Re: averaging sublists of different lengths

• To: mathgroup at smc.vnet.net
• Subject: [mg75485] Re: averaging sublists of different lengths
• From: Szabolcs <szhorvat at gmail.com>
• Date: Thu, 3 May 2007 03:38:06 -0400 (EDT)
• Organization: University of Bergen
• References: <f19fql\$4q7\$1@smc.vnet.net>

```dantimatter wrote:
> 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?
>

This is an ugly solution, but it is simple, and if your dataset is not
very large, it works works fine.

In[1]:=
dat = {{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}};

In[2]:=
maxLen = Max[Length /@ dat]

Out[2]=
18

Pad the lists with some value that can be later removed:

In[3]:=

Now we have a square matrix that can be transposed:

Select[#1, (# =!= novalue &) ]& /@ Transpose[datPadded]

Out[4]=
{{70, 70, 58, 66, 56, 60, 54, 60, 54, 66, 72},
{66, 64, 54, 56, 64, 60, 66, 60, 62, 74, 66},
{64, 64, 54, 60, 72, 70, 72, 60, 64, 70, 60},
{68, 56, 60, 64, 72, 68, 62, 62, 72, 80, 52},
{64, 66, 72, 56, 70, 60, 70, 74, 76, 54, 52},
{56, 56, 70, 62, 62, 60, 66, 80, 74, 54, 66},
{68, 62, 62, 68, 76, 50, 70, 70, 64, 66},
{78, 64, 68, 58, 76, 56, 56, 58},
{62, 66, 74, 58, 76, 60, 60}, {68, 88, 76, 58, 76, 70, 66},
{84, 54, 70, 68, 86, 62},
{72, 76, 80, 68}, {62, 100, 88}, {76, 84}, {66, 82}, {64}, {88}, {56}}

In[5]:=
Mean /@ %

Out[5]=
{686/11, 692/11, 710/11, 716/11, 714/11, 706/11, 328/5, 257/4, 456/7,
502/7, 212/3, 74, 250/3, 80,
74, 64, 88, 56}

> 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
>
>

Mean[] works fine with one data point:

In[6]:=
Mean[{1}]

Out[6]=
1

Szabolcs

```

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