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the data.
Christophe Pallier
On 7/23/07, Mike Lawrence [EMAIL PROTECTED] wrote:
Psych grad student here, R user for 3 years. Using R for experiments
is likely not advisable as it has no fine control of display timing
(ex synching stimuli to the screen refresh, etc). On recommendation
- gl(2,10)
x - rnorm(20)
x - tapply(x, group, mean)[group]
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of a within subject factor 'x' is assessed against the error term
subj:x
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Oops, sorry for the previous empty reply.
You do not tell us which Operating System you are using, so I assume it must
be Windows...
You should check the R import/export Data available from the Help/Manuals
menu
(in French: Aide/Manuels...)
Christophe
On 6/30/07, Christophe Pallier [EMAIL
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On 6/28/07, João Fadista [EMAIL PROTECTED] wrote:
I would like to take out the values from one vector that are equal to the
values in another vector.
Example:
a - c(1,2,3,4,5,6,7,8,9)
b - c(3,10,20,5,6)
b_noRepeats = c(10,20)
b[!(b %in% intersect(a,b))]
See ?intersect
--
Christophe
setdiff(b,a) is even simpler.
On 6/28/07, Christophe Pallier [EMAIL PROTECTED] wrote:
On 6/28/07, João Fadista [EMAIL PROTECTED] wrote:
I would like to take out the values from one vector that are equal to
the values in another vector.
Example:
a - c(1,2,3,4,5,6,7,8,9)
b - c
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://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
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of rows or columns in variables if there are needed
more than once.
Best,
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, to use 'points' or
'lines' to add new elements to the plot.
Also, if you want to plot several curves in one shoot, you can use
'matplot'.
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10 25 0 10_25 1 2 FALSE
7 10 25 0 10_25 2 2 FALSE
8 10 25 0 10_25 3 2 TRUE
My code is not very readable...
Yet, the 'trick' of using an helper function like 'cumcount' might be
instructive.
Christophe Pallier
On 6/22/07, Kevin E. Thorpe [EMAIL PROTECTED
',]
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)
Christophe Pallier
On 6/19/07, livia [EMAIL PROTECTED] wrote:
Hi, I am trying to write a function with the following codes and I would
like
it to return the values for alpha
beta para parab seperately. Then I would like to use this funstion for
variable with factor a and b. But the result turns out
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-help
PLEASE do read the posting guide
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'.
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back
years later, I can follow the exact flow of transformations.
Again, make (like awk or R) is not without limits and idiosyncrasies. If
someone has a simpler solution, I am interested to hear about it.
Christophe
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Christophe Pallier (http://www.pallier.org)
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Ah! I forgot to mention that it is possible to call awk from R:
a - system(awk -F'=' '/\\[/{a=$1;next}{print $1,$2,a}' example.ini,
intern=T)
z - textConnection(a)
read.table(z)
Christophe
On 6/14/07, Christophe Pallier [EMAIL PROTECTED] wrote:
On 6/14/07, Gabor Grothendieck [EMAIL
.
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, reproducible code.
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$1 - list();
next }
{ print listname [' $1 '] - $2 }
(I know, it looks cryptic... so I am shooting myself in the foot after
claiming that awk scripts are typically quite readable ;-)
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and see what scripts are more elegant (read 'short and understandable')
Best,
Christophe
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with Awk. With binary formats, it is another story...
But, again, this is my limited experience; I would like to know if there are
situations where using SAS/SPSS is really a better approach.
Christophe Pallier
On 6/8/07, Robert Wilkins [EMAIL PROTECTED] wrote:
As noted on the R-project web
, yet both are tools that one gets
attached to because they increase your productivity a lot.
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) == 1)
}
are.crossed - function (factor1,factor2)
{ all(table(factor1,factor2) 0 ) }
Christophe Pallier
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On 6/4/07, Tim Bergsma [EMAIL PROTECTED] wrote:
Is there a conventional way to test for nested factors? I.e., if 'a'
and 'b' are lists of same-length factors, does
(group)])
Christophe Pallier
On 5/22/07, John Kane [EMAIL PROTECTED] wrote:
aggregate(Measure, list(Month=Month), mean)
--- Benoit Chemineau [EMAIL PROTECTED]
wrote:
Hello,
I have a basic problem but i can't figure it out
with the
table underneath. I would like to compute monthly
tapply(Measure,Month,mean)[as.character(Month)]
--
Christophe Pallier (http://www.pallier.org)
On 5/22/07, Benoit Chemineau [EMAIL PROTECTED] wrote:
Hello,
I have a basic problem but i can't figure it out with the
table underneath. I would like to compute monthly averages.
I would
If your data is inside a data.frame names 'a' with data from each group in a
different columns,..., you can use stack and perform the anova with:
l - aov(values~ind,data=stack(a))
anova(l)
Christophe Pallier
On 5/15/07, CrazyJoe [EMAIL PROTECTED] wrote:
Thank you Guys.
