similar to: Linear Regression with Constraints

Hello! I am getting the following errors when running optim() [I tried optim() with 3 different methods as you can see]: Error in optim(c(0.66, 0.999, 0.064), pe, NULL, method = "L-BFGS-B") : objective function in optim evaluates to length 6 not 1 > out <- optim( c(0.66, 0.999, 0.064), pe, NULL, method = "Nelder-Mead") Error in optim(c(0.66, 0.999, 0.064),

Dear All, Which package/function could i use to solve following linear least square problem? A over determined system of linear equations is given. The nnls-function may would be a possibility BUT: The solving is constrained with a inequality that all unknowns are >= 0 and a equality that the sum of all unknowns is 1 The influence of the equations according to the solving process is

Dear R-users, i try to solve to following linear programm in R 0 * x_1 + 2/3 * x_2 + 1/3 * x_3 + 1/3 * x_4 = 0.3 x_1 + x_2 + x_3 + x_4 = 1 x_1, x_2, x_3, x_4 > 0, x_1, x_2, x_3, x_4 < 1 as you can see i have no objective function here besides that i use the following code. library(lpSolve) f.obj<-c(1,1,1,1) f.con<-matrix(c(0,2/3,1/3,1/3, 1,1,1,1,

I have some doubts about the validity of my procedure to estimeate linear contrasts ina a factorial design. For sake of semplicity, let's imagine a one way ANOVA with three levels. I am interested to test the significance of the difference between the first and third level (called here contrast C1) and between the first and the seconda level (called here contrast C2). I used the following

I would like to perform a regression like the one below: lm(x ~ 0 + a1 + a2 + a3 + b1 + b2 + b3 + c1 + c2 + c3, data=data) However, the data has the property that a1+a2+a3 = A, b1+b2+b3 = B, and c1+c2+c3 = C, where A, B, and C are positive constants. So there are two extra degrees of freedom, and R handles this by producing NA for two of the coefficients. Instead, I would prefer to remove the

Hi there, I am trying to use linear regression to solve the following equation - y <- c(0.2525, 0.3448, 0.2358, 0.3696, 0.2708, 0.1667, 0.2941, 0.2333, 0.1500, 0.3077, 0.3462, 0.1667, 0.2500, 0.3214, 0.1364) x2 <- c(0.368, 0.537, 0.379, 0.472, 0.401, 0.361, 0.644, 0.444, 0.440, 0.676, 0.679, 0.622, 0.450, 0.379, 0.620) x1 <- 1-x2 # equation lmFit <- lm(y ~ x1 + x2) lmFit Call:

Hello! I am trying to write a function with vector and data.frame parameters that uses the sum() function and values from the rows of the data.frame. I need to pass this function as a parameter to optim(). My starting point is: observs <- data.frame(y, x1, x2, x3) Fn <- function(par, observs) { sum( (y - (par[1] * (x1 + 1) * x2^(-par[2]) * x3^par[3])^2 ) }

hi: could someone please point me to a function that allows me to solve general non-linear functions? > irr.in <- function(r, c1, c2, c3 ) { return(c1+c2/(1+r) + c3/(1+r)^2); } > solve.nonlinear( irr.in, -100, 60, 70 ); 0.189 If someone has written an irr function, this would be helpful, too---though not difficult to write, either. thanks for any pointers. Regards, /iaw

After a few days of work, I think I nearly have it. Unfortunately, theta is unchanged after I run this (as a script from a file). I thought that theta would contain the fitted parameters. The goal here is to find the least squares fit according to the function defined as "rss" subject to the constraints defined as ui and ci. I defined ui and ci to (hopefully) force par2 and par3

Hi, I'm Pina and I'm a student in geology. I'm working with spectral profile of sand and I have to find the similarity between one spectral profile selected by hyperspectral image anche one that I created to mix different percentage of 4 mineral component. I have to find the best mix of percentage of this 4 mineral in order to have the best likeness with the spectral profile chose by

Dear All I have a constrained optimisation problem, I want to maximise the following function t(weights) %*% CovarianceMatrix %*% weights for the weights, subject to constraints on each element within the weights & the weights vector summing to 1. i.e. weights = (x1, x2, x3), where x1 is within some given range (a +b, a - b). I have tried to do this using the optim function in R,

These are supported on GK20A and GM107. Signed-off-by: Ilia Mirkin <imirkin at alum.mit.edu> --- Was a bit torn on where to place the enums... we're about to gut all the xml definitions so this seemed appropriate for now. Tested on GK20A only. src/gallium/drivers/nouveau/nv50/nv50_formats.c | 64 +++++++++++++++++++++++++ src/gallium/drivers/nouveau/nvc0/nvc0_screen.c | 10 ++++

Hi all, I have the following problem: I have a data frame (actually it is a prop.table) which I want to print as a list, e.g.: C1 C2 C3 R1 0.0 0.0 1.0 R2 1.0 0.0 0.0 R3 0.0 0.0 0.0 R4 0.0 1.0 0.0 should be printed like C1;R1;0.0 C2;R1;0.0 C3;R1;1.0 C1;R2;1.0 C2;R2;0.0 ..... Is there any existing solution out there or could somebody please give me a hint on how to

Why not fix the new names instead to be like the old names? Seems like that would be way simpler... On Feb 15, 2016 12:38 AM, "Ben Skeggs" <skeggsb at gmail.com> wrote: > From: Ben Skeggs <bskeggs at redhat.com> > > We've previously had identical naming between vertex and texture > formats, so it mostly made sense to define these together. > > However,

From: Ben Skeggs <bskeggs at redhat.com> Signed-off-by: Ben Skeggs <bskeggs at redhat.com> --- src/gallium/drivers/nouveau/nv50/g80_defs.xml.h | 279 ++++++++++++++++++++++++ 1 file changed, 279 insertions(+) create mode 100644 src/gallium/drivers/nouveau/nv50/g80_defs.xml.h diff --git a/src/gallium/drivers/nouveau/nv50/g80_defs.xml.h

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Dear all, I have a data frame that looks like this: c1 c2 c3 A B C B C A A A B and so on; I?d like to produce one single vector consisting of the columns c1,c2, c3, such that vector=("A","B","A","B","C","A","C","A","B") I guess it?s easy to do but I don?t know how...Can anyone

Dear R-experts, recently, I started to discover the world of R. I came across a problem, that I was unable to solve by myself (including searches in R-help, etc.) I have a flat table similar to key1 key2 value1 abcd_1 BP 10 abcd_1 BSMP 1A abcd_1 PD 25 abcd_2 BP 20 abcd_3 BP 80 abcd_4 IA 30 abcd_4 PD 70 abcd_4 PS N I wish to transform this table to obtain the following result: key2 key1 BP

Hi, I have a three dimensional array, e.g., my.array = array(0, dim=c(2,3,4), dimnames=list( d1=c("A1","A2"), d2=c("B1","B2","B3"), d3=c("C1","C2","C3","C4")) ) what I would like to get is then a dataframe: d1 d2 d3 value A1 B1 C1 0 A2 B1 C1 0 . . . A2 B3 C4 0 I'm sure there is one function to do

sapply() stems from S / S+ times and hence has a long tradition. In spite of that I think that it should be enhanced... As the subject mentions, sapply() produces a matrix in cases where the list components of the lapply(.) results are of the same length (and ...). However, it unfortunately "stops there". E.g., if you *nest* two sapply() calls where the inner one produces a matrix, very