Lie Ryan <[EMAIL PROTECTED]> wrote:
> That's the job of regular expression: 'import re'
>
> numbered_atom = re.compile('[A-Z][a-z]?[0-9]+')
> if numbered_atom.match('C10'):
> # this is a numbered atom
> if numbered_atom.match('C'):
> # this WON'T match
>
> read more about regular express
On Mon, 20 Oct 2008 13:16:48 +0200, Alfons Nonell-Canals wrote:
> Hello,
> I have a trouble and I don't know how to solve it. I am working with
> molecules and each molecule has a number of atoms. I obtain each atom
> spliting the molecule.
>
> Ok. It is fine and I have no problem with it.
>
> T
On Oct 20, 7:07 am, Peter Otten <[EMAIL PROTECTED]> wrote:
> Alfons Nonell-Canals wrote:
> > I have a trouble and I don't know how to solve it. I am working with
> > molecules and each molecule has a number of atoms. I obtain each atom
> > spliting the molecule.
>
> > Ok. It is fine and I have no p
atom = 'C1'
if '1' in atom:
print 'kk'
But, how can I do to identify in '1' all possibilities from 1-9, I
tried:
if '[1-9]', \d,...
You're reaching in the right direction (regexps), so just use
Python's "re" module:
import re
digit = re.compile(r'\d') # or "[0-9]"
for test in
Alfons Nonell-Canals wrote:
> I have a trouble and I don't know how to solve it. I am working with
> molecules and each molecule has a number of atoms. I obtain each atom
> spliting the molecule.
>
> Ok. It is fine and I have no problem with it.
>
> The problem is when I have to work with these
On Oct 20, 10:16 pm, Alfons Nonell-Canals <[EMAIL PROTECTED]>
wrote:
> Hello,
> I have a trouble and I don't know how to solve it. I am working with
> molecules and each molecule has a number of atoms. I obtain each atom
> spliting the molecule.
>
> Ok. It is fine and I have no problem with it.
>
>
Hello,
I have a trouble and I don't know how to solve it. I am working with
molecules and each molecule has a number of atoms. I obtain each atom
spliting the molecule.
Ok. It is fine and I have no problem with it.
The problem is when I have to work with these atoms. These atoms usually
are only
