Displaying 20 results from an estimated 11000 matches similar to: "Least Squares Restricted Estimator"
2008 Jul 25
1
Matrix from List
Hello, I have a list in which each element is a list. I want to
create a matrix indexed by the two indices of the list. I have been
using do.call, but I am not getting what I want. Let me show you:
> l.intercepts #the list that nests another list
$`1995`
$`1995`$`31`
(Intercept)
25.37164
$`1995`$`33`
(Intercept)
26.66755
$`2006`
$`2006`$`31`
(Intercept)
25.86621
$`2006`$`33`
2004 Oct 05
2
Nelson-Aalen estimator in R
Hi,
I am taking a survival class. Recently I need to do the Nelson-Aalen
estimtor in R. I searched through the R help manual and internet, but could
not find such a R function. I tried another way by calculating the
Kaplan-Meier estimator and take -log(S). However, the function only
provides the summary of KM estimator but no estimated values. Could you
please help me with this? I would
2008 Sep 04
2
Adding a legend to R graph device with several plots (no to individual plots!)
Dear Users, I already posted this question: it either went unnoticed,
or it is to basic (if this is so, please sent me a hint).
I would like to know if there is a way to add a common
legend to an arrangement of plots. In the example below, I get four
plots in my device. each one has a density for 1995 and one for 2006.
I have found that using legend or smartlegend I can add a legend to
each plot,
2004 Nov 08
2
Nonlinear weighted least squares estimation
Hi there,
I'm trying to fit a growth curve to some data and need to use a weighted least squares estimator to account for heteroscedasticity in the data. A weights argument is available in nls that would appear to be appropriate for this purpose, but it is listed as 'not yet implemented'. Is there another package which could implement this procedure?
Regards,
Robert Brown
2009 May 04
1
Nelson-Aalen estimator of cumulative hazard
Hi,
I am computing the Nelson-Aalen (NA) estimate of baseline cumulative hazard in two different ways using the "survival" package. I am expecting that they should be identical. However, they are not. Their difference is a monotonically increasing with time. This difference is probably not large to make any impact in the application, but is annoyingly non-trivial for me to just
2018 May 22
2
Nelson-Aalen Estimator in R: Error Message
Dear all,
Currently, I am doing a research project about serum sodium levels and falling. I am doing my analysis in R. I am performing the multiple imputation right now. I want to perform a survival analysis later, but therefore I need to specify the Nelson-Aalen estimator. My dataset is called DF1, the event indicator is Falls and the time variable is Time. The code that I use is as follows:
2006 Mar 07
1
breslow estimator for cumulative hazard function
Dear R-users,
I am checking the proportional hazard assumption of a cox model for a
given covariate, let say Z1, after adjusting for other relavent covariates
in the model. To this end, I fitted cox model stratified on the discrete
values of Z1 and try to get beslow estimator for the baseline cumulative
hazard function (H(t)) in each stratum. As far as i know, if the
proportionality assumption
2005 Aug 23
1
Robust M-Estimator Comparison
Hello,
I'm learning about robust M-estimators right now and had settled on the
"Huber Proposal 2" as implemented in MASS, but further reading made clear,
that at least 2 further weighting functions (Hampel, Tukey bisquare) exist.
In a post from B.D. Ripley going back to 1999 I found the following quote:
>> 2) Would huber() give me results that are similar (i.e., close
2002 Apr 09
2
Restricted Least Squares
Hi,
I need help regarding estimating a linear model where restrictions are imposed on the coefficients. An example is as follows:
Y_{t+2}=a1Y_{t+1} + a2 Y_t + b x_t + e_t
restriction
a1+ a2 =1
Is there a function or a package that can estimate the coefficient of a model like this? I want to estimate the coefficients rather than test them.
Thank you for your help
Ahmad Abu Hammour
--------------
2018 May 22
0
Nelson-Aalen Estimator in R: Error Message
Hard to tell from the info you are giving us. I assume this regards the "mice" package?
One way to proceed is to set options(error=recover) which will dump you into the browser() environment when the error occurs and you can oka around and see what the value of variables was at the point of the error. This could give you a clue about what is going on.
