Hi Marc:
Thank you for your help in this matter.
With thanks
Abou
On Tue, Aug 10, 2021, 9:28 AM Marc Schwartz wrote:
> Hi,
>
> A search would suggest that there may not be an R function/package that
> provides power/sample size calculations for the specific scenarios that
> you are describi
Hi,
A search would suggest that there may not be an R function/package that
provides power/sample size calculations for the specific scenarios that
you are describing. There may be something that I am missing, and there
is also other dedicated software such as PASS
(https://www.ncss.com/softw
Hi Marc:
First, thank you very much for your help in this matter.
Will perform an initial omnibus test of all three groups (e.g. 3 x 2
chi-square), possibly followed by
all possible 2 x 2 pairwise comparisons (e.g. 1 versus 2, 1 versus 3,
2 versus 3),
We can assume *either* the desired sample s
Hi,
You are going to need to provide more information than what you have
below and I may be mis-interpreting what you have provided.
Presuming you are designing a prospective, three-group, randomized
allocation study, there is typically an a priori specification of the
ratios of the sample s
> Thanks for the suggestion but I'm not sure that it answers my original
> question.I need to know how many samples I need to collect to collect in
> order to estimate the sample size needed to achieve a specific margin of
> error for confidence intervals for the population variance. I'm not sure
Thanks for the suggestion but I'm not sure that it answers my original
question.I need to know how many samples I need to collect to collect in order
to estimate the sample size needed to achieve a specific margin of error for
confidence intervals for the population variance. I'm not sure wheth
On Tue, 2 Jul 2019 22:23:18 + (UTC)
Thomas Subia via R-help wrote:
> Colleagues,
> Can anyone suggest a package or code which might help me calculate
> the minimum sample size required to estimate the population variance?
> I can do this in Minitab but I'd rather do this in R. Thomas Subia
Y
Hello,
Sorry for the HTML, I hadn't realised that my cellphone was not
sending in plain text.
Rui Barradas
Citando ruipbarradas :
Hello,Maybe I'm not understanding but to compute the sample variance
you need at least 2 data points? The ML estimator can be computed
with just 1 but
Hello,Maybe I'm not understanding but to compute the sample variance you need
at least 2 data points? The ML estimator can be computed with just 1 but it
will be zero.Hope this helps,Rui Barradas Enviado a partir do meu smartphone
Samsung Galaxy. Mensagem original De: Thomas Subi
Does this help?
https://www.r-bloggers.com/computing-sample-size-for-variance-estimation/
On Wed, 3 Jul 2019 at 10:23, Thomas Subia via R-help
wrote:
> Colleagues,
> Can anyone suggest a package or code which might help me calculate the
> minimum sample size required to estimate the population v
On 05/06/2019 4:34 a.m., le Gleut, Ronan wrote:
Dear R-help mailing list,
First of all, many many thanks for your great work on the R project!
I have a very small issue regarding the sample function. Depending if we
specify values for the prob argument, we don't get the same result fo
Thank you so much Marc,
that is exactly what I need. That will save me weeks of work and additionally I
learned a lot.
:-)
Have a great day!
Dagmar
Hi,
Given that your original data frame example is:
myframe <- data.frame (Timestamp=c("24.09.2012 09:00:00", "24.09.2012
10:00:00","25.09.2012
Hi,
Given that your original data frame example is:
myframe <- data.frame (Timestamp=c("24.09.2012 09:00:00", "24.09.2012
10:00:00","25.09.2012 09:00:00",
"25.09.2012 09:00:00","24.09.2012 09:00:00",
"24.09.2012 10:00:00"),
Event=c(50,60
Hi Marc,
Yes, you got it to the point! That is exactly what I want. But I do not know
how to do that. I know how to randomly pick the first day but I do not know how
to set a range of values which cover the 25 days starting from that random
value.
Best,
Dagmar
Hi,
I am confused.
As far as
Hi,
I am confused.
As far as I can tell, only the first day is selected randomly from your
dataset. The subsequent 24 days are deterministic, since they need to be
consecutive days from the first day, for a total of 25 consecutive days.
Thus, all you need to do is to randomly select 1 day fro
myfirst <- sample( seq.int( nrow(myframe)-1 ), 1 )
mysample <- myframe[seq( myfirst, myfirst+1),]
mysample
On December 7, 2018 2:24:11 AM PST, Dagmar Cimiotti
wrote:
>Hi Jim and everyone else,
>
>Mhm, no this is not what I am looking for. I think in your way I would
>randomly sample two values
Hi Jim and everyone else,
Mhm, no this is not what I am looking for. I think in your way I would
randomly sample two values of day 1 and of day 2. But I want the
opposite: I want to randomly draw two successive (!) days and put those
values in a new dataframe to continue working with them.
I
Hi Dagmar,
This will probably involve creating a variable to differentiate the
two days in each data.frame:
myframe$day<-as.Date(as.character(myframe$Timestamp),"%d.%m.%Y %H:%M:%S")
days<-unique(myframe$day)
Then just sample the two subsets and concatenate them:
myframe[c(sample(which(myframe$da
Yes.
Beating a pretty weary horse, a slightly cleaner version of my prior
offering using with(), instead of within() is:
with(dat,
dat[sampleNo[sample(var1[!var1%%2 & !sampleNo], 10, rep=FALSE)],
"sampleNo"] <- 2)
with() and within() are convenient ways to avoid having to repeatedly name
the col
Hi David,
I was about to post a reply when Bert responded. His answer is good
and his comment to use the name 'dat' rather than 'data' is instructive.
I am providing my suggestion as well because I think it may address
what was causing you some confusion (mainly to use "which", but also
the missing
For personal aesthetic reasons, I changed the name "data" to "dat".
Your code, with a slight modification:
set.seed (1357) ## for reproducibility
dat <- data.frame(var1=seq(1:40), var2=seq(40,1))
dat$sampleNo <- 0
idx <- sample(seq(1,nrow(dat)), size=10, replace=F)
dat[idx,"sampleNo"] <-1
## yi
Sounds like a homework problem. This list has a no homework policy if it is.
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Mon, Sep 4, 2017 at 10:31
Or power.t.test()
-pd
> On 12 Apr 2017, at 18:44 , Bert Gunter wrote:
>
> Search "sample size power" on rseek.org. Many useful hits, including
> "samplesize" package.
>
> -- Bert
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking thin
Search "sample size power" on rseek.org. Many useful hits, including
"samplesize" package.
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Apr 12
> On Apr 12, 2017, at 3:20 AM, Jomy Jose wrote:
>
> In R how to calculate sample size,where power,residual standard deviation
> and treatment difference is given.?
Use the non-central t-distribution. The help page has further advice:
?pt
>
> [[alternative HTML version deleted]]
Pleas
Sample without replacement and then split that sample into train and
test components.
Jim
On Fri, Dec 9, 2016 at 4:55 PM, Partha Sinha wrote:
> How to get two sets of non overlapping data?
> Regards
> Parth
__
R-help@r-project.org mailing list -- To U
df <- data.frame(x = 1:12, y = rnorm(12))
If you use sample:
RowIndex <- sample(1:nrow(df), 5)
TrainSet <- df[RowIndex, ]
TestSet <- df[-RowIndex, ]
Or with dplyr:
TrainSet <- sample_n(df, 5)
TestSet <- anti_join(TestSet, df)
HTH
Ulrik
On Fri, 9 Dec 2016, 06:56 Partha Sinha, wrote:
> How to
How to get two sets of non overlapping data?
Regards
Parth
On 8 December 2016 at 23:23, Ulrik Stervbo wrote:
> In addition to 'sample', and if you insist on dplyr, you can use
> 'sample_n'.
>
> Best,
> Ulrik
>
> On Thu, 8 Dec 2016 at 18:47 Bert Gunter wrote:
>
>> Usually we expect posters to do
In addition to 'sample', and if you insist on dplyr, you can use 'sample_n'.
