of the same variable and break
when the value changes.
On Tue, Aug 5, 2014 at 6:13 AM, William Simpson
william.a.simp...@gmail.com wrote:
This works, but it is not quite what I need:
par(mar=rep(0,4))
while(1)
{
img1-matrix(runif(2500),50,50)
dev.hold(); image(img1,useRaster=TRUE
This works, but it is not quite what I need:
par(mar=rep(0,4))
while(1)
{
img1-matrix(runif(2500),50,50)
dev.hold(); image(img1,useRaster=TRUE); dev.flush()
img2-matrix(runif(2500),50,50)
dev.hold(); image(img2,useRaster=TRUE); dev.flush()
}
I would like to do this:
while(!kbhit())
Thanks Duncan for your help.
Bill
On 1/1/12, Duncan Murdoch murdoch.dun...@gmail.com wrote:
On 12-01-01 9:05 AM, William Simpson wrote:
When using bmp() under Windows XP, I find that the saved image is a
shifted version of the correct image. Try this:
The image() function isn't designed
Today I wanted to get R help to start up in text mode under Windows
XP. I had my own notes on how to do that. The method described there
didn't work. I checked on R-help. No helpful messages on the subject.
Therefore I post here in the hope that it helps someone else (or
myself, when I forget and
...@gmail.com wrote:
On 12-01-01 9:05 AM, William Simpson wrote:
When using bmp() under Windows XP, I find that the saved image is a
shifted version of the correct image. Try this:
The image() function isn't designed to be able to do pixel-level
addressing, so it's not too surprising that some rounding
I have figured out what I wanted to do using pixmap. Pixmap writes
.pgm files which I will batch convert to .bmp using Irfanview.
Thanks for your help.
Cheers,
Bill
On 1/2/12, William Simpson william.a.simp...@gmail.com wrote:
Duncan, I checked out as.raster as you suggested. However, I can't
When using bmp() under Windows XP, I find that the saved image is a
shifted version of the correct image. Try this:
n-5
fn-01.bmp
x-matrix(runif(n*n),nrow=n)
image(x,col=gray(0:255/255),axes=F,frame.plot=F)
bmp(filename = fn,width = n, height = n, units = px)
par(mar=c(0,0,0,0),pty=s)
Suppose I have x variables x1, x2, x3 (however in general I don't know
how many x variables there are). I can do
X-cbind(x1,x2,x3)
lm(y ~ X)
This fits the no-interaction model with b0, b1, b2, b3.
How can I get lm() to fit the model that includes interactions when I
pass X to lm()? For my
Thanks for the replies.
I was just thinking that, for a two variable example, doing
X-cbind(x1,x2,x1*x2)
lm(y~X)
would work. So maybe that's what I'll do. This also allows me to pick
and choose which interactions to include.
Cheers
Bill
On Sun, Dec 5, 2010 at 8:19 PM, William Simpson
I was looking at different link functions for binomial glms recently for
the same reason as you (more zeros than ones). I did a bit of reading up
on the various link functions and IIRC, you can use the cloglog link on
your data, just turn your 0's into 1's and vice versa. This was stated
in the
I don't see why one would want to pretend that the function is
continuous. It isn't.
The x variable devices is discrete.
Moreover, the whole solution space is small: the possible solutions
are integers in the range of maybe 20-30.
Bill
On Fri, Jun 18, 2010 at 9:00 AM, José E. Lozano
Suppose I do a trigonometric regression
fit-lm(y~ cf + sf)
where cf and sf are the cos and sine components.
b-coef(fit)
I have the fitted sine component b[2] and the cos component b[3].
Doing summary(fit) gives me the p-values and SEs for b[2] and b[3].
But I want the amplitude of the fitted
On Thu, Jun 17, 2010 at 11:18 AM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
William Simpson wrote:
Suppose I do a trigonometric regression
fit-lm(y~ cf + sf)
where cf and sf are the cos and sine components.
b-coef(fit)
I have the fitted sine component b[2] and the cos component b[3
var(y)*length(y) I mean. (SSE)
Bill
On Thu, Jun 17, 2010 at 1:34 PM, William Simpson
william.a.simp...@gmail.com wrote:
Yes, I want the same test as is done for b[1] and b[2] in the summary
table -- for H0: b[]==0.
OK, do F-test on full model with cf and sf vs reduced model with intercept
Got it now. I do
anova(lm(y~ 1),lm(y~ cf+sf))
Bill
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Bill
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Thanks, Bill!
Bill
On Thu, Jun 17, 2010 at 5:12 PM, William Dunlap wdun...@tibco.com wrote:
You can also define a function that keeps the cos
and sin terms together so anova(fit) shows
one entry for the (cos,sin) pair. E.g., define
the following function
cs - function(x,
Sorry, thought you wanted to find lowest value of x that produced
maximum value of y. I see now that is not the case.
I think you have to decide on what amount of improvement per device
you judge to be 'minimal'. Then the algorithm uses the value of y that
occurs at the point where this criterion
I have the following set-up.
6 values of a continuous variable (let's say light intensity) are
presented to a system.
The input is presented as a random series of blocks lasting (say) 5 sec each.
etc
time -
The output is
[posted this at 9:25 and still hasn't appeared on the list at 13:26]
I have the following set-up.
6 values of a continuous variable (let's say light intensity) are
presented to a system.
The input is presented as a random series of blocks lasting (say) 5 sec each.
On Fri, May 14, 2010 at 1:47 PM, Thomas Levine thomas.lev...@gmail.com
wrote:
Creating the 5 indicator variables will be easy if you post your code
and sample data. This may also allow people to help with the first
problem you were having.
Here you go.
Fragment of data file with 4000 or so
I don't know why my posts aren't showing up on my email acct. I will send
again.
-- Forwarded message --
From: William Simpson william.a.simp...@gmail.com
Date: Fri, May 14, 2010 at 3:00 PM
Subject: Re: [R] Fwd: nonlinearity and interaction
To:
Cc: r-help@r-project.org
On Fri
Thanks for telling me. Sorry.
Don't know why my posts are not showing up here at gmail. Never happened
before. Thought they had been lost in the aether.
Bill
On Fri, May 14, 2010 at 4:43 PM, David Winsemius dwinsem...@comcast.netwrote:
On May 14, 2010, at 10:42 AM, William Simpson wrote:
I
On Fri, May 14, 2010 at 5:54 PM, Bert Gunter gunter.ber...@gene.com wrote:
1. As this is not an R question, this is probably not an appropriate list
for posting. You might wish to consider a list specifically devoted to
statistics and data analysis.
Well I think it is an R question to the
I would like a scatterplot matrix and a correlation matrix for the
following set-up.
The data (dataframe d) are like this:
angle resp
-90 182
-60 137
-30 ...etc
0
30
60
90
...etc
I would like each cell in the matrix to be the scatterplot of the
responses for each pair of angles (
On Sat, Jan 2, 2010 at 4:55 PM, Charles C. Berry cbe...@tajo.ucsd.edu wrote:
On Sat, 2 Jan 2010, William Simpson wrote:
I would like a scatterplot matrix and a correlation matrix for the
following set-up.
The data (dataframe d) are like this:
angle resp
-90 182
-60 137
-30
Now that we are able to help with some more detailed view of your data
(although you could have helped helping by making it easier for us to import
your data into R), the answer is:
I am preparing for the data analysis, writing the code (knowing I may
have to modify it later) while the data are
OK thanks David
Thanks very much, Uwe. I will try this (on artificial data). I think
reshape() requires a library [reshape?].
No. In fact, the reshape package does not have a reshape function.
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
I am puzzled by a filtering problem using fft(). I don't blame R.
I have a waveform y consisting of the sum of 2 sinewaves having freqs f1 and f2.
I do s = fft() of y.
Remove s's spike at freq=f2
Do inverse fft on s.
The resulting waveform still has a lot of f2 in it! But the filtering
should
28, 2009 at 9:53 AM, William Simpson
william.a.simp...@gmail.com wrote:
I am puzzled by a filtering problem using fft(). I don't blame R.
I have a waveform y consisting of the sum of 2 sinewaves having freqs f1 and
f2.
I do s = fft() of y.
Remove s's spike at freq=f2
Do inverse fft on s
Yep, that was it. Needed to remove neg freq from opposite end of s[]
coef- 1-(((1:n)8-1) ((1:n)8+1) | (((1:n)(n-6)-1) ((1:n)(n-6)+1)) )
Sorry to bother you.
