As you say, I guess that's not possible. Meanwhile I've got R's results
verified in Stata
. tabi 17 6 \ 2 1, chi2 exact
| col
row | 1 2 | Total
---+--+--
1 |17 6 |23
2 |
On Sat, Mar 21, 2009 at 5:13 PM, UBC wrote:
>
> so i am having this question
> what should i do if the give data file (.txt) has 4 columns, but different
> lengths?
> how can i read them in R?
> any idea for the following problem?
>
>
> Gas consumption (1000 cubic feet) was measured before and aft
This works with the example. If the real data is different it may not
work. To run the example below just copy and paste it into R.
To run with the real data replace textConnection(Lines) with
"insulation.txt" everywhere.
Lines <- "Before insulAfter insul.
tempgas tempgas
-0.8
Thanks berwin. See what I sent to Ben. I better start looking up my
stat textbooks or going back to class before I start using stat terms.
You guys are on top of things .
On Sat, Mar 21, 2009 at 10:50 PM, Berwin A Turlach wrote:
G'day Mark,
On Sat, 21 Mar 2009 22:08:21 -0500 (CDT)
mar
thanks ben: i think my example makes sense but my terminology of one
tailed two tailed was wrong or flipped or whatever. in fact, that's
why i gave the example. i wasn't remembering the terminology because
it's been too long since i stepped in a classroom ( 8 years ).
On Sat, Mar 21, 2009 at
One that you could write yourself:
percVar <- function(init, final) (final - init) / init * 100
On Sat, Mar 21, 2009 at 9:43 PM, Hadil Eshtayah wrote:
> Dear Members,
>
> Simple questions, I am a beginner with R. What is the function for
> percentage variance? A function that replaces: (final
verizon.net> writes:
>
> by definition, the one tailed p-value has to be <= 0.5 so there is
> still something wrong with your OpenEpi calc. Most likely it's
> calculating the
> 2 tailed p-value and then mistakenly multiplying by 2. For example:
For what it's worth, Fisher's exact test w
If the input file has a separator other than a space (e.g., tabs or
commas) then you can read it is and the missing data will be NAs and
you can decide how to handle it. If it does not have a separator,
then maybe you can read it in with read.fwf. Otherwise when you read
it in, you can tell the s
Dear Members,
Simple questions, I am a beginner with R. What is the function for percentage
variance? A function that replaces: (final value-Initial value/Initial
value)*100
Thank You.
H.
__
[[alternative HTML
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of C.H.
> Sent: Saturday, March 21, 2009 7:07 PM
> To: r-help@r-project.org
> Subject: Re: [R] Fisher test problem
>
> I tried the OpenEpi, the p-value of 1.25 is due to the fact th
so i am having this question
what should i do if the give data file (.txt) has 4 columns, but different
lengths?
how can i read them in R?
any idea for the following problem?
Gas consumption (1000 cubic feet) was measured before and after insulation
was put into
a house. We are interested in loo
by definition, the one tailed p-value has to be <= 0.5 so there is
still something wrong with your OpenEpi calc. Most likely it's
calculating the
2 tailed p-value and then mistakenly multiplying by 2. For example:
A) Suppose you are testing
Ho: u = u_0
You must go to his website because it is not on CRAN. I have built it
on mac osx by installing it with R CMD install. This should work on
other unix platforms.
Stephen Sefick
On Sat, Mar 21, 2009 at 9:26 PM, wrote:
> I cannot find sowas package by Douglas Mauran in CRAN packages list-
> On wh
Thank you all for the help!
Le
On Sat, Mar 21, 2009 at 4:23 PM, Pankaj Chopra wrote:
> data.year[j] <-
> read.table(paste("c:/data/",year[j],".csv",sep=''),header=T,sep=",")
>
>
> should do it.
>
>
>
> Le Wang wrote:
>>
>> Hi there,
>>
>> Thanks for your time in advance.
