> -Original Message-
> From: Marc Schwartz [mailto:marc_schwa...@me.com]
> Sent: Monday, August 24, 2009 9:57 AM
> To: Daniel Nordlund
> Cc: r help
> Subject: Re: [R] Combining matrices
>
>
> On Aug 24, 2009, at 11:46 AM, Marc Schwartz wrote:
>
> &
cb[[2]][rep(1:nrow(cb[[2]]), nrow(cb[[1]])), ] )
##compare m.o with all x.o where rows and columns of x.o with tied
totals are permuted##
for(i in 1:nrow(rperm)){
for(j in 1:nrow(cperm)){
if(identical(m.o,x.o[rperm[i,],cperm[j,]])) {
cat('TRUE','\n')
break
}
I updated the previously posted function for Cramer's V so that it
automatically prints Cramer's V, chi-square, the degrees of freedom, and the
significance level of Cramer's V based on the chi-square value and the
degrees of freedom with desired (user-supplied) levels of precision. An
example is
and five cities.
Age=rep(0:3,each=25)
City=rep(0:4,20)
e=rnorm(100)
#Create a dependent variable
Height=(Age-1)+(City-1)+(Age-1)*(City-1)+e
#Box whisker plot of Height by Age category for each City
bwplot(Height~factor(Age)|factor(City))
Daniel
-
cuncta stricte
asks the question how much does
group=2 with sub=b stand out from the average in its group and in its sub.
To clarify your confusion, you may want to write out the model formula.
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht
that the approach to use OLS to model your data is
fine if your error distribution (the distribution of e in the simulated
data) is normal.
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mai
ation of nested effects in R (e.g. with the nlme and lme4
libraries). So go ahead and put in the effort.
Daniel
ps: on a more general note, the R-help list is to help with the
implementation in R rather than statistical/econometric questions,
especially if these problems are exte
are
equal for both machines
#the equivalent formulation used is: is the sum of the coefficients across
machines equal to zero?
library(car)
linear.hypothesis(fm4,hypothesis.matrix=c(1,1,1,-1,-1,-1),rhs=c(0))
hope that helps,
daniel
workers=as.factor(workers)
Wen Huang-3 wrote:
>
> Hello,
The workers=as.factor(workers) codeline in my post dropped below my name. It
should be in the code before the command line for the linear model.
Daniel Malter wrote:
>
> Wen, to follow up on Thierry, your workers are nested in machines (since
> each worker only works one machine).
hen compare 2 to 1, and 3 to 2. This tells you whether including more
variables (hierarchically) makes your model better.
HTH,
Daniel
On a different note, if lme fits with "restricted maximum likelihood," I
think I remember that you cannot compare them. You have to fit them with
&quo
del with interaction effects if not all
direct effects are included. And in fact, I never see this done in analyses
in my field (business/economics).
Daniel
-
cuncta stricte discussurus
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-Ursprüngliche Nachricht-
Von: John
If you are looking for a function to standardize a variable so that it has
mean zero and unit variance:
std=function(x){if(length(which(is.na(x)))==0) (x-mean(x))/sd(x) else
(x-mean(x,na.rm=T))/sd(x,na.rm=T)
}
x=rnorm(100,3,5)
mean(x)
sd(x)
x2=std(x)
mean(x2)
sd(x2)
HTH,
Daniel
x=0.01345778543577
signif(x,3)
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von Gundala Viswanath
Gesendet: Tuesday, September 08, 2
any ideas?
Also is there a way of including the threshold in the actual model, so
that could be estimated too?
Thanks
Dan
--
**
Daniel Brewer, Ph.D.
Institute of Cancer Research
Molecular Carcinogenesis
Email: daniel.bre...@icr.ac
[i]=summary(lm(y~(x>(i-1$r.squared
tested.threshold[i]=i-1
}
#inspect r-squareds
print(data.frame(tested.threshold,r.squared))
#Should indicate the highest r-squared
#at the appropriate threshold level
HTH,
Daniel
-
cunc
Hi!
I have a need to do the ridiculous and run a batch file if certain
conditions are met within an R script. is there a simple way to do so? or
to run dos commands?
Thanks!
Daniel
[[alternative HTML version deleted]]
__
R-help@r-project.org
gt; [8,] " 8" "r"
> [9,] " 9" "s"
> [10,] "10" "t"
>
You have asked several questions about using as.matrix() on a data.frame
with colums of different types, in particular character and numeric types.
Since matrices can o
This is not a "please solve my homework for me" newsgroup. Hint: You may
want to learn about chi-square tests.
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help
ebody else do my work
for me, please. I don't know if there is any newsgroup that finds this
approach acceptable.
