Hi there,
I have a vector and would like to create a data frame, which contains
all unique combination of two elements, regardless of order.
myVec - c(1,2,3)
what expand.grid does:
1,1
1,2
1,3
2,1
2,2
2,3
3,1
3,2
3,3
what I would like to have
1,1
1,2
1,3
2,2
2,3
3,3
Can anybody help?
of a
variable?
Thanks in advance
Antje
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in (or equals the lower bin break).
In the example case, I'd like to get:
c(1,3,2,2,4,1,5)
How would you solve it?
Antje
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Thanks a lot! That's what I was looking for :-)
A
On 19 March 2011 13:56, Duncan Murdoch murdoch.dun...@gmail.com wrote:
On 11-03-19 8:18 AM, Antje Niederlein wrote:
Hi there,
probably there is a very simple solution, but I cannot think of one...
I have a vector with values:
data- c
Yes, I did send an attachment - but I forgot that attachments might be
removed, no?
Here it is:
http://rapidshare.com/files/452815636/drmData.RData
Antje
On 16 March 2011 12:19, Mike Marchywka marchy...@hotmail.com wrote:
Date: Wed, 16 Mar
Anybody who can help me with this issue?
On 15 March 2011 14:15, Antje Niederlein niederlein-rs...@yahoo.de wrote:
Hi there,
I try to model some dose response curves (drc-package). In most cases
it is fine but now I got some data which produces me the following
error:
load(drmData.RData
but the prediction fails? Anything I can do in this case?
Antje
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?
Antje
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Thanks for every helpful answer :-) !
I thought it was something easier but as long as there is a solution
it's fine for me.
Ciao,
Antje
On 21 February 2011 13:12, Martin Maechler maech...@stat.math.ethz.ch wrote:
Ted Harding ted.hard...@wlandres.net
on Mon, 21 Feb 2011 11:08:19 -
not sure whether this method
is the one to choose...
If I estimate lambda with mle2() and use the RSS as criteria to
minimize, my lambda is much smaller that with fitdistr().
I'm happy about any suggestion!
Antje
On 11 February 2011 09:16, Ingmar Visser i.vis...@uva.nl wrote:
The ML estimate
things
and I'm not sure whether I got it right how to find the ML-function of
a more complex distribution...
Antje
On 11 February 2011 10:14, Ingmar Visser i.vis...@uva.nl wrote:
Antje,
On Fri, Feb 11, 2011 at 9:58 AM, Antje Niederlein
niederlein-rs...@yahoo.de wrote:
Hi Ingmar, hi Dennis
() of stats4 package or mle2() of bbmle package, I
would have to write the function by myself which should be optimized.
But what shall I return?
-sum((y_observed - y_fitted)^2)
?
Any other suggestions or comments on my solution?
Antje
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in how
to use mle() correctly.
I think, I'll look into Bens mle2() method and figure out whether this
is a more elegant way :-)
Ciao,
Antje
On 7 February 2011 21:39, Ben Bolker bbol...@gmail.com wrote:
Antje Niederlein niederlein-rstat at yahoo.de writes:
A few day ago, I was looking
Hello,
is there somebody who can help me with my question (see below)?
Antje
On 1 February 2011 09:09, Antje Niederlein niederlein-rs...@yahoo.de wrote:
Hello,
I tried to use mle to fit a distribution(zero-inflated negbin for
count data). My call is very simple:
mle(ll)
ll() takes
Hello,
is there somebody who can help me with my question (see below)?
Antje
On 1 February 2011 09:09, Antje Niederlein niederlein-rs...@yahoo.de wrote:
Hello,
I tried to use mle to fit a distribution(zero-inflated negbin for
count data). My call is very simple:
mle(ll)
ll() takes
)
cat(x in lapply, x,\n)
fit - mle(ll)
coef(fit)
})
Can anybody help?
Antje
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)
coef(fit)
})
Can anybody help?
Antje
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? (Is there any easy way to
do it in R?)
