Dear John and R-list
Thank you for your comment. Here are my data in the dput format:
cs.not.log.bp - (
structure(c(168, 69, 16, 69, 41, 6, 148, 6, 5, 4, 7, 4, 4, 2,
7, 2, 4, 2, 4, 2, 1, 0, 2, 0), .Dim = c(4L, 6L), .Dimnames = list(
c(IUCN.Terrestrial, IUCN.Marine, National.CS.Terrestrial,
Thanks Rui and Jeff,
I thought that transposing the matrix would let me plot it the way I thought
it would but it did not. How can I take:
matrix e:
AAB BAB CCD DCD
x 1.75 2.25 3.25 3.75
y 6.25 6.75 7.25 7.75
and treat the x and y rows as the x and y axis values and plot
Dear Uwe:
Many thanks for your prompt feedback. You were right about the data.
I don't understand why but one map wasn't perfect (i.e., it didn't include
count information for each value).
If this happens the map seems alright using GIS programs because you can
still produce pretty maps based
Hi all,
I wondered how I can get the translated, rotated and scaledmatrix in a
procrustes analysis. This does not help:
##
library (vegan)
#defining the target matrix:
a= cbind (c (1,2,3,4), c( 10,12,14,16))
#defining the matrix to be rotated:
b= a; b[,2]=
In addition: the object MyNumberIs *is* created, but inside the
environment attached to the function as you evaluate it. And this
environment is gone when the function is done. (there is more to it, but
this is basically how it works). This is what Michael refers to in his
answer.
So, when
I suspect you are trying to find
your way into Circle 6 of 'The R
Inferno' but haven't yet got in.
http://www.burns-stat.com/pages/Tutor/R_inferno.pdf
Pat
On 26/04/2012 03:06, michaelyb wrote:
Hello,
I am trying to understand why the FUNCTION used in several codes, won't
create the object
Hello,
I am currently working with the betareg package, which allows the fitting of a
variable dispersion beta regression model (Simas et al. 2010, Computational
Statistics Data Analysis). I was wondering whether there is any package in R
that allows me to fit variable dispersion parameters in
I want to write to M, using row and columns taken from A and B, with values
from C. C is a lot longer than A and B, so only the first 67420 elements of
C are used in my loop.So how can I improve it to take then the next 67420
and write it to new file and so on till the 248th 67420. Many thanks
there are at least two alternatives
1) package dglm for Double generalized linear models
or probably better
2)package gamlss for Generalized Additive Models for Location Scale and
Shape. Here you can use alternative, sometimes more appropriate,
families and also you can include additive
On Wed, Apr 25, 2012 at 08:43:34PM +, Fabian Roger wrote:
sorry for cross-posting
Dear all,
I have tow (several) bivariate distributions with a known mean and
variance-covariance structure (hence a known density function) that I would
like to compare in order to get an intersect
I thought that maybe installing the RODBC package from source might fix the
problem:
install.packages(RODBC,type=source)
but to no success.
RODBC still gives the same warning with the same error code and very little
information to tackle the problem.
On Apr 25, 2012, at 10:03 AM,
Hi,
I'm having a hard time redefining the subset operation for arrays. The aim
is to be able to call a[dim2=1:5] instead of having to write a[,1:5,,,]
(assuming dimnames(a)[2]=-dim2).
I already wrote a function that does this (see bottom).
But I just can't figure out how to redefine [.array.
I
Here's an example
a=data.frame( sample(0:1,20,rep=T), sample(0:1,20,rep=T) ) # make 2 random
columns of length 20.
a.no0 = a[ rowSums(a)!=0, ]
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You could also use polyroot.
Do
x0=Re( polyroot( c(100,-0.5) - c(150,-1) ) )
In your code.
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M
[,1] [,2] [,3] [,4]
x 1.75 2.25 3.25 3.75
y 6.25 6.75 7.25 7.75
plot(t(M))
Does just fine for me oO
then again, as you write it, you actually have a data frame there.. so..
