Closed ValeryPolyakov closed 5 years ago
Hi Valery,
if you use --min-heavies-per-const-frag 3, the [:1]O[:2] >> [:1]C([:2])N transformation will not be indexed any more, if one of the rests is just a simple methyl, because that fragmentation will not be created any more. So there is no way to get back exactly this transformation during indexing.
However, for all pairs of molecules which previously had that double-cut transformation, the single-cut transformation [:1]OC >> [:1]C(C)N will still be indexed and written to the database/output.
The idea for the --min-heavies-per-frag option was that there can be a lot of use cases where having one of the two (or even more in other cases) transformations in the DB is sufficient.
Do you have a specific need to use the double-cut transformation rather than the single-cut transformation?
Bests, Christian
Thanks Christian,
Do you know why I did not see a single-cut transformation in the output? Is there a specific option during indexing to recover it?
Valery
On Tue, Mar 12, 2019 at 1:18 AM Christian Kramer notifications@github.com wrote:
Hi Valery,
if you use --min-heavies-per-const-frag 3, the [:1]O[:2] >> [:1]C([:2])N transformation will not be indexed any more, if one of the rests is just a simple methyl, because that fragmentation will not be created any more. So there is no way to get back exactly this transformation during indexing.
However, for all pairs of molecules which previously had that double-cut transformation, the single-cut transformation [:1]OC >> [:1]C(C)N will still be indexed and written to the database/output.
The idea for the --min-heavies-per-frag option was that there can be a lot of use cases where having one of the two (or even more in other cases) transformations in the DB is sufficient.
Do you have a specific need to use the double-cut transformation rather than the single-cut transformation?
Bests, Christian
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Hi Valery,
I do not think that there is a specific option about to remove/ recover this transformation during indexing. If you send me two example input SMILES that have the problem, I can try to figure out what is going on here.
Bests, Christian
Thanks. I need to think about it. Obviously, I cannot send the actual compound...
On Tue, Mar 12, 2019 at 5:28 AM Christian Kramer notifications@github.com wrote:
Hi Valery,
I do not think that there is a specific option about to remove/ recover this transformation during indexing. If you send me two example input SMILES that have the problem, I can try to figure out what is going on here.
Bests, Christian
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Hi Christian,
I which database table should I find the single and double cut smiles like this: [:1]O[:2] or [:1]OC?
Valery
Valery
On Tue, Mar 12, 2019 at 5:44 AM Valery Polyakov valery.polyakov@gmail.com wrote:
Thanks. I need to think about it. Obviously, I cannot send the actual compound...
On Tue, Mar 12, 2019 at 5:28 AM Christian Kramer notifications@github.com wrote:
Hi Valery,
I do not think that there is a specific option about to remove/ recover this transformation during indexing. If you send me two example input SMILES that have the problem, I can try to figure out what is going on here.
Bests, Christian
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Hi Valery,
that depends on what you want to do exactly. The database has a table named 'rule_smiles' which contains all the SMILES of the fragments. In SQLite3, you can check whether a given SMILES is in that table by
"select from rule_smiles where smiles like '[:1]OC';"
If you want to find all transformations where that SMILES is involved, you have to query the table named 'rule' with the ID you get from the first query. If you are interested in all transformations + environments where that SMILES is used, you have to query the 'rule_environment' table with the id you get from the query in the 'rule' table.
If you query within the DB directly, I recommend to build the DB using the --symmetric option. Otherwise, you have to use the ID for your smiles in both LHS and RHS columns.
Bests, Christian
Hi Christian,
Thanks. The query works. It is a little slow, though. Is the --symmertic option result is DB size increase?
Valery
On Wed, Mar 13, 2019 at 3:14 AM Christian Kramer notifications@github.com wrote:
Hi Valery,
that depends on what you want to do exactly. The database has a table named 'rule_smiles' which contains all the SMILES of the fragments. In SQLite3, you can check whether a given SMILES is in that table by
"select from rule_smiles where smiles like '[:1]OC';"
If you want to find all transformations where that SMILES is involved, you have to query the table named 'rule' with the ID you get from the first query. If you are interested in all transformations + environments where that SMILES is used, you have to query the 'rule_environment' table with the id you get from the query in the 'rule' table.
