Good Day All,
I am working on some xyplots using the Lattice Library. My X-axis is
the date and I am reproducing charts similar to those found in the R
Gallery (see here:
http://www.sr.bham.ac.uk/~ajrs/R/gallery/plot_midday_weather_profiles.txt)
However, the key difference is that some of
On 15.11.2011 21:34, Scott Raynaud wrote:
OK, I think I see the problem. Rather than setting method=nAGQ I need
nAGQ=1. Doing so throws the following error:
Congratulations, now you understood what R meant with its message
Argument ‘method’ is deprecated.
Warning messages:
1: glm.fit:
Looking at the help pages in the ineq package, the Theil() function
simply returns the value of the index. You can look at the source code
of the relevant functions to see what they actually compute:
library('ineq')
ineq # the main function
Theil# for the Theil index
Neither function is
On 11/16/2011 03:51 PM, Vinny Moriarty wrote:
Hello All,
Many thanks to the help I have received so far.
Here is an example data set I hope to plot
Data1
Year Data SE
1 20052 0.01
2 20064 0.01
3 20075 0.01
4 20082 0.01
5 20093 0.01
6 20106 0.01
Data2
Year
Dear all,
I was hoping someone could help with a ggplot question. I would like
to generate a faceted bar chart, but missing data are causing
problems.
g-structure(list(Date = structure(c(11322, 11687, 12052, 11322,
11687, 12052, 11322, 11687, 12052, 11322, 11687, 12052), class = Date),
Thanks a lot for your answers and reading suggestions, now I know my guess
was completely wrong.
I guess in my case it will be more informative to keep the unordered
factors. That way I can know not only that days differ in general, but also
get information on which day is differing from day 1.
Thanks to everyone who contributed to my questions. As ever, I am extremely
grateful to all those on the R-list who make it what it is.
Regards
Mike Griffiths
On Tue, Nov 15, 2011 at 5:47 PM, Joshua Wiley jwiley.ps...@gmail.comwrote:
Hi Michael,
Your strings were long so I made a bit
Hi:
Here's one way, but it puts the two countries side by side rather than
stacked (I'm not a big fan of stacked bar charts except in certain
contexts). The first version uses the original data, but one can see
immediately that there is no distinction between NA and 0:
ggplot(g, aes(x = Date, y
Try the as.dist() function -- it won't change any values, only give
you something that, upon printing, gets formatted like a dist object.
as.matrix() will reverse it.
Michael
On Wed, Nov 16, 2011 at 5:58 AM, ram basnet basnet...@yahoo.com wrote:
Dear Michael,
Thank you very much.
That what i
Hello again... sorry to be posting yet again, but I hadn't anticipated this
problem.
I am trying to now put the names found in one column in data frame 1 (lets
call it df.1[,1]) in to a list from the rows where the values in df.1[,2]
match values in a column of another dataframe (df.2[3])
I
Might I suggest you read the upstream postings in this thread you seem to have
brought back from beyond the grave and check those resources for methodological
questions (particularly CV). Once you decide what you want to do, get in touch
with SPSS support and ask them how to do it. When that
Hi,
I have a hierachical code system such as the example below (the printed data
are easiest to read). I would like to write a function that returns an
'imputed' data frame, ie. where the the parent values are calculated as the sum
of the child values. So, for instance, STAT.01.01.06 is the
Well, I could increase the sample size for my second level in hopes that my
simulation would run correctly. However, a better solution would be to assign
values of 0 to the fixed effects for this pass through the simulation. I'm
such a novice with R that I don't know if that can be done.
I'm not at a computer now, so I can't take a close look at it, but I think the
match() function can be helpful here.
I'll try to get back to you with a fuller answer later.
Michael
On Nov 16, 2011, at 8:03 AM, Rob Griffin robgriffin...@hotmail.com wrote:
Hello again... sorry to be posting
Dear Albert-Jan,
The easiest way is to create extra variables with the corresponding aggregation
level. substr() en strsplit() can be your friends. Once you have those
variables you can use aggregate() or any other aggregating function. You don't
need loops.
Best regards,
Thierry
Dear List,
I have data on a approximately 100 individuals visiting a a central logging
station over a 1000 times. I would like to be able to calculate the
distribution of inter-visit time intervals for all possible pairs am stuck
on how to code for this. Single pairs are not a problem but
On Nov 16, 2011, at 8:03 AM, Rob Griffin wrote:
Hello again... sorry to be posting yet again, but I hadn't
anticipated this problem.
I am trying to now put the names found in one column in data frame 1
(lets call it df.1[,1]) in to a list from the rows where the values
in df.1[,2] match
Hi Dennis,
that does exactly what I needed, and the treatment of missing values
is really useful.
thanks for your help.
Aidan
On Wed, Nov 16, 2011 at 12:22 PM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
Here's one way, but it puts the two countries side by side rather than
stacked (I'm not
Hi there,
I am doing some network analysis working with k-cliques and over time I
want to see what nodes are members of what cliques and how big these
cliques are. I have managed to produce a matrix which shows which k-cliques
each node is part of over the 100 time periods (slow though) but I
Hi all,
I need to check for a difference between treatment groups in the
parameter of the geometric distribution, but with a cut-off (i.e. right
censored). In my experiment I stimulated animals to see whether I got a
response, and stopped stimulating if the animal responded OR if I had
Hello, list,
I'm new to R and I'm trying to produce a chart with currency values on the y
axis.
Values should be e.g. 1,00, 1,50, 2,00, etc.
In fact they are 1,0, 1,5, 2,0, etc.
How do I get R to show two digits after the comma on that axis?
Thanks for any help,
Mario
[[alternative
Dear Researches,
I am using RF (in regression way) for analize several metrics extract from
image. I am tuning RF setting a loop using different range of mtry, tree
and nodesize using the lower value of MSE-OOB
mtry from 1 to 5
nodesize from1 to 10
tree from 1 to 500
using this paper as refery
I have a data with the sequence of events with millions records and more than
24 time stamped variables.
sample data:
1 2 3 4 5 6 7 8 9 10
A A A C C C B B D D
D D D D
Hi list!
I'm getting an error message when trying to fit an accelerated failure
time parametric model using the aftreg() function from package eha:
Error in optim(beta, Fmin, method = BFGS, control = list(trace =
as.integer(printlevel)), :
non-finite finite-difference value [2]
This only
Dear R Users, I am curious whether there is any simple solution o my problem.
My example 'series':
series - c(1,1,1,0,1,0,0,1,1,0,0,0,0,1)
My 'subset' of interest:
subset - c(0,0,0,0)
Is there any function which tells me that the subset exists in my series,
and gives me position where in series
Thanks Josh
--
View this message in context:
http://r.789695.n4.nabble.com/fitted-values-in-lm-function-tp4075426p4076397.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
I have a huge data set in the form of
V1 V2 V3 V4 V5 V6
1 2010050120100501 1.68291.38 1 0
2 20100501201005010300 0.93335.10 1 0
3 20100501201005010600 2.25 57.38 1 0
4 20100501201005010900 0.43 13.76 1 0
5
Hi
I have a data frame and i need to change the date format in it.
my dataframe
X Date
1 1/1/2009
2 2/1/2009
3 3/1/2009
I need to change it to 2009-01-01
--
View this message in context:
http://r.789695.n4.nabble.com/changing-date-format-in-a-dataframe-tp4076411p4076411.html
Sent from the R
Hello John,
Thank you. It is useful advice. I found the method which eliminates
copying the library after new installation given at by Tal at
http://www.r-statistics.com/2010/04/changing-your-r-upgrading-strategy-and-t
he-r-code-to-do-it-on-windows/ very effective. But my problem and
On Nov 16, 2011, at 7:41 AM, Mario Giesel wrote:
Hello, list,
I'm new to R and I'm trying to produce a chart with currency values
on the y axis.
Values should be e.g. 1,00, 1,50, 2,00, etc.
In fact they are 1,0, 1,5, 2,0, etc.
How do I get R to show two digits after the comma on that
Searching: time series forecasting
on the R website gives quite a few potentially useful options.
hth, Ingmar
On Wed, Nov 16, 2011 at 1:55 PM, Aher ajit.a...@cedar-consulting.comwrote:
I have a data with the sequence of events with millions records and more
than
24 time stamped variables.
In addition to getting help from others, I always find that seeking answers
through reading to be helpful. I would suggest a basic stats book and this
paper regarding issues indices like species richness:
http://www.amazon.com/Primer-Ecological-Statistics-Nicholas-Gotelli/dp/0878932690
Here
On Nov 16, 2011, at 8:56 AM, abcdef ghijk wrote:
I have a huge data set in the form of
V1 V2 V3 V4 V5 V6
1 2010050120100501 1.68291.38 1 0
2 20100501201005010300 0.93335.10 1 0
3 20100501201005010600 2.25 57.38 1 0
4
On Nov 16, 2011, at 8:25 AM, threshold wrote:
Dear R Users, I am curious whether there is any simple solution o my
problem.
My example 'series':
series - c(1,1,1,0,1,0,0,1,1,0,0,0,0,1)
My 'subset' of interest:
subset - c(0,0,0,0)
Is there any function which tells me that the subset exists
Hi:
I think you're overthinking this problem. As is usually the case in R,
a vectorized solution is clearer and provides more easily understood
code.
It's not obvious to me exactly what you want, so we'll try a couple of
variations on the same idea. Equality of floating point numbers is a
If your data is also binary like that below, it might be useful/easier
to use regular expressions tools as well:
e.g.,
series - c(1,1,1,0,1,0,0,1,1,0,0,0,0,1)
subset - c(0,0,0,0)
print(regexpr(paste(subset, collapse = ),paste(series, collapse = )))
Michael
On Wed, Nov 16, 2011 at 9:55 AM,
All right. I upped my level 2 sample size to 60. My log displays the
following:
Simulation for sample sizes of 60 macro and unbalanced micro
units
Iteration remain= 990
Iteration remain= 980
There were 27 warnings (use warnings() to see them)
Error in
Your question seems rather poorly specified to me, and you may have to
clarify it further, but I'll make a stab at it.
IF one can assume that:
1. a must occur before b, b before c etc. AND
2. sec is always increasing
THEN
z - tapply(sec, ind, FUN = min)
gives the first appearance sec value
Can you dput() your data frame? There are a different few time objects
in R and details might depend which one you have. Though, if it's
printing like that, I'd guess it's actually a string that you can
convert to a Date using as.Date(, format = XXX) where you can work out
the formatting from the
Hi,
If the dates are the same format and the same length you can try this :
dates-c(1/1/2009,2/1/2009,3/1/2009)
dates_new-as.Date(paste(substr(dates,1,3),substr(dates,7,8),sep=/),format=%d/%m/%y),
you can change the format to %m/%d/%y
Regards
M
Le 16/11/11 15:12, arunkumar a écrit :
Ok, thanks for looking in to this so far, I seem to have confused you all a
little though so I think I need to make this a bit clearer:
in the real situation:
df.1 is 271*13891, and contains (amongst others) columns with Flybase.CG,
rMF, and Affyid values.
df.2 is 14*12572 and is made from
Hi R community,
I have some data set and construct the likelihood as follows
likelihood - function(alpha,beta){
lh-1
d-0
p-0
k-NULL
data-read.table(epidemic.txt,header = TRUE)
attach(data, warn.conflicts = F)
k -which(inftime==1)
d -
On 16.11.2011 16:08, Scott Raynaud wrote:
All right. I upped my level 2 sample size to 60. My log displays the
following:
Simulation for sample sizes of 60 macro and unbalanced
micro units
Iteration remain= 990
Iteration remain= 980
There were 27 warnings (use
R -
Does anyone know of a cubic gradient descent package? I found grad.desc()
but that only allows for a 2d function. I have 3 free parameters and thus
am looking for a 3d function.
