Thanks for help, now its working.
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I know that maybe it will be stupid question.
What iswrong with it (i think that i have to do the "stop moment", but i
dont know how)
fibbonacci=function(x) {
while(x>0) {
if (x==1 || x==2) {
return(1)
} else fibbonacci(x-1)+fibbonacci(x-2)
}
}
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On Sat, Nov 10, 2012 at 7:41 AM, Haszun wrote:
> I know that maybe it will be stupid question.
>
> What iswrong with it (i think that i have to do the "stop moment", but i
> dont know how)
>
> fibbonacci=function(x) {
> while(x>0) {
> if (x==1 || x==2) {
> return(1)
> } else fibbonacci(x-1)+fibbo
On 11/09/2012 10:33 PM, Geophagus wrote:
Hi @ all,
I try to set a labeling on simple barchart.
I do it with the text function. I want to see values of the x axis
(BE_AKT$ammo) above the bars.
When I try the following code, the values are shown, but not in the correct
position.
They should be labe
Thank you for help.
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Dear all users,
I"d like to ask you how to make decision about colinearity among
categorical independent variables
when the model is multinomial logistic regression.
Any help is appreciated,
Niklas
[[alternative HTML version deleted]]
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Btw, I tried out variance inflation factor(vif)
but it works for glm models(lm) but not multinom or nnet class
Bests,
2012/11/10 Niklas Fischer
> Dear all users,
>
> I"d like to ask you how to make decision about colinearity among
> categorical independent variables
> when the model is multinomi
Dear R community,
The new version of package 'bit64' - which extends R with fast 64-bit
integers - now has fast (single-threaded) implementations of the most
important univariate algorithmic operations (those based on hashing and
sorting). Package 'bit64' now has methods for 'match', '%in%',
(cross-posted to Graphviz-devel)
Rgraphviz 2.2.1 has been released as part of the Bioconductor 2.11 release.
Rgraphviz is an R interface to the Graphviz library
(www.graphviz.org), which is used for graph
layouts. Rgraphviz is one of the most popular packages from the
Bioconductor project with m
Hi everybody!
I suceeded in extracting the sub-matrix containing just the relevant
subjects of this example data (indeed I could retrieve 190 of the 220
subjects of the data example, so that the relative kinship matrix is a
190X190). For the subjects non present in the kinship matrix case-control
k
Thanks a lot!
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Dear friends from the R-community,
I am Guy Roger and live in Germany. I am student and user of R. I wanted to do
some prediction after a linear regression fitting but could´nt because of the
following message:
> yx<-predict(lm(y~x4+x5),newdata=data.frame(X4,X5))
Warnmeldung:
In predict.lm(lm(y
Hi everybody,
I am beginer in R and I need your precious help.
I want to create a small function in R as in sas to retrieve date.
I have a file with data that import in R.
DATE PAYS nb_pays.ILI.
1 24/04/2009 usa0
2 24/04/2009
On Nov 10, 2012, at 7:56 AM, bokaha guy wrote:
> Dear friends from the R-community,
>
> I am Guy Roger and live in Germany. I am student and user of R. I wanted to
> do some prediction after a linear regression fitting but could´nt because of
> the following message:
>
>> yx<-predict(lm(y~x4+
Why it always gives me a 3?
> fun=function(x) {
+ if (x<-3) {
+ return(x)
+ } else {
+ if(x<2) {
+ return(x^2-1)
+ } else {
+ return(log(x))
+ }}}
>
> fun(-5)
[1] 3
> fun(0)
[1] 3
> fun(10)
[1] 3
> fun(-10)
[1] 3
>
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HI,
May be this helps:
dat1<-read.table(text="
DATE PAYS nb_pays.ILI.
1 24/04/2009 usa 0
2 24/04/2009 usa 0
3 24/04/2009 Mexique 0
4 24/04/2009 Mexique 0
5 26
Thank you for help.
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On Nov 10, 2012, at 8:34 AM, Haszun wrote:
> Why it always gives me a 3?
>
>> fun=function(x) {
> + if (x<-3) {
The above code assigns 3 to x.
