:47
To: Thaler,Thorn,LAUSANNE,Applied Mathematics; R-Help Mailing List
(r-help@r-project.org)
Subject: Re: [R] Automatically Remove Aliased Terms from a Model
Hi Thorn,
it is not entirely clear (at least for me) what you want to accomplish.
an easy and fail safe way of extracting used terms in a (g
Dear all,
I am trying to implement a function which removes aliased terms from a model.
The challenge I am facing is that with alias I get the aliased coefficients
of the model, which I have to translate into the terms from the model formula.
What I have tried so far:
Dear all,
I was wondering whether (or better: how) I can use R to read recursively a
directory to get all the sub-folders and files located in the root folder and
put it into a tree like structure where the leaves are files and intermediate
nodes are the directories? The idea is that I'd like
Dear all,
Is there a possibility to remove a geom from a ggplot? Background suppose I
have a function which returns a ggplot object after some data re-formatting and
aggregation. While this ggplot object is fine in 90% of the cases it turns out
that for some cases I want to suppress one of the
Dear all,
I want to use gsub to change a vector of strings. Basically, I want to replace
any dot by a space, remove the possibly appended .f and I want to capitalize
each word. I did that by chaining multiple gsubs together, but I was wondering
(for the sake of learning - maybe the current
Dear all,
Assume that I have the following data structure:
d - expand.grid(subj=1:5, time=1:3, treatment=LETTERS[1:3])
d$value - 10 ^ (as.numeric(d$treatment) + 1) + 10 * d$subj + d$time
d$value2 - 10 + d$value
where d$treatment == C stands for my control group. What I want to achieve
now
Dear all,
I want to use aggregate.formula to conveniently summarize a data.frame. I have
quiet some variables in the data.frame and thus I don't want to write all these
names by hand, but instead create them on the fly. This approach has the
advantage that if there will be even more columns in
is the root
cause?
Thanks for your help!
KR,
-Thorn
-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: Mittwoch, 31. Oktober 2012 13:15
To: Thaler,Thorn,LAUSANNE,Applied Mathematics
Cc: R help
Subject: Re: [R] aggregate.formula: formula from string
Hi,
Try this:
res
Dear all,
I produced the following graph with ggplot which is almost fine, yet I don't
like that the legend for Means and Observations includes a line, though no
line is used in the plot for those two (the line for Overall Mean on the
other hand is wanted):
library(ggplot2)
ddf - data.frame(x
Dear all,
I want to get a series of boxplots (grouped by two factors) and I want to
overlay the original observations and the following code does almost what I
want:
library(ggplot)
ddf - data.frame(x=factor(rep(LETTERS[1:4], each=30)), y = runif(120,0,10),
grp = factor(rep(rep(1:3, 10), 4)))
[mailto:ruipbarra...@sapo.pt]
Sent: Montag, 2. Juli 2012 12:20
To: Thaler,Thorn,LAUSANNE,Applied Mathematics
Cc: r-help@r-project.org
Subject: Re: [R] ggplot: dodge positions
Hello,
Though I'm not the most fluent user of ggplot, I've seen no problem
with
the graph, each subgroup of points
on the x-axis for geom_boxplot determined? Any
ideas?
Thanks for the help, anyways.
KR,
-Thorn
-Original Message-
From: John Kane [mailto:jrkrid...@inbox.com]
Sent: Montag, 2. Juli 2012 15:04
To: Thaler,Thorn,LAUSANNE,Applied Mathematics; r-help@r-project.org
Subject: RE: [R] ggplot
?
Any ideas?
Thanks for the help, anyways.
KR,
-Thorn
-Original Message-
From: John Kane [mailto:jrkrid...@inbox.com]
Sent: Montag, 2. Juli 2012 15:04
To: Thaler,Thorn,LAUSANNE,Applied Mathematics; r-help@r-project.org
Subject: RE: [R] ggplot: dodge positions
Can
...@inbox.com]
Sent: Montag, 2. Juli 2012 15:58
To: Thaler,Thorn,LAUSANNE,Applied Mathematics
Subject: RE: [R] ggplot: dodge positions
Let's hope it makes it. Just in case my version is okay let me give
you my sessionInfo in case we have some subtle difference in settings
that is having an effect
Dear all,
Sometimes I have the situation where a function takes a data.frame and
an additional argument describing come columns. For greater flexibility
I want to allow for either column names or column indices. What I
usually do then is something like the following:
-8-
Does
xyplot(y ~ seq_along(y), xlab = Index)
do what you want?
Not exactly, because it does not work once multipanel conditioning comes
into play:
xyplot(y~seq_along(y)|factor(rep(1:2, each=5)), xlab = Index)
The points in the right panel are plotted from 6:10 while the points in
the
Dear Peter,
Thanks for your concise answer, it works perfectly.
By the way, I fully agree that data or df are not good names for
data.frames and I am/was aware of that and I usually avoid those names
(not consequently though I've to admit, it is too tempting ;). However,
if one uses those evil
Dear all,
How can I make an index plot with lattice, that is plotting a vector
simply against its particular index in the vector, i.e. something
similar to
y - rnorm(10)
plot(y)
I don't want to specify the x's manually, as this could become
cumbersome when having multiple panels.
I tried
Dear all,
Quite often I have the situation that I've multiple response variables
and I create Linear Models for them in a function. The following code
illustrates my usual approach:
---8---
set.seed(123)
dat - data.frame(x = rep(rep(1:3, each = 3), 4), y = rep(1:3, 12))
Hi all,
I find myself sometimes in the situation where I lapply over a list and
in the particular function I'd like to use the name and or position of
the respective list item. What I usually do is to use mapply on the list
and the names of the list / a position list:
o - list(A=1:3, B=1:2, C=1)
Hi everybody,
I find myself quite often in the situation that I want to copy data from
Excel to R on the fly. If the source consists only of a single column, I
usually do something like
copy column in Excel
x - as.numeric(readClipboard())
If I have a matrix, I usually export this
Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: mardi 8 mars 2011 10:43
To: Thaler,Thorn,LAUSANNE,Applied Mathematics
Cc: r-help@r-project.org
Subject: Re: [R] Read data.frame from clipboard
You haven't told us your OS. But assuming Windows, why not use
read.delim(clipboard
Hi all,
Suppose we have the following matrix
m - matrix(c(1,2,3,2,1,3,3,1,2), ncol = 3, byrow=T)
where in each row each number occurs only once.
I'd like to define a permutation, e.g. 1 - 2, 2 - 1, 3 - 3 and apply
it to the matrix. Thus, the following matrix should result:
m.perm -
(values)
}
tmp - perm[as.numeric(as.factor(x))]
dim(tmp) - dim(x)
list(x = x, perm.x = tmp, perm = perm)
}
Thanks.
-Original Message-
From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
Sent: mardi 8 mars 2011 16:21
To: Thaler,Thorn,LAUSANNE,Applied Mathematics
Cc: r
Dear all,
In a function I paste a string and convert it to a formula which I pass
to lm[e]. The idea is to write a function which takes the name of the
response variable and the explanatory variable and the data frame as an
argument and calculates an lm[e]. (see example below)
This works fine,
Hi everybody,
I want an x-axis which has xlim=c(max, min) rather than xlim=c(min, max)
in order to reflect the type of the process (cooling):
library(lattice)
myprepanel - function(x,y,...) list(xlim=rev(range(x)))
x - rep(1:10, 100)
z - factor(sample(10, 1000, T))
y - rnorm(1000, x,
No (in fact that wouldn't work anyway), you can simply do
xyplot(y~x|z, xlim = rev(extendrange(x)))
The point is that in this case you have to explicitly specify a
pre-computed limit, and cannot rely on the prepanel function to give
you a nice default.
Thanks that does the trick. Because
Well, it's actually lattice:::extend.limits(range(x)), but
extendrange() does basically the same thing, and is available to the
user (i.e., exported), albeit from a different package.
Thanks again Deepayan for the insight.
A followup question though-- in another setting I'd like to have
Hi all,
In R it is possible to sum tables:
(a - table(rep(1:3, sample(10,3
1 2 3
2 5 7
a+a
1 2 3
4 10 14
Now suppose that I have a list of tables, where each table counts the
same things
k - list(a,a,a)
How can I sum all tables in k?
do.call(sum, k)
[1] 42
does not work
Perfectly, works as expected. Regarding the other questions, can anybody point
me to the right direction?
BR Thorn
From: RICHARD M. HEIBERGER [mailto:r...@temple.edu]
Sent: lundi 23 août 2010 18:36
To: Thaler,Thorn,LAUSANNE,Applied Mathematics
Cc: r-help@r-project.org
Subject: Re: [R
Hi all,
Suppose that I've two data frames, a and b say, both containing a column
'id'. While data frame 'a' contains multiple rows sharing the same id,
data frame 'b' contains just one entry per id (i.e. a 1 to n
relationship). For the ease of modeling I now want to generate a new
data frame c,
Thanks, that does the trick. Again a new command learned. Thanks.
However, any hints regarding the rownames issue?
BR Thorn
-Original Message-
From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
Sent: lundi 9 août 2010 11:07
To: Thaler,Thorn,LAUSANNE,Applied Mathematics
Cc
Hi everybody,
Suppose we have the following data structure:
ddf -data.frame(a=rep(1:4,3), b=rep(paste(p, 1:3, sep=), each=4),
c=c(1,0,0,1,1,1,0,1,1,1,1,1))
I want now to make a contingency table for each pair of values of p,
i.e. a contingency table for each of the pairs (p1,p2), (p1,p3) and
Works as expected, THX a lot.
BR thorn
-Original Message-
From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
Sent: mardi 20 juillet 2010 11:41
To: Thaler,Thorn,LAUSANNE,Applied Mathematics
Cc: r-help@r-project.org
Subject: Re: [R] apply: return list of matrices
one
Dear all,
I have a very rudimental function which takes a vector of terms and returns a
list of all possible models which can be made using the given terms. For
example for the set c(1, x, y, x:y) I'd get:
~ 1
~ x
~ y
~ x:y
~ 1 + x
~ 1 + y
~ 1 + x:y
~ x + y
~ x + x:y
~ y + x:y
~ 1 + x + y
~ 1
Try this variation of my.transform that I had posted here:
http://tolstoy.newcastle.edu.au/R/e2/help/07/09/24707.html
List - function(..., L = list()) {
f - function(){}
formals(f) - eval(substitute(as.pairlist(c(alist(...), L
body(f) - substitute(modifyList(L,
Hi everybody,
I'd like to have a list with two elements, where both elements have the same
value:
z - list(a=1, b=1)
If it happens, that I've to change the value, I've to assure that I change
both. This is error prone. Hence, a better way to achieve this is to define:
tmp - 1
z - list(a=tmp,
2010 11:47
To: Thaler,Thorn,LAUSANNE,Applied Mathematics
Cc: r-help@r-project.org
Subject: Re: [R] list() assigning the same value to two items
On 2010-06-21 3:30, Thaler, Thorn, LAUSANNE, Applied Mathematics wrote:
Hi everybody,
I'd like to have a list with two elements, where both elements have
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