[R] Links to vignettes in the base packages
Contributed packages have copies of their vignettes on CRAN (in their package page at cranmirror/web/packages/packagename). Since base packages no longer have a page here, I can't find a web link to them. I'm aware that I can find the vignette via browseVignettes() or vignette(vignettename, package = packagename). There are two use cases where a web link is more useful. If someone is asking for help and I want to direct them to a vignette, having a link to a vignette rather than getting them to open R and type a command increases the chance that they will actually bother to look at the thing. Secondly, if I want to read a vignette on a machine without R (ebook reader, phone, whatever), then a web link is more convenient. Are the base vignettes available online? If so, where are they? If not, is it feasible to make them available? -- Regards, Richie __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R Style Guide -- Was Post-hoc tests in MASS using glm.nb
LIST OF CONVENTIONS/STYLES FOR R:
[1] R coding standards in the R Internals manual
http://www.cran.r-project.org/doc/manuals/R-ints.html#R-coding-standards
[2] Bioconductor coding standards
http://wiki.fhcrc.org/bioc/Coding_Standards
[3] Google R style
[http://google-styleguide.googlecode.com/svn/trunk/google-r-style.html]
[4] R style guide by Hadley Wickham (based on [3])
http://had.co.nz/stat405/resources/r-style-guide.html
[5] R Coding Conventions (RCC) - a draft by Henrik Bengtsson
http://aroma-project.org/developers/RCC
[6] The Aroma R Coding Conventions (Aroma RCC) by Henrik Bengtsson
(based on [5])
http://aroma-project.org/developers/AromaRCC
It seems that I'm a little late to this conversation, but I'd like to add
my own guide to the list. It contains links to all the other mentioned
style guides and another one from Colin Gillespie.
http://4dpiecharts.com/r-code-style-guide/
Regards,
Richie.
Mathematical Sciences Unit
HSL
4D Pie Charts
ATTENTION:
This message contains privileged and confidential inform...{{dropped:22}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How is the relation between Frequency and Counts in hist/density defined?
In order to do this I can use the relation between count and density, but I would like to know if there is a way for me to predict it upfront. In the code for hist.default, you'll see the line dens - counts/(n * diff(breaks)) Here is an example: set.seed(242) z = rnorm(30) hist_z - hist(z) hist_z$counts / hist_z$density # the relation is 15 In your example, n is 30, and the breaks are evenly spaced every 0.5. Regards, Richie. Mathematical Sciences Unit [1]HSL [2]4D Pie Charts _ ATTENTION: This message contains privileged and confidential information intended for the addressee(s) only. If this message was sent to you in error, you must not disseminate, copy or take any action in reliance on it and we request that you notify the sender immediately by return email. Opinions expressed in this message and any attachments are not necessarily those held by the [3]Health and Safety Laboratory or any person connected with the organisation, save those by whom the opinions were expressed. Please note that any messages sent or received by the [4]Health and Safety Laboratory email system may be monitored and stored in an information retrieval system. _ Think before you print - do you really need to print this email? _ _ Scanned by MailMarshal - Marshal's comprehensive email content security solution. Download a free evaluation of MailMarshal at [5]www.marshal.com _ References 1. http://www.hsl.gov.uk/contact-us.htm 2. http://4dpiecharts.com/ 3. http://www.hsl.gov.uk/ 4. http://www.hsl.gov.uk/ 5. http://www.marshal.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] qbeta
Does any body know how I can see the code behind qbeta function?
As the code seems to be internal, you'll need to download the r-source
code and find it in there. In my copy of R it is here:
R-2.11.1/src/nmath/qbeta.c
An alternative is to view the source code online. The code for qbeta is
in
http://svn.r-project.org/R/trunk/src/nmath/qbeta.c
In general, you can find internal code by searching for
site:http://svn.r-project.org/R/trunk/src function name
Regards,
Richie.
Mathematical Sciences Unit
HSL
4D Pie Charts
ATTENTION:
This message contains privileged and confidential inform...{{dropped:22}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Implementing R's recycling rule
x - c(1, 2, 3)
n - 10
## so using the recycling rules, I would like to get from FUN(x, n)==1
## I am doing:
xRecycled - rep(x, length.out=n)[n]
This works, but it seems to me that I am missing something really basic
here
- is there more straightforward way of doing this?
x[n %% length(x)] gives you the same answer as rep(x, length.out=n)[n],
without having to create the longer vector.
Regards,
Richie.
Mathematical Sciences Unit
HSL
4D Pie Charts
ATTENTION:
This message contains privileged and confidential inform...{{dropped:22}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] table, sum, cat function
the second step in my exercice is to calculate the sum of the amout for each class et not the frequency i have this vector x y 1 100 2 1500 3 3250 4 6250 5 2000 6 450 i want to use the function table and cat to calculate the sum of the amount in each class [0-1000], [1000-3000],[ 3000] You want to use cut, not cat. Take a look at the examples on the cut help page - they will help you. Regards, Richie. Mathematical Sciences Unit [1]HSL _ ATTENTION: This message contains privileged and confidential information intended for the addressee(s) only. If this message was sent to you in error, you must not disseminate, copy or take any action in reliance on it and we request that you notify the sender immediately by return email. Opinions expressed in this message and any attachments are not necessarily those held by the [2]Health and Safety Laboratory or any person connected with the organisation, save those by whom the opinions were expressed. Please note that any messages sent or received by the [3]Health and Safety Laboratory email system may be monitored and stored in an information retrieval system. _ Think before you print - do you really need to print this email? _ _ Scanned by MailMarshal - Marshal's comprehensive email content security solution. Download a free evaluation of MailMarshal at [4]www.marshal.com _ References 1. http://www.hsl.gov.uk/contact-us.htm 2. http://www.hsl.gov.uk/ 3. http://www.hsl.gov.uk/ 4. http://www.marshal.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mixed normal distriburtion
I'm trying to draw the density function of a mixed normal distribution
in the form of:
.6*N(.4,.1)+ .4*N(.8,.1)
At first I generate a random sample with size 200 by the below code:
means = c(.4,.8)
sds = sqrt(c(.1,.1))
ind = sample(1:2, n, replace=TRUE, prob=c(.6,.4))
x=rnorm(n,mean=means[ind],sd=sds[ind])
Then I use the below code for drawing the graph:
plot(density(x))
The plot doesn't seem to be belonging to the desired distribution,
because there is just one mode in it (I've seen the real graph of this
mixed normal in a paper, it has two clear distinct modes). Even the
hist() doesn't draw a plot similar to the real graph. I think the
generation code isn't correct. Is it? (I've asked the generation code
here!)
The code is fine - the reason you can't see two peaks is that the two
distributions overlap a lot. Set
means - c(.4, 10)
to see double peaks more clearly.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:22}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] formats
what do you mean by %d-%b-%y. is it reading format or writing format.
%d-%b-%y is a date format - see the help page for strptime.
Example usage:
strptime(01-Jan-84, %d-%b-%y)
strftime(Sys.time(), %d-%b-%y)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:22}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] HELP: BRUGS/WinBUGS/RBUGS Response is a combination of random variables
Is there anyone know if BUGS language allows the combination of
variables as response
It seems doesn't work in my model. The problem is between two ##.
modelCompile(numChains=1)
multiple definitions of node bm[1]
###
bm[iter] - inprod(biomarker[iter,1:4], bta[1:4])
bm[iter] ~ dnorm(bmu[iter], sdbm.precision);
###
The problem seems to be that you've defined bm[iter] as a logical node
(-), then redefined it as a stochastic node (~). If you need more help,
try the BUGS mailing list. (See
http://www.mrc-bsu.cam.ac.uk/bugs/overview/list.shtml)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:22}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regression for loop test HELP! URGENT!
I'm new to R, and I've sent this message as a non-member, but since it's
pretty urgent, I'm sending it again now I'm on the mailing list (Thanks
Daniel for your suggestion nevertheless).
I have calculated a regression in the form of M ~ D + O + S, and I would
like to take this regression and test it with other samples, 5 sets of
M, D,
O, and S at a time(I actually have 2000 sets, so it's probably not
efficient
to make each a separate set and then index). Since I'll need to test the
regression for 400 groups, I thought a for loop might be necessary. I've
put
everything into a data frame already. Can anyone tell me how to write
the
code? I'm especially not sure about how to do the for loop.
And then how would I calculate the error of how well the test samples
fit
the original regression?
This is for my internship, so it's very urgent.
Take a deep breath, and think calm thoughts. Take a look at the posting
guide (http://www.r-project.org/posting-guide.html) - it has useful ideas
on thinking through your problem. If you can provide some code then we
can see what you want more clearly.
Show us how you've done your regression what form your data is in. Tell
us which tests you'd like to do on the samples.
If you are stuck with for loops, then take a look at section 9.2.2 in the
Intro to R guide that comes with R. (Click Help - Manuals - an
Introduction to R in RGui.)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:22}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Histograms on a log scale
I would like to be able to plot histograms/densities on a semi-log or
log-log scale.
# Get a random log-normal distribution
r - rlnorm(1000)
# Get the distribution without plotting it using tighter breaks
h - hist(r, plot=F, breaks=c(seq(0,max(r)+1, .1)))
# Plot the distribution using log scale on both axes, and use
# blue points
plot(h$counts, log=xy, pch=20, col=blue,
main=Log-normal distribution,
xlab=Value, ylab=Frequency)
This is very close to what I need, but how can I have filled rectangles
in the plot, so that it looks more like a traditional histogram?
You can use type=h to specify histogram-like plotting. You probably
also want to use a linear y-scale to make frequencies easier to compare.
Change the last line to:
plot(h$mids, h$counts, log=x, pch=20, col=blue,
main=Log-normal distribution,
xlab=Value, ylab=Frequency, type=h)
Alternatively, plot a histogram of the log data, and draw your own axes:
logr - log(r)
par(las=2)
h2 - hist(logr, axes=FALSE, breaks=seq(min(logr)-1, max(logr)+1, .5))
Axis(side=2)
Axis(side=1, at=h2$mids, labels=format(exp(h2$mids), digits=1), lty=NULL)
Or, even better, install the ggplot2 package and try something like:
qplot(r, geom=histogram) + scale_x_continuous(trans=log10)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:22}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] rle
I have an other problem, I have this vector signData with an alternation
of
1 and -1 that corrispond to the duration of two different percepts. I
extracted the durations like this:
signData- scan(dataTR10.txt)
dur-rle(signData)$length
I think that last line should be
dur-rle(signData)$lengths
Now I would like to extract only the positive duration, e.g.
signData - c(1,1,1,1,-1,-1,-1,1,1,-1,-1)
posduration - c(4,2)
If you know that the first element of signData will always be 1, then you
can simply extract the first, third, fifth etc values from signdata, like
so:
posduration - dur[c(TRUE, FALSE)]
Otherwise you need to test to see if you are extracting odd or even
elements.
if(signData[1]==1)
{
index - c(TRUE, FALSE)
} else
{
index - c(FALSE, TRUE)
}
posduration - dur[index]
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to apply a self-written function to a data frame
I have written a function in order to analyse gaze paths. It works
with the test data but when I try to apply the function to a data
frame that stores the real data in columns I receive the error
message that the
In if (pp 1) { :
condition has length 1 only the first element will be used
This means that pp is a vector (and hence pp 1 is also a vector), but
'if' requires a scalar logical value. Try, e.g.
if(1:10 5) message(foo)
for a simpler way of recreating this error message.
I interpret this error message as saying that only the first element
of pp is used. However, I'd like to analyse each row of the data
frame, row by row. Using
apply(final, 1, abst, gx,gy,tx,ty,p_pos) (with gx - p_pos being
columns of the data frame final)
By the looks of things, when you call apply(final, 1, abst,
gx,gy,tx,ty,p_pos), the value of pp (inside the abst function) is being
taken as the value of p_pos, which is all the relevant columns of the data
frame. Are you sure that this is the value you meant for pp?