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...
ooking.for,string)
##Return a boolean for each index
##whether "lorem ipsum" is found or not?
index%in%grep(i.am.looking.for,string)
##Finally, table the occurrence/absence of
##"lorem ipsum" by gender
table(index%in%grep(i.am.looking.for,stri
is rbind(a,b,c) what you mean?
Daniel
-
cuncta stricte discussurus
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-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von FMH
Gesendet: Tuesday, September 22, 2009 12:40 PM
An: r
ab[order(ab[,1]),]
-
cuncta stricte discussurus
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Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von FMH
Gesendet: Tuesday, September 22, 2009 1:20 PM
An: r-help@r-project.org
B
My guess is that you cannot keep it numeric, but that you have to convert it
to a string (the gurus may know better).
x=001
x
x="001"
x
Daniel
-
cuncta stricte discussurus
-
-Original Message-
From: r-help-boun...@r-project.org
below:
x=c(1.0001,1.0002,1.0003,1.0004,2.0001,2.0002,2.0003,3.0001,3.0002,4)
rank(x)
rank(round(x,2))
rank(x)==rank(round(x,2))
HTH,
Daniel
-
cuncta stricte discussurus
-
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help
ing down the problem? I don't
know much about ODBC in general or what RODBC is doing under the hood, or how
the SAS ODBC driver is written, but it would be useful if I could get R to
write SAS datasets, and not just read them.
Thanks,
Dan
Daniel Nordlund
Bothell, WA USA
__
this question and please apologize if my
english is not very accurate and/or the question is too simple.
Regards
Daniel.
[[alternative HTML version deleted]]
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P
ere j is in 1 to 5 and indicates the five values you want to
store.
HTH,
Daniel
-
cuncta stricte discussurus
-
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of seydahmet ercan
Sent: F
> -Original Message-
> From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
> Sent: Thursday, February 11, 2010 1:25 AM
> To: Daniel Nordlund
> Cc: r-help@r-project.org
> Subject: Re: [R] SAS and RODBC
>
> On Thu, 11 Feb 2010, Daniel Nordlund wrote:
>
>
64
> >> yHat[5]:68.67
> >>
> >> This is based on the following equation (given in the PDF): Y = 1.67 +
> X1
> >> +
> >> 9.50 X2
> >>
> >> I ran the following commands in R:
> >>
> >> data =
individual
variables (x), columns with their interactions, and one column of 1s for the
intercept. Let y be your dependent variable. Then, OLS estimates are
computed by
X'X inverse X'y
Or in R
solve(t(X)%*%X)%*%t(X)%*%y
Best,
Daniel
-
cuncta stricte d
> -Original Message-
> From: Frank E Harrell Jr [mailto:f.harr...@vanderbilt.edu]
> Sent: Saturday, February 13, 2010 5:49 AM
> To: Daniel Nordlund
> Cc: r-help@r-project.org
> Subject: Re: [R] SAS and RODBC
>
> Daniel Nordlund wrote:
> . . .
>
> >
&
> -Original Message-
> From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
> Sent: Sunday, February 14, 2010 7:09 AM
> To: Daniel Nordlund
> Cc: r-help@r-project.org
> Subject: Re: [R] SAS and RODBC
>
>
>
> On 14.02.2010 08:19, Daniel Nordlund wro
er advice can be
given. A bivariate uniform distribution is simply pairs of uniform random
numbers. The density is constant, so it is easy to get to the other quantities
that you want based on that. So what is it that you are actually wanting to do?
Dan
Daniel Nordlund
Bothell, WA USA
___
significant, it will tell you that something
about obs for which your data is missing is different than for the year for
which you have observed the independent variables. But that requires us to
figure out which model to use in the first place.
Best,
Daniel
-
cuncta
lieve some of these concerns as it would capture
unobserved heterogeneity between observed and unobserved time periods.
Daniel
-
cuncta stricte discussurus
-
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-p
Exact to the degree that real numbers can be exactly expressed. Look into
?uniroot
You want to specify the function in a one-sided way, i.e., x/(1-exp(-x))-2.2
HTH,
Daniel
-
cuncta stricte discussurus
-
-Original Message-
From: r-help
,z),]
HTH,
Daniel
-
cuncta stricte discussurus
-
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Evgenia
Sent: Friday, February 19, 2010 12:10 PM
To: r-help@r-project.org
Subject: [R
variability rather than ascending order. This is essentially what the code
further below does. After the rank transformation, a regular Mann-Whitney U
test is applied. The "manual" and code are pasted below.
