Dear friends - I'm on Win10 with R 6.3.1 and have a very simple problem with
nls which apparently gives a decent fit to the parable below, even without
starting values. But when I then think I know the meaning of the three
parameters a, b, and d it goes very wrong. I guess I am again overlooking
something easy but cannot spot it.
BW
Troels Ring,
Aalborg, Denmark
aedf <- structure(list(Flux = c(-0.141256, -0.154709, -0.215247, -0.302691,
-0.32287, -0.511211, -0.605381, -0.813901, -1.11659, -1.76906
), pH = c(7.06273, 7.11182, 7.16182, 7.18818, 7.21455, 7.23818,
7.26273, 7.28455, 7.31182, 7.34364)), class = "data.frame",
row.names = c(NA,
-10L))
m <- with(aedf,nls(Flux~a + b*pH + d*pH^2))
with(aedf,plot(pH,Flux))
with(aedf,lines(pH,fitted(m),lty=1,col="red",lwd=3))
m
# a b d
# -1630.70 457.67 -32.11
fitted(m)
# 1] -0.22087053 -0.09989926 -0.13579690 -0.21936385 -0.34761742 -0.50048799
# [7] -0.69729602 -0.90471194 -1.20692552 -1.61994657
FPG <- function(pH) -1630.70 + 457.67*pH -32.11*pH^2
FPG(aedf$pH)
# [1] -0.016359649 0.107602395 0.074773375 -0.007166685 -0.133786467
# [6] -0.285187665 -0.480463769 -0.686513537 -0.987013685 -1.398026917
# So why aren't fitted(m) and FPG(aedf$pH) not closer ("equal")?
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nls problem
8 messages · Rui Barradas, John C Nash, Troels Ring +1 more
Hello, Tip: in a formula, pH^2 = pH*pH is an interaction. pH^2 = pH*pH = pH + pH + pH:pH = pH Try I(pH^2) Hope this helps, Rui Barradas ?s 12:07 de 02/04/20, Troels Ring escreveu:
Dear friends - I'm on Win10 with R 6.3.1 and have a very simple problem with
nls which apparently gives a decent fit to the parable below, even without
starting values. But when I then think I know the meaning of the three
parameters a, b, and d it goes very wrong. I guess I am again overlooking
something easy but cannot spot it.
BW
Troels Ring,
Aalborg, Denmark
aedf <- structure(list(Flux = c(-0.141256, -0.154709, -0.215247, -0.302691,
-0.32287, -0.511211, -0.605381, -0.813901, -1.11659, -1.76906
), pH = c(7.06273, 7.11182, 7.16182, 7.18818, 7.21455, 7.23818,
7.26273, 7.28455, 7.31182, 7.34364)), class = "data.frame",
row.names = c(NA,
-10L))
m <- with(aedf,nls(Flux~a + b*pH + d*pH^2))
with(aedf,plot(pH,Flux))
with(aedf,lines(pH,fitted(m),lty=1,col="red",lwd=3))
m
# a b d
# -1630.70 457.67 -32.11
fitted(m)
# 1] -0.22087053 -0.09989926 -0.13579690 -0.21936385 -0.34761742 -0.50048799
# [7] -0.69729602 -0.90471194 -1.20692552 -1.61994657
FPG <- function(pH) -1630.70 + 457.67*pH -32.11*pH^2
FPG(aedf$pH)
# [1] -0.016359649 0.107602395 0.074773375 -0.007166685 -0.133786467
# [6] -0.285187665 -0.480463769 -0.686513537 -0.987013685 -1.398026917
# So why aren't fitted(m) and FPG(aedf$pH) not closer ("equal")?
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Hello,
Sorry, disregard my previous e-mail.
Instead of your FPG function try
FPG <- function(pH, model) {
coef(model)[1] + coef(model)[2]*pH + coef(model)[3]*pH^2
}
FPG(aedf$pH, m)
fitted(m)
Hope this helps,
Rui Barradas
?s 12:30 de 02/04/20, Rui Barradas escreveu:
Hello, Tip: in a formula, pH^2 = pH*pH is an interaction. pH^2 = pH*pH = pH + pH + pH:pH = pH Try I(pH^2) Hope this helps, Rui Barradas ?s 12:07 de 02/04/20, Troels Ring escreveu:
Dear friends - I'm on Win10 with R 6.3.1 and have a very simple
problem with
nls which apparently gives a decent fit to the parable below, even
without
starting values. But when I then think I know the meaning of the three
parameters a, b, and d it goes very wrong. I guess I am again overlooking
something easy but cannot spot it.