There is a vectorised version of if, namely the ifelse function. Perhaps
the line if (pp 1) {px = -px } should be
px - ifelse(pp 1, -px, px)
or even more simply,
px[pp 1] - -px[pp 1]
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] variable driven csv file names?
Is it possible to have variable driven csv file names? Such as:
ds.name-bob.csv
write.table(
distribution.data,
file = ~//Documents/Research/Distribution Analysis/ds.name,
sep = ,,
col.names = FALSE,
qmethod = double)
Yes. You just need to construct your filename string using paste. Replace
the file argument with
paste(~//Documents/Research/Distribution Analysis/, ds.name, sep=)
Also take a look at ?file.path for platform independent file paths.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R help
I am interested in modeling hydrological extreme events. I found
MSClaio2008 very interesting function. In this function four
criterions for choosing distributions. Can we call these criterions
as model selection techniques or goodness of fit techniques or both?
Because goodness of fit techniques are usually performed after modle
selection.
What is MSClaio2008? I don't quite understand this bit. If you provide
code examples, then questions are often easier to answer.
Can I found chi-square, kolmogrov-sminov and cramer-von mises tests
for testing goodness of fit for proposed distributions?
Yes, you can do these things in R. The functions you want are:
chisq.test for the chi-square test
ks.test for the kolmogorov-smirnoff test.
You can view the help pages for these functions by typing a question mark
before the function name, e.g. ?chisq.test.
For the Cramer-von Mises tests, I didn't know, so I typed
RSiteSearch(cramer von mises test), from which there are several
suggestions. In particular, look at the CvM2SL1Test and CvM2SL2Test
packages.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] sample unique pairs from a matrix
I have a matrix of both negative and positive values that I would like
to randomly sample with the following 2 conditions:
1. only sample positive values
2. once a cell in the matrix has been sampled the row and column of
that cell cannot be sampled from again.
#some dummy data
set.seed(101)
dataf - matrix(rnorm(36,1,2), nrow=6)
I can do this quite simply if all the values are positive by using the
sample function without replacement on the column and row indices.
samrow - sample(6,replace=F)
samcol - sample(6,replace=F)
values - numeric(6)
for(i in 1:6){
values[i] - dataf[samrow[i], samcol[i]]
}
However, I am not sure how to include the logical condition to only
include postitive values
You could create a new variable containing only the positive values of
dataf, and sample from that, e.g.
dataf - matrix(rnorm(36,1,2), nrow=6)
posdata - dataf[dataf 0]
sample(posdata, 6, replace=FALSE)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot error
I want to plot data such that the 3 time points(a,b,c) lie on the X-axis
and
the values of these times points are on Y-axis for n samples (e.g.100).
So, I have an object x, dim 100 4, it is a dataframe (when checked the
class)
x =
name a b c
10.11 1.11 0.86
2 . . .
3 . . .
.
.
.
100
so when i say:
plot(1:3, x[,2:4], type=l) - I get the error below
Error in xy.coords(x, y, xlabel, ylabel, log) :
(list) object cannot be coerced to type 'double'
However if I do:
plot(1:3, x[1,2:4], type=l) -- It works for the 1st row, and
each
individual row BUT NOT ALL ROWS
Please could someone explain what is happening here?
I wonder if I need to use 'lines' for the remaining, BUT I have another
dataset y with same dimensions as x, which I want to plot on the same
graph/plot to see the difference between x and y.
Your data looks like this:
x - data.frame(name=sample(letters, 10), a=runif(10), b=rnorm(10),
c=rlnorm(10))
The problem is that the subset x[,2:4] is also a data frame, not a matrix.
class(x[,2:4]) #[1] data.frame
The simplest thing is probably to use lines, as you say.
row - seq_len(nrow(x))
xx - x[,2:4]
plot(row, xx$a, ylim=range(xx), type=l)
lines(row, xx$b, col=blue)
lines(row, xx$c, col=green)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] r-plot
I want to make a simple plot, here is my code:
http://gist.github.com/118550
Unfortunately, the annotation of both the x- and y-axis are not correct,
as
you can see in the following picture:
http://www.nabble.com/file/p23739356/plot.png
I am not an expert of R, so maybe someone can point me to the solution
of
this problem, i.e. both of the axes should start and end at the min /
max
values of the two vectors.
From the help page on par:
'xaxs' The style of axis interval calculation to be used for the
x-axis. Possible values are 'r', 'i', 'e', 's',
'd'. The styles are generally controlled by the range of
data or 'xlim', if given. Style 'r' (regular) first extends
the data range by 4 percent at each end and then finds an
axis with pretty labels that fits within the extended range.
Style 'i' (internal) just finds an axis with pretty labels
that fits within the original data range.
You've explicitly set xaxs=r, when you really want xaxs=i. You can
also explicitly set the axis limits using xlim/ylim parameters in the call
to plot.
Compare these examples:
#Ex 1
plot(1:10) #implicitly uses par(xaxs=r, yaxs=r) unless you've
changed something
#Ex 2
oldpar - par(xaxs=i, yaxs=i)
plot(1:10)
par(oldpar)
#Ex 3
plot(1:10, xlim=c(-5, 15), ylim=c(-100, 100))
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] XML parse error
I am trying to parse XML file ( binary hex) but get an error.
Code I am using is:
xsd = xmlTreeParse(system.file(exampleData, norel.xsd, package =
XML), isSchema =TRUE) doc = xmlInternalTreeParse(system.
file(exampleData, LogCallSummary.bin, package = XML)) Start
tag expected, '' not found
xmlParse command results in same error as well:
f = system.file(exampleData, LogCallSummary.bin, package =
XML) doc = xmlParse(f)Start tag expected, '' not found
I am at beginner level with XML and will appreciate any help with
this error or general guidance.
Thanks
Kulwinder Banipal
file is:
000: 0281 0001 0201 0098 c1d5 c000
010: 000a c0a8 db35 0055 6000 00af 0001 0001 .5.U`...
020: 5f00 2200 4530 4411 2233 4455 0f08 _..E0..D.3DU..
030: 0123 4567 8901 2340 04d2 .#eg...@
040: 0002 0100 0001 0003 0303
050: 0100 6400 0100 d...
060: 6401 0300 0900 00fe fe00 012f 0001 d../
070: 0101 0001 0001 2200 0033 ..
3080: 3306 0022 1100 3...33.
090: 0033 3400 2300 0011 0001 3335 .34.#.
350a0: 0024 1100 0200 0033 3600 2500 .$.36.%.
0b0: 0011 0003 3337 0026 1100 37.
0c0: 0400 0033 3800 2700 0011 0005 .38.'...
0d0: 5504 7700 8800 0044 4406 2323 0099 U.wDD...##..
0e0: 0100 0200 0023 2400 9901 0002 .#$.
0f0: 0001 2325 0099 0100 0200 ..#%
100: 0200 0023 2600 9901 0002 0003 ...#...
110: 2327 0099 0100 0200 0400 0023 2800 #'...#(.
120: 9901 0002 0005 0102 0008 0100 0066 ...
f130: 6600 0055 5533 3400 0a35 f..UU34
5140: 0014 3600 1e37 0028 3800 67...(8.
150: 3239 003c 3a00 463b ..29...:...F;..
160: 0050 3c00 5a00 0088 8800 0077 7744 .P...Z..
wwD170: 4500 0a46 0014 4700 EF
G.180: 1e48 0028 4900 324a ...H...(I...2J..
190: 003c 4b00 464c 0050 4d00 .K...FL...PM...
1a0: 5a02 2207 7766 6604 0500 1100 0088 Z..wff.
1b0: 8800 0106 0011 8889
1c0: 0011 0700 1100 0088 8a00 2108 ..!.
1d0: 0011 888b 0031 0405 ...1
1e0: 0022 0044 0001 0600 2200 D.
1f0: 4500 1107 0022 0046 ..E...
F200: 0021 0800 2200 4700 ...!...G...
210: 3106 0001 0002 0003 0004 0005 0200 1...
220: 0033 4400 0055 6609 0101 0202 0303 0404 .3D..Uf.
230: 0505 0606 0707 0808 0909 0405 0011
240: 0044 0022 0088 0500 ...D...
250: 1200 4500 2300 8905 E...#...
260: 0013 0046 0024 008a 0500 .F...$..
270: 1400 4700 2500 8bfa
..G...%.280: ae
Um, this isn't an XML file. An XML file should look something like this:
?xml version=1.0 encoding=utf-8 ?
tag
subtagvalue/subtag
/tag
The wikipedia entry on XML gives a reasonable intro to the format.
http://en.wikipedia.org/wiki/Xml
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] [newbie] how to do a 3d plot of bivariate density?
I am new to R. Yesterday I passed the afternoon reading the
introduction and language reference, but I could'nt find a way to do a
3d plot of the density of a data table of size 2.
I am trying with:
plot(density(t(t2)))
but it mixes the two columns and calculate the density like it is a
1-dimensional casual variable.
I presume what you mean by a data table of size 2 is something like this:
data - matrix(c(3,53,36,17), nrow=2)
3D plots are almost never the best solution. (You get all sorts of issues
with perspective problems and some bits of the plot being obscured by
other bits.) Would something like this to the trick?
image(data)
If you want something different, then please provide an example of your
data (nothing too big, so we can reproduce it), and a description of what
you would like to show.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Need help on ploting Histograms
this is the command i made for a normal distribution, but when i try to
plot
the histograms, i dont know why the bars don't stick on the line...
nsamples-1000
sampsize-15
Samples-matrix(rnorm(nsamples*sampsize,0,1),nrow=nsamples)
a-apply(Samples,1,var)
NC14-a*14
x-0:40
plot(x,dchisq(x,14),type='h')
hist(NC14,freq=F,add=T)
Set the histogram bins to match the plot. Change the last line of your
code to, e.g.
hist(NC14,freq=F,add=T, breaks=x, border=transparent, col=#)
Note that the histogram contains sample values, while the plot shows a
theoretical distribution, so they won't match exactly.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] arrangement of crowded labels
I'm looking for algorithms that assist in spreading out crowded labels,
e.g.
labels of points in a scatter plot, in order to obtain a nicer visual
appearance and better legibility.
I'm probably just stuck because I didn't find the right key words for a
successful search on the R websites or in the mailing list archives.
Try thigmophobe.labels in the plotrix package.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Where to find a changelog for the survival package
since some days I try to use the versions 2.35-4 of the survival
package instead of versions 2.31, I had installed until now. Several
changes in print.survfit, plot.survfit and seemingly in the structure
of ratetabels effect some of my syntax files.
Is there somewhere a documentation of these changes, besides the code
itself?
It's in the repository on R-Forge. The latest version is here:
http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/pkg/survival/Changelog.09?rev=11234root=survivalview=markup
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to google for R stuff?
One thing I find most frustrating about R is how difficult it is to use
Google (or any other search tool) to look for answers to my R-related
questions. With languages with even slightly more distinctive names
like
Perl, Java, Python, Matlab, OCaml, etc., usually including the name of
the
language in the query is enough to ensure that the top hits are
relevant.
But this trick does not work for R, because the letter R appears by
itself
in so many pages, that the chaff overwhelms the wheat, so to speak.
There are loads of ways of finding information.
Use the function RSiteSearch, or
The R mail archive
http://www.googlesyndicatedsearch.com/u/newcastlemaths
RSeek http://www.rseek.org/
R Searchhttp://www.dangoldstein.com/search_r.html
The R Graph Gallery http://addictedtor.free.fr/graphiques/
R Help Wiki http://wiki.r-project.org/rwiki/doku.php
R manuals http://cran.r-project.org/manuals.html
FAQshttp://cran.r-project.org/faqs.html
Task Views http://cran.r-project.org/web/views/
News
http://www.r-project.org/doc/Rnews/index.html
Books
http://www.r-project.org/doc/bib/R-books.html
Cranberries http://dirk.eddelbuettel.com/cranberries/
R-Forge (http://r-forge.r-project.org/) and Bioconductor (
http://www.bioconductor.org/GettingStarted) also have their own search
tools.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] help on Nan error
When i want to do ANOSIM i get an NaN error message. What is wrong?
(lots of other code)
iwithin=rep(0,(N*(N-1)/2) )
r.w=sum(r*iwithin)/sum(iwithin)
iwithin is a vector of zeroes and so is its sum. r*iwithin is also a
vector of zeroes, and so is its sum. Thus r.w=sum(r*iwithin)/sum(iwithin)
is zero divided by zero, which is not defined.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Finding root using Newton's method
I want generate R code to determine the real root of the polynomial
x^3-2*x^2+3*x-5. Using an initial guess of 1 with Newton's method.
Homework? - see your instructor! Otherwise, provide minimal
self-contained
code and show us where you are stuck.
It gets a little suspicious when there are several people asking to
determine the root of the exact same polynomial using the same method. Is
it really so hard to RSiteSearch for an appropriate keyword?
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] decimal troubles ?
I have some trouble with the number of decimals in R (currently R
2.9.0). For instance:
options()$digits
[1] 3
let me hope that I will get three digits where useful when a number is
printed. BUT:
44.25+31.1+50
[1] 125
No way to get the right result 125.35
Can anybody tell me what's happens ?
The digits option specifies the number of significant figures, not the
number of decimal places. (The help documentation on the options page
doesn't make this clear at the moment, though it does point you to
print.default, which describes it as setting significant digits.)
Also note that the true value is being stored, so you can retrieve it with
explicit formatting, e.g.
x - 44.25+31.1+50
x # 125
print(x, digits=5) # 125.35
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] (no subject)
if i have the following function,
f - function(x) x^3-2*x^2+3*x-5
i need a simple function for the derivative of this with respect to
'x', so that i can then sub in values to the the derivative
function, and use Newtons method of finding a root for this.
You could take a look at ?deriv, but since the example you presented is
just a polymonial, you can probably learn about calculus (
http://en.wikipedia.org/wiki/Derivative#Derivatives_of_elementary_functions
) and differentiate it in your head in less time than it took you to write
this help question.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice histogram for multiple variables : adjusting x axis
I have a large data frame and I want to look at the distribution of
each variable very quickly by plotting an individual histogram for
each variable.
I'd like to do so using lattice.
Here is a small example using the iris data set:
histogram(as.formula(paste(~,paste(colnames(iris[,!sapply(iris,is.
factor)]),collapse=+))),data=iris[,!sapply(iris,is.factor)])
However in this graphic, the x-axis and the breaks are the same for
all 4 variables. I would like them to be adjusted to the data in
each panel separately. I have tried using the scales argument to no
avail:
histogram( as.formula(paste(~,paste(colnames(iris[,!sapply(iris,
is.factor)]),collapse=+))) ,data=iris[,!sapply(iris,is.factor)],
scales=list(x=list(relation=free)) )
How can I individualize the x-axis and breaks for each individual panel?
You're on the right track. If you don't specify a breaks argument, then
histogram calculates a common axis and breaks across all the panels.
Declaring breaks=NULL will force it to calculate them for each panel.
histogram(
as.formula(paste(~,paste(colnames(iris[,!sapply(iris,is.factor)]),collapse=+))),
data=iris[,!sapply(iris,is.factor)],
scales=list(x=list(relation=free)),
breaks=NULL
)
You can also specify a hist-style method, e.g. breaks=Sturges to get
what you want.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how the break is calculated by R?
As to hist,the help file says: R's default with equi-spaced breaks
(also the default) is to plot the counts in the cells defined by
breaks.
I wanna know how the break is calculated by R?
In other words: break = (max - min)/(number of group) but how the
number of group is calculated by R? Thanks!
If you look closely at the help page for hist, you'll see that in the
Usage section it says breaks=Sturges. Then in the Arguments section it
says that when breaks is a character string, it means it is the name of a
method to compute the number of cells. Finally, in the Details section,
it tells you to look at the nclass.Sturges help page for details of
Sturges' algorithm.
Wikipedia has a good description of common alorgithms for calculating the
number of breaks.
http://en.wikipedia.org/wiki/Histogram#Number_of_bins_and_width
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] xyplot: no ticks for a factor scale?
I would like to have no ticks on a scale that represents a factor. The
tick.number argument from scales does not work in such a situation, as
the help page as well as this simple (fairly stupid) code show:
require(lattice)
fact-gl(4,1,labels=LETTERS[1:4])
y-c(1,4,3,2)
xyplot(y~fact,scales=list(x=list(tick.number=0)))
I want to leave out the ticks at the x-axis (removing labels is easy,
of course). What can I do with that?
Replacing tick.number with tck will do the trick.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] working with groups of labels?
I have a graph with groups of variables. I have include the group
names as variables so that I can have them positioned correctly.
Unfortunately this means that the group names have to follow all of
the same rules as the variables within the groups. I would rather
have those group names left justified and bolded.
I know that I can do this in an mtext() function using font=2 and
adj=0, but I can never get the labels lined up exactly right.
Is there a simpler way of getting one group of labels right justified
and another group of labels left justified and still have them all
automatically positioned correctly?
mtext accepts vector inputs, so you can specify different justifications
for different labels, e.g.
plot(1:10)
mtext(c(foo, bar, baz), adj=c(0, 0.5, 1), side=1)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] What does it mean by skip=2 and skip=7?
Can anyone tell me what is skip=2, skip =7
From ?read.csv:
skip: integer: the number of lines of the data file to skip before
beginning to read data.
and %in% mean here?
%in% matches values; see ?'%in%', and example('%in%')
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] working with groups of labels?
It seems that the structure of your data is that you have two groups
(Real Bad Stuff and Other Crazy Things) which are then subdivided into
further categories. I'd be tempted to set your data up like this:
dfr - data.frame(
score=c(23, 14, 17, 8, 43, 13),
group=rep(c(Real Bad Stuff, Other Crazy Things), each=3),
variable=c(Real,Bad,Stuff,Other,Crazy,Things))
Then you can use a lattice plot to emphasise the different groups, and
avoid your mtext worries.
library(lattice)
barchart(
variable~score|group,
data=dfr,
horizontal=TRUE,
layout=c(1,2),
scales=list(y=list(relation=free)),
prepanel=function(x,y,...)
{
y - y[,drop=TRUE]
prepanel.default.bwplot(x,y,...)
},
panel=function(x,y,...)
{
y - y[,drop=TRUE]
panel.barchart(x,y,...)
}
)
(The scales, prepanel and panel arguments are just to get rid of the
variables that appear in one group but not the other.) I realise it isn't
exactly what you asked for, but I thought you might be interested in the
alternate approach.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plotting questions
1. How to plot several lines in a figure? Suppose I have several sets of
points (xi,yi), where xi and yi are equal-length vector. plot(x1,y1)
will
give a line connecting these points. Another plot(x2,y2) will erase what
plot before and plot the new line. Can I have these lines all drawn in
the
same figure?
#Draw your plot
plot(seq(0,1,length.out=20))
#Add lines to the existing plot
lines(runif(20))
#Add points to the existing plot
points(runif(20), col=red)
2. How to open another figure window? Repeating plot will redraw in the
same
window instead of opening another one.
On Windows, windows() will open a new figure window; quartz() on Mac OSX
and x11() on Linux do the same.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] I'm offering $300 for someone who know R-programming to do the assignments for me.
There are six assignments in total. It won't take you long if you were
familiar with R. For those who are interested, please send me an email
with your profile (your experience with R, how long and how often have
you been using it.) I will be paying through paypal. Thanks!
Now see, you made three mistakes here.
First, you asked for help cheating with homework on a mailing list filled
with lecturers. Second, you offered a pitiful amount to do something that
unethical. And thirdly, you violated the rules of the mailing list by not
including commented, minimal, self-contained, reproducible code.
If you can't even cheat properly, you deserve to fail.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] howto find x value where x=max(x)
fp is a data frame like this
,[ fp ]
|Frequenz AmpNorm
| 1 3322 0.0379490639
| 2 3061 0.0476033058
| 3 2833 0.0592954124
| 4 2242 0.1275510204
`
i want to find the Frequenz where AmpNorm is max.
Use which.max.
fp - data.frame(Freqenz=c(3322,3061,2833,2242),
AmpNorm=c(0.0379,0.0476,0.0592,0.1276))
index.of.max - which.max(fp$AmpNorm)
fp$Freqenz[index.of.max]
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create Pie chart from .csv file
Ive found out a way around my problem. I was trying to plaot a histogram
of
strings, but I had to change it into integers. I ran an sql query on the
original DB that I got the CSV file from and used COUNT to get the
number of
each unique item in a given column. I then used these numbers to create
a
histogram in R. It's a round about way of doing it, but it will have to
suffice.
Glad you found a solution to your problem. For future reference, an
easier way to convert strings to numbers is to use as.numeric or
as.integer.
as.numeric(1.23) # 1.23
as.integer(4) # 4
If the process of reading in your data from the CSV file converts the
strings to factors, then see the FAQ on R (
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-do-I-convert-factors-to-numeric_003f).
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Show name of dataset in graph
I?ve written a script to run several multivariate statistical analysis
automatically.
As one result a biplot and screeplot is produced.
Now I?d like to display the name of the inputdatset as part of the title
of
these graphics and I do not want to enter it each time I run the script.
How can I extract the name of a dataset?
An (shortened) extraction from the script:
Test1 - function(pri, sec, Pca, Ca, Dca){
#pri - primary Parameters (biological Dataset (abundances))
#sec ? secondary Parameters
#Pca, Ca, Dca ? Default=F, If set to TRUE these Methods are active
?several mathematical Operations?
biplot(pri, pc.biplot=T, main=paste(?PCA - ?, ???)?.
Instead of the ??? I? d like to have a variable.
For example if
Test1(dataset.typ5, sec.typ5, Pca=TRUE)
then the title of the plots should be
?PCA ? dataset.typ5?
This demonstrates the basic idea:
dfr1 - data.frame(x=1:10, y=runif(10))
dfr2 - data.frame(x=11:20, y=runif(10))
library(lattice)
myplot - function(dfr)
{
xyplot(y~x,
data=dfr,
main=as.character(substitute(dfr)))
}
myplot(dfr1)
myplot(dfr2)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create Pie chart from .csv file
I am looking to create a pie chart from a given column in a .csv file.
My class variables are as follows:
entry_type, uniquekey, types, title,url, abstract, journal,
author, month,
year, howpublished
So say I want to export a pie chart that groups together all entries
under
'types' e.g. 3 x statistics 2x education etc. Im looking to have a
piechart
represent this graphically that shows which type of entry is in most
frequently. Preferably I'd like to export to a PDF chart and while I can
do
this by typing variables directly into the R console, I cannot manage it
from a .csv file.
If you cannot help me with this specific problem, just knowing how to
create
a generic pie chart would be a great help.
Split the problem up into different stages.
1. Read in your data form the csv file.
?read.csv
2. Plot something.
?pie if you really must, but a bar chart will almost certainly be better.
See ?barplot and ?barchart (lattice package).
3. Export the graph to PDF.
?pdf
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] displaying percentage in bar plot
I have a following data
AIS LEvel
1 23
body regionA 10 15 20
B 15 25 15
Now I want to plot a barplot and in each bar (corresponding a body
region),
I need a percentage of AIS level 1 displayed in the plot. Is there an
easy
way to do this ?
Try this:
x - matrix(c(10,15,15,25,20,15), ncol=3)
pct - apply(x, 1, function(x) 100*x[1]/sum(x))
custompanelfn - function(...)
{
panel.barchart(...)
panel.text(x=c(5,5), y=c(1,2),paste(format(pct, digits=3),%, sep=))
}
library(lattice)
barchart(x, panel=custompanelfn)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plots - several pages per pdf - quality/size issue
I hope that question will not be too redundant (sorry if it is) but
i don't seem able to find the answer i need in the archives...
