])
del result[-1] # remove empty list at end

并这样做：

>>> result
[['(604)7080900', '152', 'minutes'],
['(604)8576576', '345', 'minutes', 'seconds', 'bla']]