On Mar 18, 2012, at 4:43 PM, Dajiang Liu wrote:
Dear All,
I have a seemingly very simple question, but I just cannot figure
out the answer. I attempted to run the
following:a=0.1*(1:9);which(a==0.3);it returns integer(0). But
obviously, the third element of a is equal to 0.3.
I must
This is quite a CPu consuming process. My system got hung up for the
big file I have.
Within the for loop that you have suggested, can't I have a case
statement for different value of nfields to be read and specify what
format does the variable needs to be read?
something like
case
# input format
Dear R-list,
I am trying to visualize where the dropout happens in our patient flow. We are
currently using traditional flowcharts and it bothers me that I can't visualize
both the percentage and the flow in one diagram.
The other day I came across some interesting diagrams doing exactly what
Hello,
Try
text=
fish fam length
1 a 71.46
2 a 71.06
3 a 62.94
4 b 79.46
5 b 52.38
6 b 56.78
7 b 92.08
8 c 96.86
9 d 98.09
10 d 17.23
11 d 98.35
12 d 82.43
13 e 83.85
14 e 33.92
15 e 23.16
16 e 31.39
17 e 57.08
18 e 27.05
19 f 62.38
20 f 83.21
21 f 18.72
22 f 84.32
23 g 15.99
24 h 40.33
25 h
Hello R friends,
I want to use my R script in VB to make macro in Excel.
I tried with RExcel but it seems to me that this package is just GUI API
and I still have to run(connect) R to use the script.
Google tells me there are some ways to make R script as an independent
library/module/header so
I am working on Latex and R and using following code.
echo=FALSE=
infile-read.table(test.txt,sep=\t)
Col3 - unique(infile[,3])
LCol3 - length(Col3)
for (i in 1:LCol3) {
print(paste(Column, Col3[i]))
print(infile[infile[,3]==Col3[i],-3])
}
@
I am getting following output.
1] Column C V1 V2 V4
It is working fine.
Thanks
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Col3 - unique(Msg17$V3)
LCol3 - length(Col3)
for (i in 1:LCol3) {
print(paste(Column, Col3[i]))
write.table(Msg17[Msg17$V3==Col3[i],-3], row.names=F, col.names=F,quote=F)
# If you R implementation does not accept 'F', use 'FALSE'
}
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Great it works! But how can i put space or tab between two records?
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Hi,
some sample data would be *very* helpful...
Kind regards,
Kimmo
16.03.2012 15:44, statquant2 wrote:
Hello I am looking for a special plot.
Let's suppose I have *100 days and
*each day I have a (1D)
distribution of the same variable.
I would
As to the reasons, David as given you the necessary hints.
In order to get around the issue, here is what I do:
a - round( 0.1 * ( 1:9 ), 1 )
a
[1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
which( a == 0.3 )
[1] 3
Rgds,
Rainer
Original-Nachricht
Datum: Sun, 18 Mar 2012
Thanks all of your answers and advices! They brought me some light!
I'll have a look to memois package and to tracemem function in order to
check if they can help me somehow, at least to understand and trace in
detail how memory gets consumed.
Thank you all!
David
2012/3/16 Jan T. Kim
On Sun, Mar 18, 2012 at 09:43:54PM +, Dajiang Liu wrote:
Dear All,
I have a seemingly very simple question, but I just cannot figure out the
answer. I attempted to run the following:a=0.1*(1:9);which(a==0.3);it returns
integer(0). But obviously, the third element of a is equal to 0.3.
Actually, RExcel and the StatConn DCOM connector are what you want, and this is
not the right place to discuss it. Go to http://www.statconn.com/, and read the
license carefully.
---
Jeff NewmillerThe
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 18/03/12 14:40, Uwe Ligges wrote:
On 18.03.2012 05:47, Lorenzo Cattarino wrote:
Hi R users,
Working from a PC, I am trying to install the spatstat package on a Unix
cluster. I created
the following PBS file to send a job array:
Hi,
I am using follosing code and getting the below output.
echo=FALSE=
infile-read.table(/home/manish/Desktop/test.txt,sep=\t,header=TRUE)
Col3 - unique(infile[,3])
LCol3 - length(Col3)
for (i in 1:LCol3) {
print(paste(Disease Risk:, Col3[i]),row.names=FALSE,
col.names=FALSE,quote=FALSE)
How big is the file? In the example I sent I waa using 'textConnection' to
reread the input. If the file is large, this can be slow. You will have
better luck writing the converted data outmto a temporarynfile and reading it
right back in.
