On Sep 25, 2015, at 12:54 PM, Lorenzo Isella wrote:
> Apologies for not letting this thread rest in peace.
> The small script
>
> #
> set.seed(1234)
>
> x <- rnorm(20)
> y <- rnorm(20)
>
>
> goodcls <- apply(mtxcomb , 2, function(idx) al
Apologies for not letting this thread rest in peace.
The small script
#
set.seed(1234)
x <- rnorm(20)
y <- rnorm(20)
goodcls <- apply(mtxcomb , 2, function(idx) all( dist( cbind( x[idx],
y[idx]) ) > 0.9))
mycomb <- mtxcomb [ , goodcls]
#
lt;-(dist(mm))
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thursday, September 24, 2015 6:30 PM
To: Lorenzo Isella
Cc: David L Carlson; r-help@r-project.org
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thursday, September 24, 2015 6:30 PM
To: Lorenzo Isella
Cc: David L Carlson; r-help@r-project.org
Subject: Re: [R] Sampling the Distance Matrix
On Sep 24, 2015, at 1:54 PM, Lorenzo Isella wrote:
> On
On Sep 24, 2015, at 1:54 PM, Lorenzo Isella wrote:
> On Thu, Sep 24, 2015 at 01:30:02PM -0700, David Winsemius wrote:
>>
>> On Sep 24, 2015, at 12:36 PM, Lorenzo Isella wrote:
>>
>>> Hi,
>>> And thanks for your reply.
>>> Essentially, your script gets the job done.
>>> For instance, if I run
>>
On Thu, Sep 24, 2015 at 01:30:02PM -0700, David Winsemius wrote:
On Sep 24, 2015, at 12:36 PM, Lorenzo Isella wrote:
Hi,
And thanks for your reply.
Essentially, your script gets the job done.
For instance, if I run
mm <- cbind(5/(1:5), -2*sqrt(1:5))
dst <- dist(mm)
dst2 <- as.matrix(dst)
diag
>>13 18 19
>> 13 NA 2.272407 3.606054
>> 18 2.272407 NA 1.578150
>> 19 3.606054 1.578150 NA
>>
>> -
>> David L Carlson
>> Department of Anthropology
>> Texas A&M Univers
p-boun...@r-project.org] On Behalf Of William Dunlap
Sent: Wednesday, September 23, 2015 3:23 PM
To: Lorenzo Isella
Cc: r-help@r-project.org
Subject: Re: [R] Sampling the Distance Matrix
mm <- cbind(1/(1:5), sqrt(1:5))
d <- dist(mm)
d
1 2 3 4
2 0.6492864
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of William Dunlap
Sent: Wednesday, September 23, 2015 3:23 PM
To: Lorenzo Isella
Cc: r-help@r-project.org
Subject: Re: [R] Sampling the Distance Matrix
> mm <- cbind(1/(1:5), sqrt(1:5))
> d <- dist
> mm <- cbind(1/(1:5), sqrt(1:5))
> d <- dist(mm)
> d
1 2 3 4
2 0.6492864
3 0.9901226 0.3588848
4 1.250 0.6369033 0.2806086
5 1.4723668 0.8748970 0.5213550 0.2413050
> which(as.matrix(d)>0.9, arr.ind=TRUE)
row col
3 3 1
4 4 1
5 5 1
1 1 3
1 1
Dear All,
Suppose you have a distance matrix stored like a dist object, for
instance
x<-rnorm(20)
y<-rnorm(20)
mm<-as.matrix(cbind(x,y))
dst<-(dist(mm))
Now, my problem is the following: I would like to get the rows of mm
corresponding to points whose distance is always larger of, let's say,
0
If df is the data.frame with values and you want nn samples, then this
is a slightly different approach:
# example data.frame:
df = data.frame(a1 = sample(1:20,50, replace = TRUE),
a2 = sample(seq(0.1,10,length.out =
30),50, replace = TRUE),
Hi Jean,
Thanks!
Daniel,
Yes, you are absolutely right. I want sampled vectors to be as different
as possible.
I added a little more to the earlier data set.
x1 x2 x3
[1,] 1 3.7 2.1
[2,] 2 3.7 5.3
[3,] 3 3.7 6.2
[4,] 4 3.7 8.9
[5,] 5 3.7 4.1
[6,] 1 2.9 2.1
[7,] 2 2
On 6/22/2015 9:42 AM, C W wrote:
Hello R list,
I am have question about sampling unique coordinate values.