Let say that from
')
}
It can speed up things considerably.
Christophe Pallier
On 5/15/07, Liaw, Andy [EMAIL PROTECTED] wrote:
If it's a matrix, use scan(). If the columns are not all the same type,
use the colClasses argument to read.table() to specify their types,
instead of leaving it to R to guess
The Wilcoxon paired rank sign test assumes symmetry.
(cf. http://www.basic.northwestern.edu/statguidefiles/srank_paired_alts.html
)
Christophe Pallier
On 5/10/07, Alexander Kollmann [EMAIL PROTECTED] wrote:
Hello,
given situation:
- pre / post test comparison of discrete, paired data
Have you loaded the library containing lda?
library(MASS)
Christophe Pallier
On 5/10/07, elyakhlifi mustapha [EMAIL PROTECTED] wrote:
hi,
I have this error
tr - sample(1:50, 25)
train - rbind(iris3[tr,,1], iris3[tr,,2], iris3[tr,,3])
test - rbind(iris3[-tr,,1], iris3[-tr,,2], iris3
are balanced)
Christophe Pallier
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Hello Christoph!
In the past, I used an utility called eps2wmf.
It only works under Windows though (maybe under Linux with wine?).
I believe it is available on the CTAN (Tex archives).
The nice thing is that wmf files are not bitmap and scale well.
Christophe Pallier
Christoph Lehmann wrote:
Hi
I
,
Christophe Pallier
Let me fake some `data':
set.seed(1); roiValues - rnorm(32)
subjectlabels - paste(V1:8, sep = )
options(contrasts = c(contr.helmert, contr.poly))
roi.aov - aov(roi ~ Cue*Hemisphere + Error(Subject), data=roiDataframe)
roi.aov
Call:
aov(formula = roi ~ Cue * Hemisphere + Error
getAnywhere(wilcox.test.default)
...
You will see that it uses 'psignrank' (for the one sample case).
Christophe Pallier
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It seems that you want to do is in a Anova with the within factors
factCond and factRoi, right?
If this is correct, then you should try:
summary(aov(vectdata~factCond*factRoi+Error(factCond*factRoi)))
Christophe Pallier
www.pallier.org
Bela Bauer wrote:
Hi,
I'm still trying to figure out
this helps.
Christophe Pallier
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to me that 'lme' is
useful when some assumptions of standard anova are violate (e.g. with
repeated measurements when the assumption of sphericity is false), or
when you have several random factors.
Christophe Pallier
http://www.pallier.org
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indeed use the 'aggregate' function:
aggregate(a[,1],list(a[,1],a[,2]),length)
This yields one line per unique value (but that may be what you want...)
Christophe Pallier
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]
for a while, (and I am now changing to
x-tapply(x,f,mean)[as.character(f)] because of the peculiarities of
indexing named vectors with factors )
The use of tapply(x,f,mean)[match(f,unique(f))] assumes a particular
order in the result of tapply, no? It seems a bit dangerous to me.
Christophe
We want to solve X's and Y's value!!
When we have two equation:
For example:
2X+3Y=5
X+Y=2
solve(matrix(c(2,1,3,1),2,2),c(5,2)) yields the solution.
see '?solve' (you have to know about matrix algebra to understand this
function)
Christophe Pallier
be treated as fixed-factors in the
analysis, would they not?
Any suggestion is welcome.
Christophe Pallier
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Hello Mateusz,
The 'hist' function works on the raw data.
In your data set example, you have already computed the number of data
points in each bin.
What you really want is probably a barplot of N
You could display your data:
plot(Class,N,'h')
Or
names(N)-Class
barplot(N)
Christophe Pallier
# starting point
for (i in 1:100) { x=f(x); v=append(v,x) }
There may be smarter ways...
Christophe Pallier
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the associated
MSE).
Sorry if the solution is obvious.
Christophe Pallier
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to show all the png (or use any other
diaporama system). However, I find it a dirty hack.
Is there a simpler and cleaner way to achieve this?
Christophe Pallier
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. However, I have ~30 million data points
measurements of each variable. I can read this into R from file and
produce a plot with plot(x0, x1) but as you would expect, its
not pretty
to look at and produces a postscript file of about 700MB.
Christophe Pallier
http://www.pallier.org
, not random factor.
Actually, you do not need lme for to run a repeated measure anova.
You could use the aov function:
$ summary(aov(DV~GROUP*TRIAL+Error(SUB/TRIAL)))
This, again, yields the correct results.
Hope this helps,
Christophe Pallier
http://www.pallier.org
)
factor.list=lapply(factor.names,eval)
names(factor.list)=factor.names
aggregate(d[1],factor.list,FUN)
}
Christophe Pallier
http://www.pallier.org
---
HEre is the code for aggregate.data.frame that recovers the name sof the
factors:
my.aggregate.data.frame - function (x, by, FUN
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