I try to create a file which would have 1.several pages and 2.
several plots by page. I know how to make a pdf file this way, but
my problem is that the pdf size gets way too big and cannot be open.
So my question is:
- is there a way to diminish the pdf size/quality?
- does any other file format exists that would be of smaller quality
but still be able to print plots on several pages for the same file?
You could try postscript instead. e.g.
p1 - xyplot(runif(1)~rnorm(1)|sample(letters[1:3], 1,
replace=TRUE), layout=c(1,1))
pdf(test.pdf)
print(p1)
dev.off()
# File size 623kb
postscript(test.ps)
print(p1)
dev.off()
# File size 257kb
The other alternative is to print to a raster format, e.g. png. this
creates different files for different pages, but you could always find a
way to view multiple plot files at once. Some possibilities:
1. Use, e.g., Picasa to browse your plot files.
2. Create a simple webpage containing each image
# R code
png(test%d.png)
print(p1)
dev.off()
# Web page code
!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0 Strict//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd;
html xmlns=http://www.w3.org/1999/xhtml;
head
/head
body
img alt=first plot src=test1.png /
img alt=second plot src=test2.png /
img alt=third plot src=test3.png /
/body
# File sizes 3*13kb + 1kb
Open and save this file in a word processor of your choice, and you can
probably drop the file size even further.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with a defined function which cannot access a function defined outside of the function
i have a problem with a function that i defined:
the function needs to use a function which is defined outside the
function
but
i realised that this is not working. a friend told me that this must be
a
problem with
hidden parameters.
a workaround works when i just define all parameters instead of using
the
function
itself. now the problem is that i want to integrate the function and
i'm
not
sure how to do it with this workaround since i have a plot of the
function
and
a big vektor with all points of the plot on the x-axis only.
I suggest you read up a little on the search path and how R does lexical
scoping (sections 6.3.5 and 10.7 in R-Intro). The short version of that
is that unless you specifically name the package that the function is
from, R will look inside the current function, then in the global
environment, then some other places.
If you have problems with names of things being hidden, a quick hack is to
rename the hidden item.
If you want a less abstract answer then please post a commented, minimal,
self-contained, reproducible code example.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] geometric mean to handle large number and negative values
I have created two functions to compute geometric means. Method 1 can
handle even number of negative values but not large number, vice versa
for method 2. How can I merge both functions so that both large number
and negative values can be handled ?
geometric.mean1 - function(x) prod(x)^(1/length(x))
geometric.mean2 - function(x) exp(mean(log(x)))
geometric.mean1(1:1000)
[1] Inf
geometric.mean2(1:1000)
[1] 3678798
geometric.mean1(c(-5,-4,4,5))
[1] 4.472136
geometric.mean2(c(-5,-4,4,5))
[1] NaN
Warning message:
In log(x) : NaNs produced
Geometric mean is usually restricted to positive inputs, because otherwise
the answer can have an imaginary component. If you really want the
geometric mean of negative inputs, use the second method but convert the
input to be a complex number first.
comp.x - as.complex(c(-5,-4,4,5))
geometric.mean2(comp.x)
# [1] 0+4.472136i
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] geometric mean to handle large number and negative values
geometric.mean1 - function(x) prod(x)^(1/length(x))
geometric.mean2 - function(x) exp(mean(log(x)))
geometric.mean1(c(-5,-4,4,5))
[1] 4.472136
geometric.mean2(c(-5,-4,4,5))
[1] NaN
Warning message:
In log(x) : NaNs produced
comp.x - as.complex(c(-5,-4,4,5))
geometric.mean2(comp.x)
# [1] 0+4.472136i
Obviously, there's a discrepancy between the answer of geometric.mean1 and
geometric.mean2 with complex inputs. Having thought about it a little
more, I think the problem is with my solution.
The log of a complex number decomposes as log(z) = log(abs(z)) +1i*Arg(z).
When you sum the second components, you need to take the answer modulo
2*pi, since the phase angles wrap around.
Here's an alternative geometric mean function that takes that into
account.
geometric.mean3 - function(x)
{
a - mean(log(abs(x)))
b - 1i/length(x)
c - sum(Arg(x))%%(2*pi)
exp(a+b*c)
}
geometric.mean3(comp.x)
# [1] 4.472136+0i
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extending a vector to length n
In general, how can I increase a vector of length m ( n) to length n
by padding it with m - n missing values, without losing attributes?
The two approaches I've tried, using length- and adding missings with
c, do not work in general:
a - as.Date(2008-01-01)
c(a, NA)
[1] 2008-01-01 NA
length(a) - 2
a
[1] 13879NA
b - factor(a)
c(b, NA)
[1] 1 NA
length(b) - 2
b
[1] aNA
Levels: a
You can save the attributes to a new variable, increase the length, then
reapply the attributes, e.g.
extend - function(x, n)
{
att - attributes(x)
length(x) - n
attributes(x) - att
x
}
a - as.Date(2008-01-01)
extend(a, 2)
# [1] 2008-01-01 NA
b - factor(a)
extend(b, 2)
# [1] aNA
# Levels: a
It would, perhaps, be nicer if length had an option to preserve
attributes.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice xyplot with two y axis
I have some data which needs to be plotted with lattice.
library(lattice)
cars - c(0.1, 0.3, 0.6, 0.4, 0.9)
trucks - c(0.2, 0.5, 0.4, 0.5, 0.1)
drivers-c(121,145,167,200, 210)
year-c(2005,2006,2007,2008,2009)
type-c(local,local,foreign,foreign,foreign)
xyplot(cars+trucks~year|type, data=df3, type=o)
Basically, I need to show drivers as well on a secondary y-axis. Is
this
possible with lattice and xyplot?
The trick is to use a custom y-scale component to draw the second axis.
You also need to manually make space between panels using the between
argument.
#Your data, as before
library(lattice)
cars - c(0.1, 0.3, 0.6, 0.4, 0.9)
trucks - c(0.2, 0.5, 0.4, 0.5, 0.1)
drivers-c(121,145,167,200, 210)
year-c(2005,2006,2007,2008,2009)
type-c(local,local,foreign,foreign,foreign)
#Custom y-scale component
myyscale.component - function(...)
{
ans - yscale.components.default(...)
ans$right - ans$left
foo - ans$right$labels$at
ans$right$labels$labels - as.character(200*foo) #200 is the
scale factor difference between axes, adjust as necessary
ans
}
#The plot
xyplot(cars+trucks+drivers/200~year|type,
type=o,
scales=list(
x=list(alternating=FALSE),
y=list(relation=free, rot=0)), # relation=free separates
the panels
yscale.component=myyscale.component,
between=list(x=2), #between creates more space
between the panels - you may need to adjust this value
par.settings=list(layout.widths=list(right.padding=6)), #this creates
more space on the right hand side of the plot
ylab=NULL,
key=simpleKey(text=c(cars, trucks, drivers)))
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] default print format for large numbers
Numbers like ``1239178547.653775 is inserted into a vector. I print
the vector:
route_9_80_end
[1] 1239178522 1239178526 1239178524 1239178524 1239178524
1239178523 1239178524 1239178522 1239178521 1239178565 1239178566
1239178566
[13] 1239178565 1239178566 1239178566 1239178565 1239178566 1239178566
and the decimals are suppressed. While R is wonderful and
wonderfully documented, I can only find some quasi relavent
information on options(digits) or some such as might apply here. How
can I force R to print some part of the decimal i.e. back to
1239178547.653775?
When you create the vector, each number is stored as a double-precision
floating point number. When you type the name of the variable the print
function is called, which displays the numbers on the console.
You can also manually call the print function, and set the digits argument
to a high number to show all digits.
If you want to see all the digits every time any number is displayed, use
options(digits).
For example:
x - 1239178547.653775
x
# [1] 1239178548
print(x)
# [1] 1239178548
print(x, digits=20)
# [1] 1239178547.653775
options(digits=20)
x
# [1] 1239178547.653775
Note that the accuracy of the floating point numbers is limited to
.Machine$double.eps, which is typically 2.2e-16.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R and .net/C#
There seems to be a way for calling R from .net. However, is there anyway for calling .net/C# code from R? Something similar to the RJava package for .net? To the best of my knowledge, there isn't at the moment. The latest release of MATLAB does though, so in the spirit of competitiveness, we really ought to include this feature in R. (With support for Mono, for the Linux users too, while I'm making requests.) See this video for its use in MATLAB: http://www.mathworks.com/support/2009a/matlab/7.8/demos/New-MATLAB-External- Interfacing-Features-in-R2009a.html Does anyone have any idea how much effort it would take to implement this feature? Regards, Richie. Mathematical Sciences Unit [1]HSL _ ATTENTION: This message contains privileged and confidential information intended for the addressee(s) only. If this message was sent to you in error, you must not disseminate, copy or take any action in reliance on it and we request that you notify the sender immediately by return email. Opinions expressed in this message and any attachments are not necessarily those held by the [2]Health and Safety Laboratory or any person connected with the organisation, save those by whom the opinions were expressed. Please note that any messages sent or received by the [3]Health and Safety Laboratory email system may be monitored and stored in an information retrieval system. _ _ Scanned by MailMarshal - Marshal's comprehensive email content security solution. Download a free evaluation of MailMarshal at [4]www.marshal.com _ References 1. http://www.hsl.gov.uk/contact-us.htm 2. http://www.hsl.gov.uk/ 3. http://www.hsl.gov.uk/ 4. http://www.marshal.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Difference in client vs. server graphics defaults
lawnboy34 wrote: I am having trouble with the difference between default graphic settings on my client machine and the instance of R on our company's server. I created a script locally that output graphs, but when I run it on the server the output graphs have titles running past the margins, legends improperly placed, etc. I checked the default par() settings and found differences between my machine and the server's R instance in 6 parameters. I attempted to set the server-side par() settings to mimic my client machine, but I cannot change the three read-only settings of cra, cxy, and din, and the graphs still look distorted. Is there another way to make the two sets of defaults equal each other? I am using R 2.6.1 on the server and 2.7.0 on my client computer, but I am doubtful that this would have an effect. Any ideas? Since we can't reproduce what you've done, it's hard to pinpoint the problem exactly. I suggest you start with a very simple plot, and create that on your machine and the server. Then gradually build up the plot until it looks like the one you want. That way you should be able to find the offending line of code. Also, if you write the graphs to file (using png, or whatever), then you don't need to worry about the device or character sizes, since you can explicitly define them. Finally, get your copy of R updated. You should eliminate version issues for your own sanity at least. - Regards, Richie. Mathematical Sciences Unit HSL -- View this message in context: http://www.nabble.com/Difference-in-client-vs.-server-graphics-defaults-tp22609493p22617103.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] who can give me some hint?
Hi All,
act_2
DateDtime Hour Min Second Rep
51 2006-02-22 14:52:18 14 52 18 useractivity_act
52 2006-02-22 14:52:18 14 52 18 4
55 2006-02-22 14:52:49 14 52 49 4
57 2006-02-22 14:52:51 14 52 51 4
58 2006-02-22 14:52:52 14 52 52 3
60 2006-02-22 14:54:42 14 54 42 useractivity_idle
I want to change act_2 to
DateDtime Hour Min Second Rep
51 2006-02-22 14:52:18 14 52 18 useractivity_act
52 2006-02-22 14:52:18 14 52 18 4
58 2006-02-22 14:52:52 14 52 52 3
60 2006-02-22 14:54:42 14 54 42 useractivity_idle
in other word, I want to keep 1st if there are many repeated value,
I made the program as:
rm_r-function(act_2){
dm-dim(act_2)[1]-1
for(i in 2:dm){
if(act_2$Rep[i+1]==act_2$Rep[i]){
act_2-act_2[-(i+1),]
}else{
act_2-act_2
}
}
return(act_2)
}
when it moved one row on 1st loop, i should still start 2 but it
become 3 at 2nd loop, if I add i-i-1, then i go to 1
seems not reasonbale. How should I modify it`?