Thank you,
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota
Dear list,
I'd like to access the elements of a list within another list with the
help of a variable.
dict - list( 24 = c(1,2,3,6,12,24,48,72,96,120,144,168,720),
168 = c(1,12,24,48,72,96,120,144,168,336,504,672),
8760 = c(1,24,168,730,4380,8760)
)
dict$24 works
Hello.
I recently came across the extremely strange problem (see below) of R
telling me that elements of a list do not exist...
I have tried running this small script on 2 computers (Mac OSX 10.7
Windows 7) running the latest releases or R (2.13.2 on Mac, 2.14.0 on Win).
I have tried it both
Le mercredi 16 novembre 2011 à 06:12 -0800, arunkumar a écrit :
Hi
I have a data frame and i need to change the date format in it.
my dataframe
X Date
1 1/1/2009
2 2/1/2009
3 3/1/2009
I need to change it to 2009-01-01
See ?as.Date. In your case, I think you should use
df[[2]] -
I am scraping data from a web page using XML (excellent package BTW - that's
scraping data the easy way!).
So far, I've got the code:
tables - readHTMLTable(theurl)
rhf - tables$tabResHistFull
div1 - rhf[which(rhf$V1==Div ps),]
div1
which is giving me the result:
V1 V2 V3 V4 V5
Dear Community,
I want to identify outliers in my data. I don't know how to use identify
command in the plots obtained.
I've gone through help files and use mahalanobis example for my purpose:
NormalMultivarianteComparefunc - function(x) {
Sx - cov(x)
D2 - mahalanobis(x,
HI,
I have subsetted a list of exons and put them in a vector exon_list
exon_list=levels(test5$V1)
I want to loop using that vector to create a big pdf which will contain
all my barplots and doing this I got only the last element in my
exon_list plotted rather than all of them, I guess
Thanks for the answer. I found that smth like below works
subset - c(1,0,0)
series - c(1,1,1,0,1,0,0,1,1,0,0,0,0,1)
subset1 - paste(subset, collapse='')
series1 - paste(series, collapse='')
xx - gregexpr(subset1, series1, fixed=TRUE)
as.numeric(xx[[1]]) ## yields the first element numbers in
Hi Michael,
Thanks for your response. Using the binary seems to solve partially. I am
able to install (I think!) RCurl but not able to load the library. Below is
the info you required and the error while loading RCurl.
* sessionInfo()*
R version 2.13.2 (2011-09-30)
Platform:
Dear all,
I am aiming to calculate variograms using variogram() from gstat.
The problem is that some of my data-sets are very large ( 40 points).
Running the command takes some hours of time and does not give any
error-message.
Nevertheless the result seem not to be appropriate - the first
Are you looking for something like nlm()? It does minimization in any
number of dimensions using a Newton-type algorithm;
it can be manually given gradients / Hessians, or alternatively you can
give it neither and let them approximated empirically.
Taylor
On Wed, Nov 16, 2011 at 11:00 AM, Edward
Dear Michael,
Thank you very much.
Ram
From: R. Michael Weylandt michael.weyla...@gmail.com
Sent: Wednesday, November 16, 2011 1:48 PM
Subject: Re: [R] Conversion of symmetry matrix into a vector ONE MORE QUESTION
Try the as.dist() function -- it won't
That might be an option if it weren't my most important predictor. I'm
thinking my best bet is to use MLWin for the estimation since it will properly
set fixed effects
to 0. All my other sample size simulation programs use SAS PROC IML which I
don't have/can't afford. I like R since it's
R FAQ 7.31 strikes again.
:-)
Michael
tmp[3105] == 35.52
[1] FALSE
tmp[3105] - 35.52
[1] -7.105427e-15
all.equal(tmp[3105], 35.52)
TRUE
On Wed, Nov 16, 2011 at 10:42 AM, oscar ob...@cam.ac.uk wrote:
Hello.
I recently came across the extremely strange problem (see below) of R
telling me
dict[[a]]
? `[[`
Michael
On Wed, Nov 16, 2011 at 10:16 AM, Antje Gruner n...@allesjetzt.net wrote:
Dear list,
I'd like to access the elements of a list within another list with the
help of a variable.
dict - list( 24 = c(1,2,3,6,12,24,48,72,96,120,144,168,720),
168 =
On Nov 16, 2011, at 10:16 AM, Antje Gruner wrote:
Dear list,
I'd like to access the elements of a list within another list with the
help of a variable.
dict - list( 24 = c(1,2,3,6,12,24,48,72,96,120,144,168,720),
168 = c(1,12,24,48,72,96,120,144,168,336,504,672),
8760 =
Put pdf() before your loop and dev.off() after -- as it stands now, on
each iteration you re-open the pdf and thereby wipe everything that
was previously on it.
For the axis, look at the axis() command if you want something
detailed or the xaxt, xlab arguments to boxplot if its relatively
simple.
Hello,
After I plot something how do I reactivate the console (and not the plot
window) so I don't have to click on the console each time to go to the next
command?
Example that does not work:
fun = function(x){ plot(x); dev.set(dev.prev())}
fun(1:4)
...and another that does not work:
fun =
Can you provide a dput() example of your data and put it in a workable
form -- I have no idea what that long list of numbers nor the 1 / 0 on
the end represent. I think if you do this with a real time series
class like xts you can create a new object with the interpolated
times; fill in the values
Hi all,
I would like to plot one data set as a 2d histogram and another one as a
contour. I can do it separately with the
hist2d and contour functions, but I wonder if there is a way to combine
these two plots into a single one (the ranges of the two plots are the same).
Any suggestions?
??focus ## admittedly, not the first keyword that comes to mind
?bringToTop
-- Bert
On Wed, Nov 16, 2011 at 9:07 AM, Ben quant ccqu...@gmail.com wrote:
Hello,
After I plot something how do I reactivate the console (and not the plot
window) so I don't have to click on the console each time
This is a little ugly, but I think it should work pretty robustly:
T - table(unlist(df[,-1])) # Take a close look at this to see how it
works -- it's the key to the whole thing and basically creates
something we will roughly use like a hash-table (if that term is
familiar to you) (or more
Try the add = TRUE argument to contour.
Michael
On Wed, Nov 16, 2011 at 12:35 PM, Sramkova, Anna (IEE)
anna.sramk...@iee.unibe.ch wrote:
Hi all,
I would like to plot one data set as a 2d histogram and another one as a
contour. I can do it separately with the
hist2d and contour functions,
Happy to help, but could you provide a dput() example of your data
div1 -- I'm not sure if your table elements are coded as factors or
doubles and that can make quite a difference.
Anyways, once you confirm they are doubles, something like this will do it:
apply(D, 1, function(x) all(diff(x)0))
As another potential route could I put something in to the original code
that makes df.2 (maindata2) which picks one of the AffyIds at random for the
duplicated FlybaseCG values (shown below)
maindata2-aggregate(maindata[,c(161,172,168,255,254,258,264,265,263,271)],
by = maindata[,167, drop =
Perfect, thanks! Naturally, now I need to resize the console so it doesn't
cover my new plots. I'd like to resize it on the fly (from within the
function) then reset the size to its previous size.
So, curConsoleDims() and resize.console() are a made up functions, but they
demonstrate what I am
Three things:
1) This isn't reproducible without your data file. Work out a minimal
reproducible example -- I bet you'll find your answer along the way --
and supply it.
2) What do the warnings say: they are usually pretty good at directing
you to trouble.
3) Don't use attach. Seriously -- just
Dear All,
Could anybody please advice me the way to check the goodness of fit of the
linear mixed-effects model and the suitable function in R to do so, and thank
you in advance for your assistance.
Fir
[[alternative HTML version deleted]]
Hi:
Try dict[[a]] rather than dict$a, and read the section on lists in the
Introduction to R manual to learn how to properly index them.
HTH,
Dennis
On Wed, Nov 16, 2011 at 7:16 AM, Antje Gruner n...@allesjetzt.net wrote:
Dear list,
I'd like to access the elements of a list within another
-Original Message-
From: Dennis Murphy [mailto:djmu...@gmail.com]
Sent: Tuesday, November 15, 2011 8:54 PM
To: Andy Bunn
Cc: r-help@r-project.org
Subject: Re: [R] equal spacing of the polygons in levelplot key
(lattice)
Hi:
Does this work?