>>
>> I am trying to rea
I tried the OpenEpi, the p-value of 1.25 is due to the fact that the
one tailed p-value is 0.62. The two tailed p-value then is 0.62 * 2 =
1.25. OpenEpi is not clever enough to ceiling the p-value to 1.
CH
On Sun, Mar 22, 2009 at 3:43 AM, David Winsemius wrote:
> Let me ask you: What degree of c
On 2009-March-20 , at 23:02 , johnhj wrote:
Is it possible, to use the plot() funktion and the boxplot() funktion
together ?
I will plot a simple graph and additionally to the graph on certain
places
boxplots. I have imagined to plot the graph a little bit
transparency and
show in the same
On 2009-March-20 , at 16:23 , David Winsemius wrote:
On Mar 20, 2009, at 3:55 PM, Altaweel, Mark R. wrote:
I have a problem where I have two columns of data that I can simply
plot using:
plot(wV[0:15,3],wY[0:15,3]).
Perhaps:
plot(wV[0:15,3],wY[0:15,3], col = ifelse(wY[0:15,3]>0, "blue","
I cannot find sowas package by Douglas Mauran in CRAN packages list-
On which platforms does sowas run ?
Has anybody used such a package at all ?
hHank you very much,
Maura
tutti i telefonini TIM!
[[alternative HTML version deleted]]
__
R-h
Hi Achim,
That was a very quick reply/fix that got posted in between my reading and my
response.
I have downloaded your fix from:
svn checkout svn://r-forge.r-project.org/svnroot/zoo
for the HEAD revision and confirmed that it works as expected - I tested
using the code I posted in the first m
Thank you for your reply, but I am still confused.
Let me clarify... ...my problem isn't with zoo-ness per se.
With zoo objects, there appears to be zoo-matrices and zoo-vectors.
My problem is this - I start with a zoo-matrix:
> x <- m[1,,drop=FALSE]
> x
inp
2003-02-01 5
> is.matr
On Sat, Mar 21, 2009 at 2:03 PM, joker77 wrote:
>
> Hi, I noted a discrepancy between R and openepi when I ran a fisher test with
> the same matrix. In R:
>
>> a=matrix(c(1,2,6,17), nrow=2)
>> a
> [,1] [,2]
> [1,] 1 6
> [2,] 2 17
>> fisher.test(a, conf.int=T)
>
> Fisher's Exa
On Sat, 21 Mar 2009, Ken-JP wrote:
require( zoo )
inp <- c( 5, 9, 4, 2, 1 );
m <- zoo( cbind( inp ), as.Date("2003-02-01") + (0:(length(inp)-1)));
dim( m ) # [1] 5 1
dim( m[1,,drop=FALSE] ) # [1] 1 1 - ok
dim( lag( m, -1 )) # [1] 4 1 - ok
dim( rbind( m[1,,drop=FALSE], lag(m,-1) )) # NULL - c
> library(sciplot)
> bargraph.CI(peptide, surface, group=adjunct,data = y)
Error in eval(substitute(subset), envir = data) : object "y" not found
#groan, ... why can't people offer a workable example?
> data(ToothGrowth)
# se as default
> bargraph.CI(x.factor = dose, response = len, data = To
Hi, I noted a discrepancy between R and openepi when I ran a fisher test with
the same matrix. In R:
> a=matrix(c(1,2,6,17), nrow=2)
> a
[,1] [,2]
[1,]16
[2,]2 17
> fisher.test(a, conf.int=T)
Fisher's Exact Test for Count Data
data: a
p-value = 1
alternative hypoth
Sorry about forgetting that.
Package "sciplot"
Simple example code:
This code plots the standard error on the bars; I would like to replace that
by the standard deviation
bargraph.CI(peptide, surface, group=adjunct,data = y)
David Winsemius wrote:
>
> Package name?
> Example code?
>
> --
Hi Saiwing,
If all you are asking is how to rename a factor vector, the easiest way
would be to use:
levels(Loblolly$Seed) <- c( a vector of level names you would like to use
for the factor - separated by commas)
If you are asking how to make your output look better, I am not sure I have
an idea (
hi
i need to use R to make a transfer function model. i searched
online for so long for a code. there is no result. anyone help would be
appreciated. altho the best situation is that if anyone has done the same kind
of work b4? if yes, pls upload ur work as an example.
bill
If you come across an archived thread that you would like to reply to, how
do you reply to it without starting a new thread?