Please don't repeatedly post the same question - it is irritating, and
you're not likely to get a favourable response. Try explaining your
problem more clearly. What is the condition that you want to use to keep
rows? (In your example, each of the rows is different, yet you kept some
and discarded others.)
If you just want to discard some rows form a data frame, you don't need a
loop, e.g.
dfr - data.frame(x=1:10, y=runif(10))
dfr[c(1,3,5,5),]
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with loop
I'm trying to write a loop to sum my data in the following way:
(the second - the first) + (the third - the second) + (the fourth -
the third) + ...
for each column.
This is just sum(diff(x)), or even x[length(x)] - x[1].
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help
I am trying to excess the inbuit .Fortran and .C codes of R. Can any one
help me in that. For example in kmeans clustering the algorithms are
written
in .Fortran I want to access them and see the .Fortran syntax of the
codes.
Can any one help me how can I do that?
Download the R source code from your nearest CRAN mirror. The Fortran/C
code is in the src subfolder.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Normal Probability Plot - Time series
I have some time-series data and wish to plot a normal probability plot
in
R. How do I go about this?
In R, they are known as quantile-quantile plots. Check out ?qqplot and
?qqnorm.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding text to both grid and base graphs
I generate graphs using both the grid system (with lattice) and the base
system. I'd like to be able to identify these graphs later on with a bit
of
identifying text (e.g. a date and some comments). Adding text to these
graphs cannot be done using a common system if you want to save them as
emf
files. I now use:
mtext(labelling
text,outer=TRUE,at=0.5,adj=0.5,line=-1.1,side=1,cex=0.7,col=darkgrey)
for base graphs and
grid.text(labelling
text,x=0.5,y=0.005,just=bottom,gp=gpar(fontsize=8,
col=darkgrey))
for grid graphs.
Unfortunately if I were to use mtext on a grid graph and copy to emf, I
get
the error:
Error in dev.copy(win.metafile, pltfile) : invalid graphics state
Is there a way to detect the current graphics state so I can choose
which
command to give automatically? (or is there an easier way to go about
this?)
The easier way: use win.metafile to save your plots, e.g.
win.metafile(base.emf)
plot(runif(50))
mtext(labelling
text,outer=TRUE,at=0.5,adj=0.5,line=-1.1,side=1,cex=0.7,col=darkgrey)
dev.off()
win.metafile(grid.emf)
xyplot(runif(50)~1:50)
grid.text(labelling text,x=0.5,y=0.005,just=bottom,gp=gpar(fontsize=8,
col=darkgrey))
dev.off()
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] dummy variable encoding
The best encoding depends upon which language you would like to
manipulate
the variable in. In R, genders are most naturally represented as
factors.
That means that in an external data source (like a spreadsheet of
data),
you should ideally have the gender recorded as human-understandable
text
(male and female, or M and F). Once the data is read into R,
by
default R will convert the string to factors (keeping the human
readable
labels). This way you avoid having to remember that 1 means male (or
whatever).
If you were manipulating the data in a different language that didn't
have
factors, then it might be more appropriate to use an integer. Which
integers you use doesn't matter, you need to have a look-up table to
know
what each number refers to, whatever you choose.
Yes, that's what I thought. However somebody told me that it is better
to use 1/2 rather than 0/1 for a 2 level factor such as gender, and I've
no idea why. I told them it didn't matter, but have since seen quite a
few examples where they use 1/2 (admittedly in SPSS).
The only benefit that I can see of using 1/2 instead of 0/1 is fairly
minor.
If you have cases where there are missing values, and you are working in a
language that doesn't support NA values for integers (or factors; I'm
thinking of something like C), then you could encode your genders as
0: not recorded
1: female
2: male
Then you can include logic like
if(gender)
{
do something
}
The alternative encoding of 0/1, would be something like
-1: not recorded
0: female
1: male
This makes the code slightly less pretty.
if(gender != -1)
{
do something
}
Again, none of this really applies to R, since you should be using factors
for this sort of variable.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] About warnings for non-matched items
I have many files in my directory. I want to transfer each data into
one which is readable.
They have so many possibilities, i have collected(manually and
visually) all possibilities and represent them as different numbers.
Rep[grep('context_log',log1$Remain[1:length(log1$Date)]),]-2
Rep[grep('gs',log1$Remain[1:length(log1$Date)]),]-5
Rep[grep('ClockApp',log1$Remain[1:length(log1$Date)]),]-6
Rep[grep('mce',log1$Remain[1:length(log1$Date)]),]-7
..
I manually collect all possibilities contained in all files!!!
(manually and visually: this process is so time-consuming, can i
have better ideal to collect all possibilities by computer rather
than by myself?)
The programs are fine, but each file doesn't match all
possibilities, whenever there are non-matched items with above, then
warning information come ups:
31: In max(i) ... : no non-missing arguments to max; returning -Inf
32: In max(i) ... : no non-missing arguments to max; returning -Inf
33: In max(i) ... : no non-missing arguments to max; returning -Inf
they returns non-matched items.
How to vanish those warnings?
The error means that the input you provided to the max function is missing
or NULL or of length 0. At a guess, what has happened is that you've
called grep, which didn't match anything (and so returned integer(0)),
then used that as an input to max.
To get rid of the warning, check the input to max first. e.g.
x1 - c(foo, bar, foo)
g1 - grep(foo, x1)
if(length(g1)) max(g1)
g2 - grep(baz, x1)
if(length(g2)) max(g1)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] dummy variable encoding
can anyone tell me why an encoding of 1/2 for a dummy variable for
two groups (e.g. gender) seems to be preferred over 0/1?
It's been bugging me for a while, 0/1 seems more natural, but I have
been told (without explanation) that 1/2 is better. Why?
The best encoding depends upon which language you would like to manipulate
the variable in. In R, genders are most naturally represented as factors.
That means that in an external data source (like a spreadsheet of data),
you should ideally have the gender recorded as human-understandable text
(male and female, or M and F). Once the data is read into R, by
default R will convert the string to factors (keeping the human readable
labels). This way you avoid having to remember that 1 means male (or
whatever).
If you were manipulating the data in a different language that didn't have
factors, then it might be more appropriate to use an integer. Which
integers you use doesn't matter, you need to have a look-up table to know
what each number refers to, whatever you choose.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] The Origins of R AND CALCULUS
Does any student, or teacher for that matter care whether Newton or
Leibntiz
invented calculas.
Students or teachers may not care, but Newton and Leibniz themselves were
pretty bitter about who should get credit for what.
http://en.wikipedia.org/wiki/Newton_v._Leibniz_calculus_controversy
I think this whole debate has gotten rather silly. The original article
was targetted at non-techies, and provided at reasonable summary of what R
does for that audience. The followup clarified on the history of R. As
for Ashley Vance's journalism, well, he's more qualified to write about
software than most, being the former editor of The Register. Hopefully,
this storm-in-a-teacup will blow over soon.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Equivalent of Hold On MatLab Command
Does R have a graphic command equivalent of MatLab Hold On ?
I am trying to sabve on a pdf file a composite drawing. I first
declare the canvas size, then I define the layout, finally I
generate the 4 plots according to layout order. Eventually I close
the pdf file (dev.off()).
The resulting PDF file contains 2 pages. The first one shows only
the layout. The second one shows the whole
composite drawing.
MatLab has the command Hold On to work around similar
synchronization problems.
Does R have have an equivalent command ?
If you are using base graphics (i.e. not lattice/ grid graphics), then you
can mimic MATLAB's hold on feature by using the points/lines/polygons
functions to add additional details to your plots without starting a new
plot. In the case of a multiplot layout, you can use par(mfg=...) to pick
which plot you add things to.
#open the pdf device
pdf(test.pdf)
#setup the layout
par(mfrow=c(2,2))
#plot at each location
for(i in 1:4) plot(runif(1:10))
#return to the first plot
par(mfg=c(1,1))
#add a line to this plot (without replacing anything)
lines(seq(0.1,1,0.1))
#close the device to complete the plot
dev.off()
If you have an extra page in you PDF file, it's probably because you've
called plot.new() somewhere where you didn't mean to (either explicitly or
implicitly). Without a reproducible example, it's hard to help further.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Text Outside Lattice Plot
I created the graph at the bottom using xyplot in the lattice package. I
added a title using the main=Title command in xyplot, however it is
plotted too close to the legend for my liking. To remedy this I
increased
the upper margin of the plot using plot(data, position = c(0,0,1,.9))
and
attempted to move SNA upwards and to the right. I have tried using a
variety of text functions such as:
trellis.focus(panel, 1, 1)
panel.text(x=11, y=10, labels=SNA)
trellis.unfocus()
panel.xyplot(...)
panel.text(x=11, y=10, labels=SNA)
library(grid)
ltext(grid.locator(), label='SNA')
The first two of these functions work but the text disappears once I
specify
a y coordinate ymax. The last function appears to work but requires
me to
click on the plot to specify the location (I need this to be
pre-defined).
Does anyone know how I can do this?
A simple way to get more space for the title is to change the layout
height parameter. Here's an adaptation of the OrchardSprays example in
?xyplot.
xyplot(decrease ~ treatment, OrchardSprays, groups = rowpos,
type = a,
auto.key =
list(space = top, points = FALSE, lines = TRUE),
main=Orchard sprays example,
par.settings=list(layout.heights=list(main=4)))
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] filling blanks with NA
I do have a data set with some missing values that appear as blanks. I
want
to fill these blanks with an NA. How can this be done? Thanks for your
help
Please help us to help you. What form are the data in? Are they in a
text file, or are they in R already? What do you mean by 'blanks'? If we
can reproduce your problem then we can solve it more easily.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] WinBUGS with R
I am having some problems using R with WinBUGS using the R2WinBUGS
package. Specifically, when I try to run bugs() I get the following
message.
Error in FUN(X[[1L]], ...) :
.C(..): 'type' must be real for this format
To give a little more context, my bugs() command (for a multilevel
ordinal logit similar to Gelman and Hill, Data Analysis Using
Regression and Multilevel/Hierarchical Models p. 383 is:
Wednesbury.data - list (n.judge, n, n.cut, y judge, ct,
ra, lg)
Wednesbury.inits - function(){
list(C=matrix(0,39,2))
}
Wednesbury.parameters - c(C, b1, b2, b3)
Debugging your BUGS model or dataset via R is a bit of a pain. I find
that the best way (or maybe least worst way) to weed out the problems when
you get an error like this is to find the files (model/data/inits) that
R2WinBUGS has created and open them in WinBUGS itself. Run the Model
Specification tool and you can more easily determine which part of the
file the problem lies in.
Just looking at your Wednesbury.inits variable, you don't need to define
it as a function
Wednesbury.inits - list(list(C=matrix(0,39,2)))
will do.
Also, I'm not sure if WinBUGS understands matrix data types (though I may
be wrong).
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Deleting columns where the frequency of values are too disparate
Please consider the following toy data matrix example, called x
for simplicity. There are 20 different individuals (ID), with
information about the alleles (A,T, G, C) at six different loci
(Locus1 - Locus6) for each of these 20 individuals. At any
single locus (e.g., Locus1 or Locus2, ... or Locus6), the
individuals have either one allele (from the set of A,T,C,G) or one
other allele (from the set of A,T,C, G). For example, at Locus1
individuals have have either the A or T allele only; at Locus2 the
individuals can have either C or G only; at Locus3 the individuals
can have either T or G only.
