How do I extract the standard error of the parameter estimates?
Also, if I would like to add two parameters together (x+y), can I use this
equation to calculate the new standard error?:
x = parameter 1
y = parameter 2
xSE = SE parameter 1
ySE = SE parameter 2
NewSE=(x+y)*sqrt((xSE/x)^2+(ySE/y)^2)
I extract information from list which class is nls
On Tue, Apr 26, 2011 at 2:21 PM, Schatzi <[hidden
email]> wrote:
> How do I extract the standard error of the parameter estimates?
>
> Also, if I would like to add two parameters together (x+y), can I use this
> equation to calcu
:15 AM
To: Thompson, Adele - adele_thomp...@cargill.com
Subject: Re: How can I extract information from list which class is nls
On Apr 27, 2011, at 9:28 AM, Schatzi wrote:
> Here is more information on the equation. It is a growth function:
>
> Growth = a + b*(1-exp(-k*time))
>
> w
I wanted to use an F-statistic to get p-values for treatment differences. I
have parameter estimates and standard errors. I posted about combining the
parameters in a previous post, but here I would just like to test the
parameter differences by treatments (it is a nonlinear function). I
calculated
I have a new device that takes measurements anywhere from every second, to
every 15 minutes (depending on changes). The matrix has a date, time and Y
column (Y is the measurement). For three days it is 25,000 rows. How do I
average the measurements by every 30 minutes so my matrix is 48 rows per
da
I do not want smoothing as the data should have jumps (it is weight left in
feeding bunker). I was thinking of maybe using a histogram-like function and
then averaging that. Not sure if this is possible.
-
In theory, practice and theory are the same. In practice, they are not - Albert
Einstei
In Matlab, an array can be created from 1 - 30 using the command similar to R
which is 1:30. Then, to make the array step by 0.1 the command is 1:0.1:30
which is 1, 1.1, 1.2,...,29.9,30. How can I do this in R?
-
In theory, practice and theory are the same. In practice, they are not - Albert
I can get around it by doing something like:
as.matrix(rep(1,291))*row(as.matrix(rep(1,291)))/10+.9
I was just hoping for a simple command.
Schatzi wrote:
>
> In Matlab, an array can be created from 1 - 30 using the command similar
> to R which is 1:30. Then, to make the array step b
I have times and would like to round down to the earliest 30 minute
increment. For instance, a time of
2011-04-28 09:02:00
(the as.numeric value = 1303999320)
I would like it to be rounded down to:
2011-04-28 09:00:00
(the as.numeric value = 1303999200)
Any ideas of how to do this?
-
In theo
I have a matrix where one column has a date and another column has a time. I
need to delete all times before 6am. I had combined the Date and Time column
into DateTime.
Mat1:
Weight Date Time
7.6 04/28/11 09:03
8.4 04/29/11 03:11
8.6 04/29/11 05:32
8.6 04/29/11 09:53
1.4 05/01
I have a combined date and time. I would like to separate them out into two
columns so I can do things such as take the mean by time across all dates.
meas<-runif(435)
nTime<-seq(1303975800, 1304757000, 1800)
nDateT<-as.POSIXct(nTime, origin="1970-01-01")
mat1<-cbind(nDateT,meas)
means1<- aggrega
That is wonderful. Thank you.
Adele
Ken Takagi wrote:
>
> Schatzi cargill.com> writes:
>
>>
>> I have a combined date and time. I would like to separate them out into
>> two
>> columns so I can do things such as take the mean by time across all
>> da
I have a dataset where I have missing times (11:00 and 16:00). I would like
the outputs to include the missing time so that the final time vector looks
like "realt" and has the previous time's value. Ex. If meas at time 15:30 is
0.45, then the meas for time 16:00 will also be 0.45.
meas are the mea
I am plotting 28 plots on one screen:
par(mfrow=c(4,7))
for (i in 1:28) {
a<-seq(1,3,1)
plot(a,a, ann=FALSE)
}
I want a main title for all the plots (I tried using main but that doesn't
work). I deleted the axes, but am not sure how to delete the space. There
are such large margins between plots
ose do not seem to work
or they only increase the size to a certain degree, even though I can
manually increase it to fill my screen.