> + return(x)
> + } else {
> + if(x<2) {
> + return(x^2-1)
> + } else {
> + return(log(x))
> + }}}
>>
>> fun(-5)
> [1] 3
>> fun(0)
> [1] 3
>> fun(10
Hello,
If I understand it correctly, you have a data.frame whose first column
is a date and want to extract all lines between two given dates. If so,
try the following. Note that I've added two new arguments to your function.
dat <- read.table(text="
DATE PAYS nb
Hello,
Sorry, forgot the sum part.
extraction <- function(DF, date1, date2, format = "%Y-%m-%d"){
date1 <- as.Date(date1, format)
date2 <- as.Date(date2, format)
idx <- date1 < DF[[1]] & DF[[1]] < date2
aggregate(DF[idx, 3], DF[idx, 1:2], FUN = sum)
}
Hope this helps,
Rui Barr
Thank you, Bert and Peter, for helpful responses. I'm having a little trouble
with Bert's approach because writing the lapply function is challenging when
I'm drawing from two dataframes. Peter's approach works perfectly, although it
has less "R" personality. Wrapping the model statement in
hi
could you help me to solve this issue
Question:
Using command rweibull(100,8,15), simulate n = 100 realizations from
Weibull(8; 15) distribution. Using the simulated sample, compute the sample
mean, variance and standard deviation of these observations.
I am trying like this
sim<-rweibull(100
HI Rui,
For some reason, I am not getting the result as expected.
date1
#[1] "24/04/2009"
date2
#[1] "27/04/2009"
extraction(dat,date1,date2,format="%Y-%m-%d")
#[1] DATE PAYS x
#<0 rows> (or 0-length row.names)
#Warning messages:
#1: In extraction(dat, date1, date2, format = "%Y-%m-%d") :
This is not a homework help list.
On Saturday, November 10, 2012, parvez_200207 wrote:
> hi
> could you help me to solve this issue
>
> Question:
> Using command rweibull(100,8,15), simulate n = 100 realizations from
> Weibull(8; 15) distribution. Using the simulated sample, compute the sample
>
Hello,
Arun, you're using the wrong format, "%Y-%m-%d" is the default, with
"24/04/2009" you must use
extraction(dat, date1, date2, format = "%d/%m/%Y")
#DATEPAYS x
#1 2009-04-26 Mexique 18
#2 2009-04-26 usa 100
Rui Barradas
Em 10-11-2012 18:26, arun escreveu:
HI Rui,
Fo
Hello all,
i would like to calculate the difference of all row values and the others
row values from my matrix (table 1). The output (table 2) would be a matrix
with input matrix's row names on row names and colums names, thereby the
difference values among two of the row names could be bether foun
Thank you very much, arun kirshna!
That's it! I only modified the "res1<-apply(toeplitz(dat1[,2]),1,function(x)
10-x)" for "res1<-apply(toeplitz(dat1[,2]),1,function(x) dat1[1,2]-x)" and
worked very well!
Thanks again!
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Ok, I've coerced DATE to class Date before running the function:
dat$DATE <- as.Date(dat$DATE, format = "%d/%m/%Y")
Without it the function would be:
extraction <- function(DF, date1, date2, format = "%Y-%m-%d"){
date <- as.Date(DF[[1]], format)
date1 <- as.Date(date1, format)
date
Hello,
Try the following.
# Create the dataset
Table1 <- matrix(10:6, ncol = 1)
rownames(Table1) <- letters[1:5]
Table1
t(outer(Table1[,1], Table1[,1], `-`))
Hope this helps,
Rui Barradas
Em 10-11-2012 18:32, cleberchaves escreveu:
Hello all,
i would like to calculate the difference of all r
Your code works for me, can you tell us what output you are getting, what
output you expect to see, and how they differ?
On Sat, Nov 10, 2012 at 11:23 AM, parvez_200207 wrote:
> hi
> could you help me to solve this issue
>
> Question:
> Using command rweibull(100,8,15), simulate n = 100 realiza
The package rootoned on http://r-forge.r-project.org/R/?group_id=395
has an all-R version of zeroin (the algorithm of uniroot). This should
also be in Rmpfr by Martin M., as it was set up for that use. I suspect
it can be vectorized fairly easily. However, it may be simpler to write,
or else ab
This is to all R-helpers (Sarah is just the one that I am replying to),
Have we become a little too draconian on the "not a homework help list"
issue?