Thanks Dennis.
This almost works.
On 16.11.2011 17:37, Scott Raynaud wrote:
That might be an option if it weren't my most important predictor. I'm
thinking my best bet is to use MLWin for the estimation since it will properly
set fixed effects
to 0. All my other sample size simulation programs use SAS PROC IML which I
OK, how about this instead?
# library('lattice')
levs - as.vector(quantile(volcano,c(0,0.1,0.5,0.9,0.99,1)))
levq - seq(min(levs), max(levs), length = 6)
levelplot(volcano, at = levs,
colorkey = list(at = levq,
labels = list(at = levq,
Dear All,
I am not familiar with R yet I want to use it to perform some task, hence my
posting here. I hope someone can help.
I have a set of data, genes (rows) and samples (columns). I want to do a
Pearson correlation on all the possible pairwise combinations of all the
genes (2000). Does anyone
Hello
This is probably a basic question but I am quite new to R.
I need to permute elements within rows of a binary matrix, such as
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]000010000 0
[2,]001100011
Hello,
I generate box plots from my data like this:
qplot(x=xxx,y=column,data=data,geom=boxplot) + xlab(xxx) + ylab(ylabel) +
theme_bw() + scale_y_log10() + geom_jitter(alpha=I(1/10))
The problem is that I see lot of points above the maximum at the same level as
some outliers. It looks very
-Original Message-
From: Dennis Murphy [mailto:djmu...@gmail.com]
Sent: Wednesday, November 16, 2011 11:22 AM
To: Andy Bunn
Cc: r-help@r-project.org
Subject: Re: [R] equal spacing of the polygons in levelplot key
(lattice)
OK, how about this instead?
# library('lattice')
levs
Suppose your matrix is called X.
? sample
X[sample(nrow(X)),]
Michael
On Wed, Nov 16, 2011 at 11:45 AM, Juan Antonio Balbuena balbu...@uv.es wrote:
Hello
This is probably a basic question but I am quite new to R.
I need to permute elements within rows of a binary matrix, such as
[,1]
On Wed, Nov 16, 2011 at 8:37 AM, muzz56 musah...@gmail.com wrote:
Dear All,
I am not familiar with R yet I want to use it to perform some task, hence my
posting here. I hope someone can help.
I have a set of data, genes (rows) and samples (columns). I want to do a
Pearson correlation on all
?cor
X = matrix(rnorm(400),ncol = 4)
cor(X)
Michael
On Wed, Nov 16, 2011 at 11:37 AM, muzz56 musah...@gmail.com wrote:
Dear All,
I am not familiar with R yet I want to use it to perform some task, hence my
posting here. I hope someone can help.
I have a set of data, genes (rows) and samples
OK, I'm using William Browne's MLPowSim to create an R script which will
simulate samples for estimation of sample size in mixed models. I have subjects
nested in hospitals with hospitals treated as random and all of my covariates
at level 1. My outcome is death, so it's binary and I'll have
And you got a reply on the ggplot2 list, which is why you're asked not
to cross-post.
For those who are wondering, geom_boxplot() in the ggplot2 package
will by default plot outside points along the same line as the boxplot
whiskers at their actual values. The gentleman jittered the original
Hello all,
I have a data frame that looks like this:
http://r.789695.n4.nabble.com/file/n4077622/Capture.png
I would like to know if it's possible to split a single row into two rows
when the time frame between beg and end overlaps midnight. I want to
compare the frequency of each activity for
I have csv data that extend beyond the area I want for an existing map. I want
using the boundaries of the polygon shape file as a cookie cutter so that I can
overlay the csv data on map without including anything outside the map
boundaries and create a dbf file or shapefile of the clipped data
I have the following scenario:
m - matrix(1:4, ncol=2)
m
[,1] [,2]
[1,]13
[2,]24
apply(m, 2, sum)
[1] 3 7
apply(m, 1, sum)
[1] 4 6
So I can apply to rows *or* columns. According to the documentation
(?apply)
MARGIN a vector giving the subscripts which the function
On Wed, 2011-11-16 at 14:29 -0500, R. Michael Weylandt wrote:
Suppose your matrix is called X.