Ciao,
Antje
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? (Is there any easy way to
do it in R?)
Ciao,
Antje
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from 0 to 1)? Do I have to
put it into the ll-function?
Is there any general comment on what I'm doing?
Antje
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Hallo,
hope you can help me with this question:
I have calculated a function using f-smooth.spline(data) and
approxfun(f). Now I want to calculate the mean slope of the resulting
function.
Haven't found the right R command yet, maybe you can give me a hint.
Thanx, Antje
=Riss [%],
auto.key=list(lab=levels(r),columns=3))
Thank you very much,
Antje
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a substring. At least, that is
my guess..
Ciao,
Antje
Don MacQueen wrote:
Read the help page for substr().
It says that the first argument should be a character vector.
The only one that works is the one where you gave it a character vector.
You said only third one works. But you didn't explain
to an
explanation for this behaviour?
Thanks a lot in advance!
Antje
values - factor(c(rep(abc,3), rep(bcd,3), rep(cde,3)))
substr(values,2,3) - ..
substr(as.character(values),2,3) - ..
values - as.character(values)
substr(values,2,3) - ..
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Thanks a lot for the hint!
Antje
Duncan Murdoch wrote:
On 20/11/2009 7:11 AM, Antje wrote:
Hey there,
I'm running R 2.10 on Windows XP (Professional) and I was wondering
where the HTML help window disappeared?
With earlier versions everything was fine. Now I get only this
old-fashioned
Peter Ehlers wrote:
If there's been an answer to this, I've missed it.
Here's my take.
Antje wrote:
Hi there,
I was wondering if anybody can explain to me why the boxplot ends up
with different results in the following case:
I have some integer data as a vector and I compare the stats
. But why???
Antje
,d1,d2
1,0.936585365853659,32
2,0.936585365853659,32
3,0.936585365853659,32
4,0.936585365853659,32
5,0.907317073170732,31
6,1.08292682926829,37
7,1.08292682926829,37
8,1.02439024390244,35
9,1.08292682926829,37
10,1.05365853658537,36
11,0.936585365853659,32
12,1.08292682926829,37
Hey there,
I'm running R 2.10 on Windows XP (Professional) and I was wondering
where the HTML help window disappeared?
With earlier versions everything was fine. Now I get only this
old-fashioned text windows without any links when I type
?some_function
Can anybody help me?
Antje
Hi Steve,
Steve Lianoglou wrote:
Hi,
On Nov 3, 2009, at 2:41 AM, Antje wrote:
Hi there,
currently, I've updated R on my Mac (OS X) to version 2.10. I was
wondering if I have to install all additional packages again???
In Windows, I just needed to copy the library folder of the old
Hi there,
currently, I've updated R on my Mac (OS X) to version 2.10. I was
wondering if I have to install all additional packages again???
In Windows, I just needed to copy the library folder of the old
installation but how does it work with Mac?
Can anybody give me a hint?
Antje
Karl Ove Hufthammer wrote:
In article 4addc1d0.2040...@yahoo.de, niederlein-rs...@yahoo.de
says...
In every list entry is a data.frame but the columns are only partially
the same. If I have exactly the same columns, I could do the following
command to combine my data:
do.call(rbind, myList)
-0.33022738 1.9510503 -1.7930093
30 -0.62615365 0.7330671 -0.4032405
The only thing I can think about is checking the names of each list
entry and adding NA-columns before combining them.
Is there any other way to do so?
Antje
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The following code isn't working and we can't figure out why..
letters = c(A,B,C,D,E,F,G,H,I,J)
numbers = 1:3
for(i in 1:6){ #6 letters
for (j in 1:3) { #3 numbers
for (k in -1:1) { #answer -1,right or
Hello,
I'm not very experienced with lattice and I was wondering whether I get
get some hints from you how to create a pure heatmap (using levelplot),
without any axis, title, legend, margin at all... I just want to see the
coloured squares, nothing else.
Any suggestions?
Antje
,
right.padding = 0)
)
levelplot(..., scales = list(draw = FALSE), colorkey = FALSE, xlab =
NULL, ylab = NULL, par.settings = my.padding)
But still I have a upper and lower margin (left and right margins are gone)
What else do I have to do?
Antje
Antje wrote:
Hello
,
right.padding = 0)
)
levelplot(..., scales = list(draw = FALSE), colorkey = FALSE, xlab =
NULL, ylab = NULL, par.settings = my.padding)
But still I have a upper and lower margin (left and right margins are gone)
What else do I have to do?
Antje
Antje wrote:
Hello
understand your solution with the quadratic regession
model...
Maybe my approach is completely wrong but I didn't know any other solution.
Anyhow, I've solved the issue so far with the while-construct.
Antje
bill.venab...@csiro.au wrote:
It is not at all clear what you are trying to do.
Fitting
I could also provide the dataset.
Antje
Temperaturzyklen.eps
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Hi there,
I don't have much experience with fitting at all and I'd like to get
some advice how to use the weights-argument with nls correctly.
I have created some data with a sigmoidal curve shape. Each y-Value was
generated by the mean of three values. A standard deviation was
calculated
Hi Uwe,
thanks a lot for your answer! And thanks a lot to all others helping me with
this issue!
Uwe Ligges schrieb:
Antje wrote:
Hi Uwe,
I tried to explain my problem with the given example.
I don't see any documentation which tells me that the length of
col.regions should be one less
]
...
( at[n-1],at[n] ] = color[n-1]
Please correct me if I'm wrong!!!
Antje
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- levelplot(t(mat[nrow(mat):1, ] ), at = my.at, col.regions =
my.col.regions)
print(graph)
Can anybody help me with some hints or little examples?
Antje
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they are outliers...
I've attached one graph, I've generated - maybe it helps to understand)
Any wrong assumption?
Ciao,
Antje
Uwe Ligges schrieb:
Antje wrote:
Hi there,
as I'm not sure to understand the coloring levelplot uses, I'm looking
for another easy way to create a heatmap like
'lastHeatmapColor' by the same vector
position)
I'm very confused...
Antje
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from levelplot for a certain datavalue
(3.5)
Did I explain it somehow clear now?
Antje
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whatever I put there???
With other plots it worked for me, why not in this case?
I just want to see the whole label...
I'm really sorry if I've missed a very obvious reason for this behaviour. Maybe
someone can help me?
Ciao,
Antje
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boxplot(data, las = 2)
par(mar = mar.orig) # put the original values back
Sarah
On Wed, Apr 29, 2009 at 10:06 AM, Antje niederlein-rs...@yahoo.de wrote:
Hi there,
I'm trying to solve a boxplot problem (should be simple, but I cannot find
the solution...):
data - list( long_name1_xxx
Hi Jim,
is this plotrix package version higher than 2.2-7 ?
Antje
Jim Lemon schrieb:
Antje wrote:
Hi there,
I was wondering wether it is possible to creeate plots with nested
lables like in excel?
If yes, could anyone provide me the information how to do it?
I've attached an image
Hi there,
I was wondering wether it is possible to creeate plots with nested lables like
in excel?
If yes, could anyone provide me the information how to do it?
I've attached an image of an Excel plot to show you, what I'd like to plot with
R :-)
Ciao,
Antje
inline
Hi there,
I was using Open/Save-dialogs from the package svDialogs (SciViews). But now
the package has dissapeared? How do I have to set up my R-installation to
further use these dialogs??? (beside copying my old packages to the new
installation).
Ciao,
Antje
Hi there,
I was wondering wether it's possible to generate an axis with groups (like in
Excel).
So that you can have something like this as x-axis (for example for the
levelplot-method of the lattice package):
---
| X1 | X2 | X3 | X1 | X2 |
solved by grouping... (see my next mail)
Antje schrieb:
Hi there,
I was wondering wether it's possible to generate an axis with groups
(like in Excel).