Cheers
Bill
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I have quite a complicated problem that's hard to describe.
Suppose I have a dataframe d. I want to access the vector d$var, where
var is one of the variables in d. Just take for granted that there's a
good reason I want to do this as follows.
var-DeOxyA
xx-paste(d$,var, sep=)
mean(xx)
[1] NA
,
Try this,
d - data.frame(a=1:4, b=3:6)
var - a
mean(d[var])
## or, if you are not aware of
## fortune(parse)
xx - paste(d$,var, sep=)
mean(eval(parse(text=xx)))
HTH,
baptiste
2009/11/19 William Simpson william.a.simp...@gmail.com:
I have quite a complicated problem that's hard
I am running an expt that presents a point process input x and
measures a point process output y. The times of each event are
recorded. The lengths of the data records of x and y are necessarily
different, and can be different by a factor of 10. I would like to
save these data after each
, William Simpson wrote:
I am running an expt that presents a point process input x and
measures a point process output y. The times of each event are
recorded. The lengths of the data records of x and y are necessarily
different, and can be different by a factor of 10. I would like to
save
As I understand it, they don't come in pairs anyway.
Correct.
For the same reason
a data frame is just the wrong kind of data structure. If you don't want
separate data files, you can use one file with two columns where the
second column is (say) 1 for the x and 2 for the y.
Could you
Thanks Jim. BTW the times in x and y are in ascending order (time of
occurrence).
If I do it this way, how do I actually read the data in and store in
the file? Toy code, please.
Bill
Hi Bill,
xy-list(x=1:10,y=1:100)
Note that this cheerfully ignores how you are going to figure out which
OK thanks, I look at sleep and get it
Bill
On Fri, Oct 23, 2009 at 12:21 PM, Peter Dalgaard
p.dalga...@biostat.ku.dk wrote:
William Simpson wrote:
As I understand it, they don't come in pairs anyway.
Correct.
For the same reason
a data frame is just the wrong kind of data structure. If you
I have three time series, x, y, and z, and I want to analyse the
relations between them. However, they have vastly different
resolutions. I am writing to ask for advice on how to handle this
situation in R.
x is a stimulus, and y and z are responses.
x is a rectangular pulse 4 sec long. Its
PS I think one way to get the average waveforms I want from the
analysis is using cross-correlation, but again the multiple scale
problem is present.
On Wed, Oct 21, 2009 at 9:56 AM, William Simpson
william.a.simp...@gmail.com wrote:
I have three time series, x, y, and z, and I want to analyse
I wasn't clear: x and z are pulse *trains* with irregular gaps between pulses.
x is a rectangular pulse 4 sec long. Its onset and offset are known
with sub-millisecond precision. The onset varies irregularly -- it
doesn't fall on neat 1/2 sec or sec boundaries for example.
y is a sampled
2009, William Simpson wrote:
Suppose I have two sets of (x,y) points like this:
x1-runif(n=10)
y1-runif(n=10)
A-cbind(x1,y1)
x2-runif(n=10)
y2-runif(n=10)
B-cbind(x2,y2)
I would like to measure how similar the two sets of points are.
Something like a correlation coefficient, where 0
Suppose I have two sets of (x,y) points like this:
x1-runif(n=10)
y1-runif(n=10)
A-cbind(x1,y1)
x2-runif(n=10)
y2-runif(n=10)
B-cbind(x2,y2)
I would like to measure how similar the two sets of points are.
Something like a correlation coefficient, where 0 means the two
patterns are unrelated,
Thanks everyone for your help!
Bill
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I want a function that takes an input vector, the number of columns
and returns a matrix as follows.
x- 1:5
foo(x, nc=3)
1 5 4
2 1 5
3 2 1
4 3 2
5 4 3
Thanks again for any help.
Bill
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Thanks very much Gabor
Bill
On Sun, Jul 5, 2009 at 4:44 PM, Gabor
Grothendieckggrothendi...@gmail.com wrote:
In terms of x this is:
k - 3
x - 1:5
n - length(x)
matrix(x, 2*n-1, k)[1:n, ]
On Sun, Jul 5, 2009 at 10:01 AM, Gabor
Grothendieckggrothendi...@gmail.com wrote:
Try this (ignore
Can anybody please tell me a good way to do the following?