? sample
X[sample(nrow(X)),]
That will shuffle the rows at random, not permute within the rows.
Here is an alternative, first using one of my packages (permute -
shameful promotion ;-) !:
mat -
It's the same as what you began with -- and that's because you broke
it down by columns and rows and took the sum of everything that
resulted.
I.e.,
sum(m[1,1])
sum(m[2,1])
sum(m[1,2])
sum(m[2,2])
and put them back together.
Michael
On Wed, Nov 16, 2011 at 3:13 PM, rkevinbur...@charter.net
Hi,
On Wed, Nov 16, 2011 at 3:13 PM, rkevinbur...@charter.net wrote:
I have the following scenario:
m - matrix(1:4, ncol=2)
m
[,1] [,2]
[1,] 1 3
[2,] 2 4
apply(m, 2, sum)
[1] 3 7
apply(m, 1, sum)
[1] 4 6
So I can apply to rows *or* columns. According to the
To expand on what Sarah and Michael said:
if you have a 3d array:
x-array(1:4,c(2,2,4))
x
, , 1
[,1] [,2]
[1,]13
[2,]24
, , 2
[,1] [,2]
[1,]13
[2,]24
, , 3
[,1] [,2]
[1,]13
[2,]24
, , 4
[,1] [,2]
[1,]13
[2,]
Try this: (it would have been easier if you had used 'dput' on your data)
x - read.table('/temp/example.txt', skip = 1, as.is = TRUE)
# convert to POSIXct
x$beg - as.POSIXct(paste(x$V4, x$V5))
x$end - as.POSIXct(paste(x$V6, x$V7))
# determine breaks over midnight
x$over - format(x$beg, %d) !=
Hi R users,
I try to read a data file (tab delimited format) in which some of the
entries in a particular field are missing. Is it possible to fill the
unavailable data with 'UnAvailable' string while performing read.table()
Something like
df = read.table(DataFile, header=FALSE,
Hi all,
I know that I can run an one way ANOVA on the absolute residual values
to check the assumption of equal variances:
model- lm(y~x)
summary(lm(abs(model$resid~x)))
What if I have two factors? As far as I know I have to check this
assumption on very factor/level combination. So if I have 3
Hi R users,
I try to read a data file (tab delimited format) in which some of the
entries in a particular field are missing. Is it possible to fill the
unavailable data with 'UnAvailable' string while performing read.table()
Something like
df = read.table(DataFile, header=FALSE,
I am very sorry for multiple mails.
Hi R users,
I try to read a data file (tab delimited format) in which some of the
entries in a particular field are missing. Is it possible to fill the
unavailable data with 'UnAvailable' string while performing read.table()
Something like
df =
Thanks Peter. I tried this after reading in the csv (read.csv) and
converted the data to matrix (as.matrix). But when I tried the correlation,
I keeping getting the error (x must be numeric) yet when I view the data,
its numeric.
On 16 November 2011 14:32, plangfelder [via R]
I'm just curious if Professor Koenker responded to this (e.g. in an
individual email or otherwise), as I would be interested in the answer.
--
View this message in context:
http://r.789695.n4.nabble.com/linear-against-nonlinear-alternatives-quantile-regression-tp3993327p4077713.html
Sent from
Say I have two matrixes: one is 8x28 and other 8x8
I'd like to multiply, for example, first and second column from the 8x8 and
plant them in first column of 8x28.
Then take first and third of 8x8 and plant into second column of 8x28.etc.
Any ideas?
Thanks!
[[alternative HTML
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of muzz56
Sent: Wednesday, November 16, 2011 12:28 PM
To: r-help@r-project.org
Subject: Re: [R] Pairwise correlation
Thanks Peter. I tried this after reading in the csv
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