Schatzi wrote:
>
> Thanks all for the replies. I am getting better slowly but surely. I
> imagine that I will get better at figuring out things as well s
Great. Thanks. Here is the new code:
xyplot(Y~X,groups = TRT,type="n")
text(x,y,labels=TRT)
--
View this message in context:
http://r.789695.n4.nabble.com/Plot-using-treatment-letter-as-points-tp3276743p3276808.html
Sent from the R help mailing list archive at Nabble.com.
___
I would like to create a plot of y vs x with different treatments where the
points are actually the letter of the treatment. Here is the code:
A<-as.matrix(rnorm(10,10))
B<-as.matrix(rnorm(10,9.5))
C<-as.matrix(rnorm(10,10.5))
Y<-as.matrix(rnorm(30,13))
X<-rbind(A,B,C)
nA<-matrix("A",10,1)
I am running the following nls equation. I tried it with data that excel was
fitting and got the error:
singular gradient matrix at initial parameter estimates
I thought it was due to a low number of points (6), but when I create a
dataset, I get the same problem. If I remove the parameter "a," t
I am running the following nls equation. I tried it with data that excel was
fitting and got the error:
singular gradient matrix at initial parameter estimates
I thought it was due to a low number of points (6), but when I create a
dataset, I get the same problem. If I remove the parameter "a," th
If no one is able to help with this issue, do you have any idea where I can
post this question to receive help? Thank you.
--
View this message in context:
http://r.789695.n4.nabble.com/nls-help-tp3328521p3328611.html
Sent from the R help mailing list archive at Nabble.com.
I am running the following nls equation. I tried it with data that excel was
fitting and got the error:
singular gradient matrix at initial parameter estimates
I thought it was due to a low number of points (6), but when I create a
dataset, I get the same problem. If I remove the parameter "a," t
By the way, sorry about the reposts. I subscribed but wasn't sure what was
happening. Hopefully this worked.
--
View this message in context:
http://r.789695.n4.nabble.com/nls-not-solving-tp3328647p3328659.html
Sent from the R help mailing list archive at Nabble.com.
___
I am not sure how you simplified the model to:
y = a + b(1 - exp(kl)) - b exp(-kx)
I tried simplifying it but only got to:
y = a + b - b * exp(kl) * exp(-kx)
I agree that the model must not be identifiable. That makes sense,
especially given that removing either a or l makes the model work. Can
Here is a reply by Bart:
Yes you're right (I should have taken off my glasses and looked closer).
However, the argument is essentially the same:
Suppose you have a solution with a,b,k,l. Then for any positive c, [a+b-bc]
+ [bc] + (bc) *exp(kl')exp(-kx) is also a solution, where l'
= l - log(c)/k
I am trying to do a power analysis to get the number of replicas per
treatment.
If I try to get the power it works just fine:
setn=c(2,3)
sdx=c(1.19,4.35)
power.t.test(n = setn, delta = 13.5, sd = sdx, sig.level = 0.05,power =
NULL)
If I go the other way to obtain the "n" I have problems.
sdx=c(1
"Inter ocular data"
Quite amusing :)
Thank you for the help. For some reason I was thinking that I could get the
n values for the combined test, but that doesn't make sense as there could
be an infinite number of combinations of n values.
Thanks again for the replies.
--
View this message in conte
dev.new(width=6, height=1.5,mar=c(0,0,0,0))
par(mfrow=c(1,1),mar=c(.5, .5, 1.5, .5), oma=c(.4, 0,.5, 0))
barplot(c(1,1,1,1,1,1),col=c("blue","purple","red","green","orange","yellow"),
axes = FALSE)
I have a barplot that returns six colors in a line. I would like to get the
same six color blocks in
I'm not sure this is the right location (maybe R-devel would be better).
-
In theory, practice and theory are the same. In practice, they are not - Albert
Einstein
--
View this message in context:
http://r.789695.n4.nabble.com/barplot-in-hexagram-layout-tp3807600p3807608.html
Sent from the
I will try stacking 5 barplots (with 5 bars per plot) and somehow only
showing the middle bar for the top and bottom plots and the two end bars for
the two middle plots.
