Hi,
Thanks a lot. I think this is what I want.
actuar package get more distributions.
Zero~
David Winsemius wrote:
On Sep 15, 2009, at 1:09 PM, TsubasaZero wrote:
Hi all,
I had generated 1000 random variates (u,v), and I would like to find
the
corresponding (x,y) for a
Hi all,
I want to round my values to the nearest 5.
For example, if I have a value of 11, I want R to round it to 10.
I have been searching for this command on the web and in the help file, but
I couldn't find anything useful.
Any help would be greatly appreciated.
Many thanks,
Chris
--
I have some security alert log data that I'm parsing and doing some stats on.
One of the fields is the Classtype which is the enumerated value of the
type of alert found.
classtypes = factor( alerts$Classtype )
fclass_types = table( classtypes )
fclass_types gives me a frequency table of the
The R.batch package does most of this. It's been several years, but I
used it run batch jobs on multiple (30-40) machines with different
OS:es but sharing the same file system. No communication needed
between hosts. It worked like a charm and I could process 3-4 CPU
years in a few weeks for
Hi all,
I have got a dataset like the following:
a b c
1 5
2
3 7
I am taking the minimum of each column and use it as the minimum of the
y-axis of my graphs. My scripts (simplified version) are like the following:
f-array
f[1]=a
f[2]=b
f[3]=c
for i in 1:3
name=f[i]
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal,
x5 - 5*round(x/5)
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Chris Li [chri...@austwaterenv.com.au]
Sent: 16 September 2009 09:15
To: r-help@r-project.org
Subject: [R] Rounding to the nearest 5
Hi all,
I want to
Let's say your data.frame is called xx
Then try:
apply(xx,2, function(x) {min(na.omit(x))} )
Does it work ?
On Wed, Sep 16, 2009 at 2:24 AM, Chris Li chri...@austwaterenv.com.auwrote:
Hi all,
I have got a dataset like the following:
a b c
1 5
2
3 7
I am taking the
On Tue, 15-Sep-2009 at 02:44PM -0700, William Dunlap wrote:
| perl can do more complicated processing and filtering than grep.
I once used perl to filter the useful 4 or 5 Mb out of a file that was
over 250Mb. It took about 3 lines of perl code and about 40 seconds
to run. Perl's not exactly
Hi,
I have a question to post. But I have a problem finding the appropriate
forum. Some of the FAQ on pastecs are answered by
r-h...@stat.math.ethz.chmailing list. I searched to see if this
mailing list requires registration
or not. There was no information about that. I am a registered member of
Hi All,
I need to check whether the file is empty or not, Please help me
in this regards.
Best regards,
Deepak.M.R
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PLEASE do read the posting guide
deepak m r wrote:
Hi All,
I need to check whether the file is empty or not, Please help me
in this regards.
Best regards,
Deepak.M.R
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Hello everybody,
it may be better to have sample data. I have provided data with less
levels of gene and day and only ca. 400 data points per condition.
Sample code:
small=as.data.frame(read.csv(small.csv))
small$species=factor(small$species)
small$gene=factor(small$gene)
Hi,
I'm trying to run the JGR package (in windows). I have followed all the steps
in the web page http://www.rosuda.org/JGR but when run jgr.exe it returns the
error message: Cannot find R.exe - please make sure R version 2.3.0 or higher
is installed. I have in my computer R version 2.9.0
Presumably there is no reason to attempt plotting an empty column, so best
approach is probably to remove the columns that contain no values.
Try
f.temp-f[,-which(apply(f,2,function(x)all(is.na(x]
and then run your script for f.temp instead of the original f.
Also, you may find you
Hi
I have a list which cosists out of dataframes of the same structure. Now I
want to extract one column (let's say column A) from all dataframes and
have them in a matrix (or dataframe).
At the moment I am doiong:
d - data.frame(A=runif(100), B=rnorm(100))
theList - list(e1=d, e2=d, e3=d,
Try this:
sapply(theList, '[[', 'A')
On Wed, Sep 16, 2009 at 8:34 AM, Rainer M Krug r.m.k...@gmail.com wrote:
Hi
I have a list which cosists out of dataframes of the same structure. Now I
want to extract one column (let's say column A) from all dataframes and
have them in a matrix (or
On Wed, Sep 16, 2009 at 07:11:46AM +0200, Schalk Heunis wrote:
This might be of help, first applies the formatting:print(xtable(prettyNum(d,
decimal.mark=,)))
Thanks! Your solution works for the example that I gave.
However, I failed to provide an example that really represent my
problem
sselamat wrote:
Hi,
Everytime I try to get the standard deviation of a vector and enter
command of sd(x), i get a reply
Error: could not find function sd
Can someone help me please?
Thanks.
Hmmm. Please provide more information, it's hard to imagine how this could
happen.
Perhaps with the dcolumn package ?
http://newton.ex.ac.uk/research/qsystems/people/latham/LaTeX/dcolumn.pdf
David
2009/9/16 Jakson A. Aquino jaksonaqu...@gmail.com
On Wed, Sep 16, 2009 at 07:11:46AM +0200, Schalk Heunis wrote:
This might be of help, first applies the
cls59 chuck at sharpsteen.net writes:
wesley mathew wrote:
Hello All
I have some problem for installing XML_2.6-0.tar . I am working in widows
and R version is R-2.9.1
Unfortunately, I think there are some problems with CRAN being able to build
the XML package for
Try this also;
format(coef(summary(lm.D9)), decimal.mark = ',')
or using gsub:
apply(coef(summary(lm.D9)), 2, gsub, pattern = \\., replacement = ,)
using lm.D9 object from ?lm example.
On Wed, Sep 16, 2009 at 8:50 AM, Jakson A. Aquino
jaksonaqu...@gmail.com wrote:
On Wed, Sep 16, 2009 at
On Wed, Sep 16, 2009 at 1:35 PM, Henrique Dallazuanna www...@gmail.comwrote:
Try this:
sapply(theList, '[[', 'A')
Thanks Henrique - that is much more elegant.
Rainer
On Wed, Sep 16, 2009 at 8:34 AM, Rainer M Krug r.m.k...@gmail.com wrote:
Hi
I have a list which cosists out of
Try this:
as.matrix(table(x))
On Tue, Sep 15, 2009 at 6:50 PM, esawdust lan...@360vl.com wrote:
I have some security alert log data that I'm parsing and doing some stats on.
One of the fields is the Classtype which is the enumerated value of the
type of alert found.
classtypes = factor(
Good day all, I'm trying to replicate Excel's growth trend series
interpolation, available by highlighting a number of cells and then
under Edit--Fill--Series--Type=growth and select 'Trend'.
For example, if my starting point is 1.4 and end point is 30 with the
sequence length being 11, the
On Tue, Sep 15, 2009 at 9:05 AM, Inez Campbell i...@st-andrews.ac.uk wrote:
Hi,
I am trying to find in FAQ, in the manual, everywhere, the reason and most
importantly, how to solve:
Error: protect(): protection stack overflow
It appeared after requesting a plot.
That error message comes from
exp(seq(log(1.4), log(30), length=11))
[1] 1.40 1.902074 2.584203 3.510961 4.770076 6.480741
8.804891 11.962537 16.252591 22.081162
[11] 30.00
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PLEASE
Wow :). Thanks a ton.
On Wed, Sep 16, 2009 at 2:31 PM, Richard M. Heiberger r...@temple.edu wrote:
exp(seq(log(1.4), log(30), length=11))
[1] 1.40 1.902074 2.584203 3.510961 4.770076 6.480741 8.804891
11.962537 16.252591 22.081162
[11] 30.00
On Tue, Sep 15, 2009 at 10:34 AM, Peng Yu pengyu...@gmail.com wrote:
Hi,
I run the following commands. 'A' has 3 levels and 'B' has 4 levels.
Should there be totally 3+4 = 7 coefficients (A1, A2, A3, B1, B2, B3,
B4)?
If you were to define a model matrix for those coefficients it would
be
The other package that I can think of that you might want to
investigate if you are attempting to construct bivariate distributions
is the copula package.
--
David.
On Sep 15, 2009, at 10:27 PM, TsubasaZero wrote:
Hi,
Thanks a lot. I think this is what I want.
actuar package get more
Ben Bolker wrote:
sselamat wrote:
Hi,
Everytime I try to get the standard deviation of a vector and enter
command of sd(x), i get a reply
Error: could not find function sd
Can someone help me please?
Thanks.
Hmmm. Please provide more information, it's hard to imagine how this could
Michael Bibo wrote:
cls59 chuck at sharpsteen.net writes:
wesley mathew wrote:
Hello All
I have some problem for installing XML_2.6-0.tar . I am working in widows
and R version is R-2.9.1
Unfortunately, I think there are some problems with CRAN being able to build
the XML package for
On Wed, Sep 16, 2009 at 01:59:29PM +0200, David Hajage wrote:
Perhaps with the dcolumn package ?
http://newton.ex.ac.uk/research/qsystems/people/latham/LaTeX/dcolumn.pdf
Thanks for the suggestion, but the problem isn't of alignment.
When I have commas as decimal separators, the values are
ogbos okike wrote:
Hi,
I have a question to post. But I have a problem finding the appropriate
forum. Some of the FAQ on pastecs are answered by
r-h...@stat.math.ethz.chmailing list. I searched to see if this
mailing list requires registration
or not. There was no information about that. I am
Hi,
Is there is any R package or existing codes in R to detect cycles in a
graphical model or DAG (Directed Acyclic Graph) ?
Thanks.
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
But, it seems that dcolumn can change the decimal separator too (see the
table on the first page of the pdf document).
For example:
\documentclass{article}
\usepackage{dcolumn}
\newcolumntype{d}[1]{D{.}{,}{#1}}
\begin{document}
results=tex=
x - matrix(rnorm(4), 2, 2)
library(xtable)
Hi all,
I have following ch. matrix :
mat
[,1][,2] [,3][,4]
[1,] NA0.0671746073115122 1.15281464953731
0.822277316923348
[2,] 0.113184828073156 -0.0133150789005112 0.640912657072027
-0.0667317787225847
On 9/16/2009 9:26 AM, shuva gupta wrote:
Hi,
Is there is any R package or existing codes in R to detect cycles in a
graphical model or DAG (Directed Acyclic Graph) ?
Thanks.
You might try this to search R packages:
library(sos)
findFn('Directed Acyclic Graph')
[[alternative HTML
Hi all,
I have thousands of strings like these ones:
1159_1; YP_177963; PPE FAMILY PROTEIN
1100_13; SECRETED L-ALANINE DEHYDROGENASE ALD CAA15575
1141_24; gi;2894249;emb;CAA17111.1; PROBABLE ISOCITRATE DEHYDROGENASE
and various others..
I'm interested to extract the code for the
On Sep 16, 2009, at 9:40 AM, Bogaso wrote:
Hi all,
I have following ch. matrix :
mat
[,1][,2] [,3][,4]
[1,] NA0.0671746073115122 1.15281464953731
0.822277316923348
[2,] 0.113184828073156 -0.0133150789005112
Ah whoops (and thank you). I forgot to mention an additional
constraint: I'm trying to use exactly one eval statement. Some
background information on this constraint. Some of the expressions
that are being evaluated may return an error. At the moment, I'm
wrapping each of the individual
I'd like to have something like seq() where I can pass in a length of the
desired sequence and a right limit so that the sequence goes up to the limit and
then starts again from 1.
# works now
seq(from=2, length.out=3)
[1] 2 3 4
# what I want
seq(from=2, length.out=3, rlimit=3)
[1] 2 3 1
#
Hi all,
Given the following series of numbers:
t
[1] 21.85000 30.90410 43.71000 61.82234 87.43999 123.67296
[7] 174.91997 247.40249 349.91996 494.91815 700.0
What's the simplest way of formatting them into the nearest natural
looking (rounded to nearest multiple of 10) numbers?