Here's how my data looks like
dat <- cbind(x1 = rep(1:5, 3), x2 = rep(c(3.7, 2.9, 5.2), each=5))
dat
x1 x2
[1,] 1 3.7
[2,] 2 3.7
[3,] 3 3.7
[4,] 4 3.7
[5,] 5 3.7
Mike,
There may be a more efficient way to do this, but this works on your
example.
# mix up the order of the rows
mix <- dat[order(runif(dim(dat)[1])), ]
# get rid of duplicate x1s and x2s
sub <- mix[!duplicated(mix[, "x1"]) & !duplicated(mix[, "x2"]), ]
sub
Jean
On Mon, Jun 22, 2015 at 11:42
Hello R list,
I am have question about sampling unique coordinate values.
Here's how my data looks like
> dat <- cbind(x1 = rep(1:5, 3), x2 = rep(c(3.7, 2.9, 5.2), each=5))
> dat
x1 x2
[1,] 1 3.7
[2,] 2 3.7
[3,] 3 3.7
[4,] 4 3.7
[5,] 5 3.7
[6,] 1 2.9
[7,] 2 2.9
[8,] 3 2.9
On 3/29/2015 11:10 PM, Partha Sinha wrote:
I have 1000 data points. i want to take 30 samples and find mean. I
also want to repeat this process 100 times. How to go about it?
Regards
Parth
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I have 1000 data points. i want to take 30 samples and find mean. I
also want to repeat this process 100 times. How to go about it?
Regards
Parth
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d))),]
> }
>
> Now take the samples and combine them into a single data frame:
>
> sample <- do.call(rbind, lapply(dta.list, smp))
> sample
>
> -
> David L Carlson
> Department of Anthropology
> Texas A&M University
> Colle
Since you indicated there are six more columns in the data.frame, getSample
modified below to take care of it.
> getSample
function(x)
{
sites <- unique(x$SiteID)
years <- unique(x$Year)
result <- data.frame()
x$ID <- seq(1,nrow(x))
for (i in 1:length(sites))
{
for (j in 1:length(y
Here is an implementation with function named getSample. Some modification to
the data was made so that it can be read as a table.
> fitting.set
IDbyYear SiteID Year
1 42.24 A-Airport 2006
2 42.24 A-Airport 2006
3 42.24 A-Airport 2006
4 42.
versity
College Station, TX 77840-4352
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Curtis
Burkhalter
Sent: Tuesday, March 3, 2015 3:23 PM
To: r-help@r-project.org
Subject: [R] sampling dataframe based upon number of record occurrences
Hello everyone,
I&#x
Hello everyone,
I'm having trouble performing a task that is probably very simple, but
can't seem to figure out how to get my code to work. What I want to do is
use the sample function to pick records within in a dataframe, but only if
a column attribute value is repeated more than 3 times. So if
> myProbs[ which(ntype == 0)] <- 0.75/7 # Divide so the sum will be 1.0
> myProbs[ which(ntype == 1)] <- 0.25/3
Here of course you need to divide by number of 0s and 1s, 7 and 3
were was just an example.
__
R-help@r-project.org mailing list
https://sta
If I understood correctly, you need weighted sampling. Try 'prob'
argument from 'sample'. For your example:
n <- 10
ntype <- rbinom(n, 1, 0.5)
myProbs <- rep(1/10, 10) # equally likely
myProbs[ which(ntype == 0)] <- 0.75/7 # Divide so the sum will be 1.0
myProbs[ which(ntype == 1)] <- 0.25/3
samp
I have a matrix where each entry represents a data subject's type, 1
or 0:
n <- 10
ntype <- rbinom(n, 1, 0.5)
and I'd like to sample say 3 subjects from ntype where those subjects
who are Type 1 are selected with probability say 0.75, and Type 0 with
(1-0.75). (So the sample would produce
Hi,
You may try:
dat1 <- structure(list(SubID = 1:8, CSE1 = c(6L, 6L, 5L, 5L, 5L, 5L,
3L, 3L), CSE2 = c(5L, 4L, 5L, 4L, 6L, 4L, 6L, 6L), CSE3 = c(6L,
7L, 5L, 3L, 7L, 3L, 6L, 6L), CSE4 = c(2L, 2L, 5L, 4L, 5L, 6L,
3L, 3L), WSE1 = c(6L, 6L, 5L, 4L, 6L, 4L, 6L, 6L), WSE2 = c(2L,
6L, 5L, 4L, 4L, 3L
Arguably you are looking in the wrong place (there's a special mixed-models
mailing list for R), but I can answer the question.
No.
At least, there's nothing in lme4, and I haven't done anything (since I
want a more general solution than Stata and MLWiN implement) and I'd be
surprised if someone
Perhaps I am not looking in the right place, but I am looking for a way to
use lmer() to run a multilevel model that incorporates sampling weights. I
have used the Lumley survey package to use sampling weights in the past,
but according to post I found online from Thomas Lumley in mid-2012, R is
cu
Perhaps you should read the help file for rnorm more carefully.
?rnorm
Keep in mind that the normal probability distribution is a density function, so
the smaller the standard deviation is, the greater the magnitude of the density
function is.
--
Hello R-helpers..
I want to ask about how I can sample data sets without having the infinite
numbers coming out. For example,
set.seed(1234)
a<-rnorm(15,0,1)
b<-rnorm(15,0,1)
c<-rnorm(15,0,1)
d<-rnorm(15,0,36)
After come out with the sample, I need to do a transformation (by Hoaglin,
1985) f
Hi Lorenzo,
This has the feel of a homework problem, but I will suggest to you
that this is "sampling without replacement" and there exist easy
mathematical formulas (no need to resort to R) to calculate your
desired probability.
Michael
On Sat, Dec 8, 2012 at 11:54 AM, Lorenzo Isella
wrote:
>
Dear All,
I hope this is not too off topic, but I am sure it has to be a one-liner
in R.
Suppose you have a population of size N and that you take a random sample
of n_s individuals out of this population.
This population includes a subgroup of n_i individuals.
For any individual in n_i, what
Thank you very much, Thomas!
As I need to estimate the variance components, I will most probably have to
switch from R to HLM or Mplus to apply different weights to different
levels.
Although I prefer R in general.
--
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Thank you very much, Rui!
But I am afraid that I won't be able to use this function for multilevel
analysis, as unfortunately I don't see how exactly I will combine it with
functions in the R packages for multilevel analysis .
--
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On Mon, Jun 11, 2012 at 7:00 AM, Tamara wrote:
> Dear all,
>
> I am struggling with a problem which I have been reading on the forums about
> and it did not seem to me that there is a precise answer to my question.
> However, I still hope there is one.
>
> I am working with http://timss.bc.edu/ P
Hello,
The link you've posted is to a page that does NOT have a dataset, it has
links to other pages. The proper way of posting a data example would be
# paste the output of this in a post
dput(head(yourdata, 20)) # or 30
Now, if I understand your question, function sample() does have a
we
Dear all,
I am struggling with a problem which I have been reading on the forums about
and it did not seem to me that there is a precise answer to my question.
However, I still hope there is one.
I am working with http://timss.bc.edu/ PIRLS data and trying to conduct
multilevel analysis. Ther
## recreating your data
mydata<-list(matrix(1:9, nrow=3, byrow=T),
matrix(10:15, nrow=2, byrow=T),
matrix(16:30, nrow=5, byrow=T))
## get the shortest matrix in your list
n <- min(unlist(lapply(mydata, nrow)))
## subset the list into random samples of length n
??
Something like:
lapply(mydata, function(x){
nr <- nrow(x)
x[sample(seq_len(nr),nr,rep=TRUE),]
})
maybe. The idea is to use the sampled rows as your row index.
-- Bert
On Mon, Apr 2, 2012 at 11:24 AM, Bcampbell99 wrote:
> Hi:
>
> I'm sure this seems like a rudimentary question, but I am n
Hi:
I'm sure this seems like a rudimentary question, but I am not well versed
with R syntax for lists. I have a ragged array from which I've removed
records (entire rows) with missing data. The functions I used to remove the
missing cases resulted in the generation of an R list class object, tha
Hi
I am trying to estimate bottom temperatures over a particular depth range,
based on one dataset containing synthetic temperature profiles (for set depths)
and another that contains information on bathymetry. I need to do this for
multiple regions and thus would ideally like an automated solut
Please use dput() to give a reproducible example: I can make this work
on a data frame quite easily --
x <- data.frame(1:10, letters[1:10], rnorm(10))
str(x)
print(x)
x[sample(nrow(x), 5), ]
So it's not a problem with something being a data frame or having factors.
Michael
On Thu, Mar 8, 2012 a
Thanks, but it doesn't work either, it gives me the same message error.
It works just if my first sample is taken in this way:
mysample <- sample(1:nrow(MeanA), 20, replace=FALSE)
However, in this way it sample just the number of rows:
[1] 71 24 12 36 2 39 69 62 43 38 9 44 13 54 50 63 67 66 3
>
> Hi, thank you but it does work for vectors and matrix but not
dataframes, it
> gives me this message error:
>
> MeanA <- read.csv("MeanAmf.csv",header=T)
> mysample <- MeanA[sample(1:nrow(MeanA), 20, replace=FALSE),]
Well, maybe slight correction
mysample <- sample(1:nrow(MeanA), 20, repla
Hi
I have only faint idea what was you problem as there is no context in you
message but maybe
remainder<-MeanA[-mysample, ]
could work.
Regards
Petr
>
> Hi, thank you but it does work for vectors and matrix but not
dataframes, it
> gives me this message error:
>
> MeanA <- read.csv("MeanA
Hi sarah, it is not clear to me how to do that, can you show me please?
Imagine I have a situation like this:
MeanA <- read.csv("MeanAmf.csv",header=T)
mysample <- MeanA[sample(1:nrow(MeanA), 20, replace=FALSE),]
Then?
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Hi, thank you but it does work for vectors and matrix but not dataframes, it
gives me this message error:
MeanA <- read.csv("MeanAmf.csv",header=T)
mysample <- MeanA[sample(1:nrow(MeanA), 20, replace=FALSE),]
remainder<-MeanA[-mysample]
Error in `[.default`(MeanA, -mysample) : invalid subscript ty
On Mar 7, 2012, at 11:41 AM, Oritteropus wrote:
Hi,
I need to sample randomly my dataset for 1000 times. The sample need
to be
the 80%. I know how to do that, my problem is that not only I need
the 80%,
but I also need the corresponding 20% each time. Is there any way to
do
that?
Alterna
On Wed, Mar 07, 2012 at 08:41:35AM -0800, Oritteropus wrote:
> Hi,
> I need to sample randomly my dataset for 1000 times. The sample need to be
> the 80%. I know how to do that, my problem is that not only I need the 80%,
> but I also need the corresponding 20% each time. Is there any way to do
> t
You could make a vector containing the number of TRUE values that
makes up 80% of your data, and the number of FALSE values that makes
up 20% of your data. Use sample() to reorder it, then use it to divide
your dataset.
If you had provided a reproducible example, I could write you code.
Sarah
On
Hi,
I need to sample randomly my dataset for 1000 times. The sample need to be
the 80%. I know how to do that, my problem is that not only I need the 80%,
but I also need the corresponding 20% each time. Is there any way to do
that?
Alternatively, I was thinking to something like setdiff () functio
On Wed, Jan 25, 2012 at 04:00:27AM -0800, Eliano wrote:
> Hi People,
>
> Does anyone have a good solution for this problem:
>
> a database called DB.
>
>
> index <- sample(1:nrow(DB), size=0.2*nrow(BD))
> test <- DB[index,]
> train <- DB[-index,]
>
> One of the variables in this database
Hi People,
Does anyone have a good solution for this problem:
a database called DB.
index <- sample(1:nrow(DB), size=0.2*nrow(BD))
test <- DB[index,]
train <- DB[-index,]
One of the variables in this database contais a target variable with two
values 0 and 1.
Imagine now that i want to
Hi People,
Thus anyone have a good solution for this problem:
a database called DB.
index <- sample(1:nrow(DB), size=0.2*nrow(BD))
test <- DB[index,]
train <- DB[-index,]
One of the variables in this database contais a target variable with two
values 0 and 1.