BW
Troels Ring,
Aalborg, Denmark
aedf <- structure(list(Flux = c(-0.141256, -0.154709, -0.215247,
-0.302691,
???????????? -0.32287, -0.511211, -0.605381, -0.813901, -1.11659,
-1.76906
), pH = c(7.06273, 7.11182, 7.16182, 7.18818, 7.21455, 7.23818,
?????????? 7.26273, 7.28455, 7.31182, 7.34364)), class = "data.frame",
row.names = c(NA,
-10L))
m <- with(aedf,nls(Flux~a + b*pH + d*pH^2))
with(aedf,plot(pH,Flux))
with(aedf,lines(pH,fitted(m),lty=1,col="red",lwd=3))
m
# a??????? b??????? d
# -1630.70?? 457.67?? -32.11
fitted(m)
# 1] -0.22087053 -0.09989926 -0.13579690 -0.21936385 -0.34761742
-0.50048799
# [7] -0.69729602 -0.90471194 -1.20692552 -1.61994657
FPG <- function(pH) -1630.70 + 457.67*pH -32.11*pH^2
FPG(aedf$pH)
# [1] -0.016359649? 0.107602395? 0.074773375 -0.007166685 -0.133786467
# [6] -0.285187665 -0.480463769 -0.686513537 -0.987013685 -1.398026917
# So why aren't fitted(m) and FPG(aedf$pH) not closer ("equal")?
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______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Simpler:
FPG2 <- function(x, model){
as.vector(cbind(1, x, x^2) %*% coef(model))
}
Hope this helps,
Rui Barradas
?s 12:56 de 02/04/20, Rui Barradas escreveu:
Hello,
Sorry, disregard my previous e-mail.
Instead of your FPG function try
FPG <- function(pH, model) {
? coef(model)[1] + coef(model)[2]*pH + coef(model)[3]*pH^2
}
FPG(aedf$pH, m)
fitted(m)
Hope this helps,
Rui Barradas
?s 12:30 de 02/04/20, Rui Barradas escreveu:
Hello, Tip: in a formula, pH^2 = pH*pH is an interaction. pH^2 = pH*pH = pH + pH + pH:pH = pH Try I(pH^2) Hope this helps, Rui Barradas ?s 12:07 de 02/04/20, Troels Ring escreveu:
Dear friends - I'm on Win10 with R 6.3.1 and have a very simple
problem with
nls which apparently gives a decent fit to the parable below, even
without
starting values. But when I then think I know the meaning of the three
parameters a, b, and d it goes very wrong. I guess I am again
overlooking
something easy but cannot spot it.
BW
Troels Ring,
Aalborg, Denmark
aedf <- structure(list(Flux = c(-0.141256, -0.154709, -0.215247,
-0.302691,
???????????? -0.32287, -0.511211, -0.605381, -0.813901, -1.11659,
-1.76906
), pH = c(7.06273, 7.11182, 7.16182, 7.18818, 7.21455, 7.23818,
?????????? 7.26273, 7.28455, 7.31182, 7.34364)), class = "data.frame",
row.names = c(NA,
-10L))
m <- with(aedf,nls(Flux~a + b*pH + d*pH^2))
with(aedf,plot(pH,Flux))
with(aedf,lines(pH,fitted(m),lty=1,col="red",lwd=3))
m
# a??????? b??????? d
# -1630.70?? 457.67?? -32.11
fitted(m)
# 1] -0.22087053 -0.09989926 -0.13579690 -0.21936385 -0.34761742
-0.50048799
# [7] -0.69729602 -0.90471194 -1.20692552 -1.61994657
FPG <- function(pH) -1630.70 + 457.67*pH -32.11*pH^2
FPG(aedf$pH)
# [1] -0.016359649? 0.107602395? 0.074773375 -0.007166685 -0.133786467
# [6] -0.285187665 -0.480463769 -0.686513537 -0.987013685 -1.398026917
# So why aren't fitted(m) and FPG(aedf$pH) not closer ("equal")?
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Roundoff/cancelation error: compare the following. The first is equivalent to your function, the last to fitted().
