On 2023-03-02 at 14:22:41 -0500,
jose isaias cabrera <[email protected]> wrote:

For the RegExp Gurus, consider the following python3 code:
<code>
import re
s = "pn=align upgrade sd=2023-02-"
ro = re.compile(r"pn=(.+) ")
r0=ro.match(s)

print(r0.group(1))

align upgrade
</code>

This is wrong. It should be 'align' because the group only goes up-to
the space. Thoughts? Thanks.

The bug is in your regular expression; the plus modifier is greedy.

If you want to match up to the first space, then you'll need something
like [^ ] (i.e., everything that isn't a space) instead of that dot.

--- SoupGate-Win32 v1.05
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Greetings.

For the RegExp Gurus, consider the following python3 code:
<code>
import re
s = "pn=align upgrade sd=2023-02-"
ro = re.compile(r"pn=(.+) ")
r0=ro.match(s)

print(r0.group(1))

align upgrade
</code>

This is wrong. It should be 'align' because the group only goes up-to
the space. Thoughts? Thanks.

josé

--

What if eternity is real? Where will you spend it? Hmmmm...

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Fri, 3 Mar 2023 at 06:24, jose isaias cabrera <[email protected]> wrote:

Greetings.

For the RegExp Gurus, consider the following python3 code:
<code>
import re
s = "pn=align upgrade sd=2023-02-"
ro = re.compile(r"pn=(.+) ")
r0=ro.match(s)

print(r0.group(1))

align upgrade
</code>

This is wrong. It should be 'align' because the group only goes up-to
the space. Thoughts? Thanks.

Not a bug. Find the longest possible match that fits this; as long as
you can find a space immediately after it, everything in between goes
into the .+ part.

If you want to exclude spaces, either use [^ ]+ or .+?.

ChrisA

--- SoupGate-Win32 v1.05
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On 3/2/23 12:28, Chris Angelico wrote:

On Fri, 3 Mar 2023 at 06:24, jose isaias cabrera <[email protected]> wrote:

Greetings.

For the RegExp Gurus, consider the following python3 code:
<code>
import re
s = "pn=align upgrade sd=2023-02-"
ro = re.compile(r"pn=(.+) ")
r0=ro.match(s)

print(r0.group(1))

align upgrade
</code>

This is wrong. It should be 'align' because the group only goes up-to
the space. Thoughts? Thanks.

Not a bug. Find the longest possible match that fits this; as long as
you can find a space immediately after it, everything in between goes
into the .+ part.

If you want to exclude spaces, either use [^ ]+ or .+?.

https://docs.python.org/3/howto/regex.html#greedy-versus-non-greedy

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Thu, Mar 2, 2023 at 2:32 PM <[email protected]> wrote:

On 2023-03-02 at 14:22:41 -0500,
jose isaias cabrera <[email protected]> wrote:

For the RegExp Gurus, consider the following python3 code:
<code>
import re
s = "pn=align upgrade sd=2023-02-"
ro = re.compile(r"pn=(.+) ")
r0=ro.match(s)

print(r0.group(1))

align upgrade
</code>

This is wrong. It should be 'align' because the group only goes up-to
the space. Thoughts? Thanks.

The bug is in your regular expression; the plus modifier is greedy.

If you want to match up to the first space, then you'll need something
like [^ ] (i.e., everything that isn't a space) instead of that dot.

Thanks. I appreciate your wisdom.

josé
--

What if eternity is real? Where will you spend it? Hmmmm...

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

José,

Matching can be greedy. Did it match to the last space?

What you want is a pattern that matches anything except a space (or whitespace) followed b matching a space or something similar.

Or use a construct that makes matching non-greedy.

Avi

-----Original Message-----
From: Python-list <python-list-bounces+avi.e.gross=[email protected]> On Behalf Of jose isaias cabrera
Sent: Thursday, March 2, 2023 2:23 PM
To: [email protected]
Subject: Regular Expression bug?

Greetings.