Best,
Ulrik
On Thu, 8 Dec 2016 at 18:47 Bert Gunter wrote:
> Usually we expect posters to do their homework by reading necessary R
> documentation and relevant subject matter resources (e.g. on
> clustering) and making
Usually we expect posters to do their homework by reading necessary R
documentation and relevant subject matter resources (e.g. on
clustering) and making a serious attempt to solve the problem by
offering their code to us along as part of a reproducible example of
how it failed. You have done none
lp@r-project.org
Subject: Re: [R] sample within a loop
Hi,
you can try
df1<-split(df,df$groups)
lapply(df1, function(x)
{
x<-cbind(x,entry=0)
sam <- sample(x$plotno,1)
x$entry[which(x$plotno==sam)]<-"CONTROL"
x$entry[which(!x$plotno==sam)]<-"TEST"
x
Hi,
you can try
df1<-split(df,df$groups)
lapply(df1, function(x)
{
x<-cbind(x,entry=0)
sam <- sample(x$plotno,1)
x$entry[which(x$plotno==sam)]<-"CONTROL"
x$entry[which(!x$plotno==sam)]<-"TEST"
x
}
)
Tanvir Ahamed
Göteborg, Sweden | mashra...@yahoo.com
___
I would change strategies.
Create a new variable, say,
num.in.grp <- rep(1:12, 4)
Then sample from 1:12, and add appropriate amounts so that they become row
numbers within the four sets of 12 rows
ctrls <- ssample(1:12, 4, replace=TRUE) + c(0,12,24,36)
Now that we have four random row number
Hi
> wrote:
> >Dear Sir,
> >
> >I do appreciate your views. Yes even I was also aware about the non
> >clarity in the question. Actaully, I have a large data having lots of
> >data of low magnitude and few of very high magnitude. In order to
can you explain what is low or high magnitude? Or be
R is a computing tool, and each package has implemented algorithms that have
history and books and papers that allow those algorithms to be used in a
variety if computing environments... from Fortran to Excel to Java to ... R,
and probably beyond.
>From your description I am going to hazard a g
If you have a clear idea what meaning those weights have (?) in the context of
a specific calculation (?), and you know what the weights are (?), then it is
usually trivially easy to do in R. However, your question is vague on all of
those points, so offering you a solution seems like an invita
gt; a b c d
> [1,] 10 5 5 0
> [2,] 11 4 4 1
> [3,] 12 3 3 2
> [4,] 13 2 2 3
> [5,] 14 1 1 4
> [6,] 15 0 0 5
>
> David C
>
> From: Michael Peng [mailto:michael.gang.p...@gmail.com]
> Sent: Tuesday, June 24, 2014 1:32 PM
> To: David L Carlson
> Cc: Gabor Grot
Sent: Tuesday, June 24, 2014 1:32 PM
To: David L Carlson
Cc: Gabor Grothendieck; Tahira Jamil; r-help@r-project.org
Subject: Re: [R] Sample all possible contingency tables both margin fixed
David gave a great solution. I think it is better to start from 0 to min(M)
instead of from min(M[c(1,3)]) to
--
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Gabor Grothendieck
> Sent: Tuesday, June 24, 2014 12:07 PM
> To: Tahira Jamil
> Cc: r-help@r-project.org
> Subject: Re: [R] Sample all possible contingency tables both margin fixed
>
&g
Cc: r-help@r-project.org
Subject: Re: [R] Sample all possible contingency tables both margin fixed
On Tue, Jun 24, 2014 at 10:41 AM, Tahira Jamil wrote:
> Hi,
>
> I am interested in generating all possible contingency table (2 by 2) with
> fixed row margins and column margins. Can anyone
On 24 Jun 2014, at 16:41 , Tahira Jamil wrote:
> Hi,
>
> I am interested in generating all possible contingency table (2 by 2) with
> fixed row margins and column margins. Can anyone help me.
>
It's pretty easy, just find one compatible solution and add or subtract
1 -1
-1 1
until one o
On Tue, Jun 24, 2014 at 10:41 AM, Tahira Jamil wrote:
> Hi,
>
> I am interested in generating all possible contingency table (2 by 2) with
> fixed row margins and column margins. Can anyone help me.
>
If the reason you want this is to sample them then r2dtable can do
that directly.
Hello,
At an R prompt, type
?power.t.test
Hope this helps,
Rui Barradas
Em 06-06-2013 20:58, Rebecca Greenblatt escreveu:
Looking to determine sample sizes for both my experimental and control
groups (I want only a small portion of my participants in my experimental
condition) in order to co
This is a duplicate question, right?
On Jun 6, 2013, at 12:58 PM, Rebecca Greenblatt wrote:
> Looking to determine sample sizes for both my experimental and control
> groups (I want only a small portion of my participants in my experimental
> condition) in order to compare population means. I wou
Hi,
Try:
fun1<- function(dat,Col1,Col2,number){
lst1<- split(dat,list(dat[,Col1],dat[,Col2]))
lst2<- lst1[lapply(lst1,nrow)>0]
res<- lapply(lst2,function(x) sample(x[,1],if(nrow(x)< number) nrow(x) else
number,replace=FALSE))
res}
head(fun1(Gpool,"LngtClas","SpCode",5),4)
#$`40_49.MERLMER`
#[
Thank you Sarah for this code, it's exactly what I wanted to reach.
Le 05/06/2013 16:49, Sarah Goslee a écrit :
What about using instead
size = min(5, length(Gpool$SampleNb[Gpool$LngtClas == LngtClas[[4]]&
Gpool$SpCode == SpCode[[2]]])
that would make sure your sample is either the size of t
Something like this?
> set.seed(42)
> a <- sample.int(10)
> b <- sample.int(5)
> c <- sample.int(3)
> smpl <- function(x) sample(x, ifelse(length(x)<5, length(x), 5))
> smpl(a)
[1] 4 8 1 10 2
> smpl(b)
[1] 2 3 5 1 4
> smpl(c)
[1] 2 1 3
-
David L Carlson
Ass
What about using instead
size = min(5, length(Gpool$SampleNb[Gpool$LngtClas == LngtClas[[4]] &
Gpool$SpCode == SpCode[[2]]])
that would make sure your sample is either the size of the data or 5.
Sarah
On Wed, Jun 5, 2013 at 10:27 AM, Xochitl CORMON
wrote:
> Hi all,
>
> I'm trying to randomly se
n Jones
> > Sent: Thursday, May 23, 2013 8:30 AM
> > To: r-help@r-project.org
> > Subject: Re: [R] sample(c(0, 1)...) vs. rbinom
> >
> > After a bit of playing around, I discovered that
> > sample() does something similar in other situations:
> >
&g
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Albyn Jones
> Sent: Thursday, May 23, 2013 8:30 AM
> To: r-help@r-project.org
> Subject: Re: [R] sample(c(0, 1)...) vs. rbinom
>
> After a bit of play
After a bit of playing around, I discovered that
sample() does something similar in other situations:
set.seed(105021)
sample(1:5,1,prob=c(1,1,1,1,1))
[1] 3
set.seed(105021)
sample(1:5,1)
[1] 2
set.seed(105021)
sample(1:5,5,prob=c(1,1,1,1,1))
[1] 3 4 2 1 5
set.seed(105021)
sample(1:5,5)
On May 23, 2013, at 07:01 , Jeff Newmiller wrote:
> You seem to be building an elaborate structure for testing the
> reproducibility of the random number generator. I suspect that rbinom is
> calling the random number generator a different number of times when you pass
> prob=0.5 than otherwis
You seem to be building an elaborate structure for testing the reproducibility
of the random number generator. I suspect that rbinom is calling the random
number generator a different number of times when you pass prob=0.5 than
otherwise.