Joe Boyer
Statistical Sciences
Renaissance Bldg 510, 3233-D
Mail Stop RN0320
8-275-3661
cell: (610) 209-8531
[[alternative HTML version deleted]]
___
Dear Saiwing Yeung,
You appear to be using orthogonal-polynomial contrasts (generated by
contr.poly) for Seed, which suggests that Seed is either an ordered factor
or that you've assigned these contrasts to it. Because Seed has 14 levels,
you end up fitting an degree-13 polynomial. If Seed is inde
COnsider the use of a 'list':
dataYear <- lapply(c(1940, 1950, 1960), function(.file){
read.table(paste("c:/data/", .file, '.csv', sep=''), header=TRUE, sep=',')
})
The you can access your data:
dataYear[['1940']]
or
dataYear$"1940"
On Sat, Mar 21, 2009 at 3:53 PM, Le Wang wrote:
> Hi th
Hi all,
I have a pretty basic question about categorical variables but I can't
seem to be able to find answer so I am hoping someone here can help. I
found that if the factor names are all in numbers, fitting the model
in lm would return labels that are not very recognizable.
# Example: l
The MBESS package has many sample size functions.
Hmisc has bpower and cpower .
The pwr package has some simple power functions.
ss.fromdata.pois {package::ssanv}
asypow package
I worry that beginners in statistics will get their hands on some
package or function that gives them a pleasing
data.year[j] <-
read.table(paste("c:/data/",year[j],".csv",sep=''),header=T,sep=",")
should do it.
Le Wang wrote:
Hi there,
Thanks for your time in advance.
I am trying to read in multiple files. For example,
data.1940 <- read.table("c:/data/1940.csv",header=TRUE,sep=",")
data.1950 <-
we have some initial data from our field sampling. From the means and
variances, we see your sampling field is much heterogeneous (not uniform).
We are looking for a "R" program for sampling size determination.
Right now, we are not good enough to write a whole "R" program, so please
help.
R does not directly support macros and I don't think that that is what
you meant. Also we probably want to put the data frames in a list
so we can easily operate over all of them later. (If these are time
series also see read.zoo in the zoo package.)
setwd("c:/data")
filenames <- paste(year, "cs
Sorry. Not sure what you mean by macro not " protected."
?paste
?assign
Perhaps:
> filelist <- paste("c:\\data\\", year,".csv", sep="")
> filelist
[1] "c:\\data\\1940.csv" "c:\\data\\1950.csv" "c:\\data\\1960.csv"
for (i in filelist) {
assign( paste("data.", year[i], sep="") ,
Hi again,
My bad. Below solution had a typo :(Here is the "correct" one:
year<-c(1940,1950,1960)
for (i in y) assign(paste("data.",i,sep=""),
read.table(paste("c:/data/",i,".csv",sep=""),header=TRUE,sep=","))
HTH,
Jorge
On Sat, Mar 21, 2009 at 4:01 PM, Jorge Ivan Velez
wrote:
>
> De
Dear Le,
Try this:
year<-c(1940,1950,1960)
for (i in y) assign(paste("data.",i,sep=""),
read.table("c:/data/i.csv",header=TRUE,sep=","))
data.1940
data.1950
data.1960
See ?assign for more details.
HTH,
Jorge
On Sat, Mar 21, 2009 at 3:53 PM, Le Wang wrote:
> Hi there,
Hi there,
Thanks for your time in advance.
I am trying to read in multiple files. For example,
data.1940 <- read.table("c:/data/1940.csv",header=TRUE,sep=",")
data.1950 <- read.table("c:/data/1950.csv",header=TRUE,sep=",")
data.1960 <- read.table("c:/data/1960.csv",header=TRUE
On Sat, Mar 21, 2009 at 12:01 PM, I wrote:
> I have a data frame with roughly 500 rows and 120 variables.
> I would like to generate a new data frame that will include
> one row for each PAIR of rows in the original data frame and
> will include all 120 + 120 = 240 variables from the two rows.
Let me ask you: What degree of credibility should be accorded a WWW
application that delivers a p-value of 1.25?