Imagine now that i want to constr
On Jan 24, 2012, at 20:41 , Thomas Lumley wrote:
> It's not meant for sampling weights. It's meant for precision
> weights. How best to include sampling weights in mixed models is a
> research problem at the moment, but you can rely on getting the wrong
> answer if you just use the weights= arg
On Tue, Jan 24, 2012 at 6:19 PM, Mohd masood wrote:
>
> Dear All
> I am trying to include sampling weights in multilavel regression analysis
> using packege lme4 using following codes
>
> print(fm1 <- lmer(DC~sex+age+smoker+alcohol+fruits(1|setting),
> dataset,REML = FALSE), corr = FALSE)
> prin
Dear All
I am trying to include sampling weights in multilavel regression analysis using
packege lme4 using following codes
print(fm1 <- lmer(DC~sex+age+smoker+alcohol+fruits(1|setting), dataset,REML
= FALSE), corr = FALSE)
print(fm2 <- lmer(DC~sex+age+smoker+alcohol+fruits(1|setting), d
1) Use dput() to submit data.
2) Would this work? (It requires your data are evenly spaced, but I
think that's it) d[seq(1, nrow(d), by = 3), ]
Michael
On Wed, Dec 14, 2011 at 7:17 AM, abcdef ghijk wrote:
> Good Morning ,
>
> I want to sample the following time series for every third hour. For
Good Morning ,
I want to sample the following time series for every third hour. For example at
00:00,03:00,06:00,09:00 etc.
2011-01-01 00:00:00 0.00e+00
2011-01-01 01:00:00 1.471667e+01
2011-01-01 02:00:00 1.576667e+01
2011-01-01 03:00:00 0.00e+00
2011-01-01 04:00:
Dan
Nordlund, Dan (DSHS/RDA) wrote:
>
>> -Original Message-
>> From: r-help-bounces@ [mailto:r-help-bounces@r-
>> project.org] On Behalf Of SarahJoyes
>> Sent: Tuesday, November 08, 2011 5:57 AM
>> To: r-help@
>> Subject: Re: [R] Sampling with c
-Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
>> project.org] On Behalf Of SarahJoyes
>> Sent: Tuesday, November 08, 2011 5:57 AM
>> To: r-help@r-project.org
>> Subject: Re: [R] Sampling with conditions
>>
>> Tha
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of SarahJoyes
> Sent: Tuesday, November 08, 2011 5:57 AM
> To: r-help@r-project.org
> Subject: Re: [R] Sampling with conditions
>
> That is exactly what I
That is exactly what I want, and it's so simple!
Thanks so much!
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Sent from the R help mailing list archive at Nabble.com.
__
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Sorry about being confusing, I have so many loops in loops and ifelses that I
get mixed up sometimes, it was just a typo, it was supposed to be "for(i in
1:5)" Sorry,
Thanks for you help!
SJ
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[Yet another correction -- this one is important.
I start from scratch this time]
On 07-Nov-11 22:22:54, SarahJoyes wrote:
> Hey everyone,
> I am at best, an amateur user of R, but I am stuck on how
> to set-up the following situation.
> I am trying to select a random sample of numbers from 0 t
[Correction below (I was writing too late at night ... ]
On 08-Nov-11 00:25:57, Ted Harding wrote:
> On 07-Nov-11 22:22:54, SarahJoyes wrote:
>> Hey everyone,
>> I am at best, an amateur user of R, but I am stuck on how
>> to set-up the following situation.
>> I am trying to select a random samp
On 07-Nov-11 22:22:54, SarahJoyes wrote:
> Hey everyone,
> I am at best, an amateur user of R, but I am stuck on how
> to set-up the following situation.
> I am trying to select a random sample of numbers from 0 to 10
> and insert them into the first column of a matrix (which will
> used later in
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of SarahJoyes
> Sent: Monday, November 07, 2011 2:23 PM
> To: r-help@r-project.org
> Subject: [R] Sampling with conditions
>
> Hey everyone,
> I am at
Not sure this is valid that you can have 9 random samples out of 10,
but the last one has to be fixed to meet the restraint, sum=10.
Weidong
On Mon, Nov 7, 2011 at 5:22 PM, SarahJoyes wrote:
> Hey everyone,
> I am at best, an amateur user of R, but I am stuck on how to set-up the
> following sit
Hey everyone,
I am at best, an amateur user of R, but I am stuck on how to set-up the
following situation.
I am trying to select a random sample of numbers from 0 to 10 and insert
them into the first column of a matrix (which will used later in a loop).
However, I need to have those numbers add u
Well, for 0.828324 < x[2] < Inf the probablility is roughly 0 hence not
easy to draw random numbers out there
Uwe Ligges
How is this probability roughly 0?
