similar to: curve fitting

Dear developers of R-project, I have such data height of male persons 181, 178, 182, 160, 187, 193, 184, 184, 175, 178, 184, 184, 174, 185, 175 height of female persons 164, 165, 160, 174, 167, 161, 164, 165, 169, 175, 165, 155, 172, 164, 172, 166, 160, 159, 158, 173 I want to determine if height is dependent on gender, and I need to find, as I understand, point biserial correlation. I had tried

I have the data arranged in table (in Excel, or Notepad): x y 0 100 2 100 4 80 6 40 8 0 I need to transfer these data in R file and then going to operate with x and y as with variables. I should note, that I have 2.8.1 version. Thank you for advance. With regard, Dmitry.

Hi! I have a problem of curve fitting. I use the following data : - vector of predictor data : 0 0.4 0.8 1.2 1.6 - vector of response data : 0.81954 0.64592 0.51247 0.42831 0.35371 I perform parametric fits using custom equations when I use this equation : y = yo + K *(1/(1+exp(-(a+b*ln(x))))) the fitting result is OK but when I use this more general equation : y = yo + K

Hi, It seems to me that the most suitable method in R for curve-fitting is the use of nls, which uses a Gauss-Newton (GN) algorithm, while the use of the Levenberg-Marquardt (LM) algorithm does not seem to be very stressed in R. According to this [1] by Ripley, 'Levenberg-Marquardt is hardly competitive these days' which could imply the low emphasize on LM in R. The position of LM is, to

Hi list, I have a little curve fitting problem. I would like to fit a sigmoid curve to my data using the following equation: f(x) = 1/(1 + exp(-(x-c)*b)) (or any other form for that matter) Where x is the distance/location within the dataframe, c is the shift of the curve across the dataframe and b is the steepness of the curve. I've been playing with glm() and glm.fit() but without

hello, I want to fit a curve to a simple x,y dataset - my problem is, that I want to fit it for the following term: n(1-e^x/y) - so I get the n constant for my data... can anyone help/comment on that? cheers, gregor

Dear R users, I am struggling to fit expo-linear equation to my data using "nls" function. I am always getting error message as i highlighted below in yellow color:  Theexpo-linear equation which i am interested to fit my data:       response_variable =  (c/r)*log(1+exp(r*(Day-tt))), where "Day" is time-variable my response variable rl <-

Dear R users,   I am struggling to fit expo-linear equation to my data using "nls" function. I am always getting error message as i highlighted below in yellow color:     ### Theexpo-linear equation which i am interested to fit my data:       response_variable =  (c/r)*log(1+exp(r*(Day-tt))), where "Day" is time-variable   ## my response variable   rl <-

I have the following data set, and I have to find the line of best fit using this equation, y = a*(1 - exp(-b*x)). samples = seq(1,20,by=1) species = c(5,8,9,9,11,11,12,15,17,19,20,20,21,23,23,25,25,27,27,27) plot(samples,species, main = "Accumulation Curve for Tree Species Richness", xlab = "Samples", ylab = "Number of Species") curve((y = 27*(1 -

Dear list, I'm trying to fit a chapman-richards equation to my data, only I cannot interpret the parameters a, b and d. I know that the parameter b denotes the asymptote, but for the others I couldn't figure out. But I do need to know this in order to set my starting values. Here's the model: modPoplar<- nls(Diameter ~ d*(1-exp(-b *Age))^a ,start=list(a=20,b=0.9,d=33)) I attached

Hi I'd like to fit an asymmetrical curve function to some physiological data. I've been told a weibull curve is a good place to start, but I'm having trouble specifying and fitting the function with nls and was wondering if someone could help. After some reading, I think the function specification I want is y=c*(x/a)^(b-1)*e^(-(x/a)^b) (from

I have a model Q=K*A*(R^r)*(S^s) A, R, and S are data I have and K is a curve fitting parameter. I have linearized as log(Q)=log(K)+log(A)+r*log(R)+s*log(S) I have taken the log of the data that I have and this is the model formula without the K part lm(Q~offset(A)+R+S, data=x) What is the formula that I should use? Thanks for all of your help. I can provide a subset of data if necessary.

Hello all! I have a problem with a double exponential equation. this are my data's> structure(list(proc = c(1870.52067384719, 766.789388745793, 358.701545859122, 237.113777545511, 43.2726259059654, 148.985133316262, 92.6242882655781, 88.4521557193262, 56.6404686159112, 27.0374477259404, 34.3347291080268, 18.3226992991316, 15.2196612445747, 5.31600719692165, 16.7015717397302,

Hi all, Like a lot of people I noticed that I get different results when I use nls in R compared to the exponential fit in excel. A bit annoying because often the R^2 is higher in excel but when I'm reading the different topics on this forum I kind of understand that using R is better than excel? (I don't really understand how the difference occurs, but I understand that there is a

Subject: nls, step factor 0.000488281 reduced below 'minFactor' of 0.000976563 Hi there, I'm trying to conduct nls regression using roughly the below code: nls1 <- nls(y ~ a*(1-exp(-b*x^c)), start=list(a=a1,b=b1,c=c1)) I checked my start values by plotting the relationship etc. but I kept getting an error message saying maximum iterations exceeded. I have tried changing these

Hello, I have what should be an easy question. I'm a new r user and making the transition from menus to the command line so as to do batch processing of tons of data. One of my data streams needs to be detrended. It's a vector of numbers that follows a negative exponential decay. I need to fit a curve to it and use the residuals as an object. The data looks something like this: foo.dat

I am trying to fit a curve to a cumulative mortality curve (logistic) where y is the cumulative proportion of mortalities, and t is the time in hours (see below). Asym. at 0 and 1 > y [1] 0.00000000 0.04853859 0.08303777 0.15201970 0.40995074 0.46444992 0.62862069 0.95885057 1.00000000 [10] 1.00000000 1.00000000 > t [1] 0 13 20 24 37 42 48 61 72 86 90 I tried to find starting values for

I'm trying to fit a function y=k*l^(m*x) to some data points, with reasonable starting value estimates (I think). I keep getting "singular matrix 'a' in solve". This is the code: ox <- c(-600,-300,-200,1,100,200) ir <- c(1,2.5,4,9,14,20) model <- nls(ir ~ k*l^(m*ox),start=list(k=10,l=3,m=0.004),algorithm="plinear") summary(model) plot(ox,ir) testox <-

I am having troubles in fitting functions of the form y~a*x^b+c to data, for example x<-c(0.1,0.36,0.63,0.90,1.166,1.43, 1.70, 1.96, 2.23) y<-c(8.09,9.0,9.62,10.11,10.53,10.9, 11.25, 11.56, 11.86) I tried for example with nls, which did only work with really good initial guessed values. Any suggestion, what I should use? Thanks a lot Thomas [[alternative HTML version deleted]]

Thanks a lot! This was amazing. I'm not sure I see how the conditiion pK1 < pK2 < pK3 is enforced? - it comes from the derivation via generalized Henderson-Hasselbalch but perhaps it is not really necessary. Anyway, the use of Vectorize did the trick! Best wishes Troels Den 06-11-2023 kl. 19:19 skrev Ivan Krylov: > ? Mon, 6 Nov 2023 17:53:49 +0100 > Troels Ring <tring at