Re: averaging sublists of different lengths

*To*: mathgroup at smc.vnet.net*Subject*: [mg75526] Re: averaging sublists of different lengths*From*: dimitris <dimmechan at yahoo.com>*Date*: Fri, 4 May 2007 04:12:57 -0400 (EDT)*References*: <f19fql$4q7$1@smc.vnet.net><f1c55e$har$1@smc.vnet.net>

I didn't understand what you have asked. So you can ignore my response. =CF/=C7 dimitris =DD=E3=F1=E1=F8=E5: > In[151]:= > lst = {{70, 66, 64, 68, 64, 56, 68, 78, 62, 68, 84}, {70, 64, 64, 56, > 66, 56, 62, 64, 66, 88, 54, 72}, > {58, 54, 54, 60, 72, 70, 62, 68, 74, 76, 70}, {66, 56, 60, 64, 56, > 62, 68, 58, 58, 58, 68, 76, 62, 76, 66, 64, 88, 56}, > {56, 64, 72, 72, 70, 62, 76, 76, 76, 76, 86, 80, 100}, {60, 60, > 70, 68, 60, 60, 50, 56, 60, 70, 62, 68, 88, 84, 82}, > {54, 66, 72, 62, 70, 66, 70, 56}, {60, 60, 60, 62, 74, 80, 70}, > {54, 62, 64, 72, 76, 74}, {66, 74, 70, 80, 54, 54, 64}, > {72, 66, 60, 52, 52, 66, 66, 58, 60, 66}, {3}, {0, 0}}; > > In[142]:= > Mean /@ lst > Out[142]= > {68, 391/6, 718/11, 581/9, 966/13, 998/15, 129/2, 466/7, 67, 66, > 309/5, 3, 0} > > In[144]:= > (Tr[#1]/Length[#1] & ) /@ lst > Out[144]= > {68, 391/6, 718/11, 581/9, 966/13, 998/15, 129/2, 466/7, 67, 66, > 309/5, 3, 0} > > In[145]:= > (Total[#1]/Length[#1] & ) /@ lst > Out[145]= > {68, 391/6, 718/11, 581/9, 966/13, 998/15, 129/2, 466/7, 67, 66, > 309/5, 3, 0} > > In[146]:= > (Plus @@ #1/Length[#1] & ) /@ lst > Out[146]= > {68, 391/6, 718/11, 581/9, 966/13, 998/15, 129/2, 466/7, 67, 66, > 309/5, 3, 0} > > In[148]:= > ((#1 /. {x_, y___} -> x + y)/Length[#1] & ) /@ lst > Out[148]= > {68, 391/6, 718/11, 581/9, 966/13, 998/15, 129/2, 466/7, 67, 66, > 309/5, 3, 0} > > ??? > > Dimitris > > =CF/=C7 dantimatter =DD=E3=F1=E1=F8=E5: > > Hello all, > > > > I have a list of lists like this: > > > > {{70, 66, 64, 68, 64, 56, 68, 78, 62, 68, 84}, {70, 64, 64, 56, 66, > > 56, 62, > > 64, 66, 88, 54, 72}, {58, 54, 54, 60, 72, 70, 62, 68, 74, 76, 70}, > > {66, > > 56, 60, 64, 56, 62, 68, 58, 58, 58, 68, 76, 62, 76, 66, 64, 88, > > 56}, {56, > > 64, 72, 72, 70, 62, 76, 76, 76, 76, 86, 80, 100}, {60, 60, 70, 68, > > 60, 60, > > 50, 56, 60, 70, 62, 68, 88, 84, 82}, {54, 66, 72, 62, 70, 66, > > 70, > > 56}, {60, 60, 60, 62, 74, 80, 70}, {54, 62, 64, 72, 76, 74}, {66, > > 74, 70, > > 80, 54, 54, 64}, {72, 66, 60, 52, 52, 66, 66, 58, 60, 66}} > > > > What I'd really like is the average all the 1st values, 2nd values, > > etc, but I'm having trouble figuring out how to deal with the fact > > that the lists are not all the same length. Is there a way to drop > > the sublists as they run out of points to add to the average, i.e. if > > I want the average of all the nth values, but Length[shortSublist] < > > n, can I somehow drop shortSublist and then calculate the average from > > the other sublists? > > > > Also, the Mean[] function requires more than one data point, but I'd > > still like to extract out the values for which there is only one data > > point. Is there a way to do this? > > > > Thanks! > > dan