For the RegExp Gurus, consider the following python3 code:
<code>
import re
s = "pn=align upgrade sd=2023-02-"
ro = re.compile(r"pn=(.+) ")
r0=ro.match(s)

print(r0.group(1))

align upgrade
</code>

This is wrong. It should be 'align' because the group only goes up-to the space. Thoughts? Thanks.

josé

--

What if eternity is real? Where will you spend it? Hmmmm...
--
https://mail.python.org/mailman/listinfo/python-list

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Thu, Mar 2, 2023 at 2:38 PM Mats Wichmann <[email protected]> wrote:

On 3/2/23 12:28, Chris Angelico wrote:

On Fri, 3 Mar 2023 at 06:24, jose isaias cabrera <[email protected]>

wrote:

Greetings.

For the RegExp Gurus, consider the following python3 code:
<code>
import re
s = "pn=align upgrade sd=2023-02-"
ro = re.compile(r"pn=(.+) ")
r0=ro.match(s)

print(r0.group(1))

align upgrade
</code>

This is wrong. It should be 'align' because the group only goes up-to
the space. Thoughts? Thanks.

Not a bug. Find the longest possible match that fits this; as long as
you can find a space immediately after it, everything in between goes
into the .+ part.

If you want to exclude spaces, either use [^ ]+ or .+?.

https://docs.python.org/3/howto/regex.html#greedy-versus-non-greedy

This re is a bit different than the one I am used. So, I am trying to match everything after 'pn=':

import re
s = "pm=jose pn=2017"
m0 = r"pn=(.+)"
r0 = re.compile(m0)
s0 = r0.match(s)

print(s0)

None

Any help is appreciated.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

It is a well-known fact, Jose, that GIGO.

The letters "n" and "m" are not interchangeable. Your pattern fails because you have "pn" in one place and "pm" in the other.

s = "pn=jose pn=2017"

...

s0 = r0.match(s)
s0

<re.Match object; span=(0, 15), match='pn=jose pn=2017'>



-----Original Message-----
From: Python-list <python-list-bounces+avi.e.gross=[email protected]> On Behalf Of jose isaias cabrera
Sent: Thursday, March 2, 2023 8:07 PM
To: Mats Wichmann <[email protected]>
Cc: [email protected]
Subject: Re: Regular Expression bug?

On Thu, Mar 2, 2023 at 2:38 PM Mats Wichmann <[email protected]> wrote:

On 3/2/23 12:28, Chris Angelico wrote:

On Fri, 3 Mar 2023 at 06:24, jose isaias cabrera <[email protected]>

wrote:

Greetings.

For the RegExp Gurus, consider the following python3 code:
<code>
import re
s = "pn=align upgrade sd=2023-02-"
ro = re.compile(r"pn=(.+) ")
r0=ro.match(s)

print(r0.group(1))

align upgrade
</code>

This is wrong. It should be 'align' because the group only goes up-to
the space. Thoughts? Thanks.

Not a bug. Find the longest possible match that fits this; as long as
you can find a space immediately after it, everything in between goes
into the .+ part.

If you want to exclude spaces, either use [^ ]+ or .+?.

https://docs.python.org/3/howto/regex.html#greedy-versus-non-greedy

This re is a bit different than the one I am used. So, I am trying to match everything after 'pn=':

import re
s = "pm=jose pn=2017"
m0 = r"pn=(.+)"
r0 = re.compile(m0)
s0 = r0.match(s)

print(s0)

None

Any help is appreciated.
--
https://mail.python.org/mailman/listinfo/python-list

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 02Mar2023 20:06, jose isaias cabrera <[email protected]> wrote:

This re is a bit different than the one I am used. So, I am trying to
match
everything after 'pn=':

import re
s = "pm=jose pn=2017"
m0 = r"pn=(.+)"
r0 = re.compile(m0)
s0 = r0.match(s)

`match()` matches at the start of the string. You want r0.search(s).
- Cameron Simpson <[email protected]>

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

jose isaias cabrera <[email protected]> writes:

On Thu, Mar 2, 2023 at 2:38 PM Mats Wichmann <[email protected]> wrote:

This re is a bit different than the one I am used. So, I am trying to match
everything after 'pn=':

import re
s = "pm=jose pn=2017"
m0 = r"pn=(.+)"
r0 = re.compile(m0)
s0 = r0.match(s)
>>> print(s0)
None

Assuming that you were expecting to match "pn=2017", then you probably
don't want the 'match' method. Read its documentation. Then read the documentation for the _other_ methods that a Pattern supports. Then you
will be enlightened.