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On 26.06.2011 21:26, statfan wrote:
I am trying generate a sample for a truncated multivariate normal
distribution via the rtmvnorm function in the {tmvtnorm} package.
Why does the following produce NaNs?
rtmvnorm(1, mean = rep(0, 2), matrix(c(0.06906084, -0.07463565, -0.07463565,
0.0807808
I am trying generate a sample for a truncated multivariate normal
distribution via the rtmvnorm function in the {tmvtnorm} package.
Why does the following produce NaNs?
rtmvnorm(1, mean = rep(0, 2), matrix(c(0.06906084, -0.07463565, -0.07463565,
0.08078086),2),c(-0.4316738, 0.8283240), c(Inf,
On Thu, Mar 31, 2011 at 4:01 AM, Simon Kiss wrote:
> Dear colleagues,
> I'm working with the 2008 Canada Election Studies
> (http://www.queensu.ca/cora/_files/_CES/CES2008.sav.zip), trying to construct
> a weighted national sample using the survey package.
> Three weights are included in the nat
Dear colleagues,
I'm working with the 2008 Canada Election Studies
(http://www.queensu.ca/cora/_files/_CES/CES2008.sav.zip), trying to construct a
weighted national sample using the survey package.
Three weights are included in the national survey (a household weight, a
provincial weight and a
Is anyone familiar with a way to account for sampling weights (e.g., in order
to cope with selection bias) for individual respondents using the bayesm
package (e.g., rhierMnlRwMixture)?
In the regular MNL this can easily be done in STATA using the mlogit
function with pweights option. However, I a
Hi:
A couple more approaches to consider:
# Utility function to extract two rows from a data frame
# Meant to be applied to each data subset
sampler <- function(d) if(nrow(d) > 2) d[sample(1:nrow(d), 2, replace =
FALSE), ] else d
library(plyr)
> ddply(x, 'id', sampler)
id v1 V2
1 1 2 13
2 1
On Feb 17, 2011, at 1:33 PM, andrija djurovic wrote:
This is, maybe, not the best solution but I hope it will help you:
x<-data.frame(id=c(1,1,1,2,2,2,2,3,3,3,4,4), v1=c(1:12), V2=c(12:23))
do.call("rbind",by(x,x$id,function(x) x[c(sample(nrow(x),2)),]))
Andrija
Another way (and note that
This is, maybe, not the best solution but I hope it will help you:
x<-data.frame(id=c(1,1,1,2,2,2,2,3,3,3,4,4), v1=c(1:12), V2=c(12:23))
do.call("rbind",by(x,x$id,function(x) x[c(sample(nrow(x),2)),]))
Andrija
On Thu, Feb 17, 2011 at 6:39 PM, yf wrote:
>
> But i need for each id have two data
But i need for each id have two data.
Like...
> x
id v1 V2
1 1 1 12
2 1 2 13
4 2 4 15
5 2 5 16
8 3 8 19
9 3 9 20
11 4 11 22
12 4 12 23
So should write sample( if sample id >2 ,2). I don't know how to write (if
sample id >2). Thanks.
--
View this message in context:
On Feb 16, 2011, at 11:35 PM, yf wrote:
I want to sample from the ID. For each ID, i want to have 2 set of
data. I
try the sample() function but it didn't work.
You don't say _how_ you used the sample function. You should show what
code you used when stating the _something_ "doesn't wor
Hi ,
what about split function ?
?split divided x into 2 data.frame
a<-split(x,1:2)
a[[1]] first data frame
a[[2]] second data frame
regrads
M
Le 17/02/11 05:35, yf a écrit :
> I want to sample from the ID. For each ID, i want to have 2 set of data. I
> try the sample() function but it did
I want to sample from the ID. For each ID, i want to have 2 set of data. I
try the sample() function but it didn't work.
> x<-data.frame(id=c(1,1,1,2,2,2,2,3,3,3,4,4), v1=c(1:12), V2=c(12:23))
> x
id v1 V2
1 1 1 12
2 1 2 13
3 1 3 14
4 2 4 15
5 2 5 16
6 2 6 17
7 2 7 18
8
Dear list i have a sample question
I have a dataframe of 1500 species and 13 life history traits.
small example code:
traits <- data.frame(letters[1:9],
sample(letters, 9),
sample(letters, 9),
sample(letters, 9),
sample(letters, 9),
sample(letters, 9),
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Christoph Goebel
> Sent: Friday, November 19, 2010 1:56 PM
> To: r-help@r-project.org
> Subject: [R] Sampling from multi-dimensional kernel density
Hi,
I'd like to use a three-dimensional dataset to build a kernel density and
then sample from the distribution.
