Hi Sergey, Anufriev Sergey Valentinovich wrote: > > On Wed, 13 Aug 2003, Rene Brun wrote: > Thank you for your help. I'd like to know two points more: > > 1. Is there a difference between h2.Draw() and h2->Draw() CINT allows both forms. However, we recommend to follow standard C++. Note that the new version of CINT 5.15.98 will automatically generate a warning. > > 2. To return to the previous question: if i have NTuple with the column > E_t, how can I get the TH1D I need (with the XAxis = X_t) from it. It is > not clear for me how to work with NTuple, I need an exsample. If you have a TNtuple/TTree with columns x,y,z,etc, you can do things like ntuple.Draw("x"); ntuple.Draw("sqrt(x)") ntuple.Draw("2*x/sqrt(200)") ntuple.Draw("sqrt(x):y"); ntuple.Draw("x:sqrt(y)","z<0"); etc for more information, see the documentation of TTree::Draw and the Users Guide. Rene Brun > > > Well, in this case, you have to > > -create a new histogram with the right limits in X_t > > - loop on the original histogram bins > > - getting the bin content > > - computing the new bin in the new histogram > > - call SetBinContent(newbin,bincontent) > > > > You must be careful with bin edge effects. A reason to start from the > > original data (Tree or..) instead of the above procedure. > > > > Rene Brun > > > > Anufriev Sergey Valentinovich wrote: > > > > > > But, in fact E_t is the XAxis. And I need XAxis will be X_t. I need not > > > to normalize my histogram. > > > > > > > Hi Sergey, > > > > > > > > Having TH1D* h, do > > > > h->Scale(2/sqrt(s)); > > > > > > > > see also TH1::DrawNormalized > > > > > > > > Rene Brun > > > > > > > > Anufriev Sergey Valentinovich wrote: > > > > > > > > > > Dear ROOTers! > > > > > > > > > > I have the next problem: I can draw TH1D which shows dependense CROSS > > > > > SECTION (E_transeverse), but in fact I need histogram CROSS > > > > > SECTION(X_transeverse), where X_transeverse=2*E_transeverse/SQRT(s) (just > > > > > one can say X_t=f(E_t)). Question is: how can I get it? > > > > > > > > > > Thank you in advance, Sergey. > > > > > >
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