Ooops (.03+.01+.01+.01)/(.19+.01+.09+.09) = .16 -- Ian
On Tue, Oct 4, 2011 at 12:22 PM, Ian Tickle <ianj...@gmail.com> wrote: > On Tue, Oct 4, 2011 at 11:21 AM, Adam Ralph <adam.ra...@nuim.ie> wrote: > >> Dear Brigitte, >> >> >> Looking at the formulae it could be possible to get those results. >> Take an example >> below >> >> >> Rho_cal = -0.11, 0.0, 0.05, 0.05 >> Rho_obs = -0.08, 0.01, 0.04, 0.04 >> >> >> R-fac = 0.02/0.0 = undefined >> >> >> Correl = 0.0032 - (-0.0025*0.0025) >> -------------------------------------- = 0.99 >> sqrt(0.0043 * 0.0024) >> >> Did the calculations quickly so hope they are OK. However, I designed the >> data so >> that the denominator in the R-fac is zero i.e. the sum of Rho_cal = - sum >> of Rho_obs. >> It would imply that the ATMMAP from sfall does not cover the correct set >> of grid points >> for the ligand. You expect the Fc map to be positive in this region. You >> need to generate >> a new ATMMAP for each different ligand conformation. >> >> Adam >> >> > Hi Adam > > That doesn't look right to me, the formula according to Jones et al is: > > RSR = sum(| rho_obs - rho_calc |) / sum(| rho_obs + > rho_calc |) > > So for your example we have RSR = (.03 + .01 + .01 + .01) / (.19 + .01 + > .09 + .09) = .13 which is obviously quite a reasonable number. > > If you want some numbers which will cause a zero divide you have to make > rho_obs = - rho_calc for every point so each term in the sum in the > denominator above is zero, and therefore obviously the denominator itself > would be zero. > > Here are the relevant code snippets from OVERLAPMAP: > > iave(j,i)=0 > xave(j,i)=0. > yave(j,i)=0. > > iave(jj,ii)=iave(jj,ii)+1 > xave(jj,ii)=xave(jj,ii)+xwork > yave(jj,ii)=yave(jj,ii)+ywork > > xave(jj,ii)=xave(jj,ii)/iave(jj,ii) > yave(jj,ii)=yave(jj,ii)/iave(jj,ii) > > rfac(jj,ii) = (abs(xave(jj,ii)- yave(jj,ii))) / > (abs(xave(jj,ii)+ yave(jj,ii))) > > This looks wrong to me since the absolute value is being taken after the > summation instead of before, i.e. it should be forming sums of > abs(xwork-ywork) and abs(xwork+ywork). The absolute value of a sum is not > the same as the sum of absoiute values! Note that the division throughout > by the no of points (iave(jj,ii)) has no effect on the result. > > I didn't check the formula for the correlation coefficient. > > But your broad conclusion (that the data is garbage) is very probably > correct! > > Cheers > > -- Ian >
