Re: [R] ecdf() to nls() - how to transform data?
Ok, I think I had a good idea to solve my problem and need someone who second me on that or tell me what a fool I am :-) . My problem: I went for curve-fitting with nls() and took knot()-ecdf() for collecting data for nls-basis-dataframe. I came into trouble, because my x-y-data of the data.frame() was not "equal". I computed 1000 values and there were always some ties, those ties were not returned by knot()-function. Okay, I thought about it and now I am using 100 percentile-values for the data-frame - nls() works fine - moreover 100 percentiles are always there, exceptions won't happen. Is it wise to use 100 cumulative percentile for curve-fitting-base-data instead of ecdf()-function data? Thanks in advance. -- View this message in context: http://r.789695.n4.nabble.com/ecdf-to-nls-how-to-transform-data-tp3671754p3676932.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ecdf() to nls() - how to transform data?
On Jul 17, 2011, at 8:13 PM, Jochen1980 wrote: Hi David, my first attempt to work through your code was successful, my predicted line is pretty close to the ecdf-function. I have no idea why you inverted the gumbel-function and what the advantage of this strategy is? I was advised by a person who knows way more statistics than I do that the strategy of minimizing the error around a regular step function was statistically suspect (presumably even if the x value were a random process). So I inverted the solution and put the random values on the y-axis, leaving the stair-step values on the x-axis.. I think claiming an "advantage" might be premature (especially since both methods yield very similar values) and would probably require some testing before acceptance. There is of course the fitdistr function in MASS and there is a nice vignette on "Fitting Distribution in R" some place around (yep, first google hit): http://cran.r-project.org/doc/contrib/Ricci-distributions-en.pdf I interpret your (1:100/100)-trick like this: you build a sequence of 100 Tics. In addition to this 100 is identical to the number of created points. The ecdf function object (at least as I understood from its plotted result) sets up its y-values (in the function's environment) to be a regular sequence determined by the number of original points and allows you to recover the x-values with the knots function. -- David In my real world situation, ecdf() will produce ties, same values, and this results that I do not know what number knot() will return, then this will result in a non-symmetric-data.frame()-error. R-Code: ### Tutorial 3, Gamma-Distribution-Points to Gumbel # --- create Points --- # http://r.789695.n4.nabble.com/ecdf-to-nls-how-to-transform-data-td3671754.html print( "Tutorial 3" ) p <- rgamma( 100, 2) print( p ) ep <- ecdf( p ) x <- knots(ep) hist( p, main="Tutorial 3" ) ypre <- seq( from=1, to=100, by=1 ) y <- ypre / 100 df <- data.frame( x=x, y=y ) plot( df ) res <- nls( y ~ exp( - exp( - ( x - mue ) / beta )), start=list(mue=2, beta=2), data=df, trace=TRUE ) print( summary(res) ) #print( str(res)) lines( x, predict( res, list( x=x )), lty = 1, col = "blue" ) # predicted vals Console-Output: [1] "Tutorial 3" [1] 4.75748951 5.36642043 2.03025702 1.80216440 1.40745277 0.91393251 [7] 1.30575505 2.07451301 1.18500815 0.24972503 1.36604865 2.36786796 [13] 1.37386798 1.72390843 1.93700139 0.78722468 2.92828385 1.51165415 [19] 4.17000960 2.98356424 1.20195996 1.85899533 0.94863757 1.50438053 [25] 0.50854274 3.32760075 2.21334316 0.58463817 2.26985676 2.88033389 [31] 2.00718903 1.19043470 1.62945752 0.32010478 0.54837755 1.32965131 [37] 3.32024573 0.53825226 3.30077557 0.46426513 1.00681720 0.59276030 [43] 1.31431609 2.27822419 2.56404713 0.52459218 1.05228996 