[R] Four parameter logistics.

[Victor F Seabra]

try the code below,
if you use different upper and lower limits you might get different parameter estimates
HTH, Victor

standard.curve<-data.frame(conc=c(50, 25, 12.5, 6.25, 3.125, 1.563, 0.781,50, 25, 12.5, 6.25, 3.125, 1.563, 0.781),absorb=c(1.918, 1.251, 1.104, 0.719, 0.403, 0.177, 0.083,1.918, 1.251, 1.104, 0.719, 0.403, 0.177, 0.083))
test<-nls(	absorb ~ ((a-d)/(1+(conc/c.)^b))+d, 
	data = standard.curve, 
	start = list( a=0, b=1, c.=30, d=4), 
	trace = TRUE,
	algorithm="port",
	lower = list( a=-0.45, b=-30, c.=0, d=0), 
	upper = list( a=20, b=30, c.=2000, d=20), 
	control=nls.control(maxiter = 5000, tol = 1e-05, minFactor = 1/10000))

x<-function(a, b, c., d, absorb){c.*((a-absorb)/(absorb-d))^(1/b)}
conc.fitted<-x(a=coef(test)[1],b=coef(test)[2],c.=coef(test)[3],d=coef(test)[4],seq(0,5,by=0.1))

xlim<-c(0,max(standard.curve$conc))
ylim<-c(0,max(standard.curve$absorb))
plot(standard.curve,xlim=xlim,ylim=ylim)
par(new=T)
plot(x=conc.fitted, y=seq(0,5,by=0.1), type="l", col="red",xlim=xlim,ylim=ylim, ylab="", xlab="")
text(20,0.5,expression(paste("abs = ",frac("a - d",1+(conc/c)^b))+d))
text(25,1,pos=4,paste(" a=",coef(test)[1],"\n b=",coef(test)[2],"\n c=",coef(test)[3],"\n d=",coef(test)[4]))

# calculate concentration for a measured absorbance of 1.918
measured.absorbance<-1.918
x(a=coef(test)[1],b=coef(test)[2],c.=coef(test)[3],d=coef(test)[4],measured.absorbance)



____________________________________________

Em 08/02/2011 00:53, Ramya < ramya.victory at gmail.com > escreveu:

Hi,

I decided to use the drm function for four point fitting and in the all the
examples that i see 

spinach.m1 <- drm(SLOPE~DOSE, CURVE, data = spinach, fct = LL.4())

 fit1 <- drm(wheeze~p(age>9)+smoking+cluster(id),data=wheeze,dep="B",
print=0)

what are these with '~' mean?


This is my data

 conc  il10
1 50.000 1.918
2 25.000 1.251
3 12.500 1.104
4  6.250 0.719
5  3.125 0.403
6  1.563 0.177
7  0.781 0.083

I am trying to find the co-efficients for it. 

The co-eff that i am expecting is A = -0.457, B = 0.412, C = 915, D = 9.56


Any suggestions would be appreciated.



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