The following
>>> re.sub(r'(\d+)', r'
The following
>>> re.sub(r'(\d+)', r'\1' * 2, 'test line 123')
gives
'test line 123123'
is there a way to make it give
有没有办法让它给
'test line 246'
?
float() coercion doesn't work:
float()强制不起作用:
>>> re.sub(r'(\d+)', float(r'\1') * 2, 'test line 123')
could not convert string to float: \1
nor do eval or exec.
也不是eval或exec。
2 个解决方案
#1
5
The second argument to re.sub() can also be a callable, which lets you do:
re.sub()的第二个参数也可以是可调用的,它允许您执行:
re.sub(r'(\d+)', lambda match:'%d' % (int(match.group(1))*2), 'test line 123')
BTW, there really is no reason to use float over int, as your regex does not include periods and will always be a non-negative integer
顺便说一下,确实没有理由使用float over int,因为你的正则表达式不包含句点而且总是一个非负整数
#2
4
The trick is to supply a function as the repl argument to re.sub():
诀窍是提供一个函数作为re.sub()的repl参数:
In [7]: re.sub(r'(\d+)', lambda m:'%.0f'%(float(m.group(1))*2), 'test line 123')
Out[7]: 'test line 246'
Each match is converted to float, doubled, and then converted to string using the appropriate format.
每个匹配都转换为float,doubled,然后使用适当的格式转换为字符串。
This can be simplified a little bit if the number is integer, but your question specifically mentions float, so that's what I've used.
如果数字是整数,这可以简化一点,但你的问题特别提到浮点数,所以这就是我用过的。
' * 2, 'test line 123')
>>> re.sub(r'(\d+)', r'
The following
>>> re.sub(r'(\d+)', r'\1' * 2, 'test line 123')
gives
'test line 123123'
is there a way to make it give
有没有办法让它给
'test line 246'
?
float() coercion doesn't work:
float()强制不起作用:
>>> re.sub(r'(\d+)', float(r'\1') * 2, 'test line 123')
could not convert string to float: \1
nor do eval or exec.
也不是eval或exec。
2 个解决方案
#1
5
The second argument to re.sub() can also be a callable, which lets you do:
re.sub()的第二个参数也可以是可调用的,它允许您执行:
re.sub(r'(\d+)', lambda match:'%d' % (int(match.group(1))*2), 'test line 123')
BTW, there really is no reason to use float over int, as your regex does not include periods and will always be a non-negative integer
顺便说一下,确实没有理由使用float over int,因为你的正则表达式不包含句点而且总是一个非负整数
#2
4
The trick is to supply a function as the repl argument to re.sub():
诀窍是提供一个函数作为re.sub()的repl参数:
In [7]: re.sub(r'(\d+)', lambda m:'%.0f'%(float(m.group(1))*2), 'test line 123')
Out[7]: 'test line 246'
Each match is converted to float, doubled, and then converted to string using the appropriate format.
每个匹配都转换为float,doubled,然后使用适当的格式转换为字符串。
This can be simplified a little bit if the number is integer, but your question specifically mentions float, so that's what I've used.
如果数字是整数,这可以简化一点,但你的问题特别提到浮点数,所以这就是我用过的。
' * 2,