-
In theory, practice and theory are the same. In practice, they are not - Albert
Einstein
--
View this message in context:
Here is the new code. It works just like I wanted.
dev.new(width=6, height=6.5,mar=c(0,0,0,0))
par(mfrow=c(5,1),mar=c(.5, .5, 1.5, .5), oma=c(.4, 0,.5, 0))
barplot(c(1,1,1,1,1),col=c("white","white","red","white","white"), axes =
FALSE,border=NA)
barplot(c(1,1,1,1,1),col=c("orange","white","white"
I updated the code as follows:
dev.new(width=2.5, height=3,mar=c(0,0,0,0))
par(mfrow=c(5,1),mar=c(.5, .5, 1.5, .5), oma=c(.4, 0,.5, 0))
barplot(c(1,1,1,1,1),col=c("white","white","red","white","white"), axes =
FALSE,border=NA)
barplot(c(1,1,1,1,1),col=c("orange","white","white","white","yellow"), a
I decided to go with circles instead of rectangles. Thank you for your help.
Here is the new code:
dev.new(width=2.5, height=3,mar=c(0,0,0,0))
par(mfrow=c(1,1),mar=c(0,0,0,0))
x=c(-1,1,1,-1,-3,-3,5)
y=c(1.2,0.6,-.7,-1.3,-.7,0.6,5)
plot(0,xlim=c(-4,2),ylim=c(-2,2),type="n",axes=FALSE,xlab="",ylab="
I was able to solve this problem by going back to nls and obtaining the
initial parameter estimates through optim. When I used nlsList with my
dataset, it took 2 minutes to solve and was not limited by the bounds. Now I
have the bounds working and it takes 45 seconds to solve. Here is the new
code:
Here is the code I am running:
library(nls2)
modeltest<- function(A,mu,l,b,thour){
out<-vector(length=length(thour))
for (i in 1:length(thour)) {
out[i]<-b+A/(1+exp(4*mu/A*(l-thour[i])+2))
}
return(out)
}
A=1.3
mu=.22
l = 15
b = .07
thour = 1:25
Yvals<-modeltest(A,mu,l,b,thour)-.125+runif(25)/4
Sometimes I have NA values within specific columns of a dataframe (in this
example, the first two columns can have NAs). If there are NA values, I
would like them to be removed.
I have been using the code:
y<-c(NA,5,4,2,5,6,NA)
z<-c(NA,3,4,NA,1,3,7)
x<-1:7
adata<-data.frame(y,z,x)
adata<-adata[-w
Thank you for the help and explanations. I used the "complete.cases" function
and it is working great.
adata[complete.cases(adata[,1:2]),]
-
In theory, practice and theory are the same. In practice, they are not - Albert
Einstein
--
View this message in context:
http://r.789695.n4.nabble.
I have encountered this problem on several occasions and am not sure how to
handle it. I use for-loops to cycle through datasets. When each dataset is
of equal length, it works fine as I can combine the datasets and have each
loop pick up a different column, but when the datasets are differing
leng
I adapted a selfStart function and the lower bounds are not working. The
parameter "b" is negative, whereas I would like the lower bound to be zero.
Any ideas? Thanks.
Here is my code (I am still figuring out how to easily make replicable
examples):
A<-1.75
mu<-.2
l<-2
b<-0
x<-seq(0,18,.25)
create
I tested the "optim" function and that is returning non-negative parameter
values (meaning they are bound by the lower limits), but I think those are
the starting estimates for the nlsList model which is then finding negative
values for the solution.
-
In theory, practice and theory are the sa
I ran the code again and got an error saying that the "x" was unknown. I
don't know why I hadn't seen that error before. Anyway, I made the edits to
"func1" so instead of "x", it is "xy$x."
#function to optimize
func1 <- function(value) {
A.s <- value[1]
mu.s <- value[2]
l.s <- value[3]
40 matches
Mail list logo