So that you can have something like this as x-axis (for example for the
levelplot-method of the lattice package
questions:
* I'd like to start with Y Label 1 in the upper row (that's a more general
issue, how can I have influence on the order of x,y, and groups?)
* I'd like to put the groups at the bottom
Can anybody give me some help?
Antje
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Hi there,
I'm very glad to use the R-help mailing list for R-related question but more
and more often I face general statistical problems. Does anyone know by chance
a community (mailinglist, forum, ...) where I can ask these kind of questions?
I'm glad for any link or hint :-)
Antje
not primarily bound
to this software.
Thanks a lot for all answers!
Antje
Antje schrieb:
Nobody can help with this question (or tell me where I can find help) ?
Antje schrieb:
Hi there,
I'm not sure whether this is the right mailing list to put this
question... if not, please tell me where I
Nobody can help with this question (or tell me where I can find help) ?
Antje schrieb:
Hi there,
I'm not sure whether this is the right mailing list to put this
question... if not, please tell me where I can find help.
I have a spare dual-core machine which run Windows XP Professional
(knime.org) to set up data analysis workflows. They
have some nodes to execute R on a remote machine. Is it possible to run R on
this windows machine, so that I can call it from somewhere else?
Antje
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https
.
Now, I tried the same with levelplot (instead of plot) and I see all
windows are created but the plots are only drawn in one window, one on top of
the other...
Can anybody give me a hint, why it behaves like this and what should I change?
Ciao,
Antje
I guess, I got the answer.
(http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-do-lattice_002ftrellis-graphics-not-work_003f)
Ciao,
Antje
Antje schrieb:
Hello everybody,
I just started with lattice plots and I was wondering why it behaves
different than expected.
If I generated multiple
outliers marked:
library(lattice)
my.mat - matrix(rnorm(800), nrow = 40)
threshold - c(-1,1) # should be used for the extreme colors
colorFun - colorRampPalette(c(yellow,red))
levelplot(my.mat, col.regions = colorFun(50))
I don't know how to handle the extrem values...
Can anybody help?
Antje
] + step, max(my.mat) + step, max(my.mat))
up - seq(thr[2] + step, limit + step, step)
my.at - c(lp,mp,up)
my.col.regions - c(rep(green, length(lp)), colorFun(length(mp)), rep(blue,
length(up)) )
levelplot(my.mat, at = my.at, col.regions = my.col.regions)
Antje schrieb:
Hi there,
I'd like
, can I realize it? (The list, I'm working with has not just 3 data frames
like given in my example, so I need to automize it)
Antje
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[regexpr(data, names(ALL)) 0]
names(ALL) - sub([.].*, , names(ALL))
ALL
df1 df2 df3
1 2 6 9
2 6 2 3
3 3 9 6
4 1 7 2
5 9 5 1
On Wed, Jan 21, 2009 at 3:19 AM, Antje niederlein-rs...@yahoo.de wrote:
Hi there,
I have a list of dataframes (generated by reading multiple
it for now!
Thank you!
On Wed, Jan 21, 2009 at 6:19 AM, Antje niederlein-rs...@yahoo.de
mailto:niederlein-rs...@yahoo.de wrote:
Hi there,
I have a list of dataframes (generated by reading multiple files)
and all dataframes are comparable in dimension and column names.
They also
you solve this problem in an elegant way?
Antje
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Wow, there are a lot of possibilities... thank you all very much!!!
I guess, I'll go for as.data.frame.table, because it's one line and does
exactly what I want :-)
Ciao,
Antje
Antje schrieb:
Hello,
I have a question how to reshape a given matrix to a data frame
Thanks a lot for every answer I got!
I could solve my problem!
Greg, your proposal seems to be quite useful for me :-) Thank you.
