HI,
May be this helps you:
set.seed(1)
mat1-matrix(rnorm(60,5),nrow=5,ncol=12)
colnames(mat1)-paste0(Var,1:12)
vec2-format(c(1,cor(mat1[,1],mat1[,2:12])),digits=4)
vec3-colnames(mat1)
arr2-array(rbind(vec3,vec2),dim=c(2,3,4))
Hi!
28.09.2012 08:41, Atte Tenkanen wrote:
Sorry. I should have mentioned that the order of the components is important.
So c(1,4,6) is accepted as a subvector of c(2,1,1,4,6,3), but not of
c(2,1,1,6,4,3).
How to test this?
How about this:
--- code ---
g1- c(2,1,1,4,6,3)
g2-
Hi Everyone,
I am trying a very simple task to append the Timestamp with a variable name so
something like
a_2012_09_27_00_12_30 - rnorm(1,2,1).
Tried some commands but it doesn't work out well. Hope someone has some answer
on it.
Session Info
R version 2.15.1 (2012-06-22)
Platform:
Hi!
28.09.2012 09:13, Bhupendrasinh Thakre wrote:
Statement I tried :
b - unclass(Sys.time())
b = 1348812597
c_b - rnorm(1,2,1)
Do you mean this:
--- code ---
df-data.frame(x=0,y=0)
colnames(df)
[1] x y
colnames(df)[2]-paste(b,unclass(Sys.time()),sep=_)
colnames(df)
[1] x
Hello,
Try the following:
b - unclass(Sys.time())
eval(parse(text=paste(c_,b, - rnorm(1,2,1),sep=)))
ls()
Regards,
Pascal
Le 28/09/2012 15:13, Bhupendrasinh Thakre a écrit :
Hi Everyone,
I am trying a very simple task to append the Timestamp with a variable name so
something like
On Sep 27, 2012, at 11:13 PM, Bhupendrasinh Thakre wrote:
Hi Everyone,
I am trying a very simple task to append the Timestamp with a variable name
so something like
a_2012_09_27_00_12_30 - rnorm(1,2,1).
If you want to assign a value to a character-name you need to use ... `assign`.
On Sep 27, 2012, at 9:07 PM, Ben Harrison wrote:
Hello,
I have produced some segmented regressions with the segmented package by
Viggo Mutteo. I have some example data and code below. I want to annotate
the individual segments with the slope parameter (actually it would be
nicer to
On 28-09-2012, at 07:41, Atte Tenkanen atte...@utu.fi wrote:
Sorry. I should have mentioned that the order of the components is important.
So c(1,4,6) is accepted as a subvector of c(2,1,1,4,6,3), but not of
c(2,1,1,6,4,3).
How to test this?
See this discussion for a variety of
Hi Rui,
Excellent!! This is what I was looking for. Thanks for the help.
So, now I have stored the result of the 10 regressions in summ.list
- lapply(lm.list2, summary)
And now once I enter sum.list it gives me the output for all
the 10 regressions...
I wanted to access a
Hi,
I am trying to create a scatterplot, coding each point to one of 5
populations. I was successful when I did this for one set of data, yet
when I try plotting other data a blank plot appears (although the axes are
labelled and I can fit the regression lines from each population). I have
Jessica da Silva jessica.m.dasilva at gmail.com writes:
I am trying to create a scatterplot, coding each point to
one of 5
populations. I was successful when I did this for one
set of data, yet
when I try plotting other data a blank plot appears
(although the axes are
labelled and I can
Hello, Krunal,
try
summ.list[[2]]$coefficients[2]
Note the double square brackets (as summ.list is a list)!
Hth,
Gerrit
On Fri, 28 Sep 2012, Krunal Nanavati wrote:
Hi Rui,
Excellent!! This is what I was looking for. Thanks for the help.
So, now I have stored the result of the 10
Hello,
To access list elements you need `[[`, like this:
summ.list[[2]]$coefficients
Or Use the extractor function,
coef(summ.list[[2]])
Rui Barradas
Em 28-09-2012 07:23, Krunal Nanavati escreveu:
Hi Rui,
Excellent!! This is what I was looking for. Thanks for the help.
So, now I have
On 09/27/2012 08:59 PM, Alexandra Howe wrote:
Hello,
I have data which I have arcsin transformed to analyse.
I want to plot my data with error bars however as my data is
back-transformed my standard errors are uneven.