I am not such exactly what you are asking. You can
I'm not sure I follow exactly what group of regression models you want to
create, but a good first step might be to use reshape so that each party's
vote share goes on a different row and the vote shares are all in the same
column. Then you can use plyr grouping on tipo and party to make your
Hi
This is quite a CPu consuming process. My system got hung up for the
big file I have.
Within the for loop that you have suggested, can't I have a case
statement for different value of nfields to be read and specify what
format does the variable needs to be read?
something like
case
Peter:
many thanks for your help. This is basically what I wanted to do and in
a much more elegant way.
On Mon, Mar 19, 2012 at 03:13:40AM -0700, Peter Meilstrup wrote:
I'm not sure I follow exactly what group of regression models you want to
create, but a good first step might be to use
Hi list,
I thought I was used to layouts, but today I am facing a problem I cannot
overcome :
On my R installation (Windows 7 Pro, SP1, R version 2.13.0, daily update of
packages), I am not able to put raster plots in user defined layouts :
layout.matrix-matrix(c(1,2,3,4,5,5),2,3)
Book title: Data Mining Applications with R
URL: http://www.rdatamining.com/books/book2.
Publisher: Elsevier
Chapter proposal due date: 30 April 2012
Introduction
R is one of the most widely used data mining tools in scientific and
business applications, among dozens of
Is this with SDI in Windows? I'd update to a recent version of R, and
please provide reproducible code next time.
It could be the same as this issue, now long ago fixed:
https://stat.ethz.ch/pipermail/r-devel/2011-February/059906.html
Cheers, Mike.
On Mon, Mar 19, 2012 at 9:57 PM, Olivier
Mike,
It is with SDI in Windows.
Here is reproducible code (by the way, I just add the opening of any raster and
plot it four times).
library (raster)
b- brick(system.file(external/rlogo.grd, package=raster))
layout.matrix-matrix(c(1,2,3,4,5,5),2,3,byrow=TRUE)
layout(mat=layout.matrix)
I am not sure that I understand but does something like this do what you want?
ec-1:10
vec[vec==4] - 100
vec - 1:10
vec[ vec==4 | vec==8] - 100
vec - 1:10
aa - 50
vec[vec==4] - aa
John Kane
Kingston ON Canada
-Original Message-
From: marc_...@yahoo.fr
Sent: Sun, 18 Mar 2012
Hello,
I implemented two functions reshape_long and reshape_wide (see full working
example below) to reshape data frames.
I created several small examples and the two functions seemed to work
properly. However, using the reshape_wide function
on my real data sets (about 200.000 to 300.000 rows)
Hello everyone,
I am working for a few days already on a basic algorithm, very common in
applied agronomy, that aims to determine the degree-days necessary for a
given individual to reach a given growth stade. The algorithm (and context)
is explained here:
Hi,
I'm using R to create multidimensional data, ie tensors. R, for my
work, is very good for import the data and I have seen that there are
packages to manage tensor and to factor the tensor.
I would ask a help regarding the package called tensor and tensorA. I
have seen, unfortunately, that the
Hi Ray,
Thanks for reply
/R/PlotGridded2DMap.R
/R/image.plot.fix.R
/R/image.plot.plt.fix.r
are the functions those I wrote for plotting and they work with another
data, but only I have some issue with only the codes those I provided
before.
and what do you mean by I am redefining the map()
i am new to time series
i found in help about arima
arima(x = data, order = c(p, d, q))
what is exactly p,d,q? if i not changed them,what effects will happens?
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how to cluster/classify attached time series ?(each column/time series
consider as single unit while clustering/classifying)
if my concept is wrong,tell me how to extract time series with highest
information content ?
given file is to do it
http://r.789695.n4.nabble.com/file/n4484173/rasta.txt
Hello,
I'm trying to write the sorted data in a file of a data.frame, My
question and my problem is that when I record in file adds a new column
row.name, which apparently is the original position in the file.
I wanted to write to the file without this column
Dear memberships,
I'm trying to estimate the following multivariate local regression model using
the locfit package:
BMI=m1(RCC)+m2(WCC)
where (m1) and (m2) are unknown smooth functions.
My problem is that once I get the regression done I cannot get the fitted
values of each of this
Thanks a lot for the clarification. I just find it very bizarre that if you run
a=0.1*(1:9);which(a==0.4)
it returns the right answer. Anyway, I will pay attention next time. Thanks a
lot.