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=2)))
[,1] [,2] [,3] [,4] [,5] [,6]
[,7] [,8] [,9] [,10]
[1,] -0.01635965 0.1076024 0.07477337 -0.007166685 -0.1337865 -0.2851877
-0.4804638 -0.6865135 -0.9870137 -1.398027
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=4)))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] -0.2196286 -0.09864071 -0.1345213 -0.2180792 -0.3463237 -0.4991861 -0.6959856 -0.903394 -1.205598 -1.618608
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=16)))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] -0.2208705 -0.09989926 -0.1357969 -0.2193638 -0.3476174 -0.500488 -0.697296 -0.9047119 -1.206926 -1.619947 Note that your model is linear and could be fitted with lm(data=aedf, Flux ~ pH + I(pH^2)) lm(data=aedf, Flux ~ poly(pH, 2)) The latter uses a more stable parameterization. Bill Dunlap TIBCO Software wdunlap tibco.com
On Thu, Apr 2, 2020 at 4:15 AM Troels Ring <tring at gvdnet.dk> wrote:
Dear friends - I'm on Win10 with R 6.3.1 and have a very simple problem
with
nls which apparently gives a decent fit to the parable below, even without
starting values. But when I then think I know the meaning of the three
parameters a, b, and d it goes very wrong. I guess I am again overlooking
something easy but cannot spot it.
BW
Troels Ring,
Aalborg, Denmark
aedf <- structure(list(Flux = c(-0.141256, -0.154709, -0.215247,
-0.302691,
-0.32287, -0.511211, -0.605381, -0.813901, -1.11659, -1.76906
), pH = c(7.06273, 7.11182, 7.16182, 7.18818, 7.21455, 7.23818,
7.26273, 7.28455, 7.31182, 7.34364)), class = "data.frame",
row.names = c(NA,
-10L))
m <- with(aedf,nls(Flux~a + b*pH + d*pH^2))
with(aedf,plot(pH,Flux))
with(aedf,lines(pH,fitted(m),lty=1,col="red",lwd=3))
m
# a b d
# -1630.70 457.67 -32.11
fitted(m)
# 1] -0.22087053 -0.09989926 -0.13579690 -0.21936385 -0.34761742
-0.50048799
# [7] -0.69729602 -0.90471194 -1.20692552 -1.61994657
FPG <- function(pH) -1630.70 + 457.67*pH -32.11*pH^2
FPG(aedf$pH)
# [1] -0.016359649 0.107602395 0.074773375 -0.007166685 -0.133786467
# [6] -0.285187665 -0.480463769 -0.686513537 -0.987013685 -1.398026917
# So why aren't fitted(m) and FPG(aedf$pH) not closer ("equal")?
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Yes, I was waiting to see how long before it would be noticed that this is not the sort of problem for which nls() is appropriate. And I'll beat the drum again that nls() uses a simple (and generally deprecated) forward difference derivative approximation that gets into trouble a VERY high proportion of applications. Duncan Murdoch and I published nlsr to overcome several of the weaknesses in nls(). nls() was very advanced when it first appeared, but I've always felt it was intended for use by very advanced craftsmen rather than general users. Indeed, it can do some things nlsr won't do, but usually only if handled skillfully. JN
On 2020-04-02 12:50 p.m., William Dunlap via R-help wrote:
Roundoff/cancelation error: compare the following. The first is equivalent to your function, the last to fitted().
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=2)))
[,1] [,2] [,3] [,4] [,5] [,6]
[,7] [,8] [,9] [,10]
[1,] -0.01635965 0.1076024 0.07477337 -0.007166685 -0.1337865 -0.2851877
-0.4804638 -0.6865135 -0.9870137 -1.398027
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=4)))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] -0.2196286 -0.09864071 -0.1345213 -0.2180792 -0.3463237 -0.4991861 -0.6959856 -0.903394 -1.205598 -1.618608
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=16)))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] -0.2208705 -0.09989926 -0.1357969 -0.2193638 -0.3476174 -0.500488 -0.697296 -0.9047119 -1.206926 -1.619947 Note that your model is linear and could be fitted with lm(data=aedf, Flux ~ pH + I(pH^2)) lm(data=aedf, Flux ~ poly(pH, 2)) The latter uses a more stable parameterization. Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Apr 2, 2020 at 4:15 AM Troels Ring <tring at gvdnet.dk> wrote:
Dear friends - I'm on Win10 with R 6.3.1 and have a very simple problem
with
nls which apparently gives a decent fit to the parable below, even without
starting values. But when I then think I know the meaning of the three
parameters a, b, and d it goes very wrong. I guess I am again overlooking
something easy but cannot spot it.
