Updating/Overwriting Histograms

ROOT
1

Hi all,

A part of my code involves writing the same named histograms on each other.

TH2D *myhist[56];
for(const int n=0; n<size; n++){
for (const int i=0; i<size2;i++ ) {
TFile *f = new TFile(fname);
read stuff iterated over ‘i’ and save to some buffer array;
f->Close();
TFile *fn = new TFile(“test.root”, “RECREATE”);
TString hiname = i; //because TH2D doesn’t take integer as title
myhist[i] = new TH2D(hiname,title,other parameters);
myhist[i]->Fill(n,i);
myhist[i]->Write();
fn->Close();
}}

It keeps on writing histos on themselvess and ends up with the last value of myhist[i] being written to file. When I change parameter “RECREATE” to “UPDATE”, it writes all the data but then the output takes the following format ;

hiname;1
hiname;2
.
.
hiname;n

having ‘1’ entry at each histogram (i*n histograms in total) instead of having ‘i’ histograms with n entries. After that I can combine those individual Trees but that is completely unnecessary. I can also achieve what I want by merging data by ‘htop’ but I don’t want to add extra phase to the analysis. Defining as a list and pushing ‘Addlist’ provided the same result as ‘UPDATE’ parameter option as well.

Also each iteration file I/O consumes time but myhist[i]->Write(); operates on the last file pointer by default thus reading/writing different files in the same loop is achieved like that. There should be a better way to do this.

Problems are ROOT specific, not ANSI C thus I think experienced users can save me some time on this.

Thanks.

2

Can you, please, explain what you want to do?
It’s quite difficult to find the question in your post, and the code sample looks wrong.
So, you want to update a histogram in a file? What do you mean by update - overwrite or add more data?
If you are adding more data - you do not have to re-create a hist, you have to read it from your file and add more data. But if it’s just one hist over bunch of input files … why writing/overwriting - just fill it in a loop and write it once. So, you have to clarify your problem.

3

I thought it was clear enough :slight_smile:

Let me reduce the problem;

There is a two dimensional array let’s say flatliner[i][n], I am trying to write it to a root file with ‘i’ number of histograms with each histogram containing ‘n’ number of entries.

4

If it was, I would not ask.

Well, now it’s even more confusing. This reduced description is not what you have in your original code - in your code you do not use any flatliner and do not need any input at all, since you’re just filling a bunch of 2D histograms with ‘n’ and ‘i’ indices, is it what you really want?

Anyway, I think your problem has nothing to do with ROOT (and has nothing to do “with ANSI C” or even with “ISO/IEC C++” or whatever, since it’s a problem of organising a couple of loops).

I assume that you have an input of size == size * size2 (both names are from your first code sample) and that you want to have size2 2D histograms (whatever this 2D histogram means).

Step 1: create size2 hists:

[code]TFile output(“test.root”, “recreate”);
output.cd();

for (int i = 0; i < size2; ++i) {
TString name = … somehow create a name derived from the index ‘i’.
myhist[i] = new TH2D(name, title, other parameters)
}
[/code]

So now you do not need any updates/overwrites, since you already have … all histograms you’ll be using.

step 2: iterate over your … data set (this was your outer loop with index n over a range [0, size)); it’s not clear though, if you have only one input file or size number of files.
If it’s only one, open it only once before any loop:

TFile input(fileName);
for (int i = 0; i < size; ++i) {//Here's your external loop
   //"read i-th row" (??) I suppose, 'row' must have a size == size2 == 56??
    for (int j = 0; j < size2; ++i) {
         ///do something
         myhist[j]->Fill(params);//fill it the way you want
     }
}

Now, step 3: write the histograms.

output.cd();
for (int i = 0; i < size2; ++i)
   myhist[i]->Write();

Well, if it’s still not what you want, you’ll probably have to explain why it’s not, so we can further adjust/modify this code :slight_smile: