Hi all,
Does anyone have experience estimating the sample size for beta
regression? I have three treatment groups with 0.6, 0.5 and 0.3 mean of
proportions. I tried use BetaPASS but failed,
Regards,
Greg
[[alternative HTML version deleted]]
__
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
Dear All: good morning
*Re:* Sample Size Determination to Compare Three Independent Proportions
*Situation:*
Three Binary variables (Yes, No)
Three independent populations with fixed sizes (*say:* N1 = 1500, N2 = 900,
N3 = 1350).
Power = 0.80
How to choose the sample sizes to compare th
, since there are a variety of license.
>>
>> David Winsemius, MD, (IANAL)
>>
>>
>> > On May 1, 2021, at 12:17 PM, Maja Analytics <
>> maja.analytics.b...@gmail.com> wrote:
>> >
>> > Hello!
>> >
>> > I would like to make tu
rote:
> >
> > Hello!
> >
> > I would like to make tutorials for non-data scientists in R on Medium and
> > as LinkedIn articles. Could you tell me if I can use R sample datasets (I
> > would note the acknowledgements on the end)
emius, MD, (IANAL)
>>
>>
>> > On May 1, 2021, at 12:17 PM, Maja Analytics
>> > wrote:
>> >
>> > Hello!
>> >
>> > I would like to make tutorials for non-data scientists in R on Medium and
>> > as LinkedIn articles.
cientists in R on Medium and
> as LinkedIn articles. Could you tell me if I can use R sample datasets (I
> would note the acknowledgements on the end) for that?
>
> Thanks in advance, kind regards,
> Maja
>
> [[alternative HTML version deleted]]
>
> _
Hello!
I would like to make tutorials for non-data scientists in R on Medium and
as LinkedIn articles. Could you tell me if I can use R sample datasets (I
would note the acknowledgements on the end) for that?
Thanks in advance, kind regards,
Maja
[[alternative HTML version deleted
GoodDay,
We have an urgent requirement as per attached samples & specifications. Kindly
quote your price and inquiry at the earliest .Also confirm the Terms &
Condition included, Delivery schedule for supply.
Looking forward to your valuable confirmation.
sincerely yours
--
Thanks
Ar
> 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
just 1 but it will be zero.Hope this helps,Rui Barradas Enviado
a partir do meu smartphone Samsung Galaxy. Mensagem original
De: Thomas Subia via R-help Data:
02/07/2019 23:23 (GMT+00:00) Para: r-help@r-project.org Assunto:
[R] Sample size required to estimate populatio
homas Subia via R-help
Data: 02/07/2019 23:23 (GMT+00:00) Para:
r-help@r-project.org Assunto: [R] Sample size required to estimate population
variance Colleagues,Can anyone suggest a package or code which might help me
calculate the minimum sample size required to estimate the population varianc
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
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
[[alternative HTML version deleted]]
___
t) or if
the user defines himself the probabilities (even if they are equal).
I found this post on stackoverflow which explains the reasons of this
difference (answer by Matthew Lundberg):
https://stackoverflow.com/questions/23316729/r-sample-probabilities-defaul
t-is-equal-weight-why
they are equal).
I found this post on stackoverflow which explains the reasons of this
difference (answer by Matthew Lundberg):
https://stackoverflow.com/questions/23316729/r-sample-probabilities-defaul
t-is-equal-weight-why-does-specifying-equal-weigh
I was wondering whether the sol
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
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.
In
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
Dear all,
I have data from a time span like this:
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
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
Hello everybody!
I have the following problem: I'd like to select a sample from a subsample
in a dataset. Actually, I don't want to select it, but to create a new
variable sampleNo that indicates to which sample (one or two) a case
belongs to.
Lets suppose I have a dataset containing 40 cases:
d
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
Hi
In R,how to do sample size calculation for three-way incomplete block
crossover study where within subject residual standard deviation,treatment
difference and power is given.
Thanks in advance.
Regards
Jose
[[alternative HTML version deleted]]
__
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
In R how to calculate sample size,where power,residual standard deviation
and treatment difference is given.?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/
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
I want to create two files train and test using dplyr (by random sampling
method). How to do the same same using lets say iris data.
