[R] Average of a variable against another.
David Winsemius
dwinsemius at comcast.net
Fri Feb 12 03:30:59 CET 2010
On Feb 11, 2010, at 7:32 PM, Keeeeeeee wrote:
>
> Dear helpers,
>
> FYI, I am a beginner of R, just have dealt with MATLAB or JAVA.
>
> I want to know how to solve one problem given 4 variables: year_1,
> year_2,
> tall_1, tall_2.
> The tall_1 is measured at year_1 and tall_2 at year_2.
> The tall has grown up such as uniformly 1 cm/yr.
>
> The data is like
>
> year_1 year_2 tall_1 tall_2
> 2007 2010 12 15
> 1999 2009 6 16
> 2003 2005 11 13
> 2002 2009 3 10
> . . . . .
> . . . . .
> . . . . .
> . . . . .
>
> So I need to get the average tall of the plant against year, for all
> the
> years of available data.
For an individual plant, Wouldn't this just be (tall_2 + tall_1)/2?
(Or if you wanted to do it the hard way then use seq and divide by
number of years:)
> dfp <- rd.txt("year_1 year_2 tall_1 tall_2
+ 2007 2010 12 15
+ 1999 2009 6 16
+ 2003 2005 11 13
+ 2002 2009 3 10")
> dfp$plant.avg <- apply(dfp, 1, function(x) sum(seq(x[3], x[4]) ) /
(x[2]-x[1]+1) )
> dfp
year_1 year_2 tall_1 tall_2 plant.avg
1 2007 2010 12 15 13.5
2 1999 2009 6 16 11.0
3 2003 2005 11 13 12.0
4 2002 2009 3 10 6.5
> dfp$plant.avg2 <- apply(dfp, 1, function(x) (x[3]+ x[4] ) / 2 )
> dfp
year_1 year_2 tall_1 tall_2 plant.avg plant.avg2
1 2007 2010 12 15 13.5 13.5
2 1999 2009 6 16 11.0 11.0
3 2003 2005 11 13 12.0 12.0
4 2002 2009 3 10 6.5 6.5
I don't think you need to create the display below if you want to
answer the question posed. And if this happens to be homework, be sure
that I get credit.
--
David
> The year_1 and year_2 are recorded so that a plant is alive if the
> year of a
> question is equal to or grater than the year_1 and equal to and less
> than
> the year_2.
>
> For example,
> 1999 2000 2001 2002 2003 2004 2005 2006
> 2007 2008
> 2009 2010
> 6 7 8 9 10 11 12
> 13 14 15 16
> 3 4
> 5 6
> 7 8 9 10
> 11 12 13
>
> 12 13 14 15
>
> avg
> 6 7 8 6 25/3 ...etc.
>
> Since the amount of the data is too huge, I need to use appropriate
> functions and algorithm but I am
> not good at programming R.
>
> I wish you help me out from this hell problem please.
>
> Thanks,
> Keeeeeeee
>
> --
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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