I already used the npudens function in the np package to estimate the
density and plot it:
fit<-npudens(~x+y+z)
plot(fit)
It takes some time but appears to work well.
yes, the data.frame is exactly the same as the one I posted earlier.
I was trying to see if the loop function works. And I got that message. here
below is the syntax I was trying to run, followed by the error message at
the end:
> sampleX<-function(X,nGrp1,nsamples){if(nGrp1>=4)stop("can't sampl
On 16 November 2010 16:10, wangwallace wrote:
>
> Michael, I really appreciate your help.
>
> but I got the following error message when I wan trying to run the function
> written by you:
>
> Error in out[i, ] <- apply(help[, c(grp1 + 1, grp2 + 5)], 2, sample, 1) :
> number of items to replace is
Michael, I really appreciate your help.
but I got the following error message when I wan trying to run the function
written by you:
Error in out[i, ] <- apply(help[, c(grp1 + 1, grp2 + 5)], 2, sample, 1) :
number of items to replace is not a multiple of replacement length
I am not quite sure
Hello,
Is this what you want ?
sampleX <- function(X, nGrp1, nsamples)
# X is matrix or data.frame with cols for two groups of variables
# with grp1 in cols 2:5 and grp2 in cols 6:9
#
# nGrp1 <- number of variables to sample from group 1
#
# nsamples <- number of rows in output matrix
if (nGrp
Hey,
I am hoping someone can help me with a sampling question.
I have a data frame of 8 variables (the first column is the subjects' id):
SubIDCSE1 CSE2 CSE3 CSE4 WSE1 WSE2 WSE3 WSE4
1 6 5 6 2 6 22 4
2 6 4
Hi Solafah,
You are right that two commands are equivalent when p= pnorm(a). You can
check the results by following codes.
n <- 5
a <- -1
set.seed(123456)
qnorm(runif(n,0,pnorm(a)))
p <- pnorm(a)
set.seed(123456)
qnorm(p*runif(n))
Anyway, the elements of the lower tail are not chosen equally by
Hello
If i want to resample from the lower tail of normal distribution , are these
commands equivelant??
lower tail :qnorm(runif(n,0,pnorm(a))) if a is a lower tail bound
or
lower tail:qnorm(p*runif(n)) if p is the probability of each interval(the
observations are divided to intervals)
Regard
If poproh.3 was your dataset as a data.frame (an object with row and
column dimensions), you need a comma following the row selection
(sample(...)) to indicate that you want to select those rows and all
columns:
newsample <-poprho.3[sample(1:1,100),] # note the last comma in the brackets
Gene
We'll probably need much more info, but this should get you started:
nameOfDataSet[sample(1:1, 100),]
You can replace the 1 with dim(nameOfDataSet)[1] to make it more dynamic.
Jeff.
On Tue, Oct 5, 2010 at 3:07 AM, Jumlong Vongprasert
wrote:
> Dear all.
> I have data with 2 variable x,y
Dear all.
I have data with 2 variable x,y size 1.
I want to sampling from this data with size 100.
How I can do it.
THANK.
--
Jumlong Vongprasert
Institute of Research and Development
Ubon Ratchathani Rajabhat University
Ubon Ratchathani
THAILAND
34000
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On 03/10/2010 6:38 PM, solafah bh wrote:
Hello
If i want to resampl from the tails of normal distribution , are these commans
equivelant??
upper tail:qnorm(runif(n,pnorm(b),1)) if b is an upper tail boundary
or
upper tail:qnorm((1-p)+p(runif(n)) if p is the probability of each interval
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of solafah bh
> Sent: Sunday, October 03, 2010 3:39 PM
> To: R help mailing list
> Subject: [R] sampling from normal distribution
>
> Hello
> If i wan
Hello
If i want to resampl from the tails of normal distribution , are these commans
equivelant??
upper tail:qnorm(runif(n,pnorm(b),1)) if b is an upper tail boundary
or
upper tail:qnorm((1-p)+p(runif(n)) if p is the probability of each interval
(the observatins are divided to intervals)
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