Regards
Parth
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R-help@r-project.org mailing list -- To UNSUBSCRIBE and mo
Greetings-
I am seeking to fit a model using Heckman-style selection; however the
wrinkle is that the selection is into multiple categories, not a binary
in/out. In this case, selection is into the type of higher-education
institution a student attended; the goal is to estimate post-graduation
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
Dear R users,
My data frame has four "groups" namely A1, B2, C3, & D4. Each
group has 12 rows (variable "plotno). I like to randomly sample one "plotno"
within each "groups" variable and label it as "CONTROL" and label others as
"TEST" in a new variable called "entry". I am tr
org, Sweden | mashra...@yahoo.com
From: "DIGHE, NILESH [AG/2362]"
To: "r-help@r-project.org"
Sent: Thursday, 17 March 2016, 18:18
Subject: [R] sample within a loop
Dear R users,
My data frame has four "groups" namely A1, B2,
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
Dear R Forum
I have a data.frame as
mydat =
c(6,6,5,6,4,6,8,4,6,6,6,3,4,6,5,7,7,4,3,5,5,5,3,6,7,4,4,7,4,3,4,6,4,6,5,4,4,7,6,8,5,6,5,5,8,2,3,5,7,5)
Is there any library or way in R to allocate weights to these values? Actually
I am having a large data, but for illustrative purpose, have consid
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.
Hi,
I am interested in generating all possible contingency table (2 by 2) with
fixed row margins and column margins. Can anyone help me.
Best regards,
Tahira
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://s
Hi All,
What formula can I use to determine the right sample size for clustering
analysis with 100-300 variables?
What sampling methodology can be used for k-means or hierarchical clustering on
categorical fields so that all values of the categorical fields are included in
the sample?
Thanks
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
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 would be able to
estimate standard deviation beforehand.
I'm using the bpower function from the Hmisc
ot;SpCode",5),4)
#$`40_49.MERLMER`
#[1] 34 38 143 3 19
#
#$`50_59.MERLMER`
#[1] 20 112 115 104 77
#
#$`60_69.MERLMER`
#[1] 130 136 40 128 32
#
#$`70_.MERLMER`
#[1] 44 63 96
A.K.
- Original Message -
From: Xochitl CORMON
To: r-help@r-project.org
Cc:
Sent: Wednesday, Jun
Thank you Sarah for this code, it's exactly what I wanted to reach.
Le 05/06/2013 16:49, Sarah Goslee a écrit :
What about using instead
size = min(5, length(Gpool$SampleNb[Gpool$LngtClas == LngtClas[[4]]&
Gpool$SpCode == SpCode[[2]]])
that would make sure your sample is either the size of t
---
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Xochitl CORMON
Sent: Wednesday, June 5, 2013 9
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
Hi all,
I'm trying to randomly select sample numbers for length class groups (5
per length class).
For this I'm using a loop FOR and the function sample () and specified a
size for the sampling of 5. Unfortunately, one of the length class group
does not contain 5 individuals. For me is not a
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.
-
Greetings. My wife is teaching an introductory stat class at UC Davis. The
class emphasizes the use of simulations, rather than mathematics, to get
insight into statistics, and R is the mandated tool. A student in the class
recently inquired about different approaches to sampling from a binomia
.@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$UnitName_1, FUN = length)
Hope this helps,
Rui Barradas
Em 26-04-2013 10:48, Shane Carey escreveu:
Hi,
I would like to put the sample number beside each lable in a boxplot.
How do I do this? Essentially, I need
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Shane Carey
> Sent: Friday, April 26, 2013 11:49 AM
> To: r-help@r-project.org
> Subject: [R] sample size in box plot labels
>
> Hi,
>
>
Hi,
I would like to put the sample number beside each lable in a boxplot.
How do I do this? Essentially, I need to count the sample size for each
factor, see below:
Thanks
boxplot(DATA$K_Merge~factor(DATA$UnitName_1),axes=FALSE,col=colours)
title(main=list("Tukey Boxplot by Geology:\n K(%)",cex=c
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
>
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
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
Hi all,
I have a list of genes present in 500 individuals, the individuals are the
elements:
Genes <- lapply(1:nrow(inds),function(x)
sample(1:1,inds$No_of_Genes,replace=TRUE))
(This was later written to a dataframe as well as kept as the list object:
inds2 <- data.frame(inds,Genes=I(Genes
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
Dear community
I have a dataframe and want to split it into a learn and a test partition.
However the learnset should be balanced, i.e. each class should have the
same number of cases. I tried and searched a lot, without success so far.
Maybe you can help?
Some example code
*# generate example da
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
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 value of x has equal probability to appear (1/n).
Are the following codes equivalent??
sample(x,1,replace=TRUE) and sample(x,1,replace=TRUE,prob=rep(1/n , n))
Regards
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