Given a vector, number of rows and number of columns, return a matrix
as follows. Easiest to give an example:
x=c(1,2), nrow=5, ncol=4
return the matrix:
1 0 0 0
2 1 0 0
0 2 1 0
0 0 2 1
0 0 0 2
Thanks very much for any
Thanks everyone for the help.
I should have said that I want to do this generally, not as a one-off.
So I want a function to do it. Like this
tp-function(x, nr, nc)
{
matrix( c(x,rep(0, nr-length(x)+1)), nrow=nr, ncol=nc)
}
tp(x=c(1,2), nr=5, nc=4)
This one looks good -- the warning message
, William Simpson
william.a.simp...@gmail.com wrote:
Thanks everyone for the help.
I should have said that I want to do this generally, not as a one-off.
So I want a function to do it. Like this
tp-function(x, nr, nc)
{
matrix( c(x,rep(0, nr-length(x)+1)), nrow=nr, ncol=nc)
}
tp(x=c(1,2
The following creates a point process version of a sinewave (maybe
there's a better way):
p-amp*cos(2*pi*freq*(1:n)/n ) + 0.5
as.numeric(runif(1:n)p)
I get something like this:
1ooo ooo oo o
0 ooo o o o o o ooo ooo
In case it's not
If this helps, here is what I have currently
mean.p- .5
contrast-1
amp- contrast*mean.p
freq- 3
n- 100
p-amp*cos(2*pi*freq*(1:n)/n ) + mean.p
plot(p)
dens-as.numeric(runif(1:n)p)
plot(p, type=l)
points(dens, col=red)
I would like some dots having y-value=0.5 in the middle of the plot.
These will
If I am right, the original version was a sinusoidally-modulated
(nonstationary) Bernoulli process. I want 3 possible values, not 2,
which makes it a Bernoulli scheme
http://en.wikipedia.org/wiki/Bernoulli_scheme
Not sure how to the create a sinusoidally modulated Bernoulli scheme
with values 0,
Thanks for the helpful replies.
I will try updating R and see how it goes.
MASS, nlme, multcomp are all the most recent.
Bill
On Tue, Apr 21, 2009 at 3:12 PM, William Simpson
william.a.simp...@gmail.com wrote:
I am trying to do Tukey HSD comparisons on a repeated measures expt.
I found
I would like the R console to wrap lines at 80 cols.
It does not do so, even though I have used the Rgui Configuration
Editor to set the Console cols at 80 and the Pager cols at 80.
Please tell me how to set it up so I have word wrap.
Just to be clear: in older/other R versions, console input is
I have the following between-within anova:
aovn1 - aov(amplitude ~ stereo*site*stimulus +
Error(subject/(site*stimulus)), stereon1)
This works fine. BUT I need to do Tukey HSD multiple comparisons, and
the aov() approach won't work. So I am trying the method posted on
r-help:
lmen1 -
I am trying to do Tukey HSD comparisons on a repeated measures expt.
I found the following example on r-help and quoted approvingly elsewhere.
It is broken. Can anyone please tell me how to get it to work?
I am using R 2.4.1.
require(MASS) ## for oats data set
require(nlme) ## for lme()
I would like to do as follows
plot(a,b)
points(c,d,pch=19)
Now join with a line segment point a[1], b[1] to c[1], d[1]; a[2],
b[2] to c[2], d[2] ... a[n], b[n] to c[n], d[n]
All corresponding points from the two data sets are joined by line segments.
Thanks very much for any tips on how to do
Thanks very much Rolf, Dimitris, Greg!