IDLocus1Locus2Locus3Locus4Locus5Locus6
1AGTAAC
2AGGACC
3ACGGCC
4ACGGCC
5AGGGAC
6TGGGCC
7TCGGCC
8TCGGAC
9TGGGCC
10TCGGCC
11AGGGAC
12ACGGCC
13AGGGCC
14AGGGAC
15ACGGCC
16TCGGCC
17TGGGAC
18TGGGCC
19TGGGCC
20TCGGAC
I want to delete any columns from the dataset where the rarer of the
two alleles has a frequency of ten percent or less. In other words,
I would like to delete Locus3, Locus4, and Locus6 in this data
matrix, because the frequency of the rare allele is not greater than
ten percent (and conversely, the frequency of the common allele is
not less than ninety percent). Please note that the frequency of the
rare allele in Locus6 is equal to zero (conversely, the frequency of
the common allele is equal to one hundred percent).
Would one of you know of simple way to write this sort of code? (In
my real dataset, there are 1096 loci, so this cannot be done easily by
eye.)
Most of the problem is just organising the data into a sensible form.
# read in data
data - readLines(tc - textConnection(1AGTAAC
2AGGACC
3ACGGCC
4ACGGCC
5AGGGAC
6TGGGCC
7TCGGCC
8TCGGAC
9TGGGCC
10TCGGCC
11AGGGAC
12ACGGCC
13AGGGCC
14AGGGAC
15ACGGCC
16TCGGCC
17TGGGAC
18TGGGCC
19TGGGCC
20TCGGAC)); close(tc)
# retrieve the useful bit
loci - sub([[:digit:]]{1,2}, , data)
# you may also want this
ID - grep([[:digit:]]{1,2}, data)
# find out how many of each base occurs at each locus
freqs - list()
n - length(ID)
for(i in 1:6)
{
assign(paste(locus, i, sep=), factor(substring(loci,i,i),
levels=c(A,C,G,T)))
freqs[[i]] - summary(get(paste(locus, i, sep=)))
}
freqs
# remove loci with 90% or more cases of same base
loci.to.remove - sapply(freqs, function(x) any(x0.9*n))
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] LCA (e1071 package): error
I will use the lca method in the e1071 package. But I get the following
error:
Error in pas[j, ] - drop(exp(rep(1, nvar) %*% log(mp))) :
number of items to replace is not a multiple of replacement length
Does anybody know this error and knows what this means?
The error means that you are trying to assign a variable of one size to a
variable of another fixed size. An example that recreates it is:
x - matrix(1:6, nrow=3)
x[1,] - 1:10
x[1,] cannot be resized from 3 to 10 without affecting the rest of x, so
an error is thrown.
In your example, the jth row of the matrix pas is a different size from
drop(exp(rep(1, nvar) %*% log(mp))).
Since you haven't provided a reproducible example (tut tut, read the
posting guide) you'll have to do the debugging yourself.
To get you started, type traceback() to see where the problem occurs.
Now try options(error=recover), and call the lca function again. Now you
can examine the values of nvar, mp and pas to see what is going wrong.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] R2WinBUGS stopping execution
Apologies if this isn't acceptable for the general help list.
I'm running OpenBUGS model via the R2WinBUGS package interface, under
Windows. Is it possible to terminate running models, short of using the
Windows Task Manager to forcibly exit the program?
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Saving plots as byte streams
Is it possible to save plots as byte streams? For example, if I want the
bytes for a PNG plot, I could use
#Write the plot to a PNG file
png(test.png)
plot(1:10)
dev.off()
#Read the bytes back in from the file
plotbytes - readBin(test.png, raw, n=2000)
Ideally, I'd like to avoid having to bother writing to the file in the
first place, and simply store the bytes in a variable. How do I do this?
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] for loop and if problem
I'm heaving difficulties with a dataset containing gene names and
positions
of those genes.
Not such a big problem, but each gene has multiple exons so it's hard to
say
where de gene starts and where it ends. I want the starting and ending
position of each gene in my dataset.
Attached is the dataset:
http://www.nabble.com/file/p21312449/genlistchrompos.csv
genlistchrompos.csv
Column 'B' is the gene name, 'G' is the starting position and 'H' is the
stop position.
You can load the dataset by using: data-read.csv(genlistchrompos.csv,
sep=;)
I hope someone can help me, it's giving me headaches for a week now:-((.
which(diff(as.numeric(data$Gene))!=0)
will give you a vector of the last row in each gene. The start position
is obviously the next row after the previous end.
Also take a look at
split(data, data$Gene)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Borders for rectangles in lattice plot key
Hopefully an easy question. When drawing a rectangles in a lattice plot
key, how do you omit the black borders?
Here is an example adapted from one on the xyplot help page:
bar.cols - c(red, blue)
key.list - list(
space=top,
rectangles=list(col=bar.cols),
text=list(c(foo, bar))
)
barchart(
yield ~ variety | site,
data = barley,
groups = year,
layout = c(1,6),
ylab = Barley Yield (bushels/acre),
scales = list(x = list(abbreviate = TRUE, minlength = 5)),
col=bar.cols,
border=transparent,
key=key.list
)
Notice the black borders around the rectangles in the key.
I checked to see if there was an undocumented border component for the
rectangles compoenent of key that I could set to transparent or FALSE,
but no luck. I also tried setting lwd=0 on the rectangle component but
that didn't change anything either.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with Surv
I'm trying to estimate a tobit model with survrec function. I use
the following code :
reg-survreg(Surv(crs_prod,crs_prod=1)~SOLVA+log(AF088)
+HHI+ACTIONNA,data=dat,dist=gaussian)
I get this error message with R 2.7.2
Error in survreg(Surv(crs_prod, crs_prod = 1) ~ SOLVA + log(AF088) + :
Response must be a survival object
Take a look at Surv(crs_prod, crs_prod = 1). Your example isn't
reproducible (tut tut), but if I had to guess, I'd say that this is NULL,
because crs_prod is column in dat, and Surv can't find it.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Legend and Main Title positioning
layout(matrix(c(1,2,3,4), nrow = 2, byrow = TRUE))
plot(rnorm(100))
plot(rnorm(200))
plot(rnorm(300))
plot(rnorm(400))
Now, I'd like to create a legend below each plot and generate a common
title.
How can I do that?
If you are laying plots out in grids like this then lattice graphics are
generally the way to go, but here's a solution based upon base graphics.
The trick is to include extra potting space in your layout for the
legends. The code is messy, since it requires you to manually specify
which cell of the layout to plot into, but I'm sure guven some thought you
can automate this.
#4 space for plots, 4 for legends
layout(matrix(1:8, nrow = 4, byrow = TRUE), heights=rep(c(3,1),4))
#Check the layout looks suitable
layout.show(8)
#Avoid clipping problems, and create space for your title
par(xpd=TRUE, oma=c(0,0,2,0))
#First plot
plot(rnorm(100))
#Move down and plot the first legend
par(mfg=c(2,1))
legend(0,0, legend=foo, pch=1)
#Repeat for the other plots and legends
par(mfg=c(1,2))
plot(rnorm(200))
par(mfg=c(2,2))
legend(0,0, legend=bar, pch=1)
par(mfg=c(3,1))
plot(rnorm(300))
par(mfg=c(4,1))
legend(0,0, legend=baz, pch=1)
par(mfg=c(3,2))
plot(rnorm(400))
par(mfg=c(4,2))
legend(0,0, legend=quux, pch=1)
#Title for all the plots
title(main=4 plots, outer=TRUE)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] levels update
I hope this question is not too stupid. I would like to know how to
update
levels after subsetting data from a data.frame.
df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8),
c(9,1,2,3,4))
names(df) - c(X1,X2,X3)
my.sub - subset(df, X1 == a | X1 == b)
levels(my.sub$X1)
# still gives me a,b,c, though the subset does not contain entries
with
c anymore
Two questions in one afternon; aren't I good to you!
levels(my.sub$X1[,drop=TRUE])
[1] a b
levels(factor(my.sub$X1))
[1] a b
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sink does not send graphs to sink file
I am using sink() to send the results of my analyses to a text file.
Unfortunately my graphs do not become part of the file. Is there
anyway that I can have both the text and graphic output of my
analyses appear in a file?
You can create a latex document with text, graphs and R-code using
?Sweave. If you prefer to write to Open document format , there is an
Odfweave package.
The other alternative is simply writing your graphs to files (or one file
with all the graphs), e.g.
pdf(test.pdf)
plot(1:10)
hist(rnorm(100))
dev.off()
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problems with lattice-histograms or png within loops
I have a question concerning the use of lattice plots within for-loops.
I want to create a png file containing a lattice histogram which works
out
fine (part 1).
When I loop the whole code, the graphic file does not contain anything
(part
2).
I can fix it by wrapping the histogram function into a print command
(part
3).
Why is that so? Why can I not loop it directly?
TIA,
Mark
attach(iris)
### part 1
png(filename = graphic_1.png)
histogram( ~ Sepal.Length | Species, data = iris)
dev.off()
### part 2
for (i in c(1))
{
png(filename = graphic_2.png)
histogram( ~ Sepal.Length | Species, data = iris)
dev.off()
}
### part 3
for (i in c(1))
{
png(filename = graphic_3.png)
print(histogram( ~ Sepal.Length | Species, data = iris))
dev.off()
}
This is FAQ 7.22:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-do-lattice_002ftrellis-graphics-not-work_003f
The help page on ?histogram also mentions that the function returns an
object of class 'trellis' and the 'print' method (usually called by
default) will plot it on an appropriate plotting device
It's a good idea to get into the habit of calling print any time you
create a lattice graph. If you reorder your code like
h1 - histogram( ~ Sepal.Length | Species, data = iris)
png(filename = graphic_h1.png)
print(h1)
dev.off()
then you won't end up with an open device if drawing the plot fails.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] question involving loops from intro level R programming class
a. Write a R function zerdiag.v1(m) using loop to output a square matrix
whose diagonal elements are zero and the other elements are filled in by
consecutive integers from 1 to m row-wise.
For example,
zerdiag.v1(6) = [0, 1, 2]
[3, 0, 4]
[5, 6, 0]
This function should have error checking ability. If the input m cannot
form
a square matrix, then the function will return an error message: Input
number is incorrect.
b. Write a R function zerdiag.v2(m) to produce the same output as in
part
(a) without using a loop.
c. Test your functions in part (a) and (b) using m=12 and m=14
respectively.
I'd appreciate any help with this problem... I've spent a lot of time
staring at it, and I'm still not sure where to start. Thanks!
As the posting guide (http://www.r-project.org/posting-guide.html) says
Basic statistics and classroom homework: R-help is not intended for
these.
Also, I find it a little bit cheeky that you''ve asked for help on 3 parts
of your homework, and haven't provided any evidence that you've even
attempted the problem yourself. (If you've been staring at the problem
for hours, you must have written something.) Read the help pages ?matrix
and ?for and take a look in the Intro to R manual.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] construct a vector
I have an unkown number of vectors (=2) all of the same length. Out
of these, I want to construct a new one as follows:
having vectors u,v and w, the resulting vector z should have entries:
z[1] = u[1], z[2] = v[1], z[3] = w[1]
z[4] = u[2], z[5] = v[2], z[6] = w[2]
...
i.e. go through the vector u,v,w, take at each time the 1st, 2sd, ...
elements and store them consecutively in z.
Is there an efficient way in R to do this?
u - 1:5
v - (1:5) + 0.1
w - (1:5) + 0.2
as.vector(rbind(u,v,w))
# [1] 1.0 1.1 1.2 2.0 2.1 2.2 3.0 3.1 3.2 4.0 4.1 4.2 5.0 5.1 5.2
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] basic information defining functions
i am looking from some insights to define own R functions. so far i
found most basics in documentations that are around on the web. except
for one thing:
I´d like to define some function, say:
#assume my data matrix contains vectors like data$myColumn1,data
$myColumn2 etc.
Do you really mean data matrix, or do you mean data frame?
getMyColumn - function (columnid){
x-data$MyColumn?columnid?[data$indexone=1 data$index2=5]
return(x)
}
It's not terribly clear what you want to do with this function. Question
marks don't mean anything in the middle of a statement, so I don't know
what
data$MyColumn?columnid?[data$indexone=1 data$index2=5]
is supposed to do.