BW
Troels Ring,
Aalborg, Denmark
aedf <- structure(list(Flux = c(-0.141256, -0.154709, -0.215247,
-0.302691,
-0.32287, -0.511211, -0.605381, -0.813901, -1.11659, -1.76906
), pH = c(7.06273, 7.11182, 7.16182, 7.18818, 7.21455, 7.23818,
7.26273, 7.28455, 7.31182, 7.34364)), class = "data.frame",
row.names = c(NA,
-10L))
m <- with(aedf,nls(Flux~a + b*pH + d*pH^2))
with(aedf,plot(pH,Flux))
with(aedf,lines(pH,fitted(m),lty=1,col="red",lwd=3))
m
# a b d
# -1630.70 457.67 -32.11
fitted(m)
# 1] -0.22087053 -0.09989926 -0.13579690 -0.21936385 -0.34761742
-0.50048799
# [7] -0.69729602 -0.90471194 -1.20692552 -1.61994657
FPG <- function(pH) -1630.70 + 457.67*pH -32.11*pH^2
FPG(aedf$pH)
# [1] -0.016359649 0.107602395 0.074773375 -0.007166685 -0.133786467
# [6] -0.285187665 -0.480463769 -0.686513537 -0.987013685 -1.398026917
# So why aren't fitted(m) and FPG(aedf$pH) not closer ("equal")?
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Thank you ? yes indeed, on suggestion from Rui I noticed that the problem was mainly that I took the parameters from ?m? instead of coef(m) i.e
a b d
-1630.71 457.68 -32.11
Instead of
a b d
-1630.70993 457.67555 -32.11469
And that alone saves me.
BW
Troels
Fra: William Dunlap <wdunlap at tibco.com>
Sendt: 2. april 2020 18:50
Til: Troels Ring <tring at gvdnet.dk>
Cc: r-help mailing list <r-help at r-project.org>
Emne: Re: [R] nls problem
Roundoff/cancelation error: compare the following. The first is equivalent to your function, the last to fitted().
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=2)))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] -0.01635965 0.1076024 0.07477337 -0.007166685 -0.1337865 -0.2851877 -0.4804638 -0.6865135 -0.9870137 -1.398027
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=4)))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] -0.2196286 -0.09864071 -0.1345213 -0.2180792 -0.3463237 -0.4991861 -0.6959856 -0.903394 -1.205598 -1.618608
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=16)))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] -0.2208705 -0.09989926 -0.1357969 -0.2193638 -0.3476174 -0.500488 -0.697296 -0.9047119 -1.206926 -1.619947 Note that your model is linear and could be fitted with lm(data=aedf, Flux ~ pH + I(pH^2)) lm(data=aedf, Flux ~ poly(pH, 2)) The latter uses a more stable parameterization. Bill Dunlap TIBCO Software wdunlap tibco.com <http://tibco.com>
On Thu, Apr 2, 2020 at 4:15 AM Troels Ring <tring at gvdnet.dk <mailto:tring at gvdnet.dk> > wrote:
Dear friends - I'm on Win10 with R 6.3.1 and have a very simple problem with
nls which apparently gives a decent fit to the parable below, even without
starting values. But when I then think I know the meaning of the three
parameters a, b, and d it goes very wrong. I guess I am again overlooking
something easy but cannot spot it.
BW
Troels Ring,
Aalborg, Denmark
aedf <- structure(list(Flux = c(-0.141256, -0.154709, -0.215247, -0.302691,
-0.32287, -0.511211, -0.605381, -0.813901, -1.11659, -1.76906
), pH = c(7.06273, 7.11182, 7.16182, 7.18818, 7.21455, 7.23818,
7.26273, 7.28455, 7.31182, 7.34364)), class = "data.frame",
row.names = c(NA,
-10L))
m <- with(aedf,nls(Flux~a + b*pH + d*pH^2))
with(aedf,plot(pH,Flux))
with(aedf,lines(pH,fitted(m),lty=1,col="red",lwd=3))
m
# a b d
# -1630.70 457.67 -32.11
fitted(m)
# 1] -0.22087053 -0.09989926 -0.13579690 -0.21936385 -0.34761742 -0.50048799
# [7] -0.69729602 -0.90471194 -1.20692552 -1.61994657
FPG <- function(pH) -1630.70 + 457.67*pH -32.11*pH^2
FPG(aedf$pH)