So:
On 16-Sep-09 13:40:03, Bogaso wrote:
Hi all,
I have following ch. matrix :
mat
[,1][,2] [,3][,4]
[1,] NA0.0671746073115122 1.15281464953731
0.822277316923348
[2,] 0.113184828073156 -0.0133150789005112
Try this:
library(gsubfn)
strapply(x, [A-Z]{3}[0-9]+)
On Wed, Sep 16, 2009 at 10:53 AM, Giulio Di Giovanni
perimessagg...@hotmail.com wrote:
Hi all,
I have thousands of strings like these ones:
1159_1; YP_177963; PPE FAMILY PROTEIN
1100_13; SECRETED L-ALANINE DEHYDROGENASE ALD
round(t, -1)
Uwe Ligges
Jorgy Porgee wrote:
Hi all,
Given the following series of numbers:
t
[1] 21.85000 30.90410 43.71000 61.82234 87.43999 123.67296
[7] 174.91997 247.40249 349.91996 494.91815 700.0
What's the simplest way of formatting them into the nearest natural
looking
This should do it for you:
pat - .*(\\b[A-Z]..[0-9]+).*
grep(pat, x)
[1] 1 3 5
sub(pat, '\\1', x)
[1] YP_177963 CAA15575CAA17111
On Wed, Sep 16, 2009 at 9:53 AM, Giulio Di Giovanni
perimessagg...@hotmail.com wrote:
Hi all,
I have thousands of strings like these
Hi
I'm running 32-bit R on Windows XP 64bit and the machine has 16Gb of RAM. The
help for memory.limit states:
If 32-bit R is run on some 64-bit versions of Windows the maximum value of
obtainable memory is just under 4GB.
So, using the help which states the size parameter can go up to 4095:
Try rep:
rep(2:4, lenght.out = 3, times = 10)
On Wed, Sep 16, 2009 at 11:08 AM, Jack Tanner i...@hotmail.com wrote:
I'd like to have something like seq() where I can pass in a length of the
desired sequence and a right limit so that the sequence goes up to the limit
and
then starts again
I'm was guessing that the .1 was a part of the protein code for
third example and looking at:
http://www.plosone.org/article/fetchSingleRepresentation.action?uri=info:doi/10.1371/journal.pone.0003840.s002
I see quite a few protein codes of that form. I am a complete
ignoramus with regex
On 9/16/2009 10:16 AM, michael watson (IAH-C) wrote:
Hi
I'm running 32-bit R on Windows XP 64bit and the machine has 16Gb of RAM. The
help for memory.limit states:
If 32-bit R is run on some 64-bit versions of Windows the maximum value of
obtainable memory is just under 4GB.
So, using the
Assuming the rule is an upper case alphabetic character followed by
two other characters followed by a string of digits then try this:
library(gsubfn)
strapply(x, [A-Z][^ ][^ ][0-9]+)
[[1]]
[1] YP_177963
[[2]]
[1] CAA15575
[[3]]
[1] CAA17111
If you prefer the output as one long vector of
That did not work on all three:
strapply(x, [A-Z]{3}[0-9]+)
[[1]]
NULL
[[2]]
[1] CAA15575
[[3]]
[1] CAA17111
But adding a _ to the initiation pattern and a period to the
termination pattern makes it complete:
library(gsubfn)
strapply(x, [A-Z_]{3}[0-9.]+)
[[1]]
[1] YP_177963
[[2]]
[1]
That's right, if the test is exact it is not possible to get a p-value of zero.
wilcox.test does not provide an exact p-value in the presence of ties so if
there are any ties in your data you are getting a normal approximation.
Incidentally, if there are any ties in your data set I would
Hi Duncan
Many thanks for the reply. Here are the results of the commands:
Rprofmem()
Error in Rprofmem() : memory profiling is not available on this system
memory.size(NA)
[1] 3500
memory.size()
[1] 16.19
Clearly Rprofmem() is a dead end, though if there is some way of enabling it, I
Hi all,
I'm trying to log chart but with natural looking tick marks. My
specifications are very specific -- it must indicate the lowest
number's tick as well as the maximum.
I've attached sample code and data for a particular case (and there
are a few more like this) where the bottom tickmarks
On Sep 16, 2009, at 11:13 AM, Jorgy Porgee wrote:
Hi all,
I'm trying to log chart but with natural looking tick marks. My
specifications are very specific -- it must indicate the lowest
number's tick as well as the maximum.
