[BioC] Average based on group

Moshe Olshansky olshansky at wehi.EDU.AU
Fri May 13 08:33:30 CEST 2011


You can use aggregate as below:

junk
      V1        V2       V3  V4
1  chr10 rs9971029 71916552 0.1
2  chr10 rs9971029 71916553 0.4
3  chr10 rs9971029 71916554 0.3
4  chr10 rs9971029 71916555 0.9
5  chr10 rs9971029 71916556 1.0
6  chr10 rs9971029 71916557 2.0
7  chr10 rs9971029 71916558 4.0
8  chr10 rs9971029 71916559 0.8
9  chr10 rs9971029 71916560 0.9
10 chr10 rs9971029 71916561 0.8
11 chr10 rs9971030 71916726 0.6
12 chr10 rs9971030 71916727 0.5
13 chr10 rs9971030 71916728 0.4
14 chr10 rs9971030 71916729 0.7
15 chr10 rs9971030 71916730 0.0
16 chr10 rs9971030 71916731 0.0
17 chr10 rs9971030 71916732 0.6
18 chr10 rs9971030 71916733 0.8
19 chr10 rs9971030 71916734 0.9
20 chr10 rs9971030 71916735 1.0
> colnames(junk) <- c("chr","group","someNumber","value")
> aggregate(junk$value,list(junk$group),mean)
    Group.1    x
1 rs9971029 1.12
2 rs9971030 0.55

> Dear list,
> I have dataframe, the second column is groups factor, each group has
> 10 items. The data as fellow.
> chr10   rs9971029   71916552	0.1
> chr10   rs9971029   71916553	0.4
> chr10   rs9971029   71916554	0.3
> chr10   rs9971029   71916555	0.9
> chr10   rs9971029   71916556	1
> chr10   rs9971029   71916557	2
> chr10   rs9971029   71916558	4
> chr10   rs9971029   71916559	0.8
> chr10   rs9971029   71916560	0.9
> chr10   rs9971029   71916561	0.8
> chr10   rs9971030   71916726	0.6
> chr10   rs9971030   71916727	0.5
> chr10   rs9971030   71916728	0.4
> chr10   rs9971030   71916729	0.7
> chr10   rs9971030   71916730	0
> chr10   rs9971030   71916731	0
> chr10   rs9971030   71916732	0.6
> chr10   rs9971030   71916733	0.8
> chr10   rs9971030   71916734	0.9
> chr10   rs9971030   71916735	1
>
> I want to get a average of each item based on the group factor. So at
> last I want to get a vector which length is 10.
> The value calculated as this:
>
> (0.1+0.6)/2
> (0.4+0.5)/2
> ?
> (0.8+1)/2
>
> Thank you very much in advance.
>
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