D
AAB BAB CCD DCD
x 1.75 2.25 3.25 3.75
y 6.25 6.75 7.25 7.75
plot(t(D))
Does just fine for me as well oO
You ask two questions. First, how to put points with a different y-axis on
the plot.
One possibility s like this:
plot(1:10)
plot.window(xlim=c(1,10),ylim=c(1,100))
points( seq(1,10,length=100), 100:1, col=2 )
axis(4,col.axis=2,col=2)
The command plot.window is the one that changes the
On Wed, 25 Apr 2012 20:57:31 +0200,peter dalgaard pda...@gmail.com wrote:
On Apr 25, 2012, at 20:19 , Marc Schwartz wrote:
I have not read it yet, but the acknowledgements at the end of the paper
note that:
1. It was supported by an NSF grant.
2. At least two members of R Core
Hi ,
Thanks again for helping me out.
Here is the code I am using
Ad_1 - subset(Attrition_data_1,Attrition_ind==1)
Ad_0 - subset(Attrition_data_1,Attrition_ind==0)
s1-sample(1:dim(Ad_0)[1],0.8*dim(Ad_0)[1])# 80% of the non-attrites
s2-sample(1:dim(Ad_1)[1],0.8*dim(Ad_1)[1])# 80% of attritees
s3-
Hi,
I have my data in below format.
position var1var2
2 .1 10
3 .29 89
12.56 100
425 .341234
6546 .12 21
Thank you Jorge, this did get me started and who else might be interested in
the topic, a possible code could be something like:
# Define the functions
f1 - function(x) 100-0.5*x
f2 - function(x) 150- x
# Plot the functions
par (xaxs=i, yaxs=i)
plot( 1,
Hello,
I am trying to figure out how to delete all rows which sum to zero from a
dataframe. I do not have a column with the sum of each column yet, so
perhaps that would need to be done first? Thank you for any help anyone can
offer with this.
Emily
PS sorry if this question has already been
I will be out of the office Apr 26-27, 2012 with limited or no email
access.
For immediate help, please call the JPSM main number,
301-314-7911.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
On 04/26/2012 08:02 PM, Manish Gupta wrote:
Hi,
I have my data in below format.
position var1var2
2 .1 10
3 .29 89
12.56 100
425 .34
You can run simulations to find out how large N must be so that split sample
validation yields sufficient precision to be trustworthy, in other words,
that different random splits provide the same estimate of model accuracy to
within some small tolerance. You will be surprised how large N must be
Be reminded that s1 and s2 are only the indexes on AD_0 and AD_1 of the data
which you want to keep.
therefore
traindata - rbind(s1,s2)
will not work.
you need to take data from AD_0 and AD_0 for that, similarly with what you did
with s3 and s4.
Am 26.04.2012 um 12:56 schrieb Dwaipayan
Hi
what about
rbind(a,a)
[,1] [,2] [,3] [,4]
[1,]1234
[2,]5678
[3,]1234
[4,]5678
Regards
Petr
a - matrix(1:8, 2, 4, byrow=TRUE)
a
[,1] [,2] [,3] [,4]
[1,]1234
[2,]5678
Hi,
I am sorry, could you tell me exactly how. I understand I am troubling you but
im trying to learn R since the last two weeks and havenât been able to come
to terms with the suble nuances.
Thanks again,
doy
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Dear R gurus,
I use R all the time at work, so one day a problem managing my personal arise
data made me think: Why not use R, it does everything!.