I've attached sample code and data for a particular case (and there
Hi David,
Sorry, forgot to define pxTarget. It can be anything you choose so say
pxTarget-1. I've also renamed sample to sampleData. The
following cut and paste should work:
The last tick mark (5000) is not showing for some reason... :-/
Regards,
George
sampleData-c(NA, NA, NA, NA, NA,
It shows if you lower your minimum a bit by applying a factor of 0.90
to your generation of dataMin.
--
David.
On Sep 16, 2009, at 11:27 AM, Jorgy Porgee wrote:
Hi David,
Sorry, forgot to define pxTarget. It can be anything you choose so say
pxTarget-1. I've also renamed sample to
Hi,
I have two somewhat embarassing questions about the lattice-related
plot functions:
1.) How do I make lattice (e.g. barchart) to not draw a box but only a
y-axis on the left hand side so that the plot looks like barplot with
default settings? So that the following two code snippets look more
Perfect, that works well. Thank you for the suggestion.
Henrique Dallazuanna wrote:
Try this:
as.matrix(table(x))
--
View this message in context:
http://www.nabble.com/best-method-to-format-output-of-frequency-table-tp25462448p25472239.html
Sent from the R help mailing list
Henrique Dallazuanna wwwhsd at gmail.com writes:
Try rep:
rep(2:4, length.out = 3, times = 10)
That's close, but it doesn't wrap to start at 1.
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On 2009.09.16. 16:08, Jack Tanner wrote:
I'd like to have something like seq() where I can pass in a length of the
desired sequence and a right limit so that the sequence goes up to the limit and
then starts again from 1.
Disclaimer: total R beginner here (started to learn a 1 day ago), who
On 9/16/2009 10:54 AM, michael watson (IAH-C) wrote:
Hi Duncan
Many thanks for the reply. Here are the results of the commands:
Rprofmem()
Error in Rprofmem() : memory profiling is not available on this system
memory.size(NA)
[1] 3500
memory.size()
[1] 16.19
Clearly Rprofmem() is a
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jack Tanner
Sent: Wednesday, September 16, 2009 7:08 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] a sequence that wraps around
I'd like to have something like seq() where I
Fair enough, thanks for your time.
I will wait for my Linux PCs!
-Original Message-
From: Duncan Murdoch [mailto:murd...@stats.uwo.ca]
Sent: 16 September 2009 16:42
To: michael watson (IAH-C)
Cc: r-help@r-project.org
Subject: Re: [R] Memory in R on windows running 64bit XP
On 9/16/2009
On 9/16/2009 10:08 AM, Jack Tanner wrote:
I'd like to have something like seq() where I can pass in a length of the
desired sequence and a right limit so that the sequence goes up to the limit and
then starts again from 1.
# works now
seq(from=2, length.out=3)
[1] 2 3 4
# what I want
Once one gets past the issue of the p value being extremely small,
irrespective of the test being used, the OP has asked the question of
how to report it.
Most communities will have standards for how to report p values,
covering things like how many significant digits and a minimum p value
/Dear all,///
/I am very thankful, if you could tell what is the right way to write:
mtext(paste(expression(R^2),round(marco2[1,i],digits=3), N° of
proteins:,marco3[i]),side=4,cex=.6)
in this case the output is:
R^2
I tried also in this way:
Szabolcs Horvát szhorvat at gmail.com writes:
You could use the modulo operator.
# additional examples of what I want
seq(from=2, length.out=4, rlimit=3)
[1] 2 3 1 2
seq(from=1, length.out=4) %% 3 + 1
Ah, that's so slick. You're off to a great start! Huge thanks to you and
everyone
Try this,
marco2 = matrix(rnorm(4), nrow=2, ncol=2)
marco3 = 2.12
i =1
plot.new()
mtext(bquote(R^2*.(round(marco2[1,i],digits=3))* N° of
proteins:*.(marco3[i])),side=4,cex=.6)
HTH,
baptiste
2009/9/16 Marco Chiapello marpe...@gmail.com
/Dear all,///
/I am very thankful, if you could tell
Thanks Simon,
I wasn't aware that predict.gam could be used in this situation. I
followed you advice and managed to extract the data I needed.
For people interested, I add the code with comments I used here:
#
# Full code for extracting partial
It is nice to see worked examples, but there were some errors that
relate to not including dataframe references. I've paste in code that
runs without error on my machine:
On Sep 16, 2009, at 12:39 PM, Staffan Roos wrote:
Thanks Simon,
I wasn't aware that predict.gam could be used in this
But removing the extraneous second to last line that says just:
lines
... would leave the console looking less puzzling.
--
David
On Sep 16, 2009, at 1:02 PM, David Winsemius wrote:
It is nice to see worked examples, but there were some errors that
relate to not including dataframe
On Wed, Sep 16, 2009 at 09:14:39AM -0300, Henrique Dallazuanna wrote:
Try this also;
format(coef(summary(lm.D9)), decimal.mark = ',')
or using gsub:
apply(coef(summary(lm.D9)), 2, gsub, pattern = \\., replacement = ,)
using lm.D9 object from ?lm example.
Thanks for your suggestion!
I wrote a script that I anticipating seeing a spike at 10Hz with the
function 10* sin(2*pi*10*t).
I can't figure out why my plots do not show spikes at the frequencies I
expect. Am I doing something wrong or is my expectations wrong?
require(stats)
layout(matrix(c(1,2,3), 3, 1, byrow =
On Wed, Sep 16, 2009 at 03:29:44PM +0200, David Hajage wrote:
But, it seems that dcolumn can change the decimal separator too (see the
table on the first page of the pdf document).
For example:
\documentclass{article}
\usepackage{dcolumn}
\newcolumntype{d}[1]{D{.}{,}{#1}}
Hi David,
Yes, the strange code lines was clearly a mistake from my side. Sorry.
What dataframe references did you add in your code?
/Staffan
David Winsemius wrote:
But removing the extraneous second to last line that says just:
lines
... would leave the console looking less
why not try spectrum (power spectrum)
spectrum(y, log=no)
the major spike is at 0.1 and 1/0.1 = 10Hz
On Wed, Sep 16, 2009 at 12:13 PM, delic kran...@optonline.net wrote:
I wrote a script that I anticipating seeing a spike at 10Hz with the
function 10* sin(2*pi*10*t).
I can't figure out why
Good evening to everybody. I have a data of four columns - three of the
columns represent date while the fourth is counts. Setting 4000 as my
minimum point/value, I wish to find out the number of counts that are less
or equal to 4000 and the dates they occur. I have installed pastecs and have
Your code minus the extraneous lines was:
#plot the data
plot(pp_s.x0.~x0,type=p,pch=2,cex=0.50)
sequence=order(x0)
lines(x0[sequence],pp_s.x0.[sequence])
rug(jitter(x0))
My edition was:
plot(pp_s.x0.~x0,type=p,pch=2,cex=0.50, data=f)
sequence=order(f$x0)
After extensive research I found some information hinting at my freq array
being wrong. I thought that the data spanned -F/2 to F/2. It seems that I
should only plot the first half of the fft vector? Can anyone confirm or
shed some light on this?
One other thing I thought the magnitude should be
Dear all,
I have partial data set with four colums. First column is site with three
factors (i.e., A, B, and C). From second to fourth columns (v1 ~ v3) are my
observations. In the observations of the data set, . indicates missing
value. I replaced . with NA. To replace . with NA, I used two
Yes, you're correct David, I used attach(f) earlier. Thanks for clarifying! I
should change my own code accordingly.
/Staffan
David Winsemius wrote:
Your code minus the extraneous lines was:
#plot the data
plot(pp_s.x0.~x0,type=p,pch=2,cex=0.50)
sequence=order(x0)
Hi,
On Sep 16, 2009, at 1:42 PM, ogbos okike wrote:
Good evening to everybody. I have a data of four columns - three of
the
columns represent date while the fourth is counts. Setting 4000 as my
minimum point/value, I wish to find out the number of counts that
are less
or equal to 4000 and
Hi Steve,
Here is a suggestion using your original df1:
# Create a copy -- you can avoid this
newdf1 - df1
# Process
newdf1[,2:4] - apply(newdf1[,2:4], 2, function(x) as.numeric(x))
# Removing df1
rm(df1)
# Result
newdf1
# str()
str(newdf1)
# 'data.frame': 18 obs. of 4 variables:
# $
Hi
I am using kernlab package and kpca function for dimensionality reduction of my
data. Can any body tell me that how can I get best three dimensions from output
of kpca function, which are principal component vector, eignvalues, rotated
vector, and original matrix. I want to get back my
There were some wrong NA values in the provided data set, this is now
corrected.