Is there a simple way to draw these asymmetric error bars in R?
Hi
Hi,
I have a dataframe with multiple (appr. 20) columns containing
vectors of different values (different distributions).
Now I'd like to create a crosstable
where I compare the distribution of each vector (df-column) with
each other. For the comparison I want to use the ks.test().
The result
Hello,
Try the following.
f - function(x, y, ...,
alternative = c(two.sided, less, greater), exact = NULL){
#w - getOption(warn)
#options(warn = -1) # ignore warnings
p - ks.test(x, y, ..., alternative = alternative, exact =
exact)$p.value
#options(warn = w)
p
}
Hello,
Try
names(lm.list2[[2]]$coefficient[2] )
Rui Barradas
Em 28-09-2012 11:29, Krunal Nanavati escreveu:
Ok...this solves a part of my problem
When I typelm.list2[2] ...I get the following output
[[1]]
Call:
lm(formula = as.formula(fmla), data = tryout2)
Coefficients:
(Intercept)
I would guess that if you find the bit that says pch=| and change it to
pch=1 it will solve your question, and that reading ?par will tell you why.
Sarah
On Thursday, September 27, 2012, Elaine Kuo wrote:
Hello
This is Elaine.
I am using package lattice to generate boxplots.
Using
Ok, if I'm understanding it well, you want the mean value of Price1, ,
Price5? I don't know if it makes any sense, the coefficients already are
mean values, but see if this is it.
price.coef - sapply(lm.list, function(x) coef(x)[2])
mean(price.coef)
Rui Barradas
Em 28-09-2012 12:07, Krunal
Dear Chunyan,
One possibility would be to use the harmonic mean of the person-time at risk
values. You will have to do this manually though at the moment. Here is an
example:
### let's just use the treatment group data from dat.warfarin
data(dat.warfarin)
dat - escalc(xi=x1i, ti=t1i,
Hello,
I am really new to R and it's still a challenge to me.
Currently I'm working on my Master's Thesis. My supervisor works with SAS
and is not familiar with R at all.
I want to run an Anova, a tukey-test and as a result I want to have the
tukey-grouping ( something like A - AB - B)
I came
Hello,
I'd like to know if it is Ipossible to enter in a function wich is included
in a library ?
I know how to debug function wich is in a R file (but not in a library). But
it is not the case when the function is included in a library. I want to go
step by step in this function in order to
Ok...this solves a part of my problem
When I typelm.list2[2] ...I get the following output
[[1]]
Call:
lm(formula = as.formula(fmla), data = tryout2)
Coefficients:
(Intercept) Price2 Media1 Distri1Trend
Seasonality
13491232 -5759030-1520
Hi,
Yes the thing that you provided...works finebut probably I should have
asked for some other thing.
Here is what I am trying to do
I am trying to get the mean of Price variableso I am entering the
below function:
mean(names(lm.list2[[2]]$coefficient[2] ))
but this
Ok...I am sorry for the misunderstanding
what I am trying to do is
lm.list2 - list()
for(i in seq_along(pricemedia)){
regr - paste(pricemedia[i], trendseason, sep = +)
fmla - paste(response, regr, sep = ~)
lm.list2[[i]] - lm(as.formula(fmla), data = tryout2) }
I think the package BinarySimCLF can help.
See http://cran.r-project.org/web/packages/binarySimCLF/binarySimCLF.pdf.
André Gabriel.
-Mensagem original-
De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Em
nome de Rolf Turner
Enviada em: sexta-feira, 28 de setembro
On 27/09/2012 5:15 PM, Dr. Alireza Zolfaghari wrote:
Hi List,
Would you please send me a good link to talk me through on how to write a R
package?
See the ?package.skeleton help page. After you have run it, follow the
instructions in the Read-and-delete-me file that it will create.
For
Elaine,
For panel.bwplot you see that the central dot and the outlier dots are
controlled by
the same pch argument. I initially set the pch=| to match your first
example with the horizontal
indicator for the median. I would be inclined to use the default circle
for the outliers and
therefore
HI,
I guess there is a mistake in your code. You should have used typ instead of
abun as abun is the dependent variable.
summary(fm1 - aov(breaks ~ wool + tension, data = warpbreaks))
myresults - TukeyHSD(fm1, tension, ordered = TRUE)
library(agricolae)
HSD.test(fm1,wool,group=TRUE)
Thank you Rui!
that works as I want it... :)
/Johannes
On Fri, Sep 28, 2012 at 12:30 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Try the following.
f - function(x, y, ...,
alternative = c(two.sided, less, greater), exact = NULL){
#w - getOption(warn)
I'm not able to create the proper syntax to specify a lattice bwplot() for
only one of two conditioning factors.