Ciao,
Antje
Antje schrieb:
Hi all,
I'd like to know, if I can solve this with a shorter command:
a - rnorm(100)
which(a -0.5 a 0.5)
# would give me all
)
# it's simply more human readable if I can write
which( isIn( c(-0.5, 0.5), a) )
# instead of
which( a -0.5 a 0.5 )
Thanks to baptiste! So there is no method available doing this and I have to
define this by myself. That's all I wanted to know :-)
Antje
markle...@verizon.net schrieb
to have something like:
which(within.interval(a, -0.5, 0.5))
Is there anything I could use for this purpose?
Antje
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Hi David,
thanks a lot for your proposal. I got a lot of useful hints from all of you :-)
David Winsemius schrieb:
It's not entirely clear what you are asking for, since
which(within.interval(a, -0.5, 0.5)) is actually longer than which(a
-0.5 a 0.5).
Right but in case 'a' is
Thank you very much for your help! That code does exactly what I was looking
for ( I don't have experience with lattice yet )
Ciao,
Antje
[EMAIL PROTECTED] schrieb:
layout(matrix(c(1,2,3,4), nrow = 2, byrow = TRUE))
plot(rnorm(100))
plot(rnorm(200))
plot(rnorm(300))
plot(rnorm(400))
Now
and generate a common title.
How can I do that?
Antje
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me a,b,c, though the subset does not contain entries with
c anymore
I guess, the solution is rather simple, but I cannot find it.
Antje
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Thanks a lot!!!
the drop thing was exactly what I was looking for (I already used it some
time ago but forgot about it).
Thanks to everybody else too.
Antje
Prof Brian Ripley schrieb:
On Fri, 5 Dec 2008, jim holtman wrote:
try this:
df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8),
c
Hello,
Can anybody help me to find out which colors are used automatically when
calling barplot (e.g. 3 series beside each other will get different gray values).
I want to apply a legend but I don't know the colors used...
Antje
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Hi Benjamin,
thanks for the hint.
In barplot.default I found grey.colors(nrow(height)) :-)
(I had the same problem with the help file for grey.colors)
Ciao,
Antje
Nutter, Benjamin schrieb:
I believe the defaults in barplot are found using
grey.colors{GrDevices}.
?grey.colors
(For some
-Messages in German are not easy to get a solution for...)
Antje
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Hello Annette,
that explains everything... Thank you! I just thought having an English version
of Windows would be enough but indeed my Regional and Language-Setting are set
to German...
Ciao,
Antje
Annette Heisswolf schrieb:
Hei,
in the FAQ on R for Windows it says:
3.4 I selected
Hi,
thank you both for your response.
I don't want to do anything like this - I just got some code like this from
someone else and was wondering about the result.
I would have used another approach to create a boxplot like this...
Ciao,
Antje
[EMAIL PROTECTED] schrieb:
hi: i'm not well
(l), x)
plot(df)
Thank you!
Antje
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, f2, F234562, 12345678, 234567, 912345, 333)
I'd like to get only 234567 and 912345.
Can anybody help me creating a regex for this???
For example this regex:
regexpr([^:digit:$]{6}, folders)
would match F234562, 12345678, 234567, 912345
Antje
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Hi Gabor,
it works! Thank you very much! But I still don't understand the difference
between [0-9] and [:digit:]...
Ciao,
Antje
Gabor Grothendieck schrieb:
Try this:
folders - c(folder1, f2, F234562, 12345678, 234567, 912345, 333)
grep(^[0-9]{6}$, folders, value = TRUE)
[1] 234567
Thanks a lot again to all of you!!!
Antje
Antje schrieb:
Hi Gabor,
it works! Thank you very much! But I still don't understand the
difference between [0-9] and [:digit:]...
Ciao,
Antje
Gabor Grothendieck schrieb:
Try this:
folders - c(folder1, f2, F234562, 12345678, 234567,
912345
are
no real duplicates of each other.
thanks in advance for your help!
