我有一个字符串
some_string = """
15.10.2020 102000 sometext 11112
16.11.2020 120888.89 sometext 22222
18.09.2020 1 fjjfjf sometext
15.12.2020 109555.55 sometext 22222
"""
我应该有这样的输出
15.10.2020 102000
16.11.2020 120888.89
18.09.2020 1
15.12.2020 109555.55
我尝试使用此方法，但正则表达式不匹配'1-length non-digit'
re.findall("\d{2}\.\d{2}\.\d{4}\s\d+\.{0,1}\d", some_string)
but had output
['15.10.2020 102000', '16.11.2020 120888.8', '15.12.2020 109555.5']
然后我尝试像这样使用更改它
re.findall("\d{2}\.\d{2}\.\d{4}\s\d+\.{0,1}\D", some_string)
#in this case regex matched '1-length as digit' but not as a float
['15.10.2020 102000 ',
'16.11.2020 120888.',
'18.09.2020 5 ',
'15.12.2020 109555.']
你能帮我解决我的问题吗？