residue_pair_energies

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    • #2794
      Anonymous

        Hello!

        I am trying to find the residues interacting with a particular residue of interest so as I can determine which residue-residue interaction is making the most contribution to a specific scoring term. However, I have not been able to find the residue_pair_energy function. 

        Do I need to write the function class to obtain this or is there a function that already available (e.g. to get atom-atom pairwise energies). If the latter case where is the function  located. 

        I have tried core.scoring.Energies but it does not exist there

        I am using PyRosetta4.Release.python36.mac downloaded 4 days ago

        Please advice 

         

        Thank you for your time

         

        Kamau

         

      • #13916
        Anonymous

          I’m not aware of a canned function which will allow this.

          What you’ll have to do is extract the information from the EnergyGraph object that’s stored in a Pose after a call to scoring. (Be sure to score the pose prior to extracting the data, else you might get a crash, and you’ll certainly get nonsense numbers.)

          To get the total score between to residues `resi1` and `resi2` (in the start-at-one, Pose-numbered integer form):


          pose.energies().energy_graph().find_energy_edge( resi1, resi2 ).dot( scorefxn.weights() )

          If you want the unweighted energies or the score for particular components, you can also pull that out of the particular edge.

           

          Note: For efficiency purposes, the backbone hydrogen bonds aren’t included as pairwise energies by default. To correct for this, you’ll have to set an option in the EnergyMethodOptions — Do this *before* you score the pose with the scorefunction.


          emeth_opt = scorefxn.energy_method_options().clone()
          emeth_opt.hbond_options().decompose_bb_hb_into_pair_energies( True )
          scorefxn->set_energy_method_options( emeth_opt )

           

        • #13930
          Anonymous

            Rocco

            Thanks  alot I will give this a shot 

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