Anyway, my goal is to use R to manage my personal music library, and more
precisely my playcounts. I have two XML files, one from Winamp and the
a - matrix(1:8, 2, 4, byrow=TRUE)
b-t(a)
r-rep(b,5) # can insert anything for 5
matrix(r,ncol=dim(a)[2],byrow=T)
[,1] [,2] [,3] [,4]
[1,]1234
[2,]5678
[3,]1234
[4,]5678
[5,]1234
[6,]5678
Ad_1 - subset(Attrition_data_1,Attrition_ind==1)
Ad_0 - subset(Attrition_data_1,Attrition_ind==0)
s1-sample(1:dim(Ad_0)[1],0.8*dim(Ad_0)[1])# 80% of the non-attrites
s2-sample(1:dim(Ad_1)[1],0.8*dim(Ad_1)[1])# 80% of attritees
s3- Ad_0 [-s1,]
summary(s3)
s4- Ad_1 [-s2,]
summary(s4)
s5-
Hi there,
I'm working on a very specific topic related to graph theory. In
particular I'd like to compare two graphs by a general dissimilarity
function in graph domain, named graph edit distance (GED). Do you know
if there exists a package or a script in R implementing algorithm to
Hi
I have a variable in verbatim and want to search a pattern . but i'm not
able to convert it into string.
is there anyway that i can search directly from verbatim
-
Thanks in Advance
Arun
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When using the PLM package (version 1.2-8), I encounter the probem that
calling the FGLS estimator evokes strange behavior, when choosing the
random effects model. After calling the PGGLS function to estimate FGLS,
PLM gives me a warning, stating that the random model has been replaced
with the
On 26.04.2012 01:18, Bcampbell99 wrote:
Hi:
I have a small problem that I'm not quite sure how to figure out. I've
generated some randomized (permutated) data which consists of nrows=3 by
ncols=165 (see below) per element (3 x 165 = 495 sub-elements).
[,1] [,2] [,3] [,4] [,5] [,6]
We do not know anything about the data, but given those are floting
point numbers or integers there sould not be a problem with 4500
times 3.
Uwe Ligges
On 26.04.2012 05:15, Rich Lane wrote:
Hi all,
I have an interesting project coming up, but the datasets are way bigger
than
On 26-04-2012, at 14:23, arunkumar wrote:
Hi
I have a variable in verbatim and want to search a pattern . but i'm not
able to convert it into string.
is there anyway that i can search directly from verbatim
This is utterly incomprehensible.
Please provide a reproducible example.
On Apr 26, 2012, at 4:25 AM, Amen wrote:
I want to write to M, using row and columns taken from A and B, with
values
from C. C is a lot longer than A and B, so only the first 67420
elements of
C are used in my loop.So how can I improve it to take then the next
67420
and write it to new
In a GLMM, one compares the conditional model including covariates with the
unconditional model to see whether the conditional model fits the data
better.
(1) For my unconditional model, a different random effects term fits better
(independent random effects) than for my conditional model
I'm having a problem when using heatmap. Even though the diagonal of my
matrix is all the same value, the diagonal of my heatmap is not all the same
color. Any suggestions?
Here is some reproducible code:
#
# Get data
nba -
Dear R users,
I was just wondering if anyone would be
able to advice me on how to create a confidence envelope on the graph when
plotting a pair correlation function (or mark correlation function).
Is the command qqenvl() or envl.plot() what I need or is it something
completely
hHllo,
I'm looking for an algroithm to transform an existing toeplitz matrix
(autocorrelation matrix) to the nearest positive semidefinite toeplitz
matrix.
I merely found an algorithm to transform an correlation matrix via the
function nearcor() based on the algorithm of Higham.
But as I
Peter, your solution is actually very interesting. I have never seen or heard
of before. I will look into it.