If the answer is not immediately and glaringly obvious, then tell us,
what sort of axioms of probability are you working with?
--
David Winsemius
On Mar 21, 2009, at 3:17 PM, Vij
Viju Moses,
Are you sure you got a P-value of 1.25 ?
Since P-value could only be between 0 to 1...
Tal
On Sat, Mar 21, 2009 at 9:17 PM, Viju Moses wrote:
> Hi, I noted a discrepancy between R and openepi when I ran a fisher test
> with the same matrix. In R:
>
> > a=matrix(c(1,2,6,17), nro
On Fri, Mar 20, 2009 at 8:18 PM, science! wrote:
> I am aware that it is easily possible to create var names on the fly. e.g.
> assign(paste("m",i,sep=""),j)
> but is it possible to assign dataframes to variables created on the fly?
>
> e.g.
> If I have a dataframe called master and I wanted to su
Hi, I noted a discrepancy between R and openepi when I ran a fisher test
with the same matrix. In R:
> a=matrix(c(1,2,6,17), nrow=2)
> a
[,1] [,2]
[1,]16
[2,]2 17
> fisher.test(a, conf.int=T)
Fisher's Exact Test for Count Data
data: a
p-value = 1
alternative hypothesi
On Mar 21, 2009, at 2:31 PM, Ken-JP wrote:
require( zoo )
inp <- c( 5, 9, 4, 2, 1 );
m <- zoo( cbind( inp ), as.Date("2003-02-01") + (0:(length(inp)-1)));
dim( m ) # [1] 5 1
dim( m[1,,drop=FALSE] ) # [1] 1 1 - ok
dim( lag( m, -1 )) # [1] 4 1 - ok
dim( rbind( m[1,,drop=FALSE], lag(m,-1) )) # NU
I use the CMU site. I think it is US PA #1 or some such. I
suppose the fact that I am using a Mac could affect that but I am
guessing not.
I also found a copy of the 2.14 docs and boxsize was a parameter in
that version as well.
--
David Winsemius
On Mar 21, 2009, at 2:24 PM, Gerar
require( zoo )
inp <- c( 5, 9, 4, 2, 1 );
m <- zoo( cbind( inp ), as.Date("2003-02-01") + (0:(length(inp)-1)));
dim( m ) # [1] 5 1
dim( m[1,,drop=FALSE] ) # [1] 1 1 - ok
dim( lag( m, -1 )) # [1] 4 1 - ok
dim( rbind( m[1,,drop=FALSE], lag(m,-1) )) # NULL - converted from zoo
matrix to zoo vector
I have tried several sites (2 in Ca and also Australia) and find only
version 2.14.
Which site did you pull 2.15 from?
Thanks
At 11:10 AM 3/21/2009, David Winsemius wrote:
>"Latest version" is a tad non-specific. The help page for (my) package
>rmeta, version 2.15 (download and compiled today)
J S hotmail.com> writes:
Is this sample size large enough to study differences between two groups of
the populations?
Q1: do the body temperatures differ between the two groups of the
overwintering turtles juveniles and adults?
One group (adults) has 6 turtles
Secon
Hi David,
I did a general package update, but it must not have been quite as
complete as I had hoped.
I will make sure I download 2.15 and go from there.
Thanks for you help.
Gerard
At 11:10 AM 3/21/2009, David Winsemius wrote:
>"Latest version" is a tad non-specific. The help page for (my) p
"Latest version" is a tad non-specific. The help page for (my) package
rmeta, version 2.15 (download and compiled today) says:
Usage
forestplot(labeltext, mean, lower, upper, align = NULL, is.summary =
FALSE, clip = c(-Inf, Inf), xlab = "", zero = 0, graphwidth = unit(2,
"inches"), col = m
Hi David,
I just checked to make sure I had the latest version. I see no
boxsize option in the forestplot function parameters or any other
place in the code.
function (labeltext, mean, lower, upper, align = NULL, is.summary = FALSE,
clip = c(-Inf, Inf), xlab = "", zero = 0, graphwidth =
Create an S3 generic lag.zerofill and then define methods for each class:
lag.zerofill <- function(x, k = 1, ...) UseMethod("lag.zerofill")
lag.zerofill.zoo <- function(x, k, ...) {
m <- merge(x, lag(x, k), fill = 0)
structure(m[, -(1:ncol(x))], dimnames = list(NULL
If you look at the original code (or at the help page), you should see
a boxsize parameter. If you set that to 1 in the call you get boxes
all the same size. Presumably that could be modified to suit your
needs.