Best,
Daniel
Siegel-Tukey test
Description: Non-parametric Siegel-Tukey test for e
ifelse approach
DOW.ind=ifelse(DOW=="MON",1,ifelse(DOW=="TUE",2,0))
#continue to nest ifelse statements for more days
DOW.ind
#merging approach
day=c("MON","TUE")
ind=c(1,2)
ind.frame=data.frame(day,ind)
merge(data.frame(DOW),ind.frame,by.x="DO
Greetings,
I am acquiring a new computer in order to conduct data analysis. I
currently have a 32-bit Vista OS with 3G of RAM and I consistently run into
memory allocation problems. I will likely be required to run Windows 7 on
the new system, but have flexibility as far as hardware goes. Can
ng to share? I would also be interested in other
simple examples using sem and RAM specifications. If anyone is interested, I
would also be willing to share the R code I have written for other chapters in
the Singer and Willett book.
Thanks,
Dan
Daniel Nordlund
neral or had overlooked in the
documentation for the sem package.
Again, this has been very helpful,
Dan
Daniel Nordlund
Bothell, WA USA
> -Original Message-
> From: John Fox [mailto:j...@mcmaster.ca]
> Sent: Wednesday, March 03, 2010 7:19 AM
> To: 'Chuck Cleland&#x
1 5 ...
04/15/102 5
05/15/103 NA
06/15/10NA 5
Any ideas? I have been fiddling around with plyr and reshape without success
Thanks
Dan
--
**
Daniel Brewer, Ph.D.
Institute of C
I am using curve3d in the emdbook package to graph a gaussian copula density
function generated via the copula package. Unfortunately, it appears that
emdbook masks dmvnorm from the package mvtnorm in a way that prohibits
copula from generating the gaussian copula. (Sounds very confusing!) For
exam
provide unique identification, maybe Flight and OnDate together do, which
can also be handled in merge.
Let us know if that solves the problem.
Best,
Daniel
-
cuncta stricte discussurus
-
-Original Message-
From: r-help-boun...@r
be
> equal to 1,2,3,4,5,6, corresponding to the component with lowest BIC
> Caicno # so for this, I wish to obtain 300 values of k, where k can be
> equal to 1,2,3,4,5,6, corresponding to the component with lowest CAIC
>
This is helpful, because we can now help you with R and
tween x1 and
x2, and with a level 1 random effect (intercept) only? That would give a
hint where the singularity arises.
Best,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help
26.00 10,274.00 10,326.00 10,390.00 10,570.00
> >> 10,638.00 10,728.00 10,952.00 9,854.00
> >>>
> >>
> >> Now, here's the confusion. plot(T1[,1],T1[,10]) creates a plot, but
> >> the range on both X & Y is 1-10. I want 1-10 on the X axis but ne
An Econometrician may help you with more theoretical insights, but you could
do Monte-Carlo simulations of data and analyze the effects you are
interested in.
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun
the read.xls() function from the gdata package to get
the file, something like this
library(gdata)
df <- read.xls("http://mirecords.umn.edu/miRecords/download_data.php?v=1";)
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
__
return a list of the
values
multipleoutput<-function(x)
{
squared<-x^2
cubed<-x^3
exponential<-exp(x)
factorialVal<-factorial(x)
return(list(squared=squared, cubed=cubed, exponential=exponential,
factorialVal=factorialVal))
}
Hope this is he
have you tried the pyramid function in the epicalc package?
best,
daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von Victor Manuel Garcia
Can you provide a self-contained example of what you did so far (i.e. code
for simulating some data and code for the plot)? That would greatly help to
help you find a solution.
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht
folder or
the dk folder altogether, the changes come along so to me it seems
that it's the overwrite part that doesn't work as it should - or am I
doing something wrong here?
See below for sessionInfo.
Thanks a bunch
Daniel
> sessionInfo()
R version 2.9.0 (2009-04-17)
i
TRUE, density = NULL, angle = 0,
col = c("yellow","Blue","orange"), border = par("fg"),
main = "Mujeres",
xlim =NULL, ylim = NULL, xpd = TRUE, log = "",
axisnames = FALSE,
axis(1,cex.axis=0.75))
That s
put another condition in your loop
if(all.equal(x,y)=TRUE) i=i+1 else t.test...
something in that direction.
best,
daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun
do that until I learn what I'm doing wrong.
>
> I've got more interesting questions of which the answers are far more
> important to me than this one!
>
> Thanks,
> Mark
Mark,
Take a look at the readline() function.