Meanwhile, look at this example instead:
fac-function(x){a-1
for(i in 1:x){
a-a*i
print(a)}}
The result is :
fac(5)
[1] 1
[1] 2
[1] 6
i found one clue in our forum, but it can give position for whole matrix,
n-read.table(file.choose(),header=T)
y-n[,c(1,2,3)]
my.number=1.12270420185886
z-abs(y-my.number)==min(abs(y-my.number))
which(z)
[1] 19
i have uploaded file,which
Hi all
I am looking for an add-in. I am currently working on something and I use
daily data of closing stock prices. As not all companies are traded daily
(e.g. on monday, then on thursday etc) at the stock exchange, there is
satistically a problem. There are some papers which explain the
On 26/04/2012 9:01 AM, Steven Wolf wrote:
I'm having a problem when using heatmap. Even though the diagonal of my
matrix is all the same value, the diagonal of my heatmap is not all the same
color. Any suggestions?
heatmap() rescales the matrix by default. Use
heatmap(abs(psim),
Hi,
I have a data.frame which contains timeseries from several different
locations, which I want to compare against each other for example
calculating RMSE, or normalized mean bias of each location against the
others. An example of this is the cor function where I can put in a
data.frame and
Hello,
I would like to be able to plot an array on a plot, something like:
|arg1 | arg2 | arg3
val1| 0.9| 1.1| 2.4
val2| 0.33 | 0.23 | -1.4
val3| hello| stop | test
I know Rwave is good to report but don't want to use it.
? Is there a package that allow quick and dirty plot of
On Thu, Apr 26, 2012 at 8:56 AM, michaelyb cel81009...@gmail.com wrote:
Peter, your solution is actually very interesting. I have never seen or heard
of before. I will look into it.
Meanwhile, look at this example instead:
fac-function(x){a-1
for(i in 1:x){
Thanks!
-Steve
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: Thursday, April 26, 2012 9:28 AM
To: Steven Wolf
Cc: r-help@r-project.org
Subject: Re: [R] Heatmap fidelity
On 26/04/2012 9:01 AM, Steven Wolf wrote:
I'm having a problem when using heatmap.
Hi,
you can iterate over the vertices, but it'll be probably slow. The
best solution is to create the graph directly from the data frame(s)
containing all structure and attribute data. See the
graph.data.frame() function for this.
Btw. it might be worth to post your igraph related questions to
On Apr 26, 2012, at 7:31 AM, Jim Lemon wrote:
On 04/26/2012 08:02 PM, Manish Gupta wrote:
Hi,
I have my data in below format.
position var1var2
2 .1 10
3 .29 89
12.56
On Apr 26, 2012, at 9:29 AM, Neil Davis wrote:
Hi,
I have a data.frame which contains timeseries from several different
locations, which I want to compare against each other for example
calculating RMSE, or normalized mean bias of each location against
the others. An example of this is
Perhaps PerformanceAnalytics and the CAPM.* functions.
Michael
On Thu, Apr 26, 2012 at 8:46 AM, and_mue and_muel...@bluewin.ch wrote:
Hi all
I am looking for an add-in. I am currently working on something and I use
daily data of closing stock prices. As not all companies are traded daily
Hello everybody out there using Sweave,
There are some complicated SQL queries and laborous calculations against large
data included as R code chunks using Sweave in my LaTeX document.
These code chunks create graphs that do not change most of the time, but they
are of course recompiled every
Did you read ?'-' as
Peter suggested?
Better yet, do not read ?'-' and do not use -.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Ista Zahn
Sent: Thursday, April
Hello Rui,
For the write.table, it's OK!
And for the second one (for the 2nd best correlation) seems to work great!
You're too strong ^^
I have to check a bit more to be sure, but it seems to do it!
If you come in the Alps, it will be more liqueurs such as Chartreuse or
Génépi (from mountain
The easiest way probably is to put the code that takes so long to execute, in a
separate file, such as graph1.Rnw, and get it into your master file via
SweaveInput( graph1.Rnw ).
Once you are happy with your graph1, you can comment this line out and only
compile the stuff that keeps changing.
Dear list,
I get the ifelse function to work on a data frame but don't know how to do
something similar (only more conditions) with the combination of if and else
like in the example:
A - c(a,a,b,b,c,c)
B - c(rep(2,6))
dat - data.frame(A,B)
dat$C - if(AB$A==a) {AB$B^2} else
Hi!
how do i get to the source code of kpca or even better predict.kpca(which it
tells me doesn't exist but should) ?