Now if someone just states the HW question, gives no indication that they
have done anything to try to solve it themselves, and expects us to give
On 10-11-2012, at 21:09, Greg Snow wrote:
> This is to all R-helpers (Sarah is just the one that I am replying to),
>
> Have we become a little too draconian on the "not a homework help list"
> issue?
Probably.
>
> Now if someone just states the HW question, gives no indication that they
> ha
On 10-11-2012, at 19:23, parvez_200207 wrote:
> hi
> could you help me to solve this issue
>
> Question:
> Using command rweibull(100,8,15), simulate n = 100 realizations from
> Weibull(8; 15) distribution. Using the simulated sample, compute the sample
> mean, variance and standard deviation of
I agree with much of what you said. If there is a reasonable effort to have
read the documention or otherwise to have solved the problem on their own,
and a clear question, I will frequently at least give a hint or a pointer
toward a relevant function or two. Also, I wouldn't consider that the firs
It is not always easy to discern what the instructor wants a student to get out
of an assignment. Therefore, I can't see changing the policy as it stands.
That said, it is not always easy to discern homework from self-study, and
sometimes when the question is well-constructed I don't go out of
Rolf,
Re version control: I use SVN and Git depending on the project I am working on
and what others are using. Years ago I used RCS, as you say its great for a
local repository (as is Git). The point I was making was not about version
control but that others like me might get caught out by sav
Zoosk
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Hi All
I have to run multiple stimations and to compute Likelihhod ratio.
If I compute ls function with coef and summary I can extract "outputs" that
I need.
I am not able to find something similar to log liklihood
Can you pease tell me running a ls function x on y how to extract if
posib
Thanks, Rui,
Got it right:
extraction(dat,date1,date2,format="%d/%m/%Y")
# DATE PAYS x
#1 26/04/2009 Mexique 18
#2 26/04/2009 usa 100
A.K.
- Original Message -
From: Rui Barradas
To: arun ; r-help
Cc:
Sent: Saturday, November 10, 2012 2:33 PM
Subject: Re: [R] help
Mmmm...
Actually, Rui Barradas is the right!
Arun kirshna, yours script has an error. That repeat the same set of numbers
in all columns...
Anyway, thanks for both!
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Rui and ,arun thanks you so much.
as i am a beginner, i am on that subject since two days.
Thanks,thanks so much.
It works !!!
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HI,
No problem.
You can also use ?aaply() from library(plyr)
dat1<-read.table(text="
a 10
b 9
c 8
d 7
e 6
",sep="",header=FALSE,stringsAsFactors=FALSE)
library(plyr)
res1<-aaply(dat1[1,2],1,"-",toeplitz(dat1[,2]))
res1
Once again, thanks!
MVS
-
MVS
=
Matthew Van Scoyoc
Graduate Research Assistant, Ecology
Wildland Resources Department & Ecology Center
Quinney College of Natural Resources
Utah State University
Logan, UT
=
Think SNOW!
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On Nov 10, 2012, at 11:57 AM, bgnumis bgnum wrote:
> Hi All
>
>
> I have to run multiple stimations and to compute Likelihhod ratio.
>
> If I compute ls function with coef and summary I can extract "outputs" that
> I need.
Do you mean "lm"?
>
> I am not able to find something similar to lo
I want my for loop to test for the presence of a term in a vector and return
a value to a new vector. I'm not writing it correctly though. Here's what I
have...
> testfor = letters[1:5]
> x = c("a", "b", "e", "f", "g")
> result = rep(NA, length(testfor))
>
> for (i in testfor){
+ v = any(x == t
I have this code:
IEF <- to.monthly(IEF, indexAt="endof")
SPY <- to.monthly(SPY, indexAt="endof")
I would like to use a for loop instead of separate entries,
so the only code that needs to be modified is the list
of symbols.
symbols <- c("IEF", "SPY")
for(symbol in symbols) {
symbol <- to.m
On Nov 10, 2012, at 2:07 PM, scoyoc wrote:
> I want my for loop to test for the presence of a term in a vector and return
> a value to a new vector. I'm not writing it correctly though. Here's what I
> have...
