some help

Hello,

Why answer to just me? You should keep it in the R-Help list, the odds 
of getting more (and hopefully better) answers is bigger.

Anyway, with data examples it's much easier to do something. I hope the 
code is self explanatory, except maybe for the use of scan(), not 
read.table. scan() reads in vectors, so there's no need to stack the 
columns anymore, they're just one big vector. Another thing is that 
around half of the values are zeros, and those don't show up in the 
graph you've posted, so I've filtered them out.

fdir <- "Asciis"  # change if needed
curr_dir <- setwd(fdir)  # save current directory and set working dir
fls <- list.files()
lst <- scan(fls[1], dec=",")
ndv <- scan(fls[2], dec=",")

str(lst)
str(ndv)

i0 <- ndv > 0 & lst > 0
ndv0 <- ndv[i0]
lst0 <- lst[i0]

brks <- seq(0, 1, by = 0.01)
group <- cut(ndv0, breaks = brks)

mins <- tapply(lst0, group, FUN = min) # one min per group
maxs <- tapply(lst0, group, FUN = max) # one max per group

fit.min <- lm(mins~brks[-1L])
fit.max <- lm(maxs~brks[-1L])

dev.new()
plot(ndv0, lst0, pch=".")
abline(fit.min, col = "blue")
abline(fit.max, col = "red")

#----  Now with ave(). It returns one min/max per element in lst0,
#----  so you'll have a vector as long as the original lst0, with the 
min/max
#----  corresponding values.

mins2 <- ave(lst0, group, FUN = min)  # one min per element of lst0
maxs2 <- ave(lst0, group, FUN = max)  # one max per element of lst0

fit.min2 <- lm(mins2~ndv0)
fit.max2 <- lm(maxs2~ndv0)

dev.new()
plot(ndv0, lst0, pch=".")
abline(fit.min2, col = "blue")
abline(fit.max2, col = "red")

Hope this helps,

Rui Barradas
Em 04-11-2012 10:36, Stefan M?hlbauer escreveu: