Thanks John.
I have one observation data point with a value that's exactly equal to the
predicted value, therefore the residual is 0. Would this be the reason
you mentioned below?
From: "John Fox"
To: Liang Che/US/TLS/PwC@Americas-US
Cc: "'Sanford We
Can someone please help with the below - thanks!
Warning messages:
1: Not plotting observations with leverage one:
7
2: Not plotting observations with leverage one:
7
> print(qqPlot(fit),envelop=.95);
Error in model.frame.default(formula = Y ~ X - 1, drop.unused.levels =
TRUE) :
variable lengt
Can someone please help with the error message below -- thanks!
Warning messages:
1: Not plotting observations with leverage one:
7
2: Not plotting observations with leverage one:
7
> print(qqPlot(fit),envelop=.95);
Error in model.frame.default(formula = Y ~ X - 1, drop.unused.levels =
TRUE) :
Can someone please help with the error message below?
Warning messages:
1: Not plotting observations with leverage one:
7
2: Not plotting observations with leverage one:
7
> print(qqPlot(fit),envelop=.95);
Error in model.frame.default(formula = Y ~ X - 1, drop.unused.levels =
TRUE) :
variable
For example:
How to make R write out:
Balance = 2 + 3 * IntGDP + 5 * IntUnemployment + 0.3 * d1
from the table below:
Balance Intercept IntGDP GDPNum IntUnemployment
IntInflationd1 d2 d3
3 2 3 5 0.3 0 0
will add() or drop() function work more similarly as SAS?
I understand that there are not many observation points which might cause
the problem, but why can the automated process run successfully in SAS
instead?
From: David Winsemius
To: Liang Che/US/TLS/PwC@Americas-US
Cc:
Date
one extra question
How do I export the final image (with plots and lines added) into files in
a batch mode for a bunch of regressions?
thanks
- Forwarded by Liang Che/US/TLS/PwC on 10/08/2012 08:48 PM -
From: Liang Che/US/TLS/PwC
To: Joseph Clark @INTL
Cc: r-help@r
thank you all for your answers
i have figured out the plot i wanted - it was actually pretty easy
lines(x=20:34,prd[,1],col="red",Ity=1)
lines(x=20:34,prd[,2],lty=2)
lines(x=20:34,prd[,3],lty=2)
From: Joseph Clark
To: Liang Che/US/TLS/PwC@Americas-US
Cc:
Date: 10/
Ran a bunch of variables in R and the final result of StepAIC is as below:
Why are the first 5 variables kept in the stepwise result?? Are the last
4 variables finally chosen after Stepwise? Thanks
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.315e-01 2.687e-01 0.490 0.6361
- x :
the condition has length > 1 and only the first element will be used
From: Joseph Clark
To: Liang Che/US/TLS/PwC@Americas-US
Cc:
Date: 10/08/2012 05:06 PM
Subject:RE: [R] How to use Lines function to draw the error bars?
In my example code, 'fit'
From: Joseph Clark
To: Liang Che/US/TLS/PwC@Americas-US,
Date: 10/08/2012 04:03 PM
Subject:RE: [R] How to use Lines function to draw the error bars?
I typically use the function "plotCI" from the "plotrix" package for
confidence intervals or error bars
Thanks -- I got this message:
Error in stripchart.default(x1, ...) : invalid plotting method
From: Bita Shams
To: Liang Che/US/TLS/PwC@Americas-US
Date: 10/08/2012 03:41 PM
Subject:Re: [R] How to use Lines function to draw the error bars?
try this :
plot(new,prd[,3
For a set of data showing seasonality (related to the 4th quarter), ncv
test in R shows p-value of 0.008 which rejects the null hypothesis of
constant-variance. How to apply White's standard error in R?
thanks
__
The i
fit lwrupr
1 218.4332 90.51019 346.3561
2 218.3906 90.46133 346.3198
3 218.3906 90.46133 346.3198
4 161.3982 44.85702 277.9394
5 192.4450 68.39903 316.4909
6 179.8056 56.49540 303.1158
7 219.5406 91.52707 347.5542
8 162.6761 46.65760 278.6945
9 193.8506 70.59
For a set of data showing seasonality (related to the 4th quarter),
ncv
test shows p-value of 0.008 which rejects the null hypothesis of
constant-variance. Currently a linear LM relationship is being
applied to
the data.
Should white's error be used to correct the non-cons
For example, if coefficient's p-value is less than 0.1 I want the stepwise
to automatically drop that variable. Can the stepAIC be customized to do
that? SAS seems to be able to customized stepwise function with p-value
or cooks'd.
thanks!
Can someone please help with the error message below?
thanks!
Start: AIC=-Inf
value ~ 1 + Core_CPI__ + GDP_change + Unemployment + housing +
interest + S_P + d1 + d2 + d3
Error in if (any(ch)) { : missing value where TRUE/FALSE needed
In addition: Warning message:
attempting model select
See error message below: can someone please help with this? Thanks!
Residuals:
ALL 9 residuals are 0: no residual degrees of freedom!
Residual standard error: NaN on 0 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: NaN
F-statistic: NaN on 8 and 0 DF, p-value: NA
My stepAIC function works for one set of data but not anotherone set
of data shows the steps of eliminating variables, versus another set of
data doesn't throw away any variables.
Can anyone please explain why? Thanks
__
T
hey Jim.
brilliant, very short and productive, wish that i can have such skill in the
future, i will try to learn about all functions that you used.
thanks very much for helping me, i really appreciate it.
--
View this message in context:
http://n4.nabble.com/More-than-on-loop-tp1015851p1460459
hello,
i appreciate your help, your help, comments, and suggestion really are so
helpful to develop not only my R skills, but also my programming language
sense. as i said, i am not breaking any academic rules, and this is a
softwar i have to develop to deal with my project after two months along
Here is the the written instruction as i managed to get it from my professor,
the graphs and data are attached:
The graph below shows an example of the expected outcome of this course
work. You may
procude a better one. The graph for analysing the motifs of a set of
peptides is designed
this way
yes, but the outcome graphs are almost the same, that mean it does not
calculated in a cumulative way , if you apply the following code, then run
hi(x), and then recta(x), you will see how the shape are similar to the
frequency of Amino Acid in the matrix. i am looking for a code that can do
thi
70% yes, the problem is i am trying to produce a graph similar to the one in
attachments in this message, which represents the frequency of each letter
"aminoacid" in the cleaved function and the noncleaved function. some thing
else i added to the attachments is the pattern which seemingly working
hopefully it is here, two files, one of them is .dat and the others is .txt,
just in case.
http://n4.nabble.com/file/n1290026/hiv.dat hiv.dat
http://n4.nabble.com/file/n1290026/hiv.txt hiv.txt
--
View this message in context:
http://n4.nabble.com/More-than-on-loop-tp1015851p1290026.html
Sent f
here you are the whole code, and the data is attached:
> x<-read.table("C:/Uni/502/CA2/hiv.dat", header=TRUE)
> num<-nrow(x)
> AA<-c('A','C','D','E','F','G','H','I','K','L','M','N','P','Q','R','S','T','V','W','Y')
> nc<-x$Label[61:308]
> c<-x$Label[nc]
> noncleaved<-function(x)
+ {
+ y<-matrix(0,2
can i ask your help again, please excuse my questions:
It is working perfectly now, i still have the last part which i tried a lot
with but still i can’t translate it properly for the computer through R. I
need to draw rectangular based on the frequency of each residue, actually i
found the patter
Really thanks very much, with your help i was able to write a prober code to
count the aminoacids in all the cleaved and noncleaved and then to display
the results in a matrix with 8 column, i used only two loops instead of
three. The code is working but i still have warning telling me that:
“In
help a lot to develop my
skills. Moreover i am dealing with it in a very honesty way that does not
break any academic regulations.
thanks again i will try what you sent me ..
Yours
che wrote:
>
> hello every one,
>
> How to function more than one loop in R? I have the followin
hello every one,
How to function more than one loop in R? I have the following problem to be
solved with the a method of three loops, can you help me please?
The data is attached with this message.
The data is composed of two parts, cleaved (denoted by “cleaved”) and non
cleaved (denoted by “no
ion(X)
{
y<-rep(0,20)
for(j in 1:test){
for(i in 1:nrow(x)){
res<-which(AA==substr(x$V1[i],j,j))
y[res]=y[res]+1
}
}
return(y)
}
So how to fix this code, how to give the life for the “i” and the “j” in
order to initiate the indexing. Sorry for botheri
confused between using "i" and "j" and how to iterate both of them and make
them work functionally. i attached the sequence.txt with my original
message, and i will attach it here in case. thanks for your help.
http://n4.nabble.com/file/n997087/sequence.txt sequence.txt
che wrot
may some one please help me to sort this out, i am trying to writ a R code
for calculating the frequencies of the amino acids in 9 different sequences,
i want the code to read the sequence from external text file, i used the
following code to do so:
x<-read.table("sequence.txt",header=FALSE)
then
apply a function by subjects? Thank you!
Best,
Che-hsu (Joe) Chang, Sc.D., P.T.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http
34 matches
Mail list logo