-
.@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Shane Carey
> Sent: Friday, April 26, 2013 1:09 PM
> To: Rui Barradas
> Cc: r-help@r-project.org
> Subject: Re: [R] sample size in box plot labels
>
> This works, great. Cheers
>
>
> On Fri, Apr 2
This works, great. Cheers
On Fri, Apr 26, 2013 at 12:02 PM, Rui Barradas wrote:
> Hello,
>
> To count the sample sizes for each factor try
>
> tapply(DATA$K_Merge, DATA$UnitName_1, FUN = length)
>
>
> Hope this helps,
>
> Rui Barradas
>
> Em 26-04-2013 10:48, Shane Carey escreveu:
>
> Hi,
>>
>
Hello,
To count the sample sizes for each factor try
tapply(DATA$K_Merge, DATA$UnitName_1, FUN = length)
Hope this helps,
Rui Barradas
Em 26-04-2013 10:48, Shane Carey escreveu:
Hi,
I would like to put the sample number beside each lable in a boxplot.
How do I do this? Essentially, I need
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Shane Carey
> Sent: Friday, April 26, 2013 11:49 AM
> To: r-help@r-project.org
> Subject: [R] sample size in box plot labels
>
> Hi,
>
> I would like to put the sample numbe
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 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
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
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
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
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
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
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
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
>
an then simulate mutations by changing the
> gene values, cause duplications by adding rows of duplicated genes,
> or even cause deletions by removing rows.
> Once I have this set up for the pathogen I may make a similar array
> for the host plants, then perhaps with indexing or s
indexing or some such thing I can write
functions to do the interactions and immunology and such.
Best,
Ben W.
UEA (ENV) & The Sainsbury Laboratory.
____________
From: Jean V Adams [jvad...@usgs.gov]
Sent: 07 November 2012 21:12
To: Benjamin Ward (ENV)
Cc: r-help@r-projec
Ben,
Can you provide a small example data set for
inds
so that we can run the code you have supplied?
It's difficult for me to follow what you've got and where you're trying to
go.
Jean
"Benjamin Ward (ENV)" wrote on 11/06/2012 03:29:52 PM:
>
> Hi all,
>
> I have a list of genes p
Hello,
Function caret::createDatapartition preserves the proportions of
classes, like its documentation says, so you should expected the result
to be balanced only if the original data.frame is also balanced. A
solution is to write a small function that chooses a balanced set of
indices. Note
Please don't double post.
And see my response to you here:
https://stat.ethz.ch/pipermail/r-help/2012-October/325470.html
Michael
On Sat, Oct 6, 2012 at 6:51 PM, solafah bh wrote:
> Hello
> If I have this vector x=c(5,1,2,9) and n=length(x) and I want to sample one
> value from x , and each va
Hi,
They get different results:
with the same set.seed()
x=c(3,2,6,1)
n=length(x)
set.seed(1)
sample(x,1,replace=TRUE)
#[1] 2
set.seed(1)
sample(x,1,replace=TRUE,prob=rep(1/n , n) )
#[1] 6
identical(sample(x,1,replace=TRUE),sample(x,1,replace=TRUE,prob=rep(1/n , n) ))
#[1] FALSE
A.K.
Yes and no. Same effect, but you won't get the same random numbers
because -- I believe -- a different algorithm is used. grep the source
for sample and sample2 if you're interested.
Cheers,
Michael
On Sat, Oct 6, 2012 at 5:02 PM, solafah bh wrote:
> Hello
> If I have x=c(3,2,6,1) and n=length(x
This seems more theoretical than specific to R, so you should discuss this
question in a more theoretical forum such as http://stats.stackexchange.com/.
FWIW I believe the results will be equally random either way. That doesn't say
either way will be "absolutely" random, since I don't think such
Define: "Absolutely random"; "systematic differences"
(All pseudorandom numbers are by definition generated by a
deterministic algorithm from a possibly random starting seed set).