You seem to have removed that section of the code. The two lines with
that p
Hi All,
I have been able to modify the x-axis to start at zero by adding xlow
and xhigh parameters; that was pretty simple. I have been unable to
find the location of the code that would turn off the information
weighting of the box size (I have smaller randomized trials getting
less weight t
I hacked at a bit differently than Duncan. See if these help pages and
this example point another way:
?combn
?"["
> df <- data.frame(a = 1:4, b=LETTERS[1:4])
> n <- nrow(df)
> cbind(df[combn(1:n,2)[1,],], df[combn(1:n,2)[2,],] )
a b a b
1 1 A 2 B
1.1 1 A 3 C
1.2 1 A 4 D
2 2 B 3 C
2.1
Hi R-users,
I have the following problem: I am trying to learn something about bayes
methodology and started paying around bayesm package, but could not replicate
the Conjunctive model's estimates as they appear in Rossi et al "Bayesian
Statistics and Marketing", 2005, JWS, pages 264-265, Tabl
On 21/03/2009 12:01 PM, Donald Macnaughton wrote:
I have a data frame with roughly 500 rows and 120 variables. I would like
to generate a new data frame that will include one row for each PAIR of
rows in the original data frame and will include all 120 + 120 = 240
variables from the two rows. I
Try this:
> x <- data.frame(a=1:100, b=100:1, c=sample(100))
> # assume even number of rows: bind the even/odd together
> even <- seq(nrow(x)) %% 2
> new.x <- cbind(x[even==1,], x[even==0,])
>
>
> head(new.x)
a b c a.1 b.1 c.1
1 1 100 69 2 99 60
3 3 98 24 4 97 26
5 5 96 71
I have a data frame with roughly 500 rows and 120 variables. I would like
to generate a new data frame that will include one row for each PAIR of
rows in the original data frame and will include all 120 + 120 = 240
variables from the two rows. I need only one row for each pair, not two
rows. Thu
Hi,
I need some help improving this ugly code I wrote. I would like to shift
forward a zoo object, matrix, ts, or list by "shift" items (default 1) and
fill the holes with 0's.
The code below works, but it looks ugly. I could write a function
lag.zerofill() which calls the two functions belo
Hi,
The package evd most functions that one would need for analysis of Extreme
Values so you should consider giving it a try.
By the way your vector of numerical values is not valid; there are a couple of
values with repeated decimal point separators.
Regards,
Christos Argyropoulos
University
On Mar 21, 2009, at 10:31 AM, Barry Rowlingson wrote:
On Sat, Mar 21, 2009 at 12:33 PM, Enrico R. Crema
wrote:
Dear List,
I'm trying to use different R packages for my Teaching Assistantship
classes. And I cam out to an (apparently) very simple problem. I
would
like to retrieve the verti
Thanks, David,
utils::str(pdf.options()) returns exactly the same results
as yours. The colormodel is set to rgb already.
The reason I thought this is a bug is that the same plot function
+ col option works fine when pch=0 or other numbers, except for 1 or 16 (empty
or
solid circle) or any
This came up on R-sig-geo two days ago and this is what I said:
I have the following code in ggplot2 for turning a SpatialPolygon into
a regular data frame of coordinates. You'll need to load ggplot2, and
then run fortify(yoursp).
fortify.SpatialPolygonsDataFrame <- function(shape, region = NULL
On Sat, Mar 21, 2009 at 12:33 PM, Enrico R. Crema wrote:
> Dear List,
>
> I'm trying to use different R packages for my Teaching Assistantship
> classes. And I cam out to an (apparently) very simple problem. I would
> like to retrieve the vertices coordinate of a SpatialPolygon data. I
> know this
dear all i have this dataset: x,y, datavalue
> dati[,c(1,2,5)]
[,1] [,
2] [,3]
[1,] 2.386 3.077 1.740
[2,] 2.544 1.972 1.335
[3,] 2.807 3.347
1.610
[4,] 4.308 1.933 2.150
[5,] 4.383 1.081 1.565
[6,] 3.244 4.519
1.145
[7,] 3.925 3.785 0.894
[8,] 2.116 3.498 0.525
[9,]
Hi Giuseppe,
The language of this mailing list is English.
