Your clue is... density!
Probability density is not the same as probability... you have to multiply it
by something before you can sum it.
Try typing
h
by itself and review your options.
---
Jeff Newmiller
> x <- rnorm(1000)
> h <- hist(x,plot=FALSE)
> sum(h$density)
[1] 2 - shouldn't it be 1?!
> h <- hist(x,plot=FALSE, breaks=(-4:4))
> sum(h$density)
[1] 1 - now it's 1. why?!
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (onei
Not sure I understand your question (or if there is one) and I am not
familiar with vcd::mosaic. But if you are asking is there a simpler
way ? than yes:
1. work with ?array and ?aperm
2. create the array directly in R from the original data - not excel
3. ?mosaicplot (no package required - it's in
> * David Winsemius [2012-03-13 17:53:14 -0400]:
> On Mar 13, 2012, at 5:33 PM, Sam Steingold wrote:
>> I can, of course, plot log(h$density), but then the number labels will
>> be wrong.
>
> You could try apply a log transform to the appropriate component of
> the "h" object and using barplot to
Question:
twitteR's searchTwitter() function contains a 'geocode' argument that
returns tweets from users whose location falls within a given radius.
I'm not completely familiar with the API from which twitteR pulls, but no
mechanism exists to extract location coordinates from the tweets themse
See R FAQ 7.22 -- in short, you need to print() your plot to the
graphics device -- just wrap xyplot() in print() and it should work.
Michael
On Tue, Mar 13, 2012 at 3:55 PM, Dgnn wrote:
> I am trying to write a function that generates one PDf containing plots from
> several .csv files within a
Perhaps zoo::rollapply from the zoo package can get you started.
Michael
On Tue, Mar 13, 2012 at 4:49 PM, pie' wrote:
> Hi,
>
> I would like suggestions as to how to perform rolling regressions with the
> window extended one period at a time. That is, an initial sample period is
> passed to esti
Dear chris33,
Well, actually as I said, the anova() function *will* do what you want. You can
fit multivariate linear models with lm(),
mod.1 <- lm(cbind(Y1, Y2, Y3, Y4, Y5) ~ X1*X2 +X1*X3 + X1*X4)
mod.2 <- lm(cbind(Y1, Y2, Y3, Y4, Y5) ~ X1 + X2 + X3 + X4)
and then use anova() to get multivaria
On Mar 13, 2012, at 7:02 PM, Folkes, Michael wrote:
Hello all,
I was under the (false?) assumption that an object that is class
logical, would take up less memory than an object with class integer.
Nope.
Below am I correctly showing this is not the case?
This was an attempt to reduce memor
Hi John,
Thanks for your response. The anova funtion will not work in my case,
because I have multiple response variables. In other words, I would like to
conduct an extra sums-of-squares and cross-products test between the
following models:
FULL.MODEL: (Y1, Y2, Y3, Y4, Y5) as a function of
Niroshan ucalgary.ca> writes:
> I have a question based on my research. I am analyzing reader-based
> diagnostic data set. My study involves diabetic patients who were evaluated
> for treatable diabetic retinopathy based on the presence or absence of two
> pathologies in their eyes. Pathologies
Hello all,
I was under the (false?) assumption that an object that is class
logical, would take up less memory than an object with class integer.
Below am I correctly showing this is not the case?
This was an attempt to reduce memory usage. I'm dealing with two large
arrays (could be integers).
I don't know much about sunflowerplots, but perhaps hexagonal binning
might make be worth a look in your case if you are generally looking
for a scatterplot but the point density is too high.
http://cran.r-project.org/web/packages/hexbin/index.html
Take a look at the vignettes. You can also hexb
I'm having a bit of trouble finding and understanding the correct function to
make numeric vectors to feed the sunflowerplot function. I have 33k points
to show and I want to do better than the standard scatter plot.
I gather that I need two vectors (x and y) of the same count containing the
"cent
On Tue, Mar 13, 2012 at 3:34 PM, David Winsemius wrote:
>
> When I got around to running it I was hampered by a lack of knowledge about
> what sort of data-object "price" might have been. I tried putting in a
> single number on hte theory that it would saitisfy the seq() call, and also
> got the
On Mar 13, 2012, at 5:33 PM, Sam Steingold wrote:
I have a vector x:
table(x)
2 3 4 5 6 7 8 9101112
1314
45547 11835 4692 2241 1386 820 593 425 298 239 176
158 115
1516171819202122232
For this case I would use a permutation test. Start by choosing some
statistic that represents your 4 students across the different grades,
some possibilities would be the sum of scores across grades and
students, or mean, or median, or ...
Compute the selected statistic for your 4 students and s
On Mar 13, 2012, at 4:49 PM, Berend Hasselman wrote:
On 13-03-2012, at 21:40, David Winsemius wrote:
On Mar 13, 2012, at 4:24 PM, Anna Dunietz wrote:
Hello all!
I would like to create a 3d plot, with the option price explained
by the underlying price and time. Unfortunately, I can't q
I have a vector x:
table(x)
2 3 4 5 6 7 8 91011121314
45547 11835 4692 2241 1386 820 593 425 298 239 176 158 115
15161718192021222324252627
9488766747
For Windows you can use winMenuAdd function. Type ?winAddMenu to see how...
Good luck,
Marcio
www.dsr.inpe.br/~mello
On 3/6/08 10:35 AM, Alberto Monteiro wrote:
er MIMI& piki PIKINHA wrote:
Hello, I´m spanish student, and I´m making the finish project of
computer science. I´m working in R
Dear chris33,
You can use the anova() function to compare the two multivariate linear models.
Alternatively, the Anova() function in the car package will compute "type II"
or "type III" MANOVA tests, which aren't quite what you're asking about.
I hope this helps,
John
Hi,
I would like suggestions as to how to perform rolling regressions with the
window extended one period at a time. That is, an initial sample period is
passed to estimation and that very sample is then extended one period at a
time through the remaining sample. Is there a specific package?
Thnk
I am trying to write a function that generates one PDf containing plots from
several .csv files within a directory. When I manually execute the code it
seems to work, but not when it is a function. I think I need to return()
something, but haven't had much luck figuring out what/how.
plot.isi<-fu
I would like to conduct an extra sum-of -squares test that compares a full
MANOVA model (with all 1st order interactions) to a reduced model (no
interactions) to determine if I can drop all interactions at the same time.
This is analagous to an extra sum-of-squares F-test in ANOVA, but instead
usi
On 13-03-2012, at 21:40, David Winsemius wrote:
>
> On Mar 13, 2012, at 4:24 PM, Anna Dunietz wrote:
>
>> Hello all!
>>
>> I would like to create a 3d plot, with the option price explained by the
>> underlying price and time. Unfortunately, I can't quite get it to work. I
>> would very muc
On Mar 13, 2012, at 4:24 PM, Anna Dunietz wrote:
Hello all!
I would like to create a 3d plot, with the option price explained by
the underlying price and time. Unfortunately, I can't quite get it
to work. I would very much appreciate your help!
The usual problem with lattice calls th
Hello all!
I would like to create a 3d plot, with the option price explained by
the underlying price and time. Unfortunately, I can't quite get it to
work. I would very much appreciate your help!
Thanks,
Anna
# Black-Scholes Option Graph
library(lattice)
blackscholes <- function(s, k,
On Tue, Mar 13, 2012 at 11:27 AM, aledanda wrote:
> Dear All,
>
> I hope you don't mind helping me with this small issue. I haven't been using
> R in years and I'm trying to fill in a matrix
> with the output of a function (I'm probably using the Matlab logic here and
> it's not working).
> Here i
Personally, I find it easier to fix the overall size of the "page" and
modify the margins in line size (mar and oma rather than mai and omi),
until I get the plots the way I want them. You don't specify what OS
you're using, but in windows, I would use something like
windows(h=16, w=9)
par(mfr
Suggestion below:
On Tue, Mar 13, 2012 at 1:24 PM, guillaume chaumet
wrote:
> I omit to precise that I already try to generate data based on the mean and
> sd of two variables.
>
> x=rnorm(20,1,5)+1:20
>
> y=rnorm(20,1,7)+41:60
>
> simu<-function(x,y,n) {
> simu=vector("list",length=n)
>
>
The way to represent categorical variables is with factors. See ?factor.
randomForest() will handle factors appropriately, as most modeling functions in
R.
Andy
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of abhishek
>
Is the following what you want?
> a <- c(1,2,3,4,4,4,5,5)
> b <- c(11,7,4,9,8,3,12,4)
> split(b, a)
$1
[1] 11
$2
[1] 7
$3
[1] 4
$4
[1] 9 8 3
$5
[1] 12 4
Note that your df<-cbind(a,b) produces a matrix, not the data.frame
that your df suggests you want. Use df<-da
I am working on the following database of gene sets.
database<-GSA.read.gmt( "C:/c5.all.v2.5.symbols.gmt")
I need to filter out the gene sets that contain less than 5 genes or
more than 200 genes. I would appreciate help with that matter.
Thanks in advance
[[alternative HTML version
dear all,
apologizes for bothering with a probably stupid question but I really
don' t know how to proceed.
I have a dataset which look like df
a <- c(1,2,3,4,4,4,5,5)
b <- c(11,7,4,9,8,3,12,4)
df <-cbind(a,b)
I would like to have one which looks like this:
a
1 11
2 7
3 4
4 9 8 3
5 12 4
a a
On Wed, Mar 14, 2012 at 3:41 AM, cheba meier wrote:
> Hello,
>
> I have to compute the pooled z-value and I would like to know which way is
> more appropriate
>
>
> b <- c( -0.205,1.040,0.087)
> s <- c(0.449,0.167,0.241)
> n <- c(310, 342, 348)
> z <- b/s
>
> Z <- sum(z)/sqrt(length(n))
> P <- 2*(
Dear David and Jeff,
> Only if you were going apply some sort of transformation that did not extend
> globally
Exactly, this is why the LPCM package is great, as it assigns points
to parts of a curve.
I think I pretty much got what I need - it is not perfect yet but it
should be enough to give
I omit to precise that I already try to generate data based on the mean and
sd of two variables.
x=rnorm(20,1,5)+1:20
y=rnorm(20,1,7)+41:60
simu<-function(x,y,n) {
simu=vector("list",length=n)
for(i in 1:n) {
x=c(x,rnorm(1,mean(x),sd(x)))
y=c(y,rnorm(1,mean(y),sd(y)))
Dear all,
First I would like to thank the ESS people's all the hard work. I am
watching the project closely and witnessing the improvements day by day.
Besides I found a strange situation using `ess-tracebug'. Please tell me
if I am wrong or this is a bug.
Start Emacs with "emacs -Q" and lo
No problem.
A pro-tip for future posts: the dput() function creates a plain text
representation of the data in question which is great for email and is
nicely copy-and-pasteable. It wasn't so much a thing here, but for
large or complicated data sets, the regular console printout doesn't
always rev
HI All,
I got the error : package is not installed for 'arch=x64' when building my own
package for 64bit R. How can I configure the arch ? The 'R CMD config' does not
work. Thank you very much!
The detailed error is :
** testing if installed package can be loaded
Error : package 'xxx' is not
On Mar 13, 2012, at 12:51 PM, Lik Wee Lee wrote:
Hi,
Using the multicore package and calling sample() in mclapply,
how do I get the results to be reproducible?
I know the random number is seeded by process id and so is
different for each run.
You might want to look up the thread in last mont
On 13-03-2012, at 17:51, Lik Wee Lee wrote:
> Hi,
>
> Using the multicore package and calling sample() in mclapply,
> how do I get the results to be reproducible?
> I know the random number is seeded by process id and so is
> different for each run.
Don't know about those two packages.
Have a l
> Michael
> on Mon, 12 Mar 2012 13:19:19 -0500 writes:
> The problem is: by default shouldn't it use "Huber's"?
> And it should be convex problem no?
> so when I do rlm(y~x) which is a single-beta fitting problem,
> shouldn't it always converge?
“In theory, theory
Thank you very much David. I should realize myslef that operator precedence
cause the problem. Sorry about this:-) I solved the problem.
--- On Tue, 3/13/12, David Winsemius wrote:
From: David Winsemius
Subject: Re: [R] Error " subscript out of bounds"
To: "Houhou Li"
Cc: r-help@r-project.or
Just a little typo: see below.
On Tue, Mar 13, 2012 at 1:00 PM, Haojie Yan wrote:
> Dear Michael,
>
> Thanks a lot for your hints.
>
> I have just had a try as below but still got back some error messages as
> shown:
>
> The object containing the 'date_time' data is named 'INTERVAL_END_TIME' and
Yes, the short answer is that you need to define out before running
the loop. The most effective way to do so will be to set up a matrix
with the exact right dimensions (if you know them up front); something
like out <- matrix(NA, nrow = length(input), ncol = 9)
Michael
On Tue, Mar 13, 2012 at
Hi,
Using the multicore package and calling sample() in mclapply,
how do I get the results to be reproducible?
I know the random number is seeded by process id and so is
different for each run.
Thanks,
Lik
[[alternative HTML version deleted]]
Dear All,
I hope you don't mind helping me with this small issue. I haven't been using
R in years and I'm trying to fill in a matrix
with the output of a function (I'm probably using the Matlab logic here and
it's not working).
Here is my code:
for (i in 1:length(input)){
out[i,1:3] <- MyFunc
On Tue, Mar 13, 2012 at 12:20 PM, Haojie Yan wrote:
> Dear R-user,
>
> I have read a dataset from .csv file into R. This dataset includes one
> column containing some data in 'date and time' format, e.g. 'dd/mm/
> hh:mm'.
>
> These data were automatically read and saved as 'factor' in R. When
This is just a little comment to supplement Michael's excellent
solution. If there are even a few (e.g., 5 each) repeated values,
this:
as.POSIXct(as.character(levels(x)), format = "%d/%m/%Y %H:%M")[x]
will be substantially faster, with the speed gains strongly associated
with the number of repl
On Mar 13, 2012, at 12:18 PM, Houhou Li wrote:
Hello, R-users,
I have a datafile with 37313 records and each record has 5 different
measurements on the same variables. The format looks like this:
treeID, VIG0, VIG1, VIG2, VIG3, VIG4
I was trying to convert the one row record to 5 rows reco
Michael,
Thanks for the pointer to the discussion in the ggplot list. It seems
that the reason of this behaviour of facet_grid() is already known and
being discussed by the developers of ggplot2.
facet_grid() reduces the original data frame with unique() before
applying the stats. If the data fr
Take a look at
example(HoltWinters)
Michael
On Tue, Mar 13, 2012 at 11:06 AM, sagarnikam123 wrote:
> i have data in one file below like & (i have such type of file =200,each
> file have below type of data)
>>t
> -0.15264004
> 0.056076439
> -0.07276116
> -0.00917326
> -0.02069089
> -0.00416232
res3 + t(res3)
Michael
On Tue, Mar 13, 2012 at 8:15 AM, RMSOPS wrote:
> Hello,
> With the following code get the results array
> res3<-table(df$v_source,df$v_destine)
>
> 1 2 3 4 5 6 7
> 1 0 10 0 0 0 0 0
> 2 11 0 0 0 0 0 0
> 3 0 0 18 15 0 0 0
> 4 0 0 15 11 0 0
as.POSIXct(as.character(FACTORHERE), format = "%d/%m/%Y %H:%M")
Michael
On Tue, Mar 13, 2012 at 12:20 PM, Haojie Yan wrote:
> Dear R-user,
>
> I have read a dataset from .csv file into R. This dataset includes one
> column containing some data in 'date and time' format, e.g. 'dd/mm/
> hh:mm'
Dear R-user,
I have read a dataset from .csv file into R. This dataset includes one
column containing some data in 'date and time' format, e.g. 'dd/mm/
hh:mm'.
These data were automatically read and saved as 'factor' in R. When I was
trying to produce some plots (such as time series) with the
Hello, R-users,
I have a datafile with 37313 records and each record has 5 different
measurements on the same variables. The format looks like this: treeID, VIG0,
VIG1, VIG2, VIG3, VIG4
I was trying to convert the one row record to 5 rows record with format like
this (treeID, MEASUREMENT, VIGO
:)
congratulations!
On Tue, Mar 13, 2012 at 08:59:56AM -0600, Kjetil Halvorsen wrote:
> This posting is only to celebrate that R-help-es (R-help for Spanish
> Speakers) have reached 500
> members!
>
> (and to thank Patricia for doing the bulk of admin work).
>
> Kjetil
>
>
You have a conceptual problem, as pointed out by previous helpers.
You don't have a standard error for the first level of your categorical
variable because that level's effect is not estimated.
It is being used as a reference level against which the other levels of that
categorical variable are b
Hello,
this solve my problem. table(testdata$source, testdata$destine)
Thanks
--
View this message in context:
http://r.789695.n4.nabble.com/Re-Matrix-Another-table-tp4469376p4469384.html
Sent from the R help mailing list archive at Nabble.com.
__
i have data in one file below like & (i have such type of file =200,each
file have below type of data)
>t
-0.15264004
0.056076439
-0.07276116
-0.00917326
-0.02069089
-0.00416232
-0.07225855
-0.02654577
-0.06131410
-0.09380202
0.057414014
-0.05239976
0.014397612
0.016145161
-0.00670587
0.018696335
Hello
Error: could not find function sqldf:
Hello, I'm using R Studio, and installed the option of installing the
packages sqldbf function.
But When I run the code give the next error.
install.packages("sqldf")
library("RSQLite")
require(sqldf)
x <- read.fwf(textConnection("4 - 4 5
Hi everybody,
I have a data set with a value and a status (positive or negative case) and
I want make a ROC Analysis. So, with ROCR Package, I have got the ROC curve
(True Positive Fraction [tpf] according 1-True Negative Fraction [1-tnf]).
http://r.789695.n4.nabble.com/file/n4469203/01.png
But
On Mar 13, 2012, at 9:38 AM, D_Tomas wrote:
Dear userRs,
when applied the summary function to a glm fit (e.g Poisson) the
parameter
table provides the categorical variables assuming that the first level
estimate (in alphabetical order) is 0.
Not really. It returns an estimate for the cont
Hi,
On Tue, Mar 13, 2012 at 7:51 AM, RMSOPS wrote:
> I have next table
> source destine
> 3 3
> 7 7
> 6 6
> 3 4
> 4 4
> 4 3
> 3 3
> 3 3
> 3 3
> 3 3
> 3 4
> 4 4
> 4 3
> 3 4
> 4 3
It is so much easier if you use dput to provide reproducible data, as
the posting guide asks. There's no way f
Hi,
See inline.
On Tue, Mar 13, 2012 at 6:38 AM, D_Tomas wrote:
> Dear userRs,
>
> when applied the summary function to a glm fit (e.g Poisson) the parameter
> table provides the categorical variables assuming that the first level
> estimate (in alphabetical order) is 0.
>
> What is the standard
This posting is only to celebrate that R-help-es (R-help for Spanish
Speakers) have reached 500
members!
(and to thank Patricia for doing the bulk of admin work).
Kjetil
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
Hello,
I have to compute the pooled z-value and I would like to know which way is
more appropriate
b <- c( -0.205,1.040,0.087)
s <- c(0.449,0.167,0.241)
n <- c(310, 342, 348)
z <- b/s
Z <- sum(z)/sqrt(length(n))
P <- 2*(1-pnorm(abs(Z)))
P
w <- sqrt(n)
Zw <- sum(w * z)/sqrt(sum(w^2))
Pw <- 1 -
The definition of simple in "Keep your class hierarchy simple and relevant to
your actual problem" is quite difficult. For someone who has programmed the
classes etc, it is quite simple to understand the heritance. But for example
for someone else, who has to maintain the code and the classes, it m
Dear userRs,
when applied the summary function to a glm fit (e.g Poisson) the parameter
table provides the categorical variables assuming that the first level
estimate (in alphabetical order) is 0.
What is the standard error for that variable then?
Are the standard errors calculated assuming
Thanks Josh. I'm quite new, just wondering re:factor levels?
In this example (shamelessly stolen from the internet):
*schtyp*
[1] 0 0 1 0 0 0 1 0 1 0 1 1 1 1 0 0 1 1 1 0
*schtyp.f <- factor(schtyp, labels = c("private", "public"))
schtyp.f*
[1] private private public private private private p
Hello,
With the following code get the results array
res3<-table(df$v_source,df$v_destine)
1 2 3 4 5 6 7
1 0 10 0 0 0 0 0
2 11 0 0 0 0 0 0
3 0 0 18 15 0 0 0
4 0 0 15 11 0 0 0
5 0 0 0 0 1 0 0
6 0 0 0 0 0 1 0
7 0 0 0 0 0 0 18
my
#
#
## Iris data ##
data(iris)
#
Hello,
I am sorry if there are already post that answers to this question but i
tried to find them before making this post. I did not really find relevant
posts.
I am using randomForest package for building a two class classifier. There
are categorical variables and numerical variables in my data
I have next table
source destine
3 3
7 7
6 6
3 4
4 4
4 3
3 3
3 3
3 3
3 3
3 4
4 4
4 3
3 4
4 3
I'm trying to create an array with the number of occurrences between the
source and destination. id_ap<-levels(factor(df$v_source))
num_AP<-length(levels(factor(df$v_source)))
mat<-matrix(
Hello
is the dataset that was sent to help, has over two columns the source and
destination, is the separation of position pos
POS DIF SourceDest
4 - 4 56 4 4
4 - 3 61 4 3
3 - 3 300 3 3
3 -
I have next table
source destine
33
77
6 6
3 4
4 4
4 3
3 3
3 3
3 3
3 3
3 4
4 4
4 3
3 4
4 3
I'm trying to create an array with the number of occurrences between the
source and destination.
id_ap<-level
Dear R Help Community,
I have a question and an answer (based on reading this forum and online
research), but I though I should share both since probably there's a much
better way to go about my solution. My question is specifically about how
to best visualise multiple response contingency tables.
Thank you Michael,
Your advices are truly appreciated!
Ian
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: 13 mars 2012 09:44
To: Ian Schiller
Cc: r-help@r-project.org
Subject: Re: [R] Coding C++ in R. What is faster : Using bosst external
librari
There will be ever-so-slight performance differences due to
implementation differences (I believe R's functions are just a hair
slower because they are more exact -- though that may be a comparison
with GSL I'm thinking of), but my advice would be to use the RNGs that
come with R. They are the best
Try this:
> require(sqldf)
> x <- read.fwf(textConnection("4 - 4 56
+ 4 - 3 61
+ 3 - 3 300
+ 3 - 327
+ 3 - 3 33
+ 3 - 3 87
+ 3 - 4 49
+ 4 - 4 71
+ 4 - 3 121
+ 3 - 4 138
+ 4 - 3 15"), width = c(7,8) , header = FALSE, as.is = TRUE)
> close
On 03/13/2012 04:34 AM, Alexander wrote:
Hi,
I am working on a project, which contains S4 classes and subclasses. Lets
assume the following organisation:
A: S4 Class
B,C: inherit from A
D,E,F,G: inherit from B
H,I: inherit from C
I want to define now a generic function, which returns me the name
Hi everyone,
I have built an R package and for the sake of speed I have decided to rewrite
some part of the code in C++. In my original R code I use the pnorm, qnorm,
rnorm, pgamma, dgamma, rgamma, rbeta and runif function. First I was thinking
in going with the boost libraries, but I noticed
try this:
> x <- readLines(textConnection("
+ "))
>
> closeAllConnections()
>
> # process & parse the data
> for (i in x){
+ if (grepl("^username", i)) username <- sub(".*=(.*)", '\\1', i)
+ if (grepl("^password", i)){
+ password <- sub(".*=(.*)", "\\1", i)
+ cat("found: us
On Tue, Mar 13, 2012 at 5:15 AM, sybil kennelly wrote:
> Thanks Josh. I'm quite new, just wondering re:factor levels?
>
> In this example (shamelessly stolen from the internet):
>
> schtyp
>
> [1] 0 0 1 0 0 0 1 0 1 0 1 1 1 1 0 0 1 1 1 0
>
> schtyp.f <- factor(schtyp, labels = c("private", "public"
Hello, "uday",
e.g.,
(data - 1) %/% 3 + 1
would do the job in your very specific situation, but take a look at
?findInterval
for possibly more interesting (because more general) solutions.
Hth -- Gerrit
On Tue, 13 Mar 2012, uday wrote:
I want to replace some values in my Array
e.g
data<
Hi Sybil,
You cannot turn a list into a factor. You could do:
cell_data <-c('cell1','cell2')
factor_list <- factor(cell_data)
or if you already have a list, unlist() or as.vector() may convert it
into a vector that you can then convert to a factor.
Cheers,
Josh
On Tue, Mar 13, 2012 at 4:29 A
Hi
>
> Hello
>
>I am developing a small program that to calculate the maximum,
minimum
> and average.
>
> The data.frame v is
>
> POS DIF
> 4 - 4 56
> 4 - 3 61
> 3 - 3 300
> 3 - 327
> 3 - 3 33
> 3 - 3 87
> 3 - 4 49
> 4 - 4 71
> 4
Hi,
I am working on a project, which contains S4 classes and subclasses. Lets
assume the following organisation:
A: S4 Class
B,C: inherit from A
D,E,F,G: inherit from B
H,I: inherit from C
I want to define now a generic function, which returns me the name of the
class. I can now write the function
Hello can anyone help please?
i read two words "cell1", "cell2" into a list. I want to turn this list
into a factor.
> cell_data <-list(c('cell1','cell2'))
> cell_data
[[1]]
[1] "cell1" "cell2"
> factor_list <- factor(cell_data)
Error in sort.list(y) : 'x' must be atomic for 'sort.list'
Have
I want to replace some values in my Array
e.g
data<- c(1,2,3,4,5,6,7,8,9,10,11,12)
I would like to replace 1,2,3 by 1 , 4,5,6 by 2, 7,8,9 by 3 and 10,11,12 by
4
I am expecting out put
data<- 1 1 1 2 2 2 3 3 3 4 4 4
I have tried replace function
replace(data, (data==1 |data==2| data==3),1)
Hi all,
I have a basic question concerning graphs in R. I’m using the par()
function and I’m working with multiple figures by row (mfrow) but my the
hight of my figures become compressed. I have 4 rows and 2 columns (because
I want to plot 8 histograms (freq = FALSE ) on it. I know I can adapt my
Hi, Jim
Below is the code that I try and the result I obtained:
*br<-read.table("R_beginner_starter.dat",header=TRUE,sep="\t")
library(plotrix)
barp(t(br[,c(2,4)]))
*
The result generated:
http://r.789695.n4.nabble.com/file/n4468592/ScreenHunter_01_Mar._13_17.10.jpg
Which is different with m
Hello
I am developing a small program that to calculate the maximum, minimum
and average.
The data.frame v is
POS DIF
4 - 4 56
4 - 3 61
3 - 3 300
3 - 327
3 - 3 33
3 - 3 87
3 - 4 49
4 - 4 71
4 - 3 121
3 - 4 138
4 - 3 15
Hi, I am playing around with some data and I would like to get data that is
stored in a file like this:
...
to the data.frame structure, how can I do that directly in R, currently I am
doing parse with bash, but I would like to centralize the procedure and
learn something new.
Thanks
--
View
Berend:
1. 1000 thanks for your help. It works perfectly
2. many thanks for the analysis of the expression; i will try to
understand it. Perl is really not easy to read
thanks again!
On Tue, Mar 13, 2012 at 11:51:49AM +0100, Berend Hasselman wrote:
>
> On 13-03-2012, at 11:28, Igor Sosa Mayor w
On 13-03-2012, at 11:28, Igor Sosa Mayor wrote:
> many thanks, Berend.
>
> It works well... but with a problem because i was not completely clear
> in my first email.
>
> It works with cases such as:
> Franco (El)
> Regueras (Las)
>
> but not with other cases such as:
> Fauces de San Andrés (L
many thanks, Berend.
It works well... but with a problem because i was not completely clear
in my first email.
It works with cases such as:
Franco (El)
Regueras (Las)
but not with other cases such as:
Fauces de San Andrés (Las)
any hints? or meaybe if it is very complicated, any short explanati
Hi,
Probably yes, but I think this would be more easily handled through
the command prompt or some shell. On the command prompt, I would look
at "xcopy". xcopy /? will show the arguments available. You will
need to do a bit of work to get a list of all the directories to copy
your r.css file i
Thanks Jim
For the first glance it seems to do what I want. I must go through it more
thoroughly.
Petr
>
> On 03/13/2012 03:07 AM, Petr PIKAL wrote:
> > Dear all
> >
> > with image I can plot only one set of values in one plot.
> >
> > Do somebody have any insight how to put those 2 matrices i
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