Dear all,
I have the following data in excel:
day1 y 1 2 3 2 3 x1 0.2 0.3 0.4 0.3 0.2 x2 7 3.4 2 8 6 day2 y 2 4 3
2 2 x1 0.4 0.5 0.3 0.3 0.2 x2 7 8 9.1 6 5
I have the following problems:
first of all, when I ask R to read the file (with the package xlsReadWrite
and the command read.xls) it has a problem with the fact that the left most
corner is labelled the same way, so in order for it to work I need to do:
day1
y1
1
2
3
2
3
x11
0.2
0.3
0.4
0.3
0.2
x21
7
3.4
2
8
6
day2
y2
2
4
3
2
2
x12
0.4
0.5
0.3
0.3
0.2
x22
7
8
9.1
6
5
Next I need to do the regression y~ x1+x2, for the different days.
I have converted the above data in the form of a matrix and labelled it mat.
The following loop works:
results<-c(0,0,0,0,0)
for (i in c(1,5)){
results[i]<-lm(mat[i,]~mat[i+1,]+mat[i+2,])}
results
And the results are:
Warning messages:
1: In results[i] <- lm(mat[i, ] ~ mat[i + 1, ] + mat[i + 2, ]) :
number of items to replace is not a multiple of replacement length
2: In results[i] <- lm(mat[i, ] ~ mat[i + 1, ] + mat[i + 2, ]) :
number of items to replace is not a multiple of replacement length
> results
[[1]]
(Intercept) mat[i + 1, ] mat[i + 2, ]
2.4918415 1.5151515 -0.1356220
[[2]]
[1] 0
[[3]]
[1] 0
[[4]]
[1] 0
[[5]]
(Intercept) mat[i + 1, ] mat[i + 2, ]
-0.3741935 3.4193548 0.2580645
The method does work but it is not ideal. The problem is that the real data
that I will be using will involve regressions of 20 or so variables over
3000 days. I also want to avoid finding the exact values that i should take
inside the loop.
Is there a better way of carrying out these regressions so that I do not
need to relabel everything, the results are faster and it is clear and easy
to isolate the results of the linear models for each day?
Thank you very much everyone in advance,
Regards,
Denise
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