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Hi R users, I am a new user, still learning basics of R. Is there anyway to extract y (or x) value for a known x (or y) value from ecdf (empirical cumulative distribution function) curve? Thanks in advance. Mano. [[alternative HTML version deleted]]

...pen mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Sun, Jun 18, 2017 at 9:43 AM, David Winsemius <dwinsemius at comcast.net> wrote: > >> On Jun 18, 2017, at 6:24 AM, Manoranjan Muthusamy <ranjanmano167 at gmail.com> wrote: >> >> I am using nlsLM {minpack.lm} to find the values of parameters a and b of >> function myfun which give the best fit for the data set, mydata. >> >> mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) >> &...

..., but not necessarily all the issues. 2) I used nlxb that requires an expression for the model. nlsLM can use a function because it is using derivative approximations, while nlxb actually gets a symbolic or automatic derivative if it can, else squawks. JN # Here's the script # # # Manoranjan Muthusamy <ranjanmano167 at gmail.com> # library(minpack.lm) mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) myfun=function(a,b,r,t){ prd=a*b*(1-exp(-b*r*t)) return(prd)} # and using nlsLM myfit=nlsLM(y~myfun(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05),...

...used nlxb that requires an expression for the model. nlsLM > can use a function because it is using derivative approximations, > while nlxb actually gets a symbolic or automatic derivative if > it can, else squawks. > > JN > > # Here's the script # > # > # Manoranjan Muthusamy <ranjanmano167 at gmail.com> > # > > library(minpack.lm) > mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) > > myfun=function(a,b,r,t){ > prd=a*b*(1-exp(-b*r*t)) > return(prd)} > > # and using nlsLM > > myfit=nlsLM(y~myfun(a,b,r=2,t=x),data...

> On Jun 18, 2017, at 6:24 AM, Manoranjan Muthusamy <ranjanmano167 at gmail.com> wrote: > > I am using nlsLM {minpack.lm} to find the values of parameters a and b of > function myfun which give the best fit for the data set, mydata. > > mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) > > myfun=function(a,...

I am using nlsLM {minpack.lm} to find the values of parameters a and b of function myfun which give the best fit for the data set, mydata. mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) myfun=function(a,b,r,t){ prd=a*b*(1-exp(-b*r*t)) return(prd)} and using nlsLM myfit=nlsLM(y~myfun(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05), lower = c(1000,0),

...n for the model. nlsLM >> can use a function because it is using derivative approximations, >> while nlxb actually gets a symbolic or automatic derivative if >> it can, else squawks. >> >> JN >> >> # Here's the script # >> # >> # Manoranjan Muthusamy <ranjanmano167 at gmail.com> >> # >> >> library(minpack.lm) >> mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) >> >> myfun=function(a,b,r,t){ >> prd=a*b*(1-exp(-b*r*t)) >> return(prd)} >> >> # and using nlsLM >&gt...

I need help in typing step by step commands in R software for panel data analysis saibaba

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