Python re sub：以贪婪的方式捕获组

Question

I am trying to extract digits in fallowing manner (bolded to be extracted)我正在尝试以休闲方式提取数字（粗体要提取）

0 -> 2 {2020-04-07 15:03} 0 -> 2 {2020-04-07 15:03}

0 -> 2 {2020-04-07 15:03} -> 67 {2020-04-07 15:20} 0 -> 2 {2020-04-07 15:03} -> 67 {2020-04-07 15:20}

With below code:使用以下代码：

a = '0 -> 2 {2020-04-07 15:03} -> 67 {2020-04-07 15:20}'
b = re.sub(r'^.*(\d+)\s({.+})$', r'\1', a)
print(b)
> 7

But I only get match on one (last) digit.但我只匹配一个（最后一个）数字。 Why greedeness is not working here?为什么贪婪在这里不起作用？

Answer 1

You can use:您可以使用：

re.sub(r'^.*\s(\d+)\s{.+}$', r'\1', a)

I added a \s before the capture number to make sure that it's preceded by a space.我在捕获号之前添加了一个\s以确保它前面有一个空格。