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MathGroup Archive 2007

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Re: averaging sublists of different lengths

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75512] Re: averaging sublists of different lengths
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Thu, 3 May 2007 03:52:30 -0400 (EDT)
  • References: <f19fql$4q7$1@smc.vnet.net>

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



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