- Alan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Thu, Mar 2, 2023 at 8:35 PM <[email protected]> wrote:

It is a well-known fact, Jose, that GIGO.

The letters "n" and "m" are not interchangeable. Your pattern fails because you have "pn" in one place and "pm" in the other.

It is not GIGO. pm=project manager. pn=project name. I needed search()
rather than match().

s = "pn=jose pn=2017"

...

s0 = r0.match(s)
s0

<re.Match object; span=(0, 15), match='pn=jose pn=2017'>

-----Original Message-----
From: Python-list <python-list-bounces+avi.e.gross=[email protected]> On Behalf Of jose isaias cabrera
Sent: Thursday, March 2, 2023 8:07 PM
To: Mats Wichmann <[email protected]>
Cc: [email protected]
Subject: Re: Regular Expression bug?

On Thu, Mar 2, 2023 at 2:38 PM Mats Wichmann <[email protected]> wrote:

On 3/2/23 12:28, Chris Angelico wrote:

On Fri, 3 Mar 2023 at 06:24, jose isaias cabrera <[email protected]>

wrote:

Greetings.

For the RegExp Gurus, consider the following python3 code:
<code>
import re
s = "pn=align upgrade sd=2023-02-"
ro = re.compile(r"pn=(.+) ")
r0=ro.match(s)

print(r0.group(1))

align upgrade
</code>

This is wrong. It should be 'align' because the group only goes up-to
the space. Thoughts? Thanks.

Not a bug. Find the longest possible match that fits this; as long as
you can find a space immediately after it, everything in between goes into the .+ part.

If you want to exclude spaces, either use [^ ]+ or .+?.

https://docs.python.org/3/howto/regex.html#greedy-versus-non-greedy

This re is a bit different than the one I am used. So, I am trying to match everything after 'pn=':

import re
s = "pm=jose pn=2017"
m0 = r"pn=(.+)"
r0 = re.compile(m0)
s0 = r0.match(s)

print(s0)

None

Any help is appreciated.
--
https://mail.python.org/mailman/listinfo/python-list

--

What if eternity is real? Where will you spend it? Hmmmm...

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Thu, Mar 2, 2023 at 8:30 PM Cameron Simpson <[email protected]> wrote:

On 02Mar2023 20:06, jose isaias cabrera <[email protected]> wrote:

This re is a bit different than the one I am used. So, I am trying to
match
everything after 'pn=':

import re
s = "pm=jose pn=2017"
m0 = r"pn=(.+)"
r0 = re.compile(m0)
s0 = r0.match(s)

`match()` matches at the start of the string. You want r0.search(s).
- Cameron Simpson <[email protected]>

Thanks. Darn it! I knew it was something simple.


--

What if eternity is real? Where will you spend it? Hmmmm...

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Thu, Mar 2, 2023 at 9:56 PM Alan Bawden <[email protected]> wrote:

jose isaias cabrera <[email protected]> writes:

On Thu, Mar 2, 2023 at 2:38 PM Mats Wichmann <[email protected]> wrote:

This re is a bit different than the one I am used. So, I am trying to match
everything after 'pn=':

import re
s = "pm=jose pn=2017"
m0 = r"pn=(.+)"
r0 = re.compile(m0)
s0 = r0.match(s)
>>> print(s0)
None

Assuming that you were expecting to match "pn=2017", then you probably
don't want the 'match' method. Read its documentation. Then read the documentation for the _other_ methods that a Pattern supports. Then you
will be enlightened.

Yes. I need search. Thanks.

--

What if eternity is real? Where will you spend it? Hmmmm...

--- SoupGate-Win32 v1.05
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