Displaying 5 results from an estimated 5 matches for "rnewb".
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2009 Oct 28
5
re gression with multiple dependent variables?
...1+x2+x3+x4+x5, data=data)
is it possible to run all these regs with a single command?  given that the
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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2010 Mar 11
1
how does R compute Std. Error's?
...) / (resid err)
where .[i,i] means the ith entry on the diagonal of the given matrix. 
however, doing this gives values that are off by a multiplicative 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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2009 Oct 06
1
linear model with coefficient constraints
...d by the model. 
Specifically, I want
coeff(b1) + coeff(b2) + coeff(b3) = coeff(c1) + coeff(c2) + coeff(c3) = 0.
I could 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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2009 Nov 30
1
updating subset of data.frame
...3   15  40 0
4    0  17 0
5    1  12 0
6   17  45 0
7    4  19 0
8  -13 -16 0
9   -7   5 0
10  -5  22 0
> 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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2009 Oct 03
3
getting variables based on name
...type:
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,
Rnewb
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