Displaying 20 results from an estimated 159 matches for "discussurus".
2009 Jul 09
9
Population pyramids
Hi, I hope somebody can help me with this issue: I am doing population pyramids using the barplot command, so in the left side I have male age structure and in the right side the female age structure. To plot the male age structure I put the data in negative numbers. Now, I want to change the sign in the bar plot in such way that I have no-sign numbers, both in left and right side of the graph. I
2009 Jul 24
4
CI wiskers
I have a matrix containing means and CIs (lower and upper in two columns, so
three columns for every data point) for several points. I have to build a
graph of these means accompained by the CIs (as wiskers). No problems with
making the graph of means, but I don't know how to introduce CIs.
Can anybody advise?
--
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2007 Nov 16
1
Nonparametric manova
.... This discussion is form March 2006 and
there seemed to be no package or function implemented at the time. Has this
changed? Is there a package that provides nonparametric manova as in McArdle
and Anderson (2001) and Anderson (2001) now?
Thanks,
Daniel
-------------------------
cuncta stricte discussurus
2008 Nov 20
1
sub / gsub - extracting between identical symbols
...ow to tell
"sub" between which of them to extract.
The string looks like this
12/01/03/08
The extracted variables should look like:
x1=12
x2=01
x3=03
x4=08
If anybody could help or point me to useful help, I would be greatful.
Cheers,
Daniel
-------------------------
cuncta stricte discussurus
2008 Feb 27
4
Error in cor.default(x1, x2) : missing observations in cov/cor
Hello,
I'm trying to do cor(x1,x2) and I get the following error:
Error in cor.default(x1, x2) : missing observations in cov/cor
A few things:
1. I've used cor() many times and have never encountered this error.
2. length(x1) = length(x2)
3. is.numeric(x1) = is.numeric(x2) = TRUE
4. which(is.na(x1)) = which(is.na(x2)) = integer(0) {the same goes for
is.nan()}
5. I also try
2008 Feb 26
3
OLS standard errors
...mmary(reg). The difference of 1% seems nonmarginal. Should I have
multiplied var(e)*solve(t(z)%*%z) by n and divided by n-1? But even if I do
this, I still observe a difference. Can anybody help me out what the source
of this difference is?
Cheers,
Daniel
-------------------------
cuncta stricte discussurus
2008 Jun 13
1
parsing - input buffer overflow
...t.
#This is done using:
n=100 #choose n large enough
length(which(is.na(gregexpr("Saint",x,ignore.case=TRUE)[[1]][1:n])==FALSE))
But again, if the text is large, I cannot assign it to x. I'd be grateful
for any suggestions.
Cheers,
Daniel
-------------------------
cuncta stricte discussurus
2008 Oct 09
2
Bug in ifelse
...] ## assign all x greater than a
b1 and b2 should be equal as I see it. Anybody who can explain why they are
not makes my day, especially the one who can tell me how I can get b1 equal
to b2 (without sacrificing the ifelse condition).
Thanks much,
Daniel
-------------------------
cuncta stricte discussurus
2009 Sep 20
3
statistics
The myoglobin sequence, with reference number NM_005368 in Gen bank, has the
following
frequencies of DNA nucleotides:
A C G T
237 278 309 242
Do these data provide sufficient evidence, at the 1% level of significance,
that the DNA nucleotides
have an unequal distribution, that is the DNA nucleotides are not evenly
utilised?
Clearly state your hypothesis, test statistic and conclusion.
2010 Apr 08
2
Problem using elements in a vector
Hi
So my particular problem is this:
I have a row vector of length 5200 elements - specifically created by
x<-rbinom(5200,1,0.5)
y<-matrix(x,nrow=1,ncol=5200)
y
now, each element is either a 0 or a 1 - e.g. it could be
(0,1,1,1,1,0,0,0,1,1,1) e.t.c.
when the element is a 1, i need to multiply a number (say 1000) by 1.005,
and if it is 1 again, multiply it _again_ by 1.005.
so for
2009 Nov 11
3
how to use # in a rd doc in url address
I am writing a rd doc, and need to use "#" in a url adress. This would make:
\url{http://www.xxxx.org/myfolder/#myanchor}
Of course, I suppose this will not work because # is a special character
starting a comment line in the rd dialect. I did not found a similar
example in "Writing R exentions". I am not sure bout using \dQuote{a
quotation}), and use \sQuote and \dQuote
2008 Mar 31
2
Finding a mean value of a variable holding a dummy variable fixed
I have time-series data on approval ratings of British Prime Ministers. The
prime ministers dating from MacMillan onward till today are coded as dummy
variables and the approval ratings are entered for each month. I want to
know the mean value of the approval rating of each Prime Minister in the
dataset and the approval rating during his/her first month and last month as
PM. What R code should
2009 Oct 04
1
offlist Re: AW: Urgently needed Exercise solutions related to PracticalData Analysis Using R Statisctial Software
...ines.
Interested people send their all contact details and interest at:
bhangisolutions at gmail.com
If it is possible to chat at Gtalk will be more great. MSN, Yahoo, SkyPe is
possible and also telephone.
Best regards,
Mahesh
-------------------------
cuncta stricte discussurus
-------------------------
-----Urspr?ngliche Nachricht-----
Von: David Winsemius [mailto:dwinsemius at comcast.net]
Gesendet: Saturday, October 03, 2009 11:56 PM
An: Daniel Malter
Betreff: offlist Re: AW: [R] Urgently needed Exercise solutions related to
PracticalData Analysis Using R Statisctial...
2009 Aug 05
4
for loop
I am trying to get the function "Models" to work each time there is an
instance of k. This code will stop after the first model is complete. I need
it to come back and pass the next value of c into the "Initial.State"
function. any ideas?
col<-c(23:28)
#Setup
for(k in col){
Initial.State(Response=zample[,c(k,29)],
Explanatory=zample[,variable_columns],
2008 Nov 24
3
count the cumulative for each subject
I have a data set like the following:
subject visit x1
1 1 0.5
1 2 1.2
1 3 0.7
2 1 0.4
2 2 0.6
2 3 1.0
.....
where x1 is the interval between the two visits. Now I want to calculate the
cumulative intervals since the beinging, for example
subject visit x1 cum
1 1 0.5 0.5
1 2 1.2 0.5+1.2
1 3 0.7 0.5+1.2+0.7
2 1 0.4 0.4
2 2 0.6 0.4+0.6
2 3 1.0 0.4+0.6+1.0
.....
is there an easy to generate the
2007 Dec 31
3
Survival analysis with no events in one treatment group
I'm trying to fit a Cox proportional hazards model to some hospital
admission data. About 25% of the patients have had at least one
admission, and of these, 40% have had two admissions within the 12
month period of the study. Each patients has had one of 4
treatments, and one of the treatment groups has had no admissions for
the period. I used:
2009 Sep 04
2
Nested Fixed Effects - basic questions
Hi R people,
I have a very basic question to ask - I'm sorry if it's been asked before, but I searched the archives and could not find an answer. All the examples I found were much more complicated/nuanced versions of the problem - my question is much more simple.
I have data with multiple, nested fixed effects (as I understand it, fixed effects are specified by the experimental design
2008 Aug 07
4
Obtaining the first /or last record of a subject in a longitudinal study
Dear R users,
I was wondering if anyone knows how to obtain(subset) the first and/or the
last record of a subject in a longitudinal setup.
Normally in SAS one uses first.variable1 and last.variable1. So my question
is that is there an R way of doing this.
Regards,
--
Luwis Diya, Phd student (Biostatistics),
Biostatistical Center,
School Of Public Health,
Catholic University of Leuven,
U.Z. St
2009 Sep 08
3
Omnibus test for main effects in the face ofaninteraction containing the main effects.
...with "restricted maximum likelihood," I
think I remember that you cannot compare them. You have to fit them with
"maximum likelihood." I am pointing this out because lmer with
restricted
maximum likelihood by standard, so lme might too.
-------------------------
cuncta stricte discussurus
-------------------------
-----Urspr?ngliche Nachricht-----
Von: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
Im
Auftrag von John Sorkin
Gesendet: Monday, September 07, 2009 4:00 PM
An: r-help at r-project.org
Betreff: [R] Omnibus test for main effects in the face of an...
2009 Oct 28
5
re gression with multiple dependent variables?
i have a series of regressions i need to run where everything is the same
except for the dependent variable, e.g.:
lm(y1 ~ x1+x2+x3+x4+x5, data=data)
lm(y2 ~ x1+x2+x3+x4+x5, data=data)
lm(y3 ~ x1+x2+x3+x4+x5, data=data)
is it possible to run all these regs with a single command? given that the
bulk of the work for linear regressions is inverting a matrix that depends
only on the independent