Also, you don't need the return keyword - the last line of a function is
the return value.
Do I need to use assign or eval first ? I tried to use paste to
combine something like: paste(data$MyColumn,columnid,sep=) which
did not work.
paste returns a string, not a variable. You can evaluate the contents of
the string using
eval(parse(text=paste(data$MyColumn,columnid,sep=)))
This is however a method of last resort; there is almost always a better
way to do things than using eval.
I am happy to get any help with the problem, but also thankful for
some useful link or guide on how to define own functions properly,
especially the dynamic naming and return part
You probably don't need to bother with dynamic naming, but take a look at
FAW on R, q7.21.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice contourplot background covers inward-facing ticks
I wish to have inward-pointing ticks on my contourplot graph, but the
colored background produced by the region=TRUE statement covers the
ticks up, is there any way around this? Sample code below. --Seth
library(lattice)
model - function(a,b,c,d,e, f, X1,X2) # provide model function
for contour plot
{J - a + (b*X1) + (c*X2) + (d*X1*X2) + e*(X1^2) + f*(X2^2)
pp - exp(J)/(1+exp(J))
return(pp)}
g - expand.grid(X1= seq(0.3,0.9,0.001), X2 = seq(0.3,1, 0.001))
g$z - model(-29, -14, 52, 80, -3, -56, g$X1, g$X2) # Create
variable z using gridded data, model, and variables
contourplot(z ~ X1 * X2,
data = g,
region = TRUE, # Adds color to background
cuts = 10, # Number of contour intervals...(and
color
intervals!)
scales = list(tck = c(-1,-1)) # ticks go inward
) ### END
The easiest way is to make the contours semi-transparent by adding
alpha.regions=0.7 (or whatever number you prefer) to the call to
contourplot. This has a couple of poosible disadvantages though:
1. It will change the colours you used
2. You can't print to some devices (windows metafile isn't supported)
Other solutions involve a lot of messing about. You could manually draw
ticks using grid.xaxis (and grid.yaxis) in the grid package, or stop
plotting to the edges of the panel.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculating sum of letter values
Thanks, that's almost exactly what I need...theres just a slight
difference with my requirement, in that I am looking for the actual
index value in the alphabetical sequence, so that instead of:
as.numeric(factor(unlist(strsplit(XYZ,
[1] 1 2 3
I would expect to see
[1] 24 25 26
A minor modeification of Mark's solution works in this case:
as.numeric(factor(unlist(strsplit(XYZ, )), levels=LETTERS))
# [1] 24 25 26
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R course in Scotland
pzs wrote: Several people have suggested that I just pick up R and give it a try. My reluctance to do this is that I am already very familiar with my current working method (Python + Numpy) and I worry that without a course I will work in a Python-centric way, which won't be optimal. I know this isn't what you asked, but as a sidenote, if you are used to working with Python and Numpy, then have you considered Sage ( http://sagemath.org/ http://sagemath.org/ )? It's a Python environment that you can run R commands from (amongst other things). That way you can keep your current working style but let R do the hard stats for you. - Regards, Richie. Mathematical Sciences Unit HSL -- View this message in context: http://www.nabble.com/R-course-in-Scotland-tp20601268p20623414.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dequantizing
I have some data measured with a coarsely-quantized clock. Let's say
the real data are
q- sort(rexp(100,.5))
The quantized form is floor(q), so a simple quantile plot of one
against the other can be calculated using:
plot(q,type=l); points(floor(q),col=red)
which of course shows the characteristic stair-step. I would like to
smooth the quantized form back into an approximation of the underlying
data.
The simplest approach I can think of adds a uniform random variable of
the size of the quantization:
plot(q,type=l); points(floor(q),col=red);
points(floor(q)+runif(100,0,1),col=blue)
This gives pretty good results for uniform distributions, but less
good for others (like exponential). Is there a better
interpolation/smoothing function
I'm not convinced that adding a random amount to the floor values to 'make
up' the underlying data is very meaningful.
If you know what the underlying distribution is, then you are best off
using this distribution to generate plots and extra pretend data.
If you know the distribution is exponential, then you can estimate the
rate by treating the true values as interval censored data, somewhere
between floor and floor+1.
library(survival)
q - sort(rexp(100,.5))
qq - floor(q)
qq[qq==0] - 0.1 #survreg doesn't like values that are exactly
zero
ss - Surv(qq, qq+1, type=interval2)
model - survreg(ss ~ 1, dist=exponential)
summary(model)
rate - 1/exp(model$coefficients[(Intercept)]); rate #hopefully
something close to 0.5
If you don't know the underlying distribution either, then things get
trickier. Try a histogram/kernel density plot/boxplot to see what it
looks like.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] sequencially merge multiple files in a folder
Here I have a folder with more than 50 tab-delimited files. Each
file has a few hundreds of thousands rows/subjects, and the number
of columns/variables of each file varies.The 1st row consists of all
the variable names.
Now I would like to merge all the files into one tab-delimited file
by a common column named Ident
Is there any good way to sequencially merge all of them together?
Here when I say sequencially I mean merging file_1 and file_2
first and then merge the resulting data frame and file_3, and keep
going on and on till all files are merged.
Read each of the tab-delimited files into R using read.delim
dfr1 - read.delim(file1.txt)
dfr2 - read.delim(file2.txt)
dfr3 - read.delim(file3.txt)
(If the files are nicely named then you could do this in a loop.)
It's not entirely clear what you want to do next, since you haven't
provided an example of your data. Either
1. Use rbind to concatenate the data frames if each frame has the same
columns
masterdfr - rbind(dfr1, dfr2)
masterdfr - rbind(masterdfr, dfr3)
2. Use merge to merge the data frames if they have different columns
masterdfr - merge(dfr1, dfr2)#you'll neeed to specify some other
arguments
If you want a clearer answer, you'll have to read the posting guide and
provide more details about your data.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Accessing Results from cenmle function in NADA package
I figured it out. In case anyone else ever has this question -- given
the
following output from cenmle:
fit.cen - cenmle(obs, censored, groups)
fit.cen
Value Std. Errorz p
(Intercept) 1.19473 0.0772 15.4695 5.58e-54
groups1 0.00208 0.0789 0.0263 9.79e-01
Log(scale) -0.40221 0.0168 -23.9034 2.82e-126
Scale= 0.669
Log Normal distribution
Loglik(model)= -3908.9 Loglik(intercept only)= -3908.9
Loglik-r: 0.0006265534
Chisq= 0 on 1 degrees of freedom, p= 0.98
Number of Newton-Raphson Iterations: 1
n = 1766
The p-value is (for example):
p.value - summary(fit.cen)[[9]][[11]]
Or slightly more transparently
summary(fit.cen)$table[,'p']
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Barplot Labels Problem
I’m using barplot function. I pretend to create a horizontal barplot
with
two different information (side by side) for a species list. Well I can
generate the graph easily, but the problem is that the labels with the
species names are cut by device window!! I’ve tried lots of par
functions
none seems to work properly.
Below are the script.
Height is a matrix of species and two column of numerical data. I used
par(las=) function to make the labels horizontal. When I do this they
are
cut by the device window. If I don’t do that the label stay vertical and
make nonsense with the graph.
library(xlsReadWrite)
Spp-read.xls('SppPrincipaisFVeFT.xls',sheet=1,rowNames=T)
Spp-as.matrix(Spp)
barplot(t(Spp),axis.lty='solid',horiz=T,beside=T,las=1,col=c('lightgray','bl
ack'),main='Main Speceis')
legend('topright',c('Living Fauna','Dead
Fauna'),fill=c('lightgray','black'),bty='n')
I can't recreate your code, since you haven't made the file
SppPrincipaisFVeFT.xls available. I guess it's something like this:
x = c(monkey=25, giraffe=13, rhinocerous=46, squirrel=5, sting ray=36,
cheetah=2, penguin=69, chicken=43)
par(las=1)
barplot(x)
On my machine, not all the animal names are shown when the device window
opens. You have several options to solve this.
1. Make the device bigger. Either drag the corner of the window or, if
you are writing directly to a file, try something like:
png(animal barplot.png, width=1200, height=800)
par(las=1)
barplot(x)
dev.off()
2. Make the axis text smaller, e.g.
barplot(x, cex.names=0.5)
3. Abbreviate the names, e.g.
barplot(x, names.arg=abbreviate(names(x)))
4. You can also get a little extra room by decreasing the plot margins and
the amount of space between bars, e.g.
par(las=1, mar=c(2.5,3,0.5, 0.3))
barplot(x, space=0.05)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential information intended
for the addressee(s) only. If this message was sent to you in error,
you must not disseminate, copy or take any action in reliance on it and
we request that you notify the sender immediately by return email.
Opinions expressed in this message and any attachments are not
necessarily those held by the Health and Safety Laboratory or any person
connected with the organisation, save those by whom the opinions were
expressed.
Please note that any messages sent or received by the Health and Safety
Laboratory email system may be monitored and stored in an information
retrieval system.
Scanned by MailMarshal - Marshal's comprehensive email content security
solution. Download a free evaluation of MailMarshal at www.marshal.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] counting run lengths
It works, but the for (i in ...) loop slows down the simulation a
lot.
Any suggestion on how to avoid this loop? (or in general, to speed up
this part of the simulation)
Actually, I have not specified the following: i want to consider only
the most recent sequence of zeros, that is the last part of the time
series.
That is, If I have:
[,1] [,2] [,3] [,4]
[1,]0101
[2,]1111
[3,]1101
[4,]1101
[5,]1101
[6,]0101
[7,]0101
[8,]0101
[9,]0101
I want to store the values (4 0 7 0) in unSpells. That is the values
that represent the last sequence of zeros (that is I want to ignore
other possible zeros like the ones I have inserted above).
This runs pretty quickly:
unSpells - nrow(Atr) - apply(Atr,2,function(x) max(which(x==1)))
#c(4,0,7,0)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Arrays of Trellis plots
the example below does not work. (i know it's not supposed, but it makes
it
clear what i'm trying to achieve)
par(mfrow=c(2,1))
xyplot(y~x2|x1,data=dataframe1,pch=20)
xyplot(y~x2|x1,data=dataframe2,pch=20)
i know i could probably merge the two datasets and do something like
xyplot(y~x2|x1+dataset,data=merged)
par is a base graphics command, and doesn't work with grid/lattice
graphics. While it is possible to merge grid and base graphics using for
example the gridBase package, I suspect what you want is to draw two
lattice plots on the same figure. For this, you need to read up on
viewports, and try an example like this:
pushViewport(viewport(layout=grid.layout(2,1)))
pushViewport(viewport(layout.pos.row=1))
topplot = xyplot(Sepal.Length ~ Petal.Length | Species, data = iris)
print(topplot, newpage=FALSE)
upViewport()
pushViewport(viewport(layout.pos.row=2))
bottomplot = xyplot(Sepal.Width ~ Petal.Width | Species, data = iris)
print(bottomplot, newpage=FALSE)
popViewport(2)
See also section 5.5 in Paul Murrell's book ('R Graphics').
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] convert matrix to dataframe with repeating row names
The row names on a data frame should be unique. You can try
as.data.frame(xx, row.names=FALSE) to convert zz to be a data frame.
If
you need the row name information, add it as a column in the data
frame,
e.g. mydataframe$rnames - rownames(zz). (Note to R-Core: the
documentation for as.data.frame doesn't mention the usage of
row.names=FALSE to ignore row names, but it seems to work
consistently.
Does the help page for as.data.frame need updating?)
No. row.names=FALSE is not intended to work, and did you check every
single as.data.frame() method?
It just so happens that for the matrix method invalid input for
'row.names' results in setting default row names. Other methods may
differ.
row.names=FALSE seems a natural way of supressing existing row names to
me, since it corresponds nicely to using row.names=FALSE in write.csv.