Hope this is helpful,
Dan
Daniel Nordlund
Bothell,
folder or
the dk folder altogether, the changes come along so to me it seems
that it's the overwrite part that doesn't work as it should - or am I
doing something wrong here? To me, it seems that the function
safe.dir.create (which is defined in package.skeleton never overwrit
ed in the ABOVE diagonal elements (i.e. not
including diagonal element temselves), the your loop should be
for (i in 1:9)
{
for (j in (i+1):10)
{
// do something
}
}
And you avoid your current problem.
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
__
y
#pch=16 puts a filled dot
#lcolor=NA removes the dotted lines on the chart
#now plot the confidence intervals
for(i in 1:max(CI$index)){
lines(c(CI$lower[i],CI$upper[i]),c(i,i))
}
#for each index (i.e., estimated coefficient)
#plot x-coordinates CI$upper and CI$lower
#at y-coordinate = in
example
should be fine.
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von Erin Hodgess
Gesendet: Tuesday, July 14, 2009 11:40 AM
An: R help
Bump
Anyone?
Thanks
Daniel
On 13 jul 2009, at 10.57, Daniel Klevebring wrote:
> Dear all,
>
> I am using package.skeleton to build a small packages of misc function
> for personal use. I have recently discovered that the option
> force=TRUE doesn't seem to do what is mean
It is because the nesting structure perfectly explains the data (i.e., there
is only one observation and, therefore, no variation for each Ind in each
Treatment).
e=rnorm(90,0,1)
x=rep(1:3,30)
y=rep(1:30,each=3)
z=x+y+e
ano=aov(z~factor(y)/factor(x))
ano
residuals(ano)
Best,
Daniel
Hello,
I apologize if an answer to my questions is available, or if I
submitted this question incorrectly. I have read the mailing lists, as
well as the R Project and CRAN homepages. However, I may have missed
something.
I noticed the package 'norm' has been removed. Its page
http://cran.r-projec
I appreciate the info!
I am mostly interested with ML missing data imputation. Is there
another R module which has such a function that is well regarded?
Thank you.
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PLEAS
F$d) * DF$Test + (DF$c+DF$d) * !DF$Test
>
> DF$f = with(DF, (c*d)*Test + (c+d)*!Test)
>
> DF
Mark,
Why can't you use ifelse() ?
DF$g <- with(DF, ifelse(Test,c*d,c+d))
Have I missed something in what you are doing?
Dan
Daniel Nordlund
Bothell, WA USA
__
asets...
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von Rbeginner
Gesendet: Sunday, July 19, 2009 9:49 PM
An: r-help@r-project.org
Betreff: [
prediction.
Best,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von Rbeginner
Gesendet: Monday, July 20, 2009 10:50 PM
An: r-help@r
XYZ=XYZ[XYZ$A==TRUE , ] should do. Note that this specific command
overwrites your XYZ dataframe. If you want to keep XYZ, you will want to
name the newly created frame differently.
HTH,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche
is multiple
imputation, plotting n imputations at alpha = 1/n to see where there is, or is
not, variability.)
Many thanks,
Daniel Farewell
Cardiff University
R version 2.9.1 (2009-06-26)
i386-apple-darwin8.11.1
locale:
en_GB.UTF-8/en_GB.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8
attached base package
x4)
listdata=list(data1,data2)
for(i in 1:length(listdata)){
listdata[[i]][,length(listdata[[i]])+1]=listdata[[i]][,length(listdata[[i]])
]*listdata[[i]][,length(listdata[[i]])-1]
}
HTH,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche N
vents variable given your conditions
does that work for you?
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von jimdare
Gesendet: Thursday, J
I recently answered this (or, rather, basically the same) question in a
thread with an example in it. I am trying to find it back.
Just a sec
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto
Here it is...
http://www.nabble.com/plotting-confidence-intervals-td24482119.html
Marc's answer is probably the way to go
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: Daniel Malter [mailto:dan...@um
ns
par(mfcol=c(1,2))
hist(estim[,1],main="Intercept")
hist(estim[,2],main="Effect of variable 'test'")
HTH,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [m
al variation
among the subject (in other words, respecting that some are just more likely
than others to answer yes).
But Robert has pointed out many valuable alternative approaches.
Best,
Daniel
-
cuncta stricte discussurus
-
_
V
?
> x<-c(rnorm(100), 30);
> boxplot(x)
> boxplot(x,outline=FALSE)
Dan
Daniel Nordlund
Bothell, WA USA
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
rather than an R question, but I thought one of the
many specialists in experiments might be able to help me out quickly on this
or point me to appropriate literature.
Thanks,
Daniel
---
"Who has visions should see a doctor,"
Helmut Schmi
I should have mentioned that I am using the lmer library for my analyses,
just in case other methods provide results differently.
