python正则表达式,匹配多行文本

Question

代码如下

import re
str='''
1
2
3
1
2
3
4
1
2
3
4
5
1
'''
pattern=re.compile(r'(^1\n)(^(?!1)\d.*\n)',flags=re.M)
match=pattern.findall(str)
print(match)

打印结果:[('1\n', '3\n'), ('1\n', '4\n'), ('1\n', '5\n'), ('1\n', '')]

想要打印出[('1\n', '2\n3\n'), ('1\n', '2\n3\n4\n'), ('1\n', '2\n3\n4\n5\n'), ('1\n')]
或者[('1\n', '2\n','3\n'), ('1\n', '2\n','3\n','4\n'), ('1\n', '2\n','3\n','4\n','5\n'), ('1\n')]
我应该怎么写正则表达式,

我写出的最接近的也是pattern=re.compile(r'(^1\n)((^(?!1)\d.*\n)*)',flags=re.M)
结果是[('1\n', '2\n3\n', '3\n'), ('1\n', '2\n3\n4\n', '4\n'), ('1\n', '2\n3\n5\n', '5\n'), ('1\n', '', '')]
请大神指导一下

Answer 1

import re
str='''
1
2
3
1
2
3
4
1
2
3
4
5
1
'''
pattern=re.compile(r'(^1\n)([2-9\n]*)',flags=re.M)
match=pattern.findall(str)
print(match)

Answer 2

>>> import re
>>> s='''
1
2
3
1
2
3
4
1
2
3
4
5
1
'''
>>> re.findall(r'1[^1]*',s)
['1\n2\n3\n', '1\n2\n3\n4\n', '1\n2\n3\n4\n5\n', '1\n']

#或者
>>> re.findall(r'(1\n)([^1]*)',s)
[('1\n', '2\n3\n'), ('1\n', '2\n3\n4\n'), ('1\n', '2\n3\n4\n5\n'), ('1\n', '')]

Answer 3

import re

s = '''
1
2
3
1
2
3
4
1
2
3
4
5
1
'''

result = re.findall('1\n[^1]*', s)