>
>> testfor = letters[1:5]
>> x = c("a", "b", "e", "f", "g")
>> result = rep(NA, leng
On Nov 10, 2012, at 2:36 PM, dae wrote:
> I have this code:
> IEF <- to.monthly(IEF, indexAt="endof")
> SPY <- to.monthly(SPY, indexAt="endof")
>
> I would like to use a for loop instead of separate entries,
> so the only code that needs to be modified is the list
> of symbols.
>
> symbols <- c
Hi,
Though the results from Rui, and me are similar, may be it differs in other
instances.
My result:
dat1<-read.table(text="
a 10
b 9
c 8
d 7
e 6
",sep="",header=FALSE,stringsAsFactors=FALSE)
res1<-apply(toeplitz(dat
On 11/11/2012 07:09 AM, Greg Snow wrote:
This is to all R-helpers (Sarah is just the one that I am replying to),
Have we become a little too draconian on the "not a homework help list"
issue?
...
As usual, a thoughtful comment on a problem that does not have a
straightforward solution. The ac
I want to find ML estimates of a model using mle2 in bbmle package. When I
insert new parameters (for new covariates) in model the log-likelihood value
does not change and the estimated value is exactly the initial value that I
determined. What's the problem? This is the code and the result:
As
Thanks. That got me the answer. This works:
symbols = c("IEF","SPY")
getSymbols(symbols)
for(symbol in symbols) {
assign(symbol, to.monthly(get(symbol), indexAt="endof"))
}
#end
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Hello.
How can I plot the regression curve with its confidence interval?
I use nls(y~a*x^b), then plot(x,y), add curve with
lines(x,predict(nls(y~a*x^b))). But I can't add to plot CI for my curve.
Thanks in advance,
Denis
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On Nov 10, 2012, at 6:58 PM, Jim Lemon wrote:
> On 11/11/2012 07:09 AM, Greg Snow wrote:
>> This is to all R-helpers (Sarah is just the one that I am replying to),
>>
>> Have we become a little too draconian on the "not a homework help list"
>> issue?
>> ...
>
> As usual, a thoughtful comment o
On Nov 10, 2012, at 10:17 PM, dae wrote:
> Thanks. That got me the answer.
Good. That's what the FAQ is for. You should also read the Posting Guide where
the reasons behind the request to include context for replies is laid out.
> This works:
>
> symbols = c("IEF","SPY")
>
> getSymbols(sym
On Nov 10, 2012, at 9:36 PM, Осипов Денис wrote:
> Hello.
>
> How can I plot the regression curve with its confidence interval?
> I use nls(y~a*x^b), then plot(x,y), add curve with
> lines(x,predict(nls(y~a*x^b))). But I can't add to plot CI for my curve.
Have your read:
?predict.nls # ?
-
On Nov 10, 2012, at 9:22 PM, mmosalman wrote:
> I want to find ML estimates of a model using mle2 in bbmle package. When I
> insert new parameters (for new covariates) in model the log-likelihood value
> does not change and the estimated value is exactly the initial value that I
> determined. Wha
On 11/11/2012 04:36 PM, Осипов Денис wrote:
Hello.
How can I plot the regression curve with its confidence interval?
I use nls(y~a*x^b), then plot(x,y), add curve with
lines(x,predict(nls(y~a*x^b))). But I can't add to plot CI for my curve.
Hi Denis,
If you know where you want the lines for t
Yes, but it says that arguments, that could help me, now isn't implemented.
> Hello.
>
> How can I plot the regression curve with its confidence interval?
> I use nls(y~a*x^b), then plot(x,y), add curve with
> lines(x,predict(nls(y~a*x^b))). But I can't add to plot CI for my curve.
> Have your
On Nov 10, 2012, at 11:48 PM, Осипов Денис wrote:
> Yes, but it says that arguments, that could help me, now isn't implemented.
I think the authors might have had their reasons for leaving it out. There have
been quite a few threads ofver the years discussing why it is not a
straightforward ta
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