2.23799673 [49] 1.14438962 2.41311612 1.40254244 1.20379073 0.44195457 0.95880408 [55] 1.45254027 4.49645001 2.18826796 1.07161515 1.62617544 3.15506003 [61] 2.15611491 2.43261017 1.40293461 2.79977886 3.44503201 3.25282551 [67] 0.91570531 0.70243512 5.67971774 3.55532192 1.71515046 2.97718949 [73] 0.47145131 0.32879089 0.60735231 0.92388638 4.29277857 1.73681839 [79] 0.09387383 2.81281295 1.84419058 2.63070353 1.52298124 2.89761263 [85] 1.05251987 1.24258703 3.09130229 2.66738574 3.17035060 2.15287327 [91] 2.63042751 1.69736779 3.21126033 1.56920999 2.68985477 0.82952068 [97] 3.62230865 1.65286592 2.76798783 2.08091935 Formula: y ~ exp(-exp(-(x - mue)/beta)) Parameters: Estimate Std. Error t value Pr(>|t|) mue 1.382744 0.005926 233.3 <2e-16 *** beta 0.984836 0.008816 111.7 <2e-16 *** -- View this message in context: http://r.789695.n4.nabble.com/ecdf-to-nls-how-to-transform-data-tp3671754p3674227.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ecdf() to nls() - how to transform data?
Hi David, my first attempt to work through your code was successful, my predicted line is pretty close to the ecdf-function. I have no idea why you inverted the gumbel-function and what the advantage of this strategy is? I interpret your (1:100/100)-trick like this: you build a sequence of 100 Tics. In addition to this 100 is identical to the number of created points. In my real world situation, ecdf() will produce ties, same values, and this results that I do not know what number knot() will return, then this will result in a non-symmetric-data.frame()-error. R-Code: ### Tutorial 3, Gamma-Distribution-Points to Gumbel # --- create Points --- # http://r.789695.n4.nabble.com/ecdf-to-nls-how-to-transform-data-td3671754.html print( "Tutorial 3" ) p <- rgamma( 100, 2) print( p ) ep <- ecdf( p ) x <- knots(ep) hist( p, main="Tutorial 3" ) ypre <- seq( from=1, to=100, by=1 ) y <- ypre / 100 df <- data.frame( x=x, y=y ) plot( df ) res <- nls( y ~ exp( - exp( - ( x - mue ) / beta )), start=list(mue=2, beta=2), data=df, trace=TRUE ) print( summary(res) ) #print( str(res)) lines( x, predict( res, list( x=x )), lty = 1, col = "blue" ) # predicted vals Console-Output: [1] "Tutorial 3" [1] 4.75748951 5.36642043 2.03025702 1.80216440 1.40745277 0.91393251 [7] 1.30575505 2.07451301 1.18500815 0.24972503 1.36604865 2.36786796 [13] 1.37386798 1.72390843 1.93700139 0.78722468 2.92828385 1.51165415 [19] 4.17000960 2.98356424 1.20195996 1.85899533 0.94863757 1.50438053 [25] 0.50854274 3.32760075 2.21334316 0.58463817 2.26985676 2.88033389 [31] 2.00718903 1.19043470 1.62945752 0.32010478 0.54837755 1.32965131 [37] 3.32024573 0.53825226 3.30077557 0.46426513 1.00681720 0.59276030 [43] 1.31431609 2.27822419 2.56404713 0.52459218 1.05228996 2.23799673 [49] 1.14438962 2.41311612 1.40254244 1.20379073 0.44195457 0.95880408 [55] 1.45254027 4.49645001 2.18826796 1.07161515 1.62617544 3.15506003 [61] 2.15611491 2.43261017 1.40293461 2.79977886 3.44503201 3.25282551 [67] 0.91570531 0.70243512 5.67971774 3.55532192 1.71515046 2.97718949 [73] 0.47145131 0.32879089 0.60735231 0.92388638 4.29277857 1.73681839 [79] 0.09387383 2.81281295 1.84419058 2.63070353 1.52298124 2.89761263 [85] 1.05251987 1.24258703 3.09130229 2.66738574 3.17035060 2.15287327 [91] 2.63042751 1.69736779 3.21126033 1.56920999 2.68985477 0.82952068 [97] 3.62230865 1.65286592 2.76798783 2.08091935 Formula: y ~ exp(-exp(-(x - mue)/beta)) Parameters: Estimate Std. Error t value Pr(>|t|) mue 1.382744 0.005926 233.3 <2e-16 *** beta 0.984836 0.008816 111.7 <2e-16 *** -- View this message in context: http://r.789695.n4.nabble.com/ecdf-to-nls-how-to-transform-data-tp3671754p3674227.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ecdf() to nls() - how to transform data?