The syntax that produces a box plot of each of the two conditioning
factors is:
bwplot(quant ~ param | era, data=mg.d, main='Dissolved Magnesium',
ylab='Concentration (mg/L)')
Many thanks Dr. Winsemius , Kimmo and Pascal
All of them are working and really beautiful...
Best Regards,
Bhupendrasinh Thakre
*Disclaimer :*
The information contained in this communication is confidential and may be
legally privileged. It is intended solely for the use of the individual
Jonathan,
ff has a utility function file.resize() which allows to give a new filesize
in bytes using doubles.
See ?file.resize
Regards
Jens Oehlschlägel
Gesendet: Donnerstag, 27. September 2012 um 21:17 Uhr
Von: Jonathan Greenberg j...@illinois.edu
An: r-help
Hello !
Thanks for your advice. I tried it, but the output is the same:
HSD.test(anova.typabunmit, typ, group=TRUE)
Name: typ
ds.typabunmit$typ
I don't get the values...!?!?
--
View this message in context:
http://r.789695.n4.nabble.com/Anova-and-tukey-grouping-tp4644485p4644513.html
AK: Thanks, that was very helpful. It led me to think of the function
names(base) which provided the vector of names in the correct order. I
then used the same matrix formatting and everything worked out exactly
as planned.
Dick
On 9/28/2012 1:09 AM, arun kirshna [via R] wrote:
HI,
May be
On Sep 28, 2012, at 4:35 AM, Krunal Nanavati wrote:
Ok...I am sorry for the misunderstanding
what I am trying to do is
Perhaps (and that is a really large 'perhaps'):
lm.list2 - list()
lm.means - list()
for(i in seq_along(pricemedia)){
regr - paste(pricemedia[i],
why does summary report max 27600 and not 27603?
x - c(27603, 1)
max(x)
[1] 27603
summary(x)
Min. 1st Qu. MedianMean 3rd Qu.Max.
16902 13800 13800 20700 27600
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
On 28/09/2012 12:14 PM, Sam Steingold wrote:
why does summary report max 27600 and not 27603?
x - c(27603, 1)
max(x)
[1] 27603
summary(x)
Min. 1st Qu. MedianMean 3rd Qu.Max.
16902 13800 13800 20700 27600
Because you asked for 3 digit accuracy. See
On Sep 28, 2012, at 7:49 AM, Rich Shepard wrote:
I'm not able to create the proper syntax to specify a lattice bwplot() for
only one of two conditioning factors.
Wouldn't that involve specifying the 'subset' parameter (if bwplot accepts a
subset argument) or using the 'subset' function to
Hi Everyone,
Sorry for coming back again with a new problem.
Editing question, session info and data so you don't have to scroll till
the end of page.
*Situation :*
I have a data frame and it's name is df. Now I want to add Time Stamp to
the end of *name of data Frame i.e. df_system_time*.
A small reproducible example, as requested bythe posting guide, would
have been very helpful here (if you provide one, use ?dput to provide
the data). You have also not told us what you mean by unsuccessful,
so we are left to guess what sort of problems you experienced. None
work is completely
Hi Everyone,
Sorry for coming back again with a new problem.
Editing question, session info and data so you don't have to scroll till
the end of page.
*Situation :*
I have a data frame and it's name is df. Now I want to add Time Stamp to
the end of *name of data Frame i.e. df_system_time*.
Hi,
Try this:
summary(x,digits=max(5))
# Min. 1st Qu. Median Mean 3rd Qu. Max.
# 1.0 6901.5 13802.0 13802.0 20702.0 27603.0
A.K.
- Original Message -
From: Sam Steingold s...@gnu.org
To: r-help@r-project.org
Cc:
Sent: Friday, September 28, 2012 12:14 PM
Subject: [R] max
On Fri, 28 Sep 2012, David Winsemius wrote:
Wouldn't that involve specifying the 'subset' parameter (if bwplot accepts
a subset argument) or using the 'subset' function to pass the desired rows
to the data argument if it doesn't?