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r
)))-1)))
}
}
}
Daniel Malter wrote:
>
> You can copy the code below to your R-code editor. For Yule's Q, the data
> is expected in two vectors. For cramer's phi, the data is expected in
> separate columns of a matrix or dataframe.
>
> ##Run this code
> yule.Q=function(x,y){(table
Look up the McNemar test. That sounds right...
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von mik07
Gesendet: Tuesday, July 28, 2009
Caio, check the lme4 library. The lmer function allows for fixed and random
effects.
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von
does "length" instead of "nlevels" do what you want to do?
with(Trees,tapply(SppID,PlotID,unique))
daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-b
Hi Everyone,
is there the possibility in R to assign a variable to be an alias of
another one?
Example:
x <- 17
# assign y to be an alias of x
y # returns 17
x <- 4
y # returns 4
Daniel
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https://stat.e
[1] 19
# but I am looking for the unevaluated expression (".Last.expression",
so to say), which would be expression(x * x + 3) in this case
Daniel
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PLEASE do read th
ot;D" "E" NA
> is.na(s2)
[1] FALSE FALSE FALSE FALSE FALSE TRUE
>
Notice that in string s, the characters (NA) are surrounded by quotes, and R
returns false for is.na(). But for string s2, the missing value NA is not
surrounded by quotes and is.na() returns TRUE for s2[6
Not knowing the package or the function. Could the round() function help
you? Or do you mean decimals shown in a plot?
Best,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help
Either you use
for (k%in%col)
or
for (k in min(col):max(col))
does that work for you?
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag
e posting guide. And no worries, I
have asked questions like yours before. It just takes some willingness and
effort to learn to ask better questions in order to get helpful answers.
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht
and I may quote David Winsemius at this point: "In general this falls into
the category of a request to "read my mind".
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mai
wever, without a self-contained example, it is
impossible for us to know whether the error lies here or in one of the many
other reasons (e.g. on of the functions called in your MODELS function).
Daniel
-
cuncta stricte discussurus
-
-Ursprüng
vector it gets
coerced to the more general type, in this case, numeric.
Note that "is.numeric" returns TRUE for both is.numeric(0) and
is.numeric(0L).
Daniel Gerlanc
Associate Analyst
Geode Capital Management
1 Post Office Sq, Floor 28
Boston, MA 02109
daniel.gerl...@geodecapital.com
___
you say it doesn't work, what does that mean? Do you get an error
message? Are you left with negative numbers? What? A minimal,
self-contained, reproducible example would help us help you.
> KN1 <- rnorm(100)
> KN2 <- ifelse(KN1 < 0, 0, KN1)
Works fine for fine for me.
D
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Daniel Nordlund
> Sent: Friday, August 07, 2009 2:50 PM
> To: r-help@r-project.org
> Subject: Re: [R] Seeing negative numbers to zero
>
> > -Ori
Dan
--
**
Daniel Brewer, Ph.D.
Institute of Cancer Research
Molecular Carcinogenesis
Email: daniel.bre...@icr.ac.uk
**
The Institute of Cancer Research: Royal Cancer Hospital, a charitable Company
Limited by Guarantee, Regis
nomial(logexp(days=expos))) ##Call to
glm with custom link
-
Best,
Dan Barton
--
Daniel C. Barton
PhD Candidate
USGS Montana Cooperative Wildlife Research Unit &
Program in Organismal Biology and Ecology
University of Montana
205 Natural Science
Missoula, MT 59812
daniel.ba
umeric variable and its higher
order polynomials (e.g., Rosenthal and Rosnow 1985).
Daniel
-
cuncta stricte discussurus
-
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of C.H.
Sent: Sund
mmy matrix in the first place; so the list is really the way
to go.
x1=c(1,2,3)
x2=c(4,5,6)
y1=c(0,1,2,3)
y2=c(4,5,6,7)
myList=list(x1,y1)
myList
myList[[1]]=append(myList[[1]],x2)
myList[[2]]=append(myList[[2]],y2)
myList
HTH,
Daniel
-
cuncta stricte d
DATA statement used (Total process time):
> real time 0.04 seconds
> cpu time0.01 seconds
>
> I can even concatenate windows and unix SAS datasets.
>
>
Hope this helpful,
Dan
Daniel Nordlund
Bothell, WA USA
___
nce
Dan
P.S. the reason for this is that the round end looks bad at lwd=3 or
more
Daniel Alcock
Malaria Genetics (T112)
Wellcome Trust Sanger Institute
Cambridge
CB10 1SA
UK
+44 (0)1223 834244 ext. 4994
"The most exciting phrase to hear in science, the one that heralds new
discoveries, i
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