similar to: Degradation model

Dear Colleagues, Our group is also working on implementing the use of R for pharmacokinetic compartmental analysis. Perhaps I have missed something, but > fit <- nls(noisy ~ lsoda(xstart, time, one.compartment.model, c(K1=0.5, k2=0.5)), + data=C1.lsoda, + start=list(K1=0.3, k2=0.7), + trace=T + ) Error in eval(as.name(varName), data) : Object

Hi.. i have an expression of the form: model1<-nls(y~beta1*(x1+(k1*x2)+(k1*k1*x3)+(k2*x4)+(k2*k1*x5)+(k2*k2*x6)+(k3*x7)+(k3*k4*x8)+(k3*k2*x9)+(k3*k3*x10)+ (k4*x11)+(k4*k1*x12)+(k4*k2*x13)+(k4*k3*x14)+(k4*k4*x15)+(k5*x16)+(k5*k1*x17)+(k5*k2*x18)+(k5*k3*x19)+

Hi everyone! I'm matching two samples to create one sample that have pairs of observations equal for the k1 variable. Merge() doesn't work because I dont't want to recycle the values. x <- data.frame(k1=c(1,1,2,3,3,5), k2=c(20,21,22,23,24,25)) x y <- data.frame(k1=c(1,1,2,2,3,4,5,5), k2=c(10,11,12,13,14,15,16,17)) y merge(x,y,by="k1") k1 k2.x k2.y 1 1 20

Hi, The code below is giving me this error message: Error in while (err > eps) { : missing value where TRUE/FALSE needed In addition: Warning messages: 1: In dnbinom(x, size, prob, log) : NaNs produced 2: In dnbinom(x, size, prob, log) : NaNs produced I know from the help files that for dnbinom "Invalid size or prob will result in return value NaN, with a warning", but I am not able

Error in model.frame When I run the following nls model an error message appears and I dont know how to solve that. Could you help me?? mat = c(1,2,3,4,5,6,7,8,9,12,16,24,36,48,60) for (i in 1:length(j30)) { bliss = nls(c(j[i,1:length(mat)]) ~ b0 + b1*((1-exp(-k1*mat))/(k1*mat)) + b2*(((1-exp(-k2*mat))/(k2*mat))-exp(-k2*mat)), start = list(k1=0.1993, k2=0.1993, b0= 22.0046,

HI Jim and all, I want to put one more condition. Include col2 and col3 if they are not in col1. Here is the data mydat <- read.table(textConnection("Col1 Col2 col3 K2 X1 NA Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) The desired out put would be Col1 Col2 col3 1 X1 0 0 2 K1 0 0 3 Y1 0 0 4 W1 0 0 6 K2 X1

--- v2v/convert_linux.ml | 18 +++++++++++++++--- 1 file changed, 15 insertions(+), 3 deletions(-) diff --git a/v2v/convert_linux.ml b/v2v/convert_linux.ml index f670812..39a520c 100644 --- a/v2v/convert_linux.ml +++ b/v2v/convert_linux.ml @@ -49,13 +49,14 @@ type kernel_info = { ki_modules : string list; (* The list of module names. *) ki_supports_virtio : bool; (* Kernel has

Jim has been exceedingly patient (and may well continue to be so), but this smells like "failure to launch". At what point will you start showing your (failed) attempts at solving your own problems so we can help you work on your specific weaknesses and become self-sufficient? -- Sent from my phone. Please excuse my brevity. On February 25, 2018 7:55:55 AM PST, Val <valkremk at

Hi, I have a question about how to obtain the union of several data frame rows. I'm trying to create a common key for several tests composed of different items. Here is a small scale version of the problem. These are keys for 4 different tests, not all mutually exclusive: id q1 q2 q3 q4 q5 q6 1 A C 2 B D 3 A D B 4 C D B D I would like

Hi Val, My fault - I assumed that the NA would be first in the result produced by "unique": mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) val23<-unique(unlist(mydat[,c("Col2","col3")])) napos<-which(is.na(val23)) preval<-data.frame(Col1=val23[-napos],

Ladies and Gentlemen, In MATLAB, I can write: for J=1:M Y(J+1)=Y(J)+ h * feval(f,T(J),Y(J)); ... In R, I can write above as: for (J in 2:M) { y = y + h * f(t,y) ... } ---- In MATLAB, I can write: for J=1:M k1 = feval(f,T(J),Y(J)); k2 = feval(f,T(J+1),Y(J)+ h * k1 How do I write k2 in R? k1 = f(t,y) k2 = ? Thanks, /oal

Meta: This question is somewhat long and has two parts, I would be very happy for someone just to nudge me in the right direction with the manual / tutorial, as I am somewhat lost... 1) How do I fit a curve of the form "y = k1 * x^k2" ? I want to estimate values of k1 and k2 given the x/y data I have, and I can't work out how to get R to calculate and return their estimates. 2)

Dear R-List, I would like to recode categorial variables into binary data, so that all values above median are coded 1 and all values below 0, separating each var into two equally large groups (e.g. good performers = 0 vs. bad performers =1). I have not succeeded so far in finding a nice solution to do that in R. I thought there might be a better way than ordering each column and recoding the

Dear R-List, I would like to recode categorial variables into binary data, so that all values above median are coded 1 and all values below 0, separating each var into two equally large groups (e.g. good performers = 0 vs. bad performers =1). I have not succeeded so far in finding a nice solution to do that in R. I thought there might be a better way than ordering each column and recoding the

Hi, I'm new to R and have a problem with a little test program (see below). Why doesn't <<- in function rk4 assign the new value to y so that it is seen in rktest. I thought that <<- does exactly this. But it seems that I didn't get it right. I would be very appreciative for an explanation of that behaviour of <<-. I know how to write the whole thing so that it

Thank you Jim, I read the data as you suggested but I could not find K1 in col1. rbind(preval,mydat) Col1 Col2 col3 1 <NA> <NA> <NA> 2 X1 <NA> <NA> 3 Y1 <NA> <NA> 4 K2 <NA> <NA> 5 W1 <NA> <NA> 6 Z1 K1 K2 7 Z2 <NA> <NA> 8 Z3 X1 <NA> 9 Z4 Y1 W1 On Sat, Feb 24, 2018 at 6:18 PM, Jim

Hi all, I try to assess the parameters (K1,K2) of a model that describes the adsorption of a molecule onto on adsorbent. equation: dq/dt = K1*C*(qm-q)-K2*q I know the value of 'qm' and I experimentally measure the variables 'q', 'C', and the time 't'. t C q 1 0 144.05047 0.0000000 2 565 99.71492 0.1105625 3 988 74.99426

Hi, I'm doing a function that describe two populations in competition. that's the function that i wrote: exclusao<-function(n10, n20, k1, k2, alfa, beta, t){ n1<-k1-(alfa*n20) n2<-k2-(beta*n10) if(t==0){plot(t, n10, type='b', xlim=range(c(1:t),c (1:t)), ylim=range(n10, n20), xlab='tempo', ylab='tamanho populacional') points(t, n20, type='b',

Hello, I am trying to find the analytical solution to this differential equation dR/dt = k1*(R^k2)*(1-(R/Rmax)); R(0) = Ro k1 and k2 are parameters that need to fitted, while Ro and Rmax are the baseline and max value (which can be fitted or fixed). The response (R) increases initially at an exponential rate governed by the rate constants k1 and k2. Response has a S-shaped curve as a function

Dear list, If I know the standard error for k1 and k2, is there anything I can call in R to calculate the standard error of k1/k2? Thanks. Jun [[alternative HTML version deleted]]