ah yes, my mistake was dividing by sigma instead of multiplying... thank you
very much
sincerely,
Rnewb
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factor.
the factor is the same for all coefficients, but it is not 1, and the value
varies for different data sets. what is this term?
thanks,
Rnewb
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2 -4 -5 5
> data['z']=vec
Error in `[<-.data.frame`(`*tmp*`, "z", value = c(2, -4, -5, 5)) :
replacement has 4 rows, data has 10
>
thanks,
Rnewb
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> vec
1 5 8 9
2 -4 -5 5
> data['z']=vec
Error in `[<-.data.frame`(`*tmp*`, "z", value = c(2, -4, -5, 5)) :
replacement has 4 rows, data has 10
>
thanks,
Rnewb
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thanks for the quick and accurate responses!
cheers,
Rnewb
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R
bulk of the work for linear regressions is inverting a matrix that depends
only on the independent variables, it seems like a waste to do it over and
over for each new dependent variable.
thanks,
Rnewb
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accomplish this by writing code to suitably shift the coefficients
after performing the basic regression above, but I'm hoping there's a better
way. Is there?
thanks,
Rnewb
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thanks very much for the quick and accurate response!
Rnewb
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R-help@r
:
data[1:3]
What if i don't know that the columns that start with 'a' are columns 1-3?
Is there any command that will pick out the desired columns automatically?
What if i want to match a generic regular expression instead of just looking
at the first letter?
thanks very much,
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