regex question

[Fred Mangusta]

Hi,

I would like to delete all the instances of a '.' into a number.

In other words I'd like to replace all the instances of a '.' character
with something (say nothing at all) when the '.' is representing a
decimal separator. E.g.

500.675 ---- 500675

but also

1.000.456.344 ----1000456344

I don't care about the fact the the resulting number is difficult to
read: as long as it remains a series of digits it's ok: the important
thing is to get rid of the period, because I want to keep it only where
it marks the end of a sentence.

I was trying to do like this

s=re.sub("[(\d+)(\.)(\d+)]","... ",s)

but I don't know much about regular expressions, and don't know how to
get the two groups of numbers and join them in the sub. Moreover doing
like this I only match things like "345.000" and not "1.000.000" .

What's the correct approach?

Thanks
F.

[Marc 'BlackJack' Rintsch]

Re: regex question

On Tue, 05 Aug 2008 11:39:36 +0100, Fred Mangusta wrote:

In other words I'd like to replace all the instances of a '.' character
with something (say nothing at all) when the '.' is representing a
decimal separator. E.g.
>
500.675 ---- 500675
>
but also
>
1.000.456.344 ----1000456344
>
I don't care about the fact the the resulting number is difficult to
read: as long as it remains a series of digits it's ok: the important
thing is to get rid of the period, because I want to keep it only where
it marks the end of a sentence.
>
I was trying to do like this
>
s=re.sub("[(\d+)(\.)(\d+)]","... ",s)
>
but I don't know much about regular expressions, and don't know how to
get the two groups of numbers and join them in the sub. Moreover doing
like this I only match things like "345.000" and not "1.000.000" .
>
What's the correct approach?

In [13]: re.sub(r'(\d)\. (\d)', r'\1\2', '1.000.456.344' )
Out[13]: '1000456344'

Ciao,
Marc 'BlackJack' Rintsch

[Jeff]

Re: regex question

On Aug 5, 7:10 am, Marc 'BlackJack' Rintsch <bj_...@gmx.net wrote:

On Tue, 05 Aug 2008 11:39:36 +0100, Fred Mangusta wrote:

In other words I'd like to replace all the instances of a '.' character
with something (say nothing at all) when the '.' is representing a
decimal separator. E.g.

>

500.675  ----      500675

>

but also

>

1.000.456.344 ----1000456344

>

I don't care about the fact the the resulting number is difficult to
read: as long as it remains a series of digits it's ok: the important
thing is to get rid of the period, because I want to keep it only where
it marks the end of a sentence.

>

I was trying to do like this

>

s=re.sub("[(\d+)(\.)(\d+)]","... ",s)

>

but I don't know much about regular expressions, and don't know how to
get the two groups of numbers and join them in the sub. Moreover doing
like this I only match things like "345.000" and not "1.000.000" .

>

What's the correct approach?

>
In [13]: re.sub(r'(\d)\. (\d)', r'\1\2', '1.000.456.344' )
Out[13]: '1000456344'
>
Ciao,
        Marc 'BlackJack' Rintsch

Even faster:

'1.000.456.344' .replace('.', '') ='1000456344'

[Chris]

Re: regex question

On Aug 5, 2:23 pm, Jeff <jeffo...@gmail .comwrote:

On Aug 5, 7:10 am, Marc 'BlackJack' Rintsch <bj_...@gmx.net wrote:
>
>
>

On Tue, 05 Aug 2008 11:39:36 +0100, Fred Mangusta wrote:

In other words I'd like to replace all the instances of a '.' character
with something (say nothing at all) when the '.' is representing a
decimal separator. E.g.

>

500.675  ----      500675

>

but also

>

1.000.456.344 ----1000456344

>

I don't care about the fact the the resulting number is difficult to
read: as long as it remains a series of digits it's ok: the important
thing is to get rid of the period, because I want to keep it only where
it marks the end of a sentence.

>

I was trying to do like this

>

s=re.sub("[(\d+)(\.)(\d+)]","... ",s)

>

but I don't know much about regular expressions, and don't know how to
get the two groups of numbers and join them in the sub. Moreover doing
like this I only match things like "345.000" and not "1.000.000" .

>

What's the correct approach?

>

In [13]: re.sub(r'(\d)\. (\d)', r'\1\2', '1.000.456.344' )
Out[13]: '1000456344'

>

Ciao,
        Marc 'BlackJack' Rintsch

>
Even faster:
>
'1.000.456.344' .replace('.', '') ='1000456344'

Doesn't work for his use case as he wants to keep periods marking the
end of a sentence.

[Fred Mangusta]

Re: regex question

Chris wrote:

Doesn't work for his use case as he wants to keep periods marking the
end of a sentence.

Exactly. Thanks to all of you anyway, now I have a better understanding
on how to go on :)

F.

[MRAB]

Re: regex question

On Aug 5, 11:39 am, Fred Mangusta <a...@bbb.itwro te:

Hi,
>
I would like to delete all the instances of a '.' into a number.
>
In other words I'd like to replace all the instances of a '.' character
with something (say nothing at all) when the '.' is representing a
decimal separator. E.g.
>
500.675  ----      500675
>
but also
>
1.000.456.344 ----1000456344
>
I don't care about the fact the the resulting number is difficult to
read: as long as it remains a series of digits it's ok: the important
thing is to get rid of the period, because I want to keep it only where
it marks the end of a sentence.
>
I was trying to do like this
>
s=re.sub("[(\d+)(\.)(\d+)]","... ",s)
>
but I don't know much about regular expressions, and don't know how to
get the two groups of numbers and join them in the sub. Moreover doing
like this I only match things like "345.000" and not "1.000.000" .
>
What's the correct approach?
>

I would use look-behind (is it preceded by a digit?) and look-ahead
(is it followed by a digit?):

s = re.sub(r'(?<=\d )\.(?=\d)', '', s)

[Tobiah]

Re: regex question

On Tue, 05 Aug 2008 15:55:46 +0100, Fred Mangusta wrote:

Chris wrote:
>

>Doesn't work for his use case as he wants to keep periods marking the
>end of a sentence.

Doesn't it? The period has to be surrounded by digits in the
example solution, so wouldn't periods followed by a space
(end of sentence) always make it through?



** Posted from http://www.teranews.com **