-pd
> On 22 May 2018, at 15:29
2009 Jan 22
2
Standard errors of least squares adjusted means
Hello,
I have the following model:
lm.7 <- lm(Y ~ F + C1 + C2 , data = EM4)
F is a 4-level factor, the rest are covariates centered at their mean (Y
is a two-column matrix).
I have tried to find functions to give the model-adjusted means
(adjusted at the covariates'means) and their standard deviations for each.
(That is, what I believe is called in SAS "least square or LS-means,
2007 Feb 28
3
Packages in R for least median squares regression and computing outliers (thompson tau technique etc.)
Hi
I am looking for suitable packages in R that do
regression analyses using least median squares method
(or better). Additionally, I am also looking for
packages that implement algorithms/methods for
detecting outliers that can be discarded before doing
the regression analyses.
Although some websites refer to "lms" method under
package "lps" in R, I am unable to find such a
2003 Sep 26
1
least squares regression using (inequality) restrictions
Dear R Users,
I would like to make a lesast squares regression similar to that what is
done by the command "lm". But additionally, I would like to impose some
restrictions:
1) The sum of all regression coefficients should be equal to 1.
2) Each coefficient should assume a value between 0 and 1. (inequality
restrictions)
Which command is the best to use in order to solve this problem
2007 May 28
1
monthly least squares estimation
Hi R-programmers !
I would like to perform a linear model regression month by month using the
'lm' function and i don't know how to do it.
The data is organised as below:
Month ExcessReturn Return STO
8 0.047595875 0.05274292 0.854352503
8 0.016134874 0.049226941 4.399372005
8 -0.000443869 0.004357305 -1.04980297
9 0.002206554 -0.089068828 0.544809429
9 0.021296551
2009 May 30
0
improve efficiency of a loop
Dear All:
I need advice about efficient looping/vectorization. I am trying to
bootstrap a regression model with one lag of the dependent variable in
the RHS. Specifically, let error^b_(t) be the bootstrapped error of
the regression y_(t) = gamma y_(t-1) + beta x +error_(t) at time (t),
y_(t) is the original dependent variable, and y^b_(t) the bootstraped
y_(t) using parameter estimates gamma and
2009 Jul 01
1
Iteratively Reweighted Least Squares of nonlinear regression
Dear all,
When doing nonlinear regression, we normally use nls if e are iid normal.
i learned that if the form of the variance of e is not completely known,
we can use the IRWLS (Iteratively Reweighted Least Squares )
algorithm:
for example, var e*i =*g0+g1*x*1
1. Start with *w**i = *1
2. Use least squares to estimate b.
3. Use the residuals to estimate g, perhaps by regressing e^2 on
2009 Mar 19
0
Restrained least squares fitting
Hi All,
I've found a few references in the mailing list and documentation to
constrained least squares fitting, but little on restrained least squares.
To clarify what I mean, a constraint might limit a parameter to a particular
value (e.g. x=5.0, or exactly within the bounds 4.9 - 5.1), whereas a
restraint adds some further information to the problem about the certainty
of the starting point
2008 Mar 27
1
Significance of confidence intervals in the Non-Linear Least Squares Program.
I am using the non-linear least squares routine in "R" -- nls. I have a
dataset where the nls routine outputs tight confidence intervals on the
2 parameters I am solving for.
As a check on my results, I used the Python SciPy leastsq module on the
same data set and it yields the same answer as "R" for the
coefficients. However, what was somewhat surprising was the the
2011 Apr 26
1
Least Squares Method
Hi everyone,
I am running the 'gls' command (least squares method) for a number of data
out of which many are zeros. I strongly believe that the output is wrong and
I think that this is due to the large number of zero values included in my
dataset.
I would like to ask if there is a command that would allow me to run the gls
command disregarding all the zero values?
Thank you in
2012 Feb 04
1
least squares solution to linear system
Dear all
I am having a linear system of the form
A*X=B and I want to find the X by using least squares.
For example my A is of dimension [205,3] and my B is of dimension[205,1]
I am looking for the X matrix which is of the size [3,1]. In the matlab I was doing that by the function
X = LSCOV(A,B) returns the ordinary least squares solution to the
linear system of equations A*X = B, i.e., X