Date: Mon, 19 Mar 2012 08:59:59 +0100
From: rainer.schuerm...@gmx.net
Subject: Re: [R] a very simple
Hello,
I want to determine the regression relationship between a proportion (y)
and a continuous variable (x).
Reading a number of sources (e.g. The R Book, Quick R,help), I believe I
should be able to designate the model as:
model-glm(formula=proportion~x, family=binomial(link=logit))
this runs
Hi everyone,
I'm trying to reduce the font size in the Y exe in this plot:
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=150
Anyone knows how to do it?
I have checked the argument lab.cex and cex, but any of these works!
if you want to check us this code:
### read the data
On 12-03-19 7:42 AM, Letnichev wrote:
Hello everyone,
I am working for a few days already on a basic algorithm, very common in
applied agronomy, that aims to determine the degree-days necessary for a
given individual to reach a given growth stade. The algorithm (and context)
is explained here:
The logit link requires a binary response variable, not a proportion. Better
bet is a beta regression. You can also do some stuff with linear regression if
you do some transformations, but linear regression assumes the outcome is any
number on the real number line bounded between -Inf and Inf.
On 19-03-2012, at 12:42, Letnichev wrote:
Hello everyone,
I am working for a few days already on a basic algorithm, very common in
applied agronomy, that aims to determine the degree-days necessary for a
given individual to reach a given growth stade. The algorithm (and context)
is
On 19-03-2012, at 13:47, Dajiang Liu wrote:
Thanks a lot for the clarification. I just find it very bizarre that if you
run a=0.1*(1:9);which(a==0.4)
it returns the right answer. Anyway, I will pay attention next time. Thanks a
lot.
Look at
a = 0.1*(1:4)
a - 0.4
[1] -0.3 -0.2
Your response variable is not binomial, it's a proportion.
Try the betareg function in the betareg package, which more correctly assumes
that your response variable is Beta distributed (but beware that 1 and 0 are
not allowed). The syntax is the same as in a glm.
HTH
Ruben
-Mensaje
Hi Georgiana,
Take a look at the betareg package at
http://cran.r-project.org/web/packages/betareg/index.html
HTH,
Jorge.-
On Mon, Mar 19, 2012 at 10:05 AM, Georgiana May wrote:
Hello,
I want to determine the regression relationship between a proportion (y)
and a continuous variable (x).
Hi,
You're not following the algorithm as given. The asin step shouldn't
be done for all values, but only for the ones that don't meet the
previous conditions. You're trying to calculate that step for ALL
values, then only use certain ones. You must instead subset the
values, THEN calculate that
It doesn't have anything to do with your use of order(). Those are the
row names of your data frame. You can disable writing them with the
row.names=FALSE argument to write.table().
Sarah
On Mon, Mar 19, 2012 at 8:16 AM, MSousa ricardosousa2...@clix.pt wrote:
Hello,
I'm trying to write
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Georgiana May
Sent: 19 March 2012 14:06
To: r-help@r-project.org
Subject: [R] regression with proportion data
I understand that the binomial function concerns successes
https://en.wikipedia.org/wiki/Autoregressive_integrated_moving_average
You may also be interested in forecast:::auto.arima
Michael
On Mon, Mar 19, 2012 at 6:54 AM, sagarnikam123 sagarnikam...@gmail.com wrote:
i am new to time series
i found in help about arima
arima(x = data, order = c(p, d,
To view the source of (most) functions, simply type funcname without
parentheses: here, you get
dmvt
function (x, delta, sigma, df = 1, log = TRUE, type = shifted)
{
if (df == 0)
return(dmvnorm(x, mean = delta, sigma = sigma, log = log))
if (is.vector(x)) {
x - matrix(x,
Once again, R has been accepted as an organization for the Google Summer of
Code (2012).
We invite students interested in this program to learn more about it. A good
starting
point is http://rwiki.sciviews.org/doku.php?id=developers:projects:gsoc2012.
The Google
GSOC home page is
On 19-Mar-2012 Dajiang Liu wrote:
Thanks a lot for the clarification. I just find it very bizarre
that if you run
a=0.1*(1:9);which(a==0.4)
it returns the right answer. Anyway, I will pay attention next time.
Thanks a lot.
