Thomas,
The resulting histogram must have the same number of bins than the
operand histograms.
Rene Brun
On Thu, 18 Mar 2004, Thomas Bretz wrote:
> Hi Rene,
>
> why does h4 need the same number of bins? It is enough if h4 'takes' the
> binning from either h3 or h2, isn't it?
>
> Thomas.
>
> Rene Brun wrote:
> > Your two projections h2 and h3 have the same number of bins,
> > eg you can do h3->Divide(h2);
> > However, you call the default constructor for h4. h4 does not have the
> > same number of bins as h2 and h3.
> > Instead of
> > TH2D *h4 = new TH2D;
> > h4->Divide(h3, h2);
> > do
> > TH2D *h4 = (TH2D*)h2->Clone();
> > h4->Divide(h3, h2);
> >
> > Rene Brun
> >
> > Thomas Bretz wrote:
> >
> >>Hi,
> >>
> >>I have a
> >>Double_t x[5] = { 0, 1, 2, 3, 4 };
> >>TH3D fHist("", "", 4, x, 4, x, 4, x);
> >>and do:
> >>fHist.GetZaxis()->SetRange(3, 4);
> >>TH1 *h2 = fHist.Project3D("xy_off"); //, -1, 9999, "E");
> >>
> >>fHist.GetZaxis()->SetRange(1,2);
> >>TH1 *h3 = fHist.Project3D("xy_on"); //, -1, 9999, "E");
> >>
> >>TH2D *h4 = new TH2D;
> >>h4->Divide(h3, h2);
> >>
> >>And I get
> >>Attempt to divide histograms with different number of bins...
> >>
> >>What am I doing wrong? I would expect to have two projections to have
> >>the same binning...
> >>
> >>Thomas.
> >
> >
>
This archive was generated by hypermail 2b29 : Sun Jan 02 2005 - 05:50:06 MET