On Jul 16, 2009, at 5:53 PM, KARSTEN G HOLMQUIST wrote:
Hello,
I have a series of questions that I hope will be simple to answer.
Basically I would like a code to do the following so that I can
compute the distribution free test for the slope of a postulated
regression line (Theil test). As I am testing the null hypothesis
that slope = 0 against the general alternative the slope does not
equal 0, it should be pretty straight-forward.
I have a data frame as follows;
index Score Mean_Temp SD_Mean_Temp
i 10 3.7 0.65
i+1 12 7.5 2.11
i+2 6 10.3 0.85
... ... ... ...
n 8 12.6 4.2
The first question is how can I compute the n(n-1)/2 differences
between Score i and Score j where i is from 1 to (n-1) and j is from
(i+1) to n? I would then like R to spit out the differences in a
file something like the following;
i j Difference
1 2 2
1 3 -4
... .. ...
(n-1) n D(n-1,n)
Without an example, I am going to suggest something:
dfg<-data.frame(index=1:10, Score=sample(1:20,
10),Mean_Temp=10*abs(rnorm(10)), SD_Mean_Temp=abs(rnorm(10)) )
The trick is to get the differences and here is one method:
mapply(FUN="-", dfg$Mean_Temp[combn(dfg$index,2)[1,] ], dfg
$Mean_Temp[combn(dfg$index,2)[2,] ])
Should be pretty obvious how to line up the indices since they are
just the arguments to "[".
The second question is for each indexed sample from i to n, I would
like to use something like rnorm(n, mean = 0, sd = 1) so that I
could generate 1000 random draws from the distribution specified so
that the arguments for rnorm () are as follows; mean = the value of
Mean_Temp & sd = SD_Mean_Temp for indexed values i to n. I would,
of course, then like R to spit out the massive table with 1000
columns of randomly generated temperature values for each index i to
n.
You can get 1000 such values with rnorm(1000, Mean_Temp, SD_Mean_Temp)
Perhaps something along these lines:
Mtx <- matrix( , ncol=1000, nrow=nrow(dfg))
Mtx[1:nrow(dfg), ] <- apply(dfg, 1, function(x) rnorm(1000, x[3],
x[4]) )
Finally I would like to compute the slope estimator associated with
the Theil statistic. All I need to do is compute the n(n-1)/n
^^^^^^^^^^==n, is that what you meant?
Or is it n(n-1)/2?
individual slope values; Sij = (Yj -Yi)/(Xj-Xi) where the Ys are the
Scores for samples i to n and the Xs are the set of randomly
generated temperature values, as before, 1 less than or equal to i
less than j less than or equal to n. The median is then the
estimator of Beta (the slope). I would like to compute the slopes
for all 1000 sets of randomly generated temperature values.
There is a package, mblm, that computes the Theil estimator. I don't
see why we should reinvent that wheel.
(Found via tsearching on theil statistic)
http://search.r-project.org/cgi-bin/namazu.cgi?query=theil+statistic&max=100&result=normal&sort=score&idxname=functions&idxname=Rhelp08&idxname=views
...and following what seemed like promising links
library(mblm)
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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