On Wed, Aug 15, 2012 at 9:11 AM, wrote:
> Hello,
>
> Vector y is an alphabetically sorted version of vector x. Wil
On Sun, Feb 12, 2012 at 01:57:18PM -0500, SUPAKORN LAOHAPITAKVORN wrote:
> This is what I got:
>
> > sessionInfo()
> R version 2.14.1 (2011-12-22)
> Platform: x86_64-pc-mingw32/x64 (64-bit)
>
> locale:
> [1] LC_COLLATE=English_United States.1252
> [2] LC_CTYPE=English_United States.1252
> [3] LC_
Supakorn,
Try:
rm(sample)
#then
sample(x)
Etienne
2012/2/12 SUPAKORN LAOHAPITAKVORN
> This is what I got:
>
> > sessionInfo()
> R version 2.14.1 (2011-12-22)
> Platform: x86_64-pc-mingw32/x64 (64-bit)
>
> locale:
> [1] LC_COLLATE=English_United States.1252
> [2] LC_CTYPE=English_United States.
This is what I got:
> sessionInfo()
R version 2.14.1 (2011-12-22)
Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
Hi,
The following is what I get:
> x =1:12
> x
[1] 1 2 3 4 5 6 7 8 9 10 11 12
> sample(x)
[1] 9 6 12 5 3 4 1 11 8 7 10 2
> sample(x, size = 2)
[1] 9 4
What's the output of sessionInfo() and ls() ? Perhaps you have a
different "sample" function in your workspace?
HTH,
Jorge
sessionInfo()?
Can you replicate this behavior in a R --vanilla session? This seems
very odd and I presume you've overwritten sample() somewhere in your
workspace.
Michael
On Sun, Feb 12, 2012 at 12:52 PM, SUPAKORN LAOHAPITAKVORN
wrote:
> Hi,
> Can anyone help me with the sample () in R?
>
> If
I suggest you ask this question on r-sig-geo.
And it would be best if you could create a small self-contained example.
For example, what class of object is dataset2?
Is there any reason to expect that the coordinates of the sample will be
exactly the same as any of the coordinates in dataset2?
P
From: https://stat.ethz.ch/pipermail/r-help/2011-November/294329.html
> I'm trying to compute sample size requirements for a binomial exact test.
> we want to show that the proportion is at least 90% assuming that it is
> 95%, with 80% power so any asymptotic approximations are out of the
> quest
2011/9/6 Jean-Christophe BOUËTTÉ :
you could tapply
function(x) if(length(x)==1) x else sample(x)
or something like this
>
> 2011/9/6 Jean-Christophe BOUËTTÉ :
>> Hi there,
>> in the third case you get sample(5) which is exactly what you asked for.
>>
>> from ?sample:
>> "If x has length 1, is nu
Hi there,
in the third case you get sample(5) which is exactly what you asked for.
from ?sample:
"If x has length 1, is numeric (in the sense of is.numeric) and x >=
1, sampling via sample takes place from 1:x. Note that this
convenience feature may lead to undesired behaviour when x is of
varying
On 11-09-06 8:13 PM, Jack Siegrist wrote:
I want to sample within groups, and when a group has only one associated
number to just return that number.
If I use this code:
groups<- c(1, 2, 2, 2, 3)
numbers<- 1:5
tapply(numbers, groups, FUN = sample)
I get the following output:
groups<- c(1,
On Sep 6, 2011, at 8:13 PM, Jack Siegrist wrote:
I want to sample within groups, and when a group has only one
associated
number to just return that number.
And what we supposed to do when it has more than one value?