The basic explanation is that, for an integer r (0r10), what is
Your description was too general for me to know exactly what you want but
perhaps this will help you solve your own problem
set.seed(123)
evtlist = sample(c('fwd','rev'),100,replace=TRUE)
evtlist
[1] fwd rev fwd rev rev fwd rev rev rev fwd rev
fwd rev
[14] rev fwd rev fwd fwd fwd rev rev rev
On Mon, Mar 19, 2012 at 7:56 AM, Jose Bustos Melo jbustosm...@yahoo.es wrote:
Hi everyone,
I'm trying to reduce the font size in the Y exe in this plot:
dotplot( bank ~ MV2007 + MV2009 , data = d, horiz = T,
par.settings = list( superpose.symbol = list( pch = 21, fill = c(
lightblue,
On Mon, Mar 19, 2012 at 12:47:12PM +, Dajiang Liu wrote:
Thanks a lot for the clarification. I just find it very bizarre that if you
run a=0.1*(1:9);which(a==0.4)
it returns the right answer. Anyway, I will pay attention next time. Thanks a
lot.
Hi.
Yes, these things are bizarre
Dear Ieva,
plm(.., model=within) (which is the default for plm()) estimates a
within model on time-demeaned data, which is equivalent to using the
LSDV estimator. Therefore any time-constant dummy variable you add by
hand will be discarded because of perfect collinearity.
What kind of dummies
On Mar 19, 2012, at 03:32 , Dominic Comtois wrote:
Say I fit a logistic model and want to calculate an odds ratio between 2
sets of predictors. It is easy to obtain the difference in the predicted
logodds using the predict() function, and thus get a point-estimate OR. But
I can't see how to
Hello,
I am trying to fit my histogram to a smooth Gaussian curve(the data
closely resembles one except a few bars).
This is my code :
#!/usr/bin/Rscript
out_file = irc_20M_opencl_test.png
png(out_file)
scan(my.csv) - myvals
hist(myvals, breaks = 50, main = My Distribution,xlab = My Values)
Thanks
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Dear R-help,
I am trying to rename the variables in a dataframe, called 'T1A' here.
Seems renaming was successful, but when I call one of the variable I got
error message and I wanted to know why.
The data frame contains 365 rows and 49 columns. I would like to name the
first column `DATE` and
Hello,
you're totally right, I tried first to control the flow with if
(MaxDailyTemp k MinDailyTemp k){statement} but it was a bit messy.
Then ifelse() was supposed to help me out, but it didn't.
Thank you for your time, your code works exactly as I want :)
P.
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Yes, with the same input I had two different outputs with Excel and R. When
printing a debug report of Excel, it showed no anomalies and I am certain it
didn't calculate odd values (such as NaNs).
The way I coded was wrong, as Sarah said, I didn't follow completely the
algorithm. The solution she
Use the eol=\n\n option. The records should have a 2 line space.
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Thanks for your advice. I actually meant to ask about the pmvt for the
distribution function. Viewing the source code pmvt uses the function
mvt which uses the function probval which sources the fortran code:
Fortran(mvtdst, N = as.integer(n), NU = as.integer(df),
LOWER =
Hello.
I'm allready this far. I have a function which is calculating the lower (l)
and upper (u) limit for a confidence interval for the odds ratio.
For example for 5 simulated 2x2 tables the upper and lower limits are:
u
[1] 2.496141 7.436524 8.209161 4.313587 3.318612
l
[1] -0.9718608
Hi hubinho,
You are almost there. Try this slightly modification of your function:
# theta, u and l are vectors of the same length
foo - function(theta, u, l) mean(theta = l theta = u, na.rm = TRUE)
foo(theta, u, l)
HTH,
Jorge.-
On Mon, Mar 19, 2012 at 12:55 PM, hubinho wrote:
Hello.
Dear colleagues,
I had a question wrt the car package. How do I evaluate whether a
simpler multivariate regression model is adequate?
For instance, I do the following:
ami - read.table(file =
http://www.public.iastate.edu/~maitra/stat501/datasets/amitriptyline.dat;,
col.names=c(TCAD, drug,
On 19-03-2012, at 17:39, HJ YAN wrote:
Dear R-help,
I am trying to rename the variables in a dataframe, called 'T1A' here.
Seems renaming was successful, but when I call one of the variable I got
error message and I wanted to know why.
The data frame contains 365 rows and 49 columns.
On 19-03-2012, at 16:54, statfan wrote:
Thanks for your advice. I actually meant to ask about the pmvt for the
distribution function. Viewing the source code pmvt uses the function
mvt which uses the function probval which sources the fortran code:
No it doesn't source. It call a
hey,
thnks a lot. I got exactly what I wanted.
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Hello there,
I am new to using regression in R. I wanted to solve a simple regression
problem where I have 2 equations and 2 unknowns.