Ciao a tutti ho appena iniziato ad utilizzare R per ora per attuare un'analisi
geostatistica di dati. Volevo sapere come poter creare un oggetto gstat
partendo da un file testo(che ho gia importato con read.table)e che contiene 3
col
Package name?
Example code?
--
On Mar 21, 2009, at 4:22 AM, herwig wrote:
Hi there,
I am a beginner.
I would like to change the error bars in the bargraph.CI function
from the
default (se) to (sd). The help file says
ci.fun= function(x) c(fun(x)-se(x), fun(x)+se(x))
Is there a simple way
If it is in an S4 slot then there should be an extraction function,
possibly with an obvious name like coords(). Failing that you could
offer the name of the package that you are using. Perhaps that would
contain an example to be used for illustration.
--
David Winsemius
On Mar 21, 2009,
Ciao a tutti ho appena iniziato ad utilizzare R per ora per attuare un'analisi
geostatistica di dati. Volevo sapere come poter creare un oggetto gstat
partendo da un file testo(che ho gia importato con read.table)e che contiene 3
colonne: x,y,value. Mi servirebbe far questo per costruire un var
Dear List,
I'm trying to use different R packages for my Teaching Assistantship
classes. And I cam out to an (apparently) very simple problem. I would
like to retrieve the vertices coordinate of a SpatialPolygon data. I
know this is stored in the "coords" slot, but I can't get access to
it
I've found where the problem was and a way to solve this problem:
One dataset was encoded (and read) as UTF-8
and the other one was encoded (en read) as latin3
In this case, even if at the terminal you see
the same characters, R states that the two
elements are not equal.
Don't know if this is th
Hi there,
I am a beginner.
I would like to change the error bars in the bargraph.CI function from the
default (se) to (sd). The help file says
ci.fun= function(x) c(fun(x)-se(x), fun(x)+se(x))
Is there a simple way of telling the function what (x) precisely is - I
already define in in the of the
On Sat, Mar 21, 2009 at 1:35 PM, Umesh Srinivasan
wrote:
> Hi,
>
> I am trying to simulate animal movement in a gridded landscape made up of
> cells. At each time step (iteration), the animal moves from one cell to
> another in a random fashion.
> This is how I am simulating movement, where a and
Hi,
I am trying to simulate animal movement in a gridded landscape made up of
cells. At each time step (iteration), the animal moves from one cell to
another in a random fashion.
This is how I am simulating movement, where a and b are the x,y co-ordinates
of the animal at the previous time step:
Andrew Ellis wrote:
You cannot have \label in an S chunk, as R will not know what to do
with it. Neither can you have \label in a verbatim environment, as it
will simply be typeset verbatim.
Actually, with your definition of Sinput
\DefineVerbatimEnvironment{Sinput}{Verbatim} {
commentchar
mau...@alice.it wrote:
Now that I have my list of flags with theri respective values (thanks to all
those who posted their suggestions):
Flags Values
1 TrendOff 0
2 MOdwt 1
3ZeroPadding 1
4 Step1HSOff 1
5 Step1NumHSOff 4
6 Step1NumHSOff
Hi there.
Try the "add parameter" in boxplot:
add logical, if true *add* boxplot to current plot.
On Sat, Mar 21, 2009 at 5:02 AM, johnhj wrote:
>
> Hii,
>
> Is it possible, to use the plot() funktion and the boxplot() funktion
> together ?
> I will plot a simple graph and additionally to t
Whenever I try to load the Matrix package, I get the following error
message:
libRlapack.so: cannot open shared object file: No such file or directory
A file with that name is indeed not on the hard disk.
I am using the R version which comes with Ubuntu Hardy Heron LTS. Here
is the output of R.Ver
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