On Jul 17, 2011, at 11:07 AM, David Winsemius wrote: On Jul 17, 2011, at 4:12 AM, Jochen1980 wrote: Thanks David and Peter! I tried to improve my R-script to get closer to my goal. I guess I have to use nls(), because later I want to work with Levenberg-Marquardt-Algorithm and when I got it right, LM-Algorithm uses least squares as well, fitdistr() instead uses Maximum Likelihoods. Anyway, I am wondering, how to create the appropriate data.frame for nls()? In my opinion all data must be there: I got the absolute numbers in my dataset from the internet. I got the histogram/density function right, as you can see in the graph number 3 after running the following example. I got the cumulative density right as well - so how to find x and y for the data-frame in line 116? The trick I used (which has been criticised by my statistical betters) was to recognize that the y values for the ecdf function were simply (1:100)/100 df <- dataframe(y=(1:100)/100, x=knots(ecdf) # worked for me res <- nls( y ~ exp( - exp( - ( x - mue ) / beta ), start=list(mue=2, beta=3), data=df) The expression should have been: y ~ exp( - exp( - ( x - mue ) / beta )) Using: gdta <- rgamma(100, 2) ecdfgr <- ecdf(gdta) dfecdf <- data.frame(knots=knots(ecdfgr), Fn =(1:100)/100) > res <- nls( Fn ~ exp( - exp( - ( knots - mue ) / beta )), start=list(mue=2, beta=2), data=dfecdf) > res Nonlinear regression model model: Fn ~ exp(-exp(-(knots - mue)/beta)) data: dfecdf mue beta 1.395 1.068 residual sum-of-squares: 0.04979 The concern statistically was that the nls function is minimizing error around y which has no error. Whether one can invert the function and thereby avoid the concern about minimizing error around a regular series ... I don't know. You could certainly try both, since the inverted function is so easy to construct: res <- nls( x ~ beta(- log(- log(y))) - mue, start=list(mue=2, beta=3), data=df) I think I got the sign of mue wrong and I get an Inf error with the 100th value for the dataframe but leaving it off gave me a comparable answer: > res <- nls( knots ~ - beta1*(- log(- log(Fn))) - mue, start=list(mue=2, beta1=2), data=dfecdf[1:99, ]) > res <- nls( knots ~ beta1*(- log(- log(Fn))) + mue, start=list(mue=2, beta1=2), data=dfecdf[1:99, ]) > res Nonlinear regression model model: knots ~ beta1 * (-log(-log(Fn))) + mue data: dfecdf[1:99, ] mue beta1 1.455 1.004 residual sum-of-squares: 3.041 Number of iterations to convergence: 1 Achieved convergence tolerance: 3.295e-09 (None of this code was tested, although the first part is very similar if not identical to code I tested yesterday. I was concerned this was a homework problem and was less specific than I might have been. I now see that you have at least made efforts, although I am in a hurry to get to a sailing and have not gone through it.) -- David. Is it necessary/easier to get this example worked with absolute numbers? The two easy tutorials for nls() worked out as well, so that can't be that hard I guess, but I am sitting here for two days and got no idea what to do now!? Thanks in advance for a little more input guys. # ### R-Skript Example Fit ### # cat( " * \n" ) cat( " Script curvefit - from histogram to function \n" ) cat( " \n" ) # Preparation setwd( "/home/joba/jobascripts/rproject/" ) rm( list=ls( all=TRUE ) ) # Organise X11, x Rows, y Graphs par( mfrow=c( 2, 2 ) ) ### tutorial 1 # Build data.frame() x <- seq( 0.0, 0.30, length=10 ) # seq( from, to, by ) y <- c( 0.04, 0.06, 0.12, 0.14, 0.15, 0.16, 0.18, 0.19, 0.20, 0.26 ) # vals of target function df <- data.frame( x=x, y=y ) #print( df ) plot( df, main="Tutorial 1" ) # Formula: y = x^a , look for a res <- nls( y ~ I(x^exponent), data = df, start = list( exponent = 1 ), trace = T ) print( round( summary(res)$coefficients[1], 3 )) lines( x, predict( res, list( x = x ) ), lty = 1, col = "blue" ) # Formula: y = x^a + b, look for a and b res2 <- nls( y ~ I( x^exponent + gamma ), data = df, start = list( exponent=1, gamma=0 ), trace = TRUE ) print( round( summary(res2)$coefficients[1], 3 )) lines( x, predict( res2, list( x = x ) ), lty = 1, col = "red" ) ### tutorial 2 len <- 24 x <- runif( len ) # 24 random values between 0 and 1 y <- x^3 + rnorm( len, 0, 0.06 ) # compute y-values and add some noise ds <- data.frame( x = x, y = y ) # build data.frame #str( ds ) # Compute prediction s <- seq( 0, 1, length = 100 ) # compute real-data, to get y-values m <- nls( y ~ I(x^power), data = ds, start = list( power = 1 ), trace = T ) #print( class(m) ) print( summary(m) ) # plot f
Re: [R] ecdf() to nls() - how to transform data?
On Jul 17, 2011, at 4:12 AM, Jochen1980 wrote:
Thanks David and Peter!
I tried to improve my R-script to get closer to my goal.
I guess I have to use nls(), because later I want to work with
Levenberg-Marquardt-Algorithm and when I got it right, LM-Algorithm
uses
least squares as well, fitdistr() instead uses Maximum Likelihoods.
Anyway,
I am wondering, how to create the appropriate data.frame for nls()?
In my
opinion all data must be there:
I got the absolute numbers in my dataset from the internet.
I got the histogram/density function right, as you can see in the
graph
number 3 after running the following example.
I got the cumulative density right as well - so how to find x and y
for the
data-frame in line 116?
The trick I used (which has been criticised by my statistical betters)
was to recognize that the y values for the ecdf function were simply
(1:100)/100
df <- dataframe(y=(1:100)/100, x=knots(ecdf) # worked for me
res <- nls( y ~ exp( - exp( - ( x - mue ) / beta ), start=list(mue=2,
beta=3), data=df)
The concern statistically was that the nls function is minimizing
error around y which has no error. Whether one can invert the
function and thereby avoid the concern about minimizing error around a
regular series ... I don't know.
You could certainly try both, since the inverted function is so easy
to construct:
res <- nls( x ~ beta(- log(- log(y))) - mue, start=list(mue=2,
beta=3), data=df)
(None of this code was tested, although the first part is very similar
if not identical to code I tested yesterday. I was concerned this was
a homework problem and was less specific than I might have been. I now
see that you have at least made efforts, although I am in a hurry to
get to a sailing and have not gone through it.)
--
David.
Is it necessary/easier to get this example worked with absolute
numbers?
The two easy tutorials for nls() worked out as well, so that can't
be that
hard I guess, but I am sitting here for two days and got no idea
what to do
now!? Thanks in advance for a little more input guys.
#
### R-Skript Example Fit ###
#
cat( " *
\n" )
cat( " Script curvefit - from histogram to function \n" )
cat( " \n" )
# Preparation
setwd( "/home/joba/jobascripts/rproject/" )
rm( list=ls( all=TRUE ) )
# Organise X11, x Rows, y Graphs
par( mfrow=c( 2, 2 ) )
### tutorial 1
# Build data.frame()
x <- seq( 0.0, 0.30, length=10 ) # seq(
from, to, by )
y <- c( 0.04, 0.06, 0.12, 0.14, 0.15, 0.16, 0.18, 0.19, 0.20, 0.26 )
# vals
of target function
df <- data.frame( x=x, y=y )
#print( df )
plot( df, main="Tutorial 1" )
# Formula: y = x^a , look for a
res <- nls( y ~ I(x^exponent), data = df, start = list( exponent =
1 ),
trace = T )
print( round( summary(res)$coefficients[1], 3 ))
lines( x, predict( res, list( x = x ) ), lty = 1, col = "blue" )
# Formula: y = x^a + b, look for a and b
res2 <- nls( y ~ I( x^exponent + gamma ), data = df, start = list(
exponent=1, gamma=0 ), trace = TRUE )
print( round( summary(res2)$coefficients[1], 3 ))
lines( x, predict( res2, list( x = x ) ), lty = 1, col = "red" )
### tutorial 2
len <- 24
x <- runif( len ) # 24 random values between 0 and 1
y <- x^3 + rnorm( len, 0, 0.06 ) # compute y-values and add some noise
ds <- data.frame( x = x, y = y ) # build data.frame
#str( ds )
# Compute prediction
s <- seq( 0, 1, length = 100 ) # compute real-data, to get y-values
m <- nls( y ~ I(x^power), data = ds, start = list( power = 1 ),
trace = T )
#print( class(m) )
print( summary(m) )
# plot fitted curve and known curve
power <- round( summary(m)$coefficients[1], 3 ) # get power from
summary
power.se <- round( summary(m)$coefficients[2], 3 ) # get failure from
summary
plot( y ~ x, main = "Tutorial 2", sub="Blue: fit; green: known" )
lines( s, s^3, lty = 2, col = "green" ) # known real data
lines( s, predict( m, list( x = s )), lty = 1, col = "blue" ) #
predicted
vals
text( 0, 0.5, paste( "y =x^ (", power, " +/- ", power.se, ")",
sep=""),
pos=4 )
### real data #
# --- Create Points ---
# Real data and density
hgd <- read.csv(
file="http://www.jochen-bauer.net/downloads/curve-fitting-tutorial/data01.txt
",
sep=",", head=FALSE)
#print( hgd$V1 )
print( summary( hgd$V1 ) )
denspoints <- density( hgd$V1 )
# define variables for later use
msa <- "MSA-Name"
c1 <- "col1"
c2 <- "col2"
histtitle <- paste( msa, "Spalten:", c1, ",", c2 )
#
# --- Plot ---
# Organise X11, x Rows, y Graphs
#par( mfrow=c( 1, 2 ) )
# Histogram
histo <- hist( hgd$V1, freq=FALSE, main=histtitle, xlab="U-Value",
Re: [R] ecdf() to nls() - how to transform data?
Thanks David and Peter!
I tried to improve my R-script to get closer to my goal.
I guess I have to use nls(), because later I want to work with
Levenberg-Marquardt-Algorithm and when I got it right, LM-Algorithm uses
least squares as well, fitdistr() instead uses Maximum Likelihoods. Anyway,
I am wondering, how to create the appropriate data.frame for nls()? In my
opinion all data must be there:
I got the absolute numbers in my dataset from the internet.
I got the histogram/density function right, as you can see in the graph
number 3 after running the following example.
I got the cumulative density right as well - so how to find x and y for the
data-frame in line 116?
Is it necessary/easier to get this example worked with absolute numbers?
The two easy tutorials for nls() worked out as well, so that can't be that
hard I guess, but I am sitting here for two days and got no idea what to do
now!? Thanks in advance for a little more input guys.
#
### R-Skript Example Fit ###
#
cat( " * \n" )
cat( " Script curvefit - from histogram to function \n" )
cat( " \n" )
# Preparation
setwd( "/home/joba/jobascripts/rproject/" )
rm( list=ls( all=TRUE ) )
# Organise X11, x Rows, y Graphs
par( mfrow=c( 2, 2 ) )
### tutorial 1
# Build data.frame()
x <- seq( 0.0, 0.30, length=10 )# seq(
from, to, by )
y <- c( 0.04, 0.06, 0.12, 0.14, 0.15, 0.16, 0.18, 0.19, 0.20, 0.26 )# vals
of target function
df <- data.frame( x=x, y=y )
#print( df )
plot( df, main="Tutorial 1" )
# Formula: y = x^a , look for a
res <- nls( y ~ I(x^exponent), data = df, start = list( exponent = 1 ),
trace = T )
print( round( summary(res)$coefficients[1], 3 ))
lines( x, predict( res, list( x = x ) ), lty = 1, col = "blue" )
# Formula: y = x^a + b, look for a and b
res2 <- nls( y ~ I( x^exponent + gamma ), data = df, start = list(
exponent=1, gamma=0 ), trace = TRUE )
print( round( summary(res2)$coefficients[1], 3 ))
lines( x, predict( res2, list( x = x ) ), lty = 1, col = "red" )
### tutorial 2
len <- 24
x <- runif( len )# 24 random values between 0 and 1
y <- x^3 + rnorm( len, 0, 0.06 ) # compute y-values and add some noise
ds <- data.frame( x = x, y = y ) # build data.frame
#str( ds )
# Compute prediction
s <- seq( 0, 1, length = 100 ) # compute real-data, to get y-values
m <- nls( y ~ I(x^power), data = ds, start = list( power = 1 ), trace = T )
#print( class(m) )
print( summary(m) )
# plot fitted curve and known curve
power <- round( summary(m)$coefficients[1], 3 ) # get power from summary
power.se <- round( summary(m)$coefficients[2], 3 ) # get failure from
summary
plot( y ~ x, main = "Tutorial 2", sub="Blue: fit; green: known" )
lines( s, s^3, lty = 2, col = "green" ) # known real data
lines( s, predict( m, list( x = s )), lty = 1, col = "blue" ) # predicted
vals
text( 0, 0.5, paste( "y =x^ (", power, " +/- ", power.se, ")", sep=""),
pos=4 )
### real data #
# --- Create Points ---
# Real data and density
hgd <- read.csv(
file="http://www.jochen-bauer.net/downloads/curve-fitting-tutorial/data01.txt";,
sep=",", head=FALSE)
#print( hgd$V1 )
print( summary( hgd$V1 ) )
denspoints <- density( hgd$V1 )
# define variables for later use
msa <- "MSA-Name"
c1 <- "col1"
c2 <- "col2"
histtitle <- paste( msa, "Spalten:", c1, ",", c2 )
#
# --- Plot ---
# Organise X11, x Rows, y Graphs
#par( mfrow=c( 1, 2 ) )
# Histogram
histo <- hist( hgd$V1, freq=FALSE, main=histtitle, xlab="U-Value",
ylab="Density")
# Density-Line
lines( denspoints, col="GREEN", lwd=1 )
# Plot cumulatives
plot( ecdf( hgd$V1 ), col="GREEN", lwd=1, main="", xlab="U-Value",
ylab="Probability" ) # real data
#
# --- Fit ---
# Matching Gumbel distribution as tip from expert
# Gumbel-Dist-Function, cumulative
# ( - ( x - mue ) / beta )
# ( -e )^
# F(x) = e^
# => exp( - exp( - ( x - mue ) / beta ) ) # Thanks to Peter Dalgaard :-)
# Starting guesses: mue = medianOfData, beta = sdevOfData
mueStart <- median( hgd$V1 )
betaStart <- sd( hgd$V1 )
print( paste( "mueStart: ", mueStart ))
print( paste( "betaStart: ", betaStart ))
# Build data.frame()
# x = min of uvals to max of uvals
# y = probabilities, which add to 1, normalized histogramms could help ?
### XXX # X #
x <- seq( 0.0, 0.30, length=10 )# seq(
from, to, by )
y <- c( 0.04, 0.06, 0.12, 0.14, 0.15, 0.16, 0.18, 0.19, 0.20, 0.26 )# vals
of target function
df <- data.frame( x=x, y=y )
print( df )
# Formula: y = x^a , l
Re: [R] ecdf() to nls() - how to transform data?
On Jul 16, 2011, at 14:17 , Jochen1980 wrote: > Hi, > > I am using ecdf-function and want to use the ecdf()-data-points for nls() as > data-parameter. > nls() expects 'list' or 'environment' as a data-type, knots(ecdf(mydata)) > gives me 'numeric'. > What should I do now? Consider using fitdistr() from the MASS package. What you're trying to do is just wrong! > > Thanks in advance - Jochen > > Here is the code: > # > # --- Fit --- > # Gumbel-Dist-Function, cumulative, > http://en.wikipedia.org/wiki/Gumbel_distribution > # ( - ( x - mue ) / beta ) > # ( -e )^ > # F(x) = e^ > # formula for fitting-function > # formula: y ~ I( exp(1) ^ ( - exp(1) ) ^ ( - ( x - mue ) / beta ) ) Ouch Do teach yourself what exp() means. I believe that wants to be exp( - exp( - ( x - mue ) / beta ) ) notice that fitdistr() needs the density, though. I.e. the derivative of the above with respect to x. -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ecdf() to nls() - how to transform data?
On Jul 16, 2011, at 8:17 AM, Jochen1980 wrote:
Hi,
I am using ecdf-function and want to use the ecdf()-data-points for
nls() as
data-parameter.
nls() expects 'list' or 'environment' as a data-type,
knots(ecdf(mydata))
gives me 'numeric'.
If you put them into 'df' with appropriate name and add an appropriate
'y' value you should get success. See below. And perhaps you should
plot your ecdf so you can figure out what you y-values should be.
What should I do now?
Thanks in advance - Jochen
Here is the code:
#
# --- Fit ---
# Gumbel-Dist-Function, cumulative,
http://en.wikipedia.org/wiki/Gumbel_distribution
# ( - ( x - mue ) / beta )
# ( -e )^
# F(x) = e^
# formula for fitting-function
# formula: y ~ I( exp(1) ^ ( - exp(1) ) ^ ( - ( x - mue ) / beta ) )
# data: ecdf( hgd$V1 )
# start: list( mue=0.1, beta=0.025 )
gpfunction <- ecdf( hgd$V1 )
gplistrange <- seq( 0.0, 1, by=0.001 )
gplist <- gpfunction( gplistrange )
print( gplist )
print( class( gplist ) )
print("---")
#res <- nls( y ~ I( exp(1) ^ ( - exp(1) ) ^ ( - ( x - mue ) /
beta ) ), df,
list( mue=0.1, beta=0.025 ), trace=TRUE )
This may or may not work. I didn't try it because I was pretty sure
that the I() was unnecessary and I knew that exp(1)^(.) was just a
convoluted way of doing exp(.), so it can be considerably simplified:
The 'df' object needs to be constructed so that the variables inside
the function match column names in df, so you need df to have columns
'y' and 'x'.
Try it out with a function like rgamma. Generate a random sample,
rgamma(100, shape=2), apply ecdf, construct a 'df' argument and use
nls to solve for y ~ pgamma(knots, shape). Then substitute in your
(hopefully) simplified Gumbel function.
#print( summary( res ) )
David Winsemius, MD
West Hartford, CT
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