# [1] -0.016359649 0.107602395 0.074773375 -0.007166685 -0.133786467
# [6] -0.285187665 -0.480463769 -0.686513537 -0.987013685 -1.398026917
# So why aren't fitted(m) and FPG(aedf$pH) not closer ("equal")?
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If you will be copying the printed coefficients into your function (instead of just using fitted() or predict()), then use dput(coef(m)) to get them printed to full precision. Also, if you regress on pH-7 instead of pH you don't have to worry so much about the roundoff or cancellation error. This is akin to what poly(pH,degree) does. Bill Dunlap TIBCO Software wdunlap tibco.com
On Thu, Apr 2, 2020 at 9:44 PM Troels Ring <tring at gvdnet.dk> wrote:
Thank you ? yes indeed, on suggestion from Rui I noticed that the problem
was mainly that I took the parameters from ?m? instead of coef(m) i.e
a b d
-1630.71 457.68 -32.11
Instead of
a b d
-1630.70993 457.67555 -32.11469
And that alone saves me.
BW
Troels
*Fra:* William Dunlap <wdunlap at tibco.com>
*Sendt:* 2. april 2020 18:50
*Til:* Troels Ring <tring at gvdnet.dk>
*Cc:* r-help mailing list <r-help at r-project.org>
*Emne:* Re: [R] nls problem
Roundoff/cancelation error: compare the following. The first is
equivalent to your function, the last to fitted().
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=2)))
[,1] [,2] [,3] [,4] [,5] [,6]
[,7] [,8] [,9] [,10]
[1,] -0.01635965 0.1076024 0.07477337 -0.007166685 -0.1337865 -0.2851877
-0.4804638 -0.6865135 -0.9870137 -1.398027
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=4)))
[,1] [,2] [,3] [,4] [,5] [,6]
[,7] [,8] [,9] [,10]
[1,] -0.2196286 -0.09864071 -0.1345213 -0.2180792 -0.3463237 -0.4991861
-0.6959856 -0.903394 -1.205598 -1.618608
with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=16)))
[,1] [,2] [,3] [,4] [,5] [,6]
[,7] [,8] [,9] [,10]
[1,] -0.2208705 -0.09989926 -0.1357969 -0.2193638 -0.3476174 -0.500488
-0.697296 -0.9047119 -1.206926 -1.619947
Note that your model is linear and could be fitted with
lm(data=aedf, Flux ~ pH + I(pH^2))
lm(data=aedf, Flux ~ poly(pH, 2))
The latter uses a more stable parameterization.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Apr 2, 2020 at 4:15 AM Troels Ring <tring at gvdnet.dk> wrote:
Dear friends - I'm on Win10 with R 6.3.1 and have a very simple problem
with
nls which apparently gives a decent fit to the parable below, even without
starting values. But when I then think I know the meaning of the three
parameters a, b, and d it goes very wrong. I guess I am again overlooking
something easy but cannot spot it.
BW
Troels Ring,
Aalborg, Denmark
aedf <- structure(list(Flux = c(-0.141256, -0.154709, -0.215247,
-0.302691,
-0.32287, -0.511211, -0.605381, -0.813901, -1.11659, -1.76906
), pH = c(7.06273, 7.11182, 7.16182, 7.18818, 7.21455, 7.23818,
7.26273, 7.28455, 7.31182, 7.34364)), class = "data.frame",
row.names = c(NA,
-10L))
m <- with(aedf,nls(Flux~a + b*pH + d*pH^2))
with(aedf,plot(pH,Flux))
with(aedf,lines(pH,fitted(m),lty=1,col="red",lwd=3))
m
# a b d
# -1630.70 457.67 -32.11
fitted(m)
# 1] -0.22087053 -0.09989926 -0.13579690 -0.21936385 -0.34761742
-0.50048799
# [7] -0.69729602 -0.90471194 -1.20692552 -1.61994657
FPG <- function(pH) -1630.70 + 457.67*pH -32.11*pH^2
FPG(aedf$pH)
# [1] -0.016359649 0.107602395 0.074773375 -0.007166685 -0.133786467
# [6] -0.285187665 -0.480463769 -0.686513537 -0.987013685 -1.398026917
# So why aren't fitted(m) and FPG(aedf$pH) not closer ("equal")?
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