Bill
On Thu, Feb 26, 2009 at 8:56 PM, Rolf Turner r.tur...@auckland.ac.nz wrote:
On 27/02/2009, at 9:46 AM, William Simpson wrote:
I would like to do as follows
plot(a,b)
points(c,d,pch=19)
Now join with a line segment point a[1], b[1] to c[1], d[1
I have data in a format like this:
namessexsex viewnum rating rt
ahl4f m f 56 -1082246
ahl4f m f 74 85 1444
ahl4f m f 52 151 1595
ahl4f m f 85 1 1447
I wish to write a new link function for a GLM. R's glm routine does
not supply the loglog link. I modified the make.link function adding
the code:
}, loglog = {
linkfun - function(mu) -log(-log(mu))
linkinv - function(eta) exp(-exp(-eta))
mu.eta - function(eta)
I would like to do an R glm() with
family = binomial(link=loglog)
Right now, the cloglog link exists, which is nice when the data have a
heavy tail to the left. I have the opposite case and the loglog link
is what I need. Can someone suggest how to add the loglog link onto
glm()? It would be
I have a dataframe like this (toy example):
x y z
a a 0
a b 1
a c 2
b a .9
b b 0
b c 1.3
c a 2.2
c b 1.1
c c 0
The observations are from a matrix like this:
c 2.2 1.1 0.0
b 0.9 0.0 1.3
a 0.0 1.0 2.0
ab
On re-reading ?cmdscale I see that I can also use cmdscale(d) on a
full matrix rather than just the lower triangle.
d: a distance structure such as that returned by 'dist' or a
full symmetric matrix containing the dissimilarities.
So how to get a full symmetric matrix from a dataframe
Dallazuanna [EMAIL PROTECTED] wrote:
Try this:
x$zMean - ave(x$z,
apply(x[,1:2], 1, function(x)paste(sort(x),
collapse = )),
FUN = mean)
xtabs(zMean ~ x + y, data = x)
On Wed, Dec 10, 2008 at 11:16 AM, William Simpson
[EMAIL PROTECTED] wrote
regression problem (note that b is unknown to you but is constant).
--- On Tue, 2/9/08, William Simpson [EMAIL PROTECTED] wrote:
From: William Simpson [EMAIL PROTECTED]
Subject: [R] intercept of 3D line? (Orthogonal regression)
To: r-help@r-project.org
Received: Tuesday, 2 September, 2008, 4:53
I posted before recently about fitting 3D data x, y, z where all have
error attached.
I want to predict z from x and y; something like
z = b0 + b1*x + b2*y
But multiple regression is not suitable because all of x, y, and z have errors.
I have plotted a 3D scatterplot of some data using rgl. I see
I want to fit something like:
z = b0 + b1*x + b2*y
Since x, y,and z all have measurement errors attached, the proper way
to do the fit is with principal components analysis, and to use the
first component (called loadings in princomp output).
My dumb question is: how do I convert the princomp
Thanks everyone for the help.
Greg, I posted the equation from Prof Brillinger's paper. To me it is
not clear how or why to subtract a vector from a matrix (since they
have diff dimensions). That's why I posted my question to the experts
here!
Anyway, I have some answers now.
Cheers
Bill
On
Thanks Phipp very much for your help. I had meant, given that I'd
computed the matrix f[x,y] and the vector e[x], how to take the
difference. What is confusing is how to subtract a vector from a
matrix. I don't want the recycling rule.
Cheers
Bill
On Tue, Apr 22, 2008 at 9:53 AM, Philipp Pagel
Hi everyone,
The following is mysterious to me. David Brillinger (famous
statistician at Berkeley) has an equation in a paper that is
essentially
g(x,y) = f(x,y) - e(x)- e(y)
These are continuous functions. I am not sure how to do this with the
discrete equivalents in R.
Please tell me how to do
Hi Everyone,
I am running into a problem with matrices. I use R version 2.4.1 and
an older version.
The problem is this:
m-matrix(ncol=3,nrow=4)
m[,1:3]-runif(n=4)
That does what I expect; it fills up the rows of the matrix with the
data vector
m
[,1] [,2] [,3]
[1,]
Hi everyone,
I would like to do the following.
Given matrix m and matrix n, I would like to compute mn[i,,j]= m[i,,j]
+ n[i,,j] if either of these elements is 0. (In other words, whichever
number is nonzero.)
Else I want mn[i,,j]=(m[i,,j] + n[i,,j])/2
I need a fast method.
Thanks very much for
Thanks Julien Rolf for your help. The double commas were just typos.
Bill
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Thanks very much Petr and Rold for your helpful replies.
Cheers
Bill
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Dear R experts,
I have been given data files in the following configuration and have
been puzzling about how to read them in. First I will give a snippet
of the beginning of file:
Data File: W
Para File: GABOR_0.gor v 10.6,
Date : 29/10/2007
Time : 13:33
3.00
5.000
Noise SD(deg):
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