(And if anyone has too much time:
Now if i got that right, the @pcv attribute consists of the principal
components, and for kpca, these are defined as projections of some
I guess there is not a steep learning curve if you want to use cache
in knitr or cacheSweave -- just use the chunk option cache=TRUE and
you are all set.
heavy-chunk, cache=TRUE=
# time-consuming computation here
@
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web:
Hi Jessica,
On Thu, Apr 26, 2012 at 11:59 AM, Jessica Streicher
j.streic...@micromata.de wrote:
Hi!
how do i get to the source code of kpca or even better predict.kpca(which it
tells me doesn't exist but should) ?
Probably you have to do kernlab:::predict.kpca from your R workspace,
but why
Hello,
Your example code has a bug, there's no fata.frame called 'AB'. It's
corrected below.
pannigh wrote
Dear list,
I get the ifelse function to work on a data frame but don't know how to do
something similar (only more conditions) with the combination of if and
else like in the
Dear all,
I have a series of variables that looks roughly like the sample data below and
I'm trying to conduct a factor analysis. I've omitted cases with missing
values for the factor analysis, but now I'd like to use the scores on each
component as new variables in the *original* data set for
On Apr 26, 2012, at 9:26 AM, statquant2 wrote:
Hello,
I would like to be able to plot an array on a plot, something like:
|arg1 | arg2 | arg3
val1| 0.9| 1.1| 2.4
val2| 0.33 | 0.23 | -1.4
val3| hello| stop | test
I know Rwave is good to report but don't want to use it.
? Is
Hello R users,
Hope everyone is doing great.
I have a dataset that is in .csv format and consists of two columns: one
named Period (which contains dates in the format _mm) and goes from
1995_10 to 2007_09 and the second column named pcumsdry which is a
volumetric measure and has been
Thanks a lot, totally forgot cran there.
Hm.. so they're multiplying some specifically computed Kernelmatrix with the
pcv's.. interesting.. too tired to check the math there, guess i'll just accept
its possible and go to sleep.
Am 26.04.2012 um 18:10 schrieb Steve Lianoglou:
Hi Jessica,
Dear All,
I am recently working on neural network using nnet package. The network has
4 hidden layers and 1 output layer, the target output 1 or 0.
The model I use is as follows:
nn-nnet(target~f1+f2+f3+f4+f5+f6+f7+f8+f9+f10,data=train,size=4,linout=FALSE,decay=0.025,maxit=800)
It works well
At 00:55 26/04/2012, Steven Orzack wrote:
I am looking for an R package with which one can calculate an effect
size for a set of contrasts given an omnibus chi-square test
statistic (more than 1 degree of freedom). Is there such a package?
Presumably, it would implement the procedure (or
You could also use the grconvertX function to convert from the middle
of the plotting region (from='npc') to user coordinates (to='user'),
or many other combinations.
On Wed, Apr 25, 2012 at 5:53 AM, Ramon Ovelar ramon.ove...@gmail.com wrote:
Many many thanks for the tip and for authoring this
On 25/04/12 14:02, Mabille, Geraldine wrote:
Hi, I am working with gam models in the mgcv library. My response
variable (Y) is binary (0/1), and my dataset contains repeated
measures over 110 individuals (same number of 0/1 within a given
individual: e.g. 345-zero and 345-one for individual A,
Any solution for that type of problem?
I did read the ?-, and seems very similar to the assign function, if I
am not mistaken
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Hello,
ok for xtable, but how will I print the latex code generated in a plot ?
Cheers
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Hi , R is a new language for me so sorry in advance if this error is to basic
for posting. I have tried the R manual and search online for quite a few, if
anyone could help i would be very thankful.
