• Q: I defined an "userD3PDBranchesToFill" branch as "meff". While looping over the meff values and channels I got following output:

  here the source code:
  -------------------------
  for (Int_t iMeff=0; iMeff< meff4jValues->size(); ++iMeff) {
      cout << "\n\nMeff= " << meff4jValues->at(iMeff) << endl;
      for (Int_t iCh=0; iCh< meff4jChannels->at(iMeff).size(); ++iCh) {
        cout << " " << meff4jChannels->at(iMeff).at(iCh) << endl;
      }
    }
  
  Here is the output:
  --------------------
  
  Meff= 693016
     4
     j
     1
     l
     e
     p
     C
     S
     C
       

• A: The error is in the reading code. meff4jValues is a vector of floats. And meff4jChannels is a vector of strings. So what you do in your code above is printing the character of the strings.

For particles, e.g. for a jet, you get a jetChannels branch, formatted as vector of vector of strings, because you have many particles in each event. But for branches defined in the "=userD3PDBranchesToFill=" dictionary you get only one entry per event, and thus you have a vector of strings to store the channels name whose sets of particles were used to compute the quantity. Below a python example of the reading code is provided:

Source code
----------------
for val in enumerate(t.meff4jValues):  # --> it actually stores only one value
        print "val: ", val
        for chan in t.meff4jChannels:
            print "chan: ", chan



Output
--------

val:  315487.78125
chan:  4j1lepCSC
chan:  4j0lepCSC
chan:  3j1lepMediumCutsMTSmaller100GeV
chan:  3j0lepMediumCuts
chan:  DEFAULT
chan:  3j1lepMediumCutsMTBigger100GeV
chan:  2j0lepCSC
chan:  3j0lepCSC
chan:  1lep

Instead, the example below reproduces the error:

Source Code
-----------------
for i,val in enumerate(t.meff4jValues):
        print "val: ", val
        for pr in t.meff4jChannels[i]:
            print "pr: ", pr


Output
---------
val:  315487.78125
pr:  1
pr:  l
pr:  e
pr:  p

-- RiccardoMariaBianchi - 04 Mar 2009