So lets say -
y1 = alpha1*A + beta1*B
y2 = alpha2*A + beta2*B
y1 - runif(10, 0,1)
y2 - runif(10,0,1)
alpha1 - 0.6
alpha2 - 0.75
beta1 - 1-alpha1
beta2
On 2012-03-19 09:39, HJ YAN wrote:
Dear R-help,
I am trying to rename the variables in a dataframe, called 'T1A' here.
Seems renaming was successful, but when I call one of the variable I got
error message and I wanted to know why.
The data frame contains 365 rows and 49 columns. I would like
Dear all,
I have made a function that given a number of list elements plot them to the
same window.
The first element is plotted by using plot and all the rest are plotted under
the
same window by using lines.
I have below a small and simple reproducible example.
x1-c(1:10)
plot(x1)
Dear Ranjan,
As you no doubt noticed, the Manova() function in the car package, or the
Anova() function for which Manova() is an alias, produces type II or III tests
for a multivariate linear model. To compare two nested multivariate linear
models, as you wish to do, you can use the standard R
Hello R community,
I am processing raw Affymetrix CEL files and I am using the Michigan custom
CDF library hgu133plus2hsentrezgprobe. I have been looking for
documentation on the function that it contains...I am specifically
interested in converting probe names to gene symbols. Does anybody know
If I understand you correctly, a univariate Gaussian distribution is
uniquely determined by its first two moments so you can just fit those
directly (using sample mean for population mean and sample variance
with Besel's correction for population variance) and get the best
Gaussian (in a ML
I don't believe this is possible in base graphics: you need to plan
your graphics ahead with something like plot(, ylim = range(x1, x2,
x3)). There's a pen-and-paper approach which means once something is
there, it's on the device permanently (unless you write over it).
Perhaps an interactive
I see, that could be an option, however isn't there a fitting function
which would do that on given data?
On 19 March 2012 19:49, R. Michael Weylandt michael.weyla...@gmail.com wrote:
If I understand you correctly, a univariate Gaussian distribution is
uniquely determined by its first two
Take a look at fitdistr in the MASS package.
fitdistr(x, normal)
I don't think you need to supply start values for the normal since its
loglikelihood function is nicely behaved. You may need to for harder
distributions.
Michael
On Mon, Mar 19, 2012 at 2:54 PM, Vihan Pandey
1. Homework assignment? We don't do homework here.
2. If not, a mixture model of some sort? I suggest you state the
context of the problem more fully. R has several packages to do
mixture modeling, if that's what you're trying to do.
3. In any case, this cannot be done with lm() (at least
I'll check it out, thanks a million Micheal!
On 19 March 2012 19:59, R. Michael Weylandt michael.weyla...@gmail.com wrote:
Take a look at fitdistr in the MASS package.
fitdistr(x, normal)
I don't think you need to supply start values for the normal since its
loglikelihood function is nicely
Perhaps matplot()?
matplot(cbind(x1, x2, x3), type = 'l')
See ?matplot for more information.
HTH,
Jorge.-
On Mon, Mar 19, 2012 at 2:40 PM, Alaios wrote:
Dear all,
I have made a function that given a number of list elements plot them to
the same window.
The first element is plotted by
Thanks for the immediate answer.
is ther any alternative for the matplot? There might few limitations with
matplot in my case. I will post again if needed when I will be at office
tomorrow.
Regards
Alex
From: Jorge I Velez jorgeivanve...@gmail.com
Cc: R
On 2012-03-19 07:35, S Ellison wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Georgiana May
Sent: 19 March 2012 14:06
To: r-help@r-project.org
Subject: [R] regression with proportion data
I understand that the binomial
Dear all,
While I am executing my code I receive the error below
Error in sort.int(x, na.last = na.last, decreasing = decreasing, ...) :
'x' must be atomic
the weird thing that I am not calling anywhere sort function nor do I rely on
anyh sorting.
How I can discover where this comes from
I call upon the great and mighty Google (hallowed be its name) to discover:
traceback()
and its more powerful cousin
options(error = recover)
Michael
On Mon, Mar 19, 2012 at 3:22 PM, Alaios ala...@yahoo.com wrote:
Dear all,
While I am executing my code I receive the error below
Error in
Call traceback() after seeing the error message.
E.g.,
factor(list(1, 2:3, 4:6))
Error in sort.list(y) : 'x' must be atomic for 'sort.list'
Have you called 'sort' on a list?
traceback()
3: stop('x' must be atomic for 'sort.list'\nHave you called 'sort' on a
list?)
2: sort.list(y)
?traceback
?options ## consider changing error option to recover
?debug
Search on debugging in R to find more possibilities. R is a
programming language. You need to learn how to debug code if you wish
to program in R.
-- Bert
On Mon, Mar 19, 2012 at 12:22 PM, Alaios ala...@yahoo.com wrote:
Or look at the xlim and ylim arguments to plot. E.g.,
x1 - 1:10 ; x2 - 11:17 ; x3 - 21:23
plot(NA, NA, xlim=range(1, length(x1), length(x2), length(x3)),
ylim=range(x1, x2, x3), type=n, xlab=, ylab=)
points(x1, type=b)
lines(x2)
points(x3)
title(xlab=The X Values, ylab=The Y Values)
Hello Bert,
This is definitely not for a homework problem. I am trying to estimate
frequencies of mutations in different groups. The mutation frequencies can
be modeled as a linear relation in cases of mixtures. So I have a lot of
populations that follow the relationship -
y = alpha*A + beta*B
hi
I think You Can Use solve function to solve the equations.
___
Niloofar.Javanrouh
MSc Student Of BioStatistics
Mashad University Of Medical Sciences
From: Bert Gunter
Thank you very much. This was, was i needed. Unfortunately I have one futher
problem with this Code. I don't only need the coverage probability for one
but for a range of different odds ratios. (for example [1;30]). I tried it
with a loop but I get an error. I think again, that I'm almost there
We are pleased to announce version 0.8 of the acs package for R, now
available on CRAN
(http://cran.r-project.org/web/packages/acs/index.html.
The package provides a general toolkit for managing, analyzing, and
presenting data from the U.S. Census American Community Survey
(ACS). Confidence
Note that your equations can be written:
y = alpha*A + (1-alpha)*B, which is equivalent to
y = (A-B) * alpha + B , i.e. of form
y = C*alpha + B a simple linear equation in alpha
You have two different values of alpha at which y was measured, so
just stack up all your results into a single
Hi hubinho,
You need to initialize the for() loop and then store the u and l values
properly:
# parameters
n1 - 10
n2 - 10
y - 100
alpha - 1
z-1.96
# creating B 2x2 tables
B - 50
u - l - vector('numeric', B)
for (i in 1:B){
theta - i
x1 - exp(alpha +theta)/ (1+ exp(alpha +theta))
Hello all,
I need to figure out a way to lag a variable in by a number of days
without using the zoo package. I need to use a remote R connection
that doesn't have the zoo package installed and is unwilling to do so.
So that is, I want a function where I can specify the number of days
to lag a
Thank you very much again. But in this case I get the coverage probability as
an average over all values for the odds ratio.
I need a coverage probability for every value for the odds ratio.
So the coverage probability for odds ratio = 1, than for odds ratio = 2 and
so on.
Sorry to bother you
Hi,
Although the following statements work individually in R, they produce an error
if placed inside a function as below:
fsubt - function(a) {
b - 1:length(a)
b-a
}
The error message is:
Error: unexpected input in:
b - 1:length(a)
b-
Any insight would be greatly appreciated.
Thanks,
Jack
I think you'll need to provide a reproducible example, because your
code works for me:
fsubt - function(a) {
+ b - 1:length(a)
+ b-a
+ }
fsubt(1:5)
[1] 0 0 0 0 0
fsubt(sample(1:10))
[1] -8 -6 1 1 -1 5 3 1 4 0
fsubt(2)
[1] -1
On Mon, Mar 19, 2012 at 4:01 PM, Schryver, Jack C.
Hi Eleni
Question like this are better served on the bioconductor mailing list.
Nonetheless try this
ALL - topTable(fit2, coef=1, number=Inf)
ALL$SYMBOL - unlist(mget(ALL$ID, hgu133plus2hsentrezgSYMBOL, ifnotfound=NA))
Here ALL is the output from limma for differential expression (ALL$ID is
The OP's error suggests (to me) that there's a line break error
somewhere so it may be a funny quirk of encoding/OS incompatibility if
it's from a source()'d script.
Incidentally, the OP could also write the body of his function as a
one liner with:
seq_along(a) - a
Michael
On Mon, Mar 19,
Hi hubinho,
This starts to look as homework to me so this will be my last try in
helping you.
The general strategy would be along the lines of (1) write a function that
does what you want for a value of theta and (2) sapply() that function to
the vector of theta values you would like to
I think the most likely explanation is that something in
the input string has had the effect of inserting an invisible
character between the - and the a in b-a, and a
possible suspect is pollution by UTF8: see the discussion at
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