Question

我有一个字符串：

st = "Dee Fee, MD is a good person. Kas Egre, MD came here"

我想将", MD"替换为"Name:"并将其放在名称之前，因此新字符串应为：

new = "Name: Dee Fee is a good person. Name: Kas Egre came here"

我编写以下代码，它可以工作，但它没有给出我想要的东西。结果如下：

Name: Dee Fee, MD is a good person. Name: Dee Fee, MD came here    
Name: Kas Egre, MD is a good person. Name: Kas Egre, MD came here

这是我的代码：

rename = re.compile(r"""([A-Z][a-z]+\s[A-Z\s]*[A-Z][a-z]+)(,\s)(MD)""")
match = rename.search(st)
for match in rename.finditer(st):
    if match.group(3) == 'MD':
        new = rename.sub("Name: %s"%(match.group(0)),st)
        print new

如何修复我的代码？谢谢你的帮助

Answer 1

稍微简化你的正则表达式......

>>> re.sub(r'(\w+\s+\w+),\s*MD', lambda x: 'Name: '+ x.group(1), st)
'Name: Dee Fee is a good person. Name: Kas Egre came here'

实际上，我们甚至不需要这里的函数，因为我们可以插入\N组（其中N是组号）......

>>> re.sub(r'(\w+\s+\w+),\s*MD', r'Name: \1', st)
'Name: Dee Fee is a good person. Name: Kas Egre came here'

Answer 2

将re.sub与捕获组一起使用：

>>> import re
>>> st = "Dee Fee, MD is a good person. Kas Egre, MD came here"
>>> rename = re.compile(r'([A-Z][a-z]+\s+[A-Z][a-z]+),\s*\bMD\b')
>>> rename.sub(r'Name: \1', st)
'Name: Dee Fee is a good person. Name: Kas Egre came here'

替换字符串中的

\1将替换为第一个捕获的组（名称部分）。

用名字替换标题

2 个答案: