questions regarding the reproduction of your test results
pengzhangzhi opened this issue · comments
Hi. I am trying to reproduce your test results about generating antibody CDRs (sequence-structure co-design) using the DiffAb model.
Using the design_testset.py
script, index 10 (pdb 7bwj_H_L_E
), and codesign_single
ckpt, the results on CDR-H3 are unacceptably bad.
The following table is the rmsd-ca between generated structure and native structure.
H_CDR1 H_CDR2 H_CDR3 L_CDR1 L_CDR2 L_CDR3
mean 1.428283 1.659423 52.084544 1.605134 0.385445 3.645147
min 0.861669 0.935878 28.523428 0.730051 0.208125 1.603799
max 2.923090 3.224749 153.499802 2.118108 0.752684 6.416183
the rmsd-ca is calculated by the following code.
generate_flags = variant['data']['generate_flag']
native_atom_positions = variant['data']['pos_heavyatom'][...,BBHeavyAtom.CA,:][generate_flags]
# native_atom_positions = native_atom_positions[mask_ha[generate_flags]]
pred_atom_positions = pos_ha[...,BBHeavyAtom.CA,:][generate_flags]
# pred_atom_positions = pred_atom_positions[mask_ha[generate_flags]]
rmsd = ((native_atom_positions - pred_atom_positions)**2).sum(-1).mean()
If this case has such a high rmsd, I doubt that the testset rmsd reported in your paper, Table 1 would also high.
No offense, I am trying to find out what is wrong with my reproduction.
Let me know if you want more details about my reproduction.
rmsd = ((native_atom_positions - pred_atom_positions)**2).sum(-1).mean()
R.M.S.D means Root-mean-square deviation, where is the square-root?
oops, My bad! Thank you sooooo much~