Currently it seems that if a matrix has duplicate row names, then
converting it to be a data frame requires
rnames - rownames(mymatrix)
rownames(mymatrix) - NULL
as.data.frame(mymatrix)
rownames(mymatrix) - rnames
Ideally, three of these lines of code shouldn't really need to be there.
If you disagree that allowing row.names=FALSE is a good idea, or you don't
want to change the function interface, then perhaps having as.data.frame
check for duplicates and throwing a warning (rather than an error) would
be preferable behaviour. I do realise that there are dozens of
as.data.frame methods, and the documentation does state that Few of the
methods check for duplicated row names, but it would be beneficial from a
user standpoint.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] convert matrix to dataframe with repeating row names
I have a matrix x with repeating row names.
zz-matrix(0,4,4)
rownames(zz)=c(a,a,b,b)
data.frame(zz) (?)
The row names on a data frame should be unique. You can try
as.data.frame(xx, row.names=FALSE) to convert zz to be a data frame. If
you need the row name information, add it as a column in the data frame,
e.g. mydataframe$rnames - rownames(zz). (Note to R-Core: the
documentation for as.data.frame doesn't mention the usage of
row.names=FALSE to ignore row names, but it seems to work consistently.
Does the help page for as.data.frame need updating?)
lm(as.formula(paste(final_dat[,5]~,paste(colnames(x),collapse=+))),x
)
this gives me a error
Error in model.frame.default(formula =
as.formula(paste(final_dat[,5]~, :
'data' must be a data.frame, not a matrix or an array
I suspect that if you try class(x), it will be a matrix, not the requisite
data frame.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Using key.opts in Ecdf/labcurve (Hmisc package)
I'm presumably missing something very obvious, but how does one use the
key.opts argument in labcurve (via Ecdf)?
In this example, I want the key to be big and have a blue background, but
it isn't and doesn't.
ch - rnorm(1000, 200, 40)
sex - factor(sample(c('female','male'), 1000, TRUE))
Ecdf(~ch, group=sex, label.curves=list(keys=c(f, m),
key.opts=list(cex=3, background=blue)))
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to store lme/lmer fit result
I am building a hierarchical model on a large data set. It can take
quite
some time to finish one fit, I was just wondering whether it is possible
to
store the fit object (the result) to a file for later (offline)
analysis.
See ?save.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Space between bars in barplot
I played around with your suggestions to change the appearance of my
graph,
but there is still a problem I could not fix.
The vector, which I want to plot, contains 60 entries. After 3 bars I
want
to have a large gap
between the next 3 bars. But what always happens is, that R took at the
beginning just 2 bars
and then always 3 bars ending up with one bar at the end.
So I have created a new vector v - c(0,0,vectorToPrint) plotting this
vector with the following
command: barplot(v, space=rep(c(0,0,2)) ). It works but getting always
some
warning messages.
Instead of 0,0 I tried as well NA, NA but same result. So is there a way
to
achieve that without
warning messages?
Seeing as it's late on a friday afternoon, could you give us a fighting
chance of helping you by
1) providing enough code to reproduce the problem, and
2) telling us what the warning said.
(As per the posting guide.)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Space between bars in barplot
with the space parameter it is possible to change the gap / distance
between the bars, but is it also
possible to change the space after each 6th bar?
So for example you have bars from 1 to 6 then a large gap and then the
next
six bars from 7 to 12
Try, for example
y - runif(13)
y[7] - NA
barplot(y)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] histogram loses top row with alpha transparency under Windows
k.ponting wrote:
Hello all.
Trying to use transparency for overlaid histogram plots I have come
across an interesting inconsistency, possibly a bug when running under
Windows. Originally noticed in R 2.7.1, it is still there in 2.8.0 beta.
library(lattice)
zz - function(n,alpha)
{
ranges - NULL
for(ds in 1:n){
ranges -
rbind(ranges,data.frame(confidence=c(0,100),dataset=as.character(ds),cor
rect=c(FALSE,TRUE)))
}
panel.colhist = function(x, group.number, col, ...) {
panel.histogram(x, col=group.number+1, ...)
}
x - histogram(~confidence|dataset,data=ranges,alpha=alpha,
panel=panel.superpose,panel.groups=panel.colhist,groups=correct)
print(x)
}
zz(12,1) # works as expected, 12 identical plots
zz(12,0.5) # top row of plots has no bars at all, lower rows are as
expected
zz(1,1) # two bars fine
zz(1,0.5) # no bars at all
The rectangles being drawn extend higher than the top of the panel. (Your y
axis ranges from 0 to 50, but the bars go up to 100.)
In the top row of plots, depending upon the shape of your device window, the
bars can extend beyond the range of the device window. For some reason,
(take a look in panel.rect), when you specify alpha less than 1, this
prevents the bar being drawn.
You need to add type=count to the call to histogram, or rescale the bar
heights.
-
Regards,
Richie.
Mathematical Sciences Unit
HSL
--
View this message in context:
http://www.nabble.com/histogram-loses-top-row-with-alpha-transparency-under-Windows-tp19879687p19897501.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Exporting symnum() result from cor()
Michael Just wrote: Hello, I am trying to export the results from symnum() while maintain their readability. I tried using sink to text file and also copying and pasting but the results end up looking like this: symnum(c5.s) bC bED bEN bLP bLS bPA bPD bPR p bbContag 1 bbED + 1 bbENN_MN + B 1 bbLPI , , , 1 bbLSI + B B , 1 bbPAFRAC , * * , * 1 bbPD , B B , B B 1 bbPROX_MN + B B , B * * 1 pfor 1 attr(,legend) Is there a way I can get these out of R with their readability intact so I can share them with non-R users? Is it possible to export the way things look in View() ? write.csv works, e.g. write.csv(symnum(c5.s), test.csv) - Regards, Richie. Mathematical Sciences Unit HSL -- View this message in context: http://www.nabble.com/Exporting-symnum%28%29-result-from-cor%28%29-tp19895075p19897840.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] vectorization instead of using loop
I've sent this question 2 days ago and got response from Sarah. Thanks
for
that. But unfortunately, it did not really solve our problem. The main
issue
is that we want to use our own (manipulated) covariance matrix in the
calculation of the mahalanobis distance. Does anyone know how to
vectorize
the below code instead of using a loop (which slows it down)?
I'd really appreciate any help on this, thank you all in advance!
Cheers,
Frank
This is what I posted 2 days ago:
We have a data frame x with n people as rows and k variables as columns.
Now, for each person (i.e., each row) we want to calculate a distance
between him/her and EACH other person in x. In other words, we want to
create a n x n matrix with distances (with zeros in the diagonal).
However, we do not want to calculate Euclidian distances. We want to
calculate Mahalanobis distances, which take into account the covariance
among variables.
Below is the piece of code we wrote (covmat in the function below is
the
variance-covariance matrix among variables in Data that has to be fed
into
mahalonobis function we are using).
mahadist = function(x, covmat) {
dismat = matrix(0,ncol=nrow(x),nrow=nrow(x))
for (i in 1:nrow(x)) {
dismat[i,] = mahalanobis(as.matrix(x), as.matrix(x[i,]),
covmat)^.5
}
return(dismat)
}
This piece of code works, but it is very slow. We were wondering if it's
at
all possible to somehow vectorize this function. Any help would be
greatly
appreciated.
You can save a substantial time by calling as.matrix before the loop, e.g.
x - data.frame(runif(1000), runif(1000), runif(1000))
covmat - cov(x)
mahadist = function(x, covmat) #yours
{
dismat = matrix(0,ncol=nrow(x),nrow=nrow(x))
for (i in 1:nrow(x))
{
dismat[i,] = mahalanobis(as.matrix(x), as.matrix(x[i,]),
covmat)^.5
}
return(dismat)
}
mahadist2 - function(x, covmat) #my modification
{
n - nrow(x)
dismat - matrix(0,ncol=n,nrow=n)
matx - as.matrix(x)
for (i in 1:n)
{
dismat[i,] - mahalanobis(matx, matx[i,], covmat)^.5
}
dismat
}
system.time(mahadist(x, covmat))
# user system elapsed
# 2.820.062.95
system.time(mahadist2(x, covmat))
# user system elapsed
# 1.390.041.45
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] vectorization instead of using loop
Frank said:
This piece of code works, but it is very slow. We were wondering if
it's
at
all possible to somehow vectorize this function. Any help would be
greatly
appreciated.
Richie said:
You can save a substantial time by calling as.matrix before the loop
Patrick said:
One thing that would speed it up is if you
inverted 'covmat' once and then used
'inverted=TRUE' in the call to 'mahalanobis'.
The timings before:
system.time(mahadist(x, covmat))
# user system elapsed
# 2.820.062.95
system.time(mahadist2(x, covmat))
# user system elapsed
# 1.390.041.45
With Patrick's modification, and moving the square root out of the loop:
mahadist3 - function(x, covmat) #patrick's modification
{
n - nrow(x)
dismat - matrix(0,ncol=n,nrow=n)
matx - as.matrix(x)
icovmat - chol2inv(chol(covmat))
for (i in 1:n)
{
dismat[i,] - mahalanobis(matx, matx[i,], icovmat, inverted=TRUE)
}
dismat^.5
}
system.time(mahadist3(x, covmat))
# user system elapsed
# 0.800.000.85
Not bad - a better than threefold speed up, without worrying about
vectorization.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] question from Braun/Murdoch book
I am looking at the Braun/Murdoch book, A First Course in
Statistical Programming in R, and I have a question about a function
there. It's on page 52, Example 4.5; the sieve of Erastosthenes.
There is a line:
primes - c()
Is there a difference between using that and
primes - NULL
please?
When you put in primes - c(), primes comes back as NULL.
They return the same thing
identical(c(), NULL)#TRUE
Is one more efficient or is this just a matter of programming style,
please?
system.time(for(i in 1:100) c())
# user system elapsed
# 0.630.020.64
system.time(for(i in 1:100) NULL)
# user system elapsed
# 0.280.000.28
Using NULL appears to be quicker on my machine, but given that you can do
a million of these assignments in a fraction of a second, it makes no
practical difference. NULL is perhaps more intuitive than c() for
demonstrating that the variable is empty.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fitting weibull, exponential and lognormal distributions to left-truncated data.
I have several datasets, all left-truncated at x=1, that I am
attempting
to fit distributions to (lognormal, weibull and exponential). I had
been using fitdistr in the MASS package as follows:
A possible solution is to use the survreg() in the survival package
without specifying the covariates, i.e.
library(survival)
survreg(Surv(..)~1, dist=weibull)
where Surv(..) accepts information about times, censoring/truncation
variables and dist allows to specify alternative distributions.
See ?Surv e ?survreg
The survival package is mostly targeted at right-censored data. The NADA
package provides wrappers for many of the survival routines so they work
with left-censored data.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] horizontal boxplot + xlim
I get a strange behaviour of a boxplot with the following code.
There seems to
be a problem with the xlim-parameter. Did I do anything wrong? What
else can I
do to force the boxplot to have a defined x-range?
x - rnorm(100)
boxplot(x, notch=TRUE, xlab=parameter, xlim - c(-4,4), horizontal =
TRUE)
Did you mean xlim = c(-4,4) in that statement, rather than '-'?
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] horizontal boxplot + xlim
True, I made a mistake here. Still, I have problems to visualize my data
(not
the example code I used). I just see a flat line instead a proper
plot...
Another example code with creating a strange plot:
x - rnorm(100) + 100
maxval - max(x)
boxplot(x, notch=TRUE, xlim = c(0,maxval), horizontal = TRUE)
When you use horizontal=TRUE, the meanings of the xlim and ylim arguments
to boxplot are swapped. Thus xlim affects the range of the vertical axis.
Take a look at
boxplot(x, notch=TRUE, ylim = c(0,maxval))
and
boxplot(x, notch=TRUE, ylim = c(0,maxval), horizontal = TRUE)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