If I use this code:
groups <- c(1, 2, 2, 2, 3)
numbers <- 1:5
tapp
but I am confused a little
Karl
- Ursprüngliche Mail
Von: David Winsemius
An: Karl Knoblick
CC: Greg Snow ; "r-h...@stat.math.ethz.ch"
Gesendet: Samstag, den 13. August 2011, 2:18:37 Uhr
Betreff: Re: [R] Sample size AUC for ROC curves
On Aug 11, 2011, at 5:50 AM, Kar
- Ursprüngliche Mail
Von: Greg Snow
An: Karl Knoblick ; "r-h...@stat.math.ethz.ch"
Gesendet: Dienstag, den 9. August 2011, 19:45:12 Uhr
Betreff: RE: [R] Sample size AUC for ROC curves
If you know how to generate random data that represents your null
hypothesis
(chance, auc=0.5) and
Von: Greg Snow
An: Karl Knoblick ; "r-h...@stat.math.ethz.ch"
Gesendet: Dienstag, den 9. August 2011, 19:45:12 Uhr
Betreff: RE: [R] Sample size AUC for ROC curves
If you know how to generate random data that represents your null hypothesis
(chance, auc=0.5) and how to do your anal
If you know how to generate random data that represents your null hypothesis
(chance, auc=0.5) and how to do your analysis, then you can do this by
simulation, simulate a dataset at a given sample size, analyze it, repeat a
bunch of times and see if that sample size is about the right size. If
Well, you can have exactly 70:30%, i.e. 70% 1s and 30% 0s, but
in random order. For example:
Popn <- c(rep(1,70),rep(0,30))
Samp <- sample(Pop)
(see '?sample' for this usage -- the result of sample(x) is a
random permutation of the elements of x).
In probabilistic terms, this is a "condition
On 05/07/2011 2:25 PM, Joshua Wiley wrote:
Hi Ana,
Look at the documentation for ?sample, specifically, the "prob"
argument. In your case this should work:
sample(c(0,1), 100, replace = TRUE, prob = c(.3, .7))
note that you may not have *exactly* 70% 1 and 30%, in any given sample.
And if y
Hi Ana,
Look at the documentation for ?sample, specifically, the "prob"
argument. In your case this should work:
sample(c(0,1), 100, replace = TRUE, prob = c(.3, .7))
note that you may not have *exactly* 70% 1 and 30%, in any given sample.
HTH,
Josh
On Tue, Jul 5, 2011 at 11:21 AM, Ana Kolar
Thanks so much. I really appreciate it.
Carlos
On 5/11/2011 3:18 PM, Thomas Lumley-2 [via R] wrote:
> On Thu, May 12, 2011 at 2:43 AM, jour4life <[hidden email]
> > wrote:
> > I have a follow up question. When using svyglm, it does not matter
> that I am
> > not using survey design and only wei
On Thu, May 12, 2011 at 2:43 AM, jour4life wrote:
> I have a follow up question. When using svyglm, it does not matter that I am
> not using survey design and only weights?
> In other words,
>
> fit<-svyglm(y~x1+x2+...xk,data=dataset,weights=weightvariable)
>
> Or am I going to have to construct a
I have a follow up question. When using svyglm, it does not matter that I am
not using survey design and only weights?
In other words,
fit<-svyglm(y~x1+x2+...xk,data=dataset,weights=weightvariable)
Or am I going to have to construct a survey design variable, using only the
weight variable?
Than
On 5/10/2011 3:12 PM, Thomas Lumley-2 [via R] wrote:
> On Tue, May 10, 2011 at 2:50 PM, jour4life <[hidden email]
> > wrote:
>
> > Hello all,
> >
> > I am wondering if there is a way to specify sampling weights for an ols
> > model using sample weights.
> >
> > For instance, right now, my code is:
On Tue, May 10, 2011 at 2:50 PM, jour4life wrote:
> Hello all,
>
> I am wondering if there is a way to specify sampling weights for an ols
> model using sample weights.
>
> For instance, right now, my code is:
>
> fit.ex<-lm(y~x1+x2+x3+...xk,data=dataset,weights=weightvariable.)
> summary(fit.ex)
Awesome! Thanks, David and Dennis! And now I know how to search for
packages more effectively.
Tom
On Mon, Apr 4, 2011 at 9:38 PM, Dennis Murphy wrote:
> Start here:
>
> library(sos) # install first if necessary
> findFn('sample size survey')
>
> I got 238 hits, many of which could be relev
Start here:
library(sos) # install first if necessary
findFn('sample size survey')
I got 238 hits, many of which could be relevant.
HTH,
Dennis
On Mon, Apr 4, 2011 at 6:05 PM, Thomas Levine wrote:
> Hi,
>
> Is there an R package for estimating sample size requirements for
> parameter esti
1 - 100 of 206 matches
Mail list logo
