Re: [Rdkit-discuss] Can't kelulize

2014-12-09 Thread Wong, Sergio E. 
Dear Ling;

   Thank you for pointing out the issue with the lactam ring.  I manually 
changed the bond types in the mol2 file and now the error is gone.  The 
MolFromMol2File function can sanitize the molecule.  However, I still have a 
problem with the output.  Again, my code is:

   mol=Chem.MolFromMol2File(%s.lig.%d.mol2%(name, i), sanitize = True, 
removeHs = False)

   aromatic_6=[c,n]1[c,n][c,n][c,n][c,n][c,n]1
   aromatic_5=[c,n]1[c,n][c,n][c,n][c,n]1

   pattern6=Chem.MolFromSmarts(aromatic_6)
   pattern5=Chem.MolFromSmarts(aromatic_5)

   print Pattern 6 
   lar = mol.GetSubstructMatch(pattern6)
   print lar
   print Pattern 5 
   lar = mol.GetSubstructMatch(pattern5)
   print lar

The output is:

   Pattern 6
   (0, 1, 3, 5, 7, 9)
   Pattern 5
   (30, 31, 32, 41, 42)

So for some reason, the pattern match for an aromatic six-membered ring returns 
the conjugated lactam ring, but fails to recognize the other two (all-carbon) 
aromatic rings in the system.  Interesting, it correctly recognizes the 
five-membered ring system.  Do you have any idea's on how to address the issue? 
 I am attaching the hand-edited mol2 file.

Thanks!
-Sergio



From: S.L. Chan [slch...@yahoo.com]
Sent: Monday, December 08, 2014 8:26 PM
To: Wong, Sergio E.; rdkit-discuss@lists.sourceforge.net
Subject: Re: [Rdkit-discuss] Can't kelulize

Dear Sergio,
The lactam ring (atoms 1 2 4 6 8 10) is not really aromatic. The bonds 4-6, 
6-8, 8-10 should all be single rather than aromatic in the mol2 file. The 
remaining three bonds in the ring should be double or single rather than 
aromatic.

Ling


From: Wong, Sergio E. wong...@llnl.gov
To: rdkit-discuss@lists.sourceforge.net rdkit-discuss@lists.sourceforge.net
Sent: Monday, December 8, 2014 3:27 PM
Subject: Re: [Rdkit-discuss] Can't kelulize

Dear RDKit users:

I tried reading a mol2 file using the function MolFromMol2 ().  The goal of my 
script is to read the molecule and find 5 or 6 membered aromatic rings.  First 
I got the following error:

Can't Kekulize mol

The code I used is as follows:

mol=Chem.MolFromMol2File(%s.lig.%d.mol2%(name, i), sanitize = True, removeHs 
= False)

As a work-around I tried removing the sanitize flag and did the following


   mol=Chem.MolFromMol2File(%s.lig.%d.mol2%(name, i), sanitize = False, 
removeHs = False)

   aromatic_6=[c,n]1[c,n][c,n][c,n][c,n][c,n]1
   aromatic_5=[c,n]1[c,n][c,n][c,n][c,n]1

   pattern6=Chem.MolFromSmarts(aromatic_6)
   pattern5=Chem.MolFromSmarts(aromatic_5)

   print Pattern 6 
   lar = mol.GetSubstructMatch(pattern6)
   print lar
   print Pattern 5 
   lar = mol.GetSubstructMatch(pattern5)
   print lar

The output should 3 aromatic six-membered rings and 1 aromatic five-membered 
ring.  Instead I get only the first six-membered ring and no listing of the 
five-membered ring.:

   Pattern 6
   (0, 1, 3, 5, 7, 9)
   Pattern 5
   ()

So basically, I can not get around the kekulize function.  I looked the mol2 
file (attached) and it correctly lists the bond types as aromatic for all of 
the rings.  Is there a way to use the bond information from the mol2 file to 
assign aromaticity?

Thanks!!

-Sergio





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Description: MIZOLASTINE.lig.999.mol2
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Re: [Rdkit-discuss] Can't kelulize

2014-12-09 Thread Christos Kannas 
Sergio,

You have to use GetSubstructMatches.
Look at my sample here
http://nbviewer.ipython.org/gist/CKannas/5a762b97c52e389d492e.

Best,

Christos

Christos Kannas

Researcher
Ph.D Student

Mob (UK): +44 (0) 7447700937
Mob (Cyprus): +357 99530608

[image: View Christos Kannas's profile on LinkedIn]
http://cy.linkedin.com/in/christoskannas

On 9 December 2014 at 18:48, Wong, Sergio E. wong...@llnl.gov wrote:

  Dear Ling;

Thank you for pointing out the issue with the lactam ring.  I manually
 changed the bond types in the mol2 file and now the error is gone.  The
 MolFromMol2File function can sanitize the molecule.  However, I still have
 a problem with the output.  Again, my code is:

mol=Chem.MolFromMol2File(%s.lig.%d.mol2%(name, i), sanitize = True,
 removeHs = False)

aromatic_6=[c,n]1[c,n][c,n][c,n][c,n][c,n]1
aromatic_5=[c,n]1[c,n][c,n][c,n][c,n]1

pattern6=Chem.MolFromSmarts(aromatic_6)
pattern5=Chem.MolFromSmarts(aromatic_5)

print Pattern 6 
lar = mol.GetSubstructMatch(pattern6)
print lar
print Pattern 5 
lar = mol.GetSubstructMatch(pattern5)
print lar

 The output is:

Pattern 6
(0, 1, 3, 5, 7, 9)
Pattern 5
(30, 31, 32, 41, 42)

 So for some reason, the pattern match for an aromatic six-membered ring
 returns the conjugated lactam ring, but fails to recognize the other two
 (all-carbon) aromatic rings in the system.  Interesting, it correctly
 recognizes the five-membered ring system.  Do you have any idea's on how to
 address the issue?  I am attaching the hand-edited mol2 file.

 Thanks!
 -Sergio



 *From:* S.L. Chan [slch...@yahoo.com]
  *Sent:* Monday, December 08, 2014 8:26 PM
 *To:* Wong, Sergio E.; rdkit-discuss@lists.sourceforge.net

 *Subject:* Re: [Rdkit-discuss] Can't kelulize

   Dear Sergio,
  The lactam ring (atoms 1 2 4 6 8 10) is not really aromatic. The bonds
 4-6, 6-8, 8-10 should all be single rather than aromatic in the mol2 file.
 The remaining three bonds in the ring should be double or single rather
 than aromatic.

  Ling

   --
 *From:* Wong, Sergio E. wong...@llnl.gov
 *To:* rdkit-discuss@lists.sourceforge.net 
 rdkit-discuss@lists.sourceforge.net
 *Sent:* Monday, December 8, 2014 3:27 PM
 *Subject:* Re: [Rdkit-discuss] Can't kelulize

   Dear RDKit users:

 I tried reading a mol2 file using the function MolFromMol2 ().  The goal
 of my script is to read the molecule and find 5 or 6 membered aromatic
 rings.  First I got the following error:

 Can't Kekulize mol

 The code I used is as follows:

 mol=Chem.MolFromMol2File(%s.lig.%d.mol2%(name, i), sanitize = True,
 removeHs = False)

 As a work-around I tried removing the sanitize flag and did the following


mol=Chem.MolFromMol2File(%s.lig.%d.mol2%(name, i), sanitize = False,
 removeHs = False)

aromatic_6=[c,n]1[c,n][c,n][c,n][c,n][c,n]1
aromatic_5=[c,n]1[c,n][c,n][c,n][c,n]1

pattern6=Chem.MolFromSmarts(aromatic_6)
pattern5=Chem.MolFromSmarts(aromatic_5)

print Pattern 6 
lar = mol.GetSubstructMatch(pattern6)
print lar
print Pattern 5 
lar = mol.GetSubstructMatch(pattern5)
print lar

 The output should 3 aromatic six-membered rings and 1 aromatic
 five-membered ring.  Instead I get only the first six-membered ring and no
 listing of the five-membered ring.:

Pattern 6
(0, 1, 3, 5, 7, 9)
Pattern 5
()

 So basically, I can not get around the kekulize function.  I looked the
 mol2 file (attached) and it correctly lists the bond types as aromatic for
 all of the rings.  Is there a way to use the bond information from the mol2
 file to assign aromaticity?

 Thanks!!

 -Sergio






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Re: [Rdkit-discuss] Can't kelulize

2014-12-08 Thread Wong, Sergio E. 
Dear RDKit users:

I tried reading a mol2 file using the function MolFromMol2 ().  The goal of my 
script is to read the molecule and find 5 or 6 membered aromatic rings.  First 
I got the following error:

Can't Kekulize mol

The code I used is as follows:

mol=Chem.MolFromMol2File(%s.lig.%d.mol2%(name, i), sanitize = True, removeHs 
= False)

As a work-around I tried removing the sanitize flag and did the following


   mol=Chem.MolFromMol2File(%s.lig.%d.mol2%(name, i), sanitize = False, 
removeHs = False)

   aromatic_6=[c,n]1[c,n][c,n][c,n][c,n][c,n]1
   aromatic_5=[c,n]1[c,n][c,n][c,n][c,n]1

   pattern6=Chem.MolFromSmarts(aromatic_6)
   pattern5=Chem.MolFromSmarts(aromatic_5)

   print Pattern 6 
   lar = mol.GetSubstructMatch(pattern6)
   print lar
   print Pattern 5 
   lar = mol.GetSubstructMatch(pattern5)
   print lar

The output should 3 aromatic six-membered rings and 1 aromatic five-membered 
ring.  Instead I get only the first six-membered ring and no listing of the 
five-membered ring.:

   Pattern 6
   (0, 1, 3, 5, 7, 9)
   Pattern 5
   ()

So basically, I can not get around the kekulize function.  I looked the mol2 
file (attached) and it correctly lists the bond types as aromatic for all of 
the rings.  Is there a way to use the bond information from the mol2 file to 
assign aromaticity?

Thanks!!

-Sergio






MIZOLASTINE.lig.30.mol2
Description: MIZOLASTINE.lig.30.mol2
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Re: [Rdkit-discuss] Can't kelulize

2014-12-08 Thread S.L. Chan 
Dear Sergio,
The lactam ring (atoms 1 2 4 6 8 10) is not really aromatic. The bonds 4-6, 
6-8, 8-10 should all be single rather than aromatic in the mol2 file. The 
remaining three bonds in the ring should be double or single rather than 
aromatic.
Ling
 From: Wong, Sergio E. wong...@llnl.gov
 To: rdkit-discuss@lists.sourceforge.net 
rdkit-discuss@lists.sourceforge.net 
 Sent: Monday, December 8, 2014 3:27 PM
 Subject: Re: [Rdkit-discuss] Can't kelulize
   
 #yiv2379857403 P {margin-top:0;margin-bottom:0;}Dear RDKit users:

I tried reading a mol2 file using the function MolFromMol2 ().  The goal of my 
script is to read the molecule and find 5 or 6 membered aromatic rings.  First 
I got the following error:

Can't Kekulize mol

The code I used is as follows:   

mol=Chem.MolFromMol2File(%s.lig.%d.mol2%(name, i), sanitize = True, removeHs 
= False)

As a work-around I tried removing the sanitize flag and did the following


   mol=Chem.MolFromMol2File(%s.lig.%d.mol2%(name, i), sanitize = False, 
removeHs = False)
 
   aromatic_6=[c,n]1[c,n][c,n][c,n][c,n][c,n]1
   aromatic_5=[c,n]1[c,n][c,n][c,n][c,n]1

   pattern6=Chem.MolFromSmarts(aromatic_6)
   pattern5=Chem.MolFromSmarts(aromatic_5)

   print Pattern 6 
   lar = mol.GetSubstructMatch(pattern6)
   print lar
   print Pattern 5 
   lar = mol.GetSubstructMatch(pattern5)
   print lar

The output should 3 aromatic six-membered rings and 1 aromatic five-membered 
ring.  Instead I get only the first six-membered ring and no listing of the 
five-membered ring.:

   Pattern 6
   (0, 1, 3, 5, 7, 9)
   Pattern 5
   ()

So basically, I can not get around the kekulize function.  I looked the mol2 
file (attached) and it correctly lists the bond types as aromatic for all of 
the rings.  Is there a way to use the bond information from the mol2 file to 
assign aromaticity?

Thanks!!

-Sergio
 




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