The data can be read in as
small=read.csv(small.csv,colClasses=c(character,rep(integer,2),rep(factor,5)))
The high number of residual df can be seen using the nlme package (can
it be seen in the lme4 package, too
Try this:
sapply(x, format, decimal.mark = ',')
On Wed, Sep 16, 2009 at 2:06 PM, Jakson A. Aquino
jaksonaqu...@gmail.com wrote:
On Wed, Sep 16, 2009 at 09:14:39AM -0300, Henrique Dallazuanna wrote:
Try this also;
format(coef(summary(lm.D9)), decimal.mark = ',')
or using gsub:
On Sep 16, 2009, at 1:42 PM, ogbos okike wrote:
Good evening to everybody. I have a data of four columns - three of
the
columns represent date while the fourth is counts. Setting 4000 as my
minimum point/value, I wish to find out the number of counts that
are less
or equal to 4000
You
On Wed, Sep 16, 2009 at 03:16:12PM -0300, Henrique Dallazuanna wrote:
Try this:
sapply(x, format, decimal.mark = ',')
Yes, this works as I need, regarding the replacement of dots with
commas.
Thanks!
On Wed, Sep 16, 2009 at 2:06 PM, Jakson A. Aquino
jaksonaqu...@gmail.com wrote:
On
Hi,
I have the precision values of a system on two different data sets.
The snippets of these results are as shown:
sample1: (total 194 samples)
0.600238
0.800119
0.600238
0.200030
0.600238
...
...
sample2: (total 188 samples)
0.8001
0.2000
0.8001
0.
Robert,
We unfortunately do not have enough information to help you interpret the
results, and this is not really an R question at all, but general statistical
advice. You will probably have much better understanding and confidence in
your results by consulting a local statistical consultant
Hi,
I was just going to send this when I saw Erik's post. He's right -- we
can't say anything about your data, but we can say something about
using a t-test.
I'm not a real statistician, so this answer isn't very rigorous, but
might be helpful.
On Sep 16, 2009, at 2:55 PM, Robert Hall
Greetings,
I am attempting to run a function, which produces a vector and
requires two input variables, across two nested factor levels. I can
do this using by(X, list(factor1, factor2), function), however I
haven't found a simple way to extract the list output into an
organized vector
Hello,
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Jon Loehrke
Sent: Wednesday, September 16, 2009 2:23 PM
To: r-help@r-project.org
Subject: [R] apply function across two variables by mult factors
Greetings,
I am
One correction below,
---snip---
# example data frame
testDF-data.frame(
x=rnorm(12),
y=rnorm(12),
f1=gl(3,4),
f2=gl(2,2,12))
Try this using lapply, split, mapply? Maybe it is in a nicer output
object for you?
testFun2 - function(x, y) {
X - abs(x);
Hi Jon,
Here is a suggestion:
foo - function(x) as.numeric( paste( abs( round( x ) ), collapse = . ) )
testDF$value - apply( testDF[,1:2], 1, foo )
testDF
HTH,
Jorge
On Wed, Sep 16, 2009 at 3:22 PM, Jon Loehrke jloeh...@umassd.edu wrote:
Greetings,
I am attempting to run a function, which
Try this:
transform(testDF, value = as.numeric(paste(round(abs(x)),
round(abs(y)), sep = .)))
On Wed, Sep 16, 2009 at 3:22 PM, Jon Loehrke jloeh...@umassd.edu wrote:
Greetings,
I am attempting to run a function, which produces a vector and
requires two input variables, across two nested
I am loathe to expound basic statistics here ... but, at the considerable
risk of pedantry, I must note that Steve's reply below contains fundamental
errors, which I feel should not be left on this list unremarked: t-tests do
**not** test for differences in **sample** means; they test for
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