If you query within the DB directly, I recommend to build the DB using the --symmetric option. Otherwise, you have to use the ID for your smiles in both LHS and RHS columns.
Bests, Christian
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Hi Valery,
yes, it results in an almost 2 fold increase of the DB size.
Christian
Hi Christian,
Can you tell me what SQL commands go into the following query:
python mmpdb predict --smiles "smiles1" --reference "smiles2" --property "propName" --save-details --prefix noOptions master_full.mmpdb > noOptions.txt
obviously, there are real smiles under smiles1 and smiles2.
Thanks a lot,
Valery
On Thu, Mar 14, 2019 at 2:44 AM Christian Kramer notifications@github.com wrote:
Hi Valery,
yes, it results in an almost 2 fold increase of the DB size.
Christian
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In mmpdblib/schema.py change the method MMPDatabase.execute from
if 0:
import time
print("EXECUTE")
print(sql)
print(repr(args))
to
if 1:
import time
print("EXECUTE")
print(sql)
print(repr(args))
That is, change the "0" to a "1". If I recall correctly, that will print out all of the SQL calls and their parameters.
The --symmetric
flag roughly doubles the database size but reduces the number of required SQL queries.
Thanks. I will try that.
On Tue, Mar 19, 2019, 1:45 AM Andrew Dalke notifications@github.com wrote:
In mmpdblib/schema.py change the method MMPDatabase.execute from
if 0: import time print("EXECUTE") print(sql) print(repr(args))
to
if 1: import time print("EXECUTE") print(sql) print(repr(args))
That is, change the "0" to a "1". If I recall correctly, that will print out all of the SQL calls and their parameters.
The --symmetric flag roughly doubles the database size but reduces the number of required SQL queries.
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Hi Andrew,
I did that, but the sql statements are not being printed...
Valery R. Polyakov
On Tue, Mar 19, 2019 at 1:45 AM Andrew Dalke notifications@github.com wrote:
In mmpdblib/schema.py change the method MMPDatabase.execute from
if 0: import time print("EXECUTE") print(sql) print(repr(args))
to
if 1: import time print("EXECUTE") print(sql) print(repr(args))
That is, change the "0" to a "1". If I recall correctly, that will print out all of the SQL calls and their parameters.
The --symmetric flag roughly doubles the database size but reduces the number of required SQL queries.
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I am surprised at that. As far as I can tell, the analysis routines all use something like self.mmpa_db.execute
, which goes through the method I asked you to modify.
I don't have the time to figure out where those specific calls are being done.
Are you sure that you've modified the right code? That is, sometimes it's hard to figure out if Python is using a modified file vs. the installed package file.
All I can suggest is that you trace through the code to find out where those SQL calls are being done, and add the print statements in the correct places. This can be done with the debugger (including the graphical IDE "IDLE" which comes as part of the distribution), among other methods.
Hi Valery,
is this issue still open? If yes, could you post an example that I can use to trace down the problem in the code?
Thank you, Christian
Christian and Andrew,
I was able to print the statements. Thanks.
By the way, is there any way for me to close the issue?
Valery R. Polyakov
On Sun, May 19, 2019 at 11:28 PM Christian Kramer notifications@github.com wrote:
Hi Valery,
is this issue still open? If yes, could you post an example that I can use to trace down the problem in the code?
Thank you, Christian
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Hi Valery,
I don't know whether you can close it. I will close this issue. If there are still questions open regarding this issue, we can open it again.
Best regards, Christian
Thanks
On Mon, May 20, 2019, 1:00 PM Christian Kramer notifications@github.com wrote:
Hi Valery,
I don't know whether you can close it. I will close this issue. If there are still questions open regarding this issue, we can open it again.
Best regards, Christian
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When using --min-heavies-per-const-frag 3 option during the fragmentation stage, I noticed that I am loosing the following transformation: [:1]O[:2] to [:1]C([:2])N in which one of the Rs is a simple methyl group. Is it possible to somehow loosing this transformation by playing with any option on during the indexing step.