Here is my code.
kappa = 1.1
theta = 0.1
sigma = 0.4
rho = -0.6
v0 = 0.2
r = 0.05
T = 0.5
s0 = 1
Hello,
I am using constrOptim to maximize a likelihood function (the values of my
parameter vector must be between zero and one and must sum up to =1). I am
getting the error 'initial value in 'vmmin' is not finite'. I've tracked it
down to a problem in the function 'R' defined within the
The phrase does not work is not very helpful, it can mean quit a few
things including:
* Your computer exploded.
* No explosion, but smoke is pouring out the back and microsoft's
NoSmoke utility is not compatible with your power supply.
* The computer stopped working.
* The computer sits around
Yes, don't do it.
I know this can seem appealing if you're coming from another language,
but playing fast and loose with globals, non-local effects, and
superassignment is almost-never a good idea.
Michael
On Thu, Apr 26, 2012 at 1:32 PM, michaelyb cel81009...@gmail.com wrote:
Any solution for
Just what it says:
You define g but refer to a variable g. in the next line.
Just get rid of the typo.
Sarah
On Thu, Apr 26, 2012 at 1:43 PM, Guaramy guar...@hotmail.com wrote:
Hi , R is a new language for me so sorry in advance if this error is to basic
for posting. I have tried the R
On Thu, Apr 26, 2012 at 1:32 PM, michaelyb cel81009...@gmail.com wrote:
Any solution for that type of problem?
Solution for what type of problem?
I did read the ?-, and seems very similar to the assign function, if I
am not mistaken
Bottom line: don't use - until you know what you are
Ista,
Since you seem to know your stuff very well, how would you get 120 out of a
function that gives you the factorial of 5, without using factorial(5)?
Meanwhile, look at this example instead:
fac-function(x){a-1
for(i in 1:x){
a-a*i
Simply return it like a function is supposed to:
fac - function(x, loud = TRUE){
a - 1
for(i in seq_len(x)) { # seq_len is faster and more robust
a - a * i
if(loud) print(a)
}
return(a)
}
fac3 - fac(3)
print(fac3) # As desired
Michael
On Thu, Apr 26, 2012 at
On Apr 26, 2012, at 2:16 PM, michaelyb wrote:
Ista,
Since you seem to know your stuff very well, how would you get 120
out of a
function that gives you the factorial of 5, without using
factorial(5)?
Meanwhile, look at this example instead:
fac-function(x){a-1
for(i in
In R, the preferred method is to assign the result to a new object:
fac - function(x) {
a-1
for(i in 1:x){
a-a*i
print(a)
}
a # need to explicitly state what the function should return
}
myresult - fac(5)
myresult
Sarah
On Thu, Apr 26, 2012 at 2:16 PM, michaelyb
Dear Rainer
On Wed, Apr 25, 2012 at 5:34 PM, Rainer Schuermann
rainer.schuerm...@gmx.net wrote:
chunk_name_1,eval=FALSE,echo=FALSE=
I like the 'eval=FALSE' trick.
SweaveInput( setup.Rnw )
and from here, I can suse the named chunks almost like function calls, as you
you describe below.
the solution is to write functions that return the data you want changed, and
let the caller of the function decide where to put the answer. If you want to
return multiple answers, collect them in a vector or list and return that.
Sorry I took so long getting back to this, but the paying job needs to
take priority.
The regular expression (?!un)(?!non)muta looks for a string that
matches muta then looks at the characters immediately before it to
see if they match either un or non in which case it makes it a not
match.
Hello,
gives you 120, but you cannot access it after the end of execution.
Because you're just printing the final value of 'a', not returning it.
fac - function(x){
a - 1
for(i in 1:x) a - a*i
a
}
The return value must be the last instruction in a function.
Then, if
The latest version of fortunes (from R-forge, not sure about CRAN)
have fortunes up to number 317 which is from just a couple of days ago
(and 312 is the quoted one from February). For some reason some of
the instances of R on my computer stop at 291, others go up to 317
(running a new instance
Nice, thank you !
I particularly appreciated the list of possible explanations for it does
not work ! :-)
Arnaud
Le 26 avril 2012 13:49, Greg Snow 538...@gmail.com a écrit :
The phrase does not work is not very helpful, it can mean quit a few
things including:
* Your computer exploded.
*
Dear R users.
I'm trying to know how to create a function. I have developed a big code that
use a file and makes many process and then creates a table that is exported as
txt.
The point is to make something like a function. I need to make it easy for
others, so they can run it. This has a
Hello R-users,
I am having a problem with the 'break' command in R. I am wondering if
anyone can help me out with this. My program is similar to the following.
a=rep(NA,5)
a[1]=0
for(i in 2:5)
{
a[i]=a[i-1]+runif(1,0,3)
if(a[i]5)
{
i=2
break
}
}
What exactly I am
On Apr 26, 2012, at 12:49 PM, Greg Snow wrote:
The phrase does not work is not very helpful, it can mean quit a few
things including:
* Your computer exploded.
* No explosion, but smoke is pouring out the back and microsoft's
NoSmoke utility is not compatible with your power supply.
*
Hi there,
I wish to merge a common variable between a list and a data.frame return
rows via the data.frame where there is NO match. Here are some details:
The list, where the variable/col.name = CLAIM_NO
CLAIM_NO
20
83
1440
4439
7002
...
dim(hrc78_clm_no)
[1] 66781
The data.frame, where
On 26-04-2012, at 21:30, cassie jones wrote:
Hello R-users,
I am having a problem with the 'break' command in R. I am wondering if
anyone can help me out with this. My program is similar to the following.
a=rep(NA,5)
a[1]=0
for(i in 2:5)
{
a[i]=a[i-1]+runif(1,0,3)
if(a[i]5)
Hi,
To increase the chances of you getting help on this one, please give
example data (a small data.frame, a small list) that you are trying to
do this on, and also show the desired output. Whip these variables up
in your R workspace and paste the output of `dput` for each into your
follow up
Thanks Berend, you are right. The break command would not work here. But
the while loop is taking time to generate the desired.
On Thu, Apr 26, 2012 at 2:39 PM, Berend Hasselman b...@xs4all.nl wrote:
On 26-04-2012, at 21:30, cassie jones wrote:
Hello R-users,
I am having a problem with
David -
My question to you may sound (actually, it really is) silly, but please do
take your time to answer it.
What is the difference between:
fac-function(x){a-1
for (i in 1:x){
a-a*i
}a}
and:
fac-function(x){a-1
for (i
On Thu, Apr 26, 2012 at 02:30:09PM -0500, cassie jones wrote:
Hello R-users,
I am having a problem with the 'break' command in R. I am wondering if
anyone can help me out with this. My program is similar to the following.
a=rep(NA,5)
a[1]=0
for(i in 2:5)
{
a[i]=a[i-1]+runif(1,0,3)
Hi Steve,
Thanks for replying. Here's a small piece of the data.frame:
bestPartAreadmin[1:5,1:6]
DESY_SORT_KEY PRVDR_NUM CLM_THRU_DT CLAIM_NO
NCH_NEAR_LINE_REC_IDEN_CD NCH_CLM_TYPE_CD
1 10193 290003 20090323 20
Dear R-community,
I am using R (V 2.14.1) on Windows 7. I have a dataset which consists of 19
variables for 91 individuals or rows. Two of my variables are Age
(adult/chick, with no NA values) and Sex (0 for females/1 for females, with
quite a few NA values). The sex of many adult birds is
On Apr 26, 2012, at 3:40 PM, michaelyb wrote:
David -
My question to you may sound (actually, it really is) silly, but
please do
take your time to answer it.
What is the difference between:
fac-function(x){a-1
for (i in 1:x){
a-a*i
}a}
and:
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