Antje
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(h = gridlines, col=lightgray, lty=2)
abline(h = 1, col=red, lwd=3)
par(new=TRUE)
boxplot(mydata[i], pars= bpars, main = title)
dev.off()
})
Ciao,
Antje
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)
})
ciao,
Antje
Dieter Menne schrieb:
Antje niederlein-rstat at yahoo.de writes:
I want to create some boxplots (as png) within an lapply method. To get
nice gridlines behind the boxplot, I plotted it twice and therefore I
set par(new=TRUE).
This works nicely for the first plot but the second
), horizontal = TRUE)
Antje
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and provide commented, minimal, self-contained, reproducible code.
a strange plot:
x - rnorm(100) + 100
maxval - max(x)
boxplot(x, notch=TRUE, xlim = c(0,maxval), horizontal = TRUE)
Any idea?
Antje
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged
of an array???
Antje
Alain Guillet schrieb:
Hi,
If all your matrices have the same size, you should work with an array
and not with a list. Then you can use dimnames to set the names of the
rows, columns, and so on..
Alain
Antje wrote:
Hello,
I have another stupid question. I hope you can
elements to a single string.
# Example code (how it should look like):
t1 - c(1,2,3)
t2 - c(3.4,5.5,1.1)
paste(t1,t2, sep=\t)
# and now how the data is available
tl - list(t1,t2)
??? what do I have to do to get the same output ???
Can anybody help me?
Antje
Great! That's exactly what I was looking for.
(I see, I still have to learn a lot...)
Thank you!
Antje
Dimitris Rizopoulos schrieb:
try this:
t1 - c(1, 2, 3)
t2 - c(3.4, 5.5, 1.1)
tl - list(t1, t2)
do.call(paste, c(tl, sep = \t))
I hope it helps.
Best,
Dimitris
Antje wrote:
Hello
:50, nrow=5)
# ... there can be much more than two matrices
l - list()
l[[1]] - list(m1,m2)
r_names - LETTERS[1:5]
c_names - LETTERS[6:10]
? how can I apply these names to any number of matrices within this list-list ?
Ciao,
Antje
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R-help@r
= function(ex) { value})
it works pretty nice. but within the lapply construct I always end up with
strings.
Does anybody have an idea why? (I can try to create a simple testcode for this
but maybe there is already someone who knows what's wrong?
Ciao,
Antje
Hi there,
does anybody know how to return the xmlPath from a node?
For example, at several location in the xml file, I have nodes with the same
name and I'd like to process only the nodes from a certain path.
Any idea?
Antje
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R-help@r
but does not change the root?
So looking for a subnode at all levels below my current node is not possible
with the xPath syntax? (search on all levels starting from root is possible
with //nodename)
Antje
Duncan Temple Lang schrieb:
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Antje
Duncan Temple Lang schrieb:
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Hash: SHA1
Antje wrote:
Hi Duncan,
thanks a lot for your explanations.
I tried the following now to understand a bit more:
data - getNodeSet(doc, //Data)
xmlName(data[[1]])
xmlName(xmlRoot(data[[1]]))
xpathApply(data[[1
/val
val i=t2 11 /val
/data
/root
Now, I'd like to do something like this in R. Most important would be to
retrieve a node just by its name, not by the whole path. How is it possible?
Can anybody help me with this issue?
Antje
__
R-help@r
the possibilities to
access data as I was used to.
Antje schrieb:
Hi there,
I try to rewrite some Java-code with R. It deals with reading XML files.
I started with the XML package. In Java, I had a very useful method
which gave me a node by using:
name of the node
index of appearance
start point
Hello everybody,
I don't understand the warning I get when executing the following code:
x - rnorm(100)
h - hist(x, plot= FALSE, freq = FALSE)
If I don't use the plot-argument, I don't get a warning...
Any explanations for me?
Antje
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R-help@r
. Anyway, I had a strange problem but currently I
cannot reproduce, which led me somehow to this warning (I had problems to get a
customized y-range).
Thanks a lot!
Antje
Regards,
Richie.
Mathematical Sciences Unit
HSL
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