David,
That's what I tried:
bwplot(quant ~ param | era,
On 28-09-2012, at 18:40, Bhupendrasinh Thakre vickytha...@gmail.com wrote:
Hi Everyone,
Sorry for coming back again with a new problem.
Editing question, session info and data so you don't have to scroll till
the end of page.
*Situation :*
I have a data frame and it's name is df. Now
Rui:
Quick follow-up -- it looks like seek does do what I want (I see Simon
suggested it some time ago) -- what do mean by trash your disk? What I'm
trying to accomplish is getting parallel, asynchronous writes to a large
binary image (just a binary file) working. Each node writes to a
On Sep 28, 2012, at 9:56 AM, Rich Shepard wrote:
On Fri, 28 Sep 2012, David Winsemius wrote:
Wouldn't that involve specifying the 'subset' parameter (if bwplot accepts
a subset argument) or using the 'subset' function to pass the desired rows
to the data argument if it doesn't?
David,
On Fri, 28 Sep 2012, David Winsemius wrote:
bwplot(quant ~ param , data=mg.d, main='Magnesium', ylab='Concentration
(mg/L)', subset= era=='Pre-mining' )
David, Don:
Thank you. I tried subset= and era== separately, not together.
Now I know.
Much appreciated,
Rich
Yes. Now I understand what was wanted.
1. the subset argument is certainly documented on the Help page:
subset
An expression that evaluates to a logical or integer indexing vector.
Like groups, it is evaluated in data. Only the resulting rows of data
are used for the plot. If subscripts is
On 28.09.2012 00:32, Duncan Murdoch wrote:
On 12-09-27 2:53 PM, Anju R wrote:
Sometimes when I try to install certain packages I get a warning message.
For example, I tried to install the package Imtest on windows R version
2.15.1 and got the following message:
Warning message:
package
Hi,
As I mentioned earlier, these are just guess work until you provide a subset of
your data with dput(). Also, please check the structure of the data with str().
A.K.
- Original Message -
From: Landi ent-ar...@gmx.de
To: r-help@r-project.org
Cc:
Sent: Friday, September 28,
On 28.09.2012 14:22, Duncan Murdoch wrote:
On 27/09/2012 5:15 PM, Dr. Alireza Zolfaghari wrote:
Hi List,
Would you please send me a good link to talk me through on how to
write a R
package?
See the ?package.skeleton help page. After you have run it, follow the
instructions in the
Thanks a ton Berend. That worked like a charm..
R comes with thousands of Sweet Surprises everyday
Bhupendrasinh Thakre
On Sep 28, 2012, at 12:00 PM, Berend Hasselman b...@xs4all.nl wrote:
On 28-09-2012, at 18:40, Bhupendrasinh Thakre vickytha...@gmail.com wrote:
Hi Everyone,
On Fri, Sep 28, 2012 at 11:15 AM, Bhupendrasinh Thakre
vickytha...@gmail.com wrote:
Thanks a ton Berend. That worked like a charm..
R comes with thousands of Sweet Surprises everyday
-- Not for those who read the docs. :-o
-- Bert
Bhupendrasinh Thakre
On Sep 28, 2012, at 12:00
Thank you!
___
Lähettäjä: Berend Hasselman [b...@xs4all.nl]
Lähetetty: 28. syyskuuta 2012 10:47
Vastaanottaja: Atte Tenkanen
Cc: R help
Aihe: Re: [R] How to test if there is a subvector in a longer vector
On 28-09-2012, at 07:41, Atte Tenkanen atte...@utu.fi wrote:
Sorry. I should
Hi Ankur,
I am running into the exact same issue you have described above. Were you
able to find out why it didn't work on your data set and resolve it? If yes,
could you share?
Much thanks regards,
Alice
--
View this message in context:
Hello,
I've written a function to try to answer to your op request, but I've
run into a problem. See in the end.
In the mean time, inline.
Em 28-09-2012 17:44, Jonathan Greenberg escreveu:
Rui:
Quick follow-up -- it looks like seek does do what I want (I see Simon
suggested it some time ago)
On Fri, Sep 28, 2012 at 6:57 AM, Richard M. Heiberger r...@temple.eduwrote:
Elaine,
For panel.bwplot you see that the central dot and the outlier dots are
controlled by
the same pch argument.
??? I don't think so...
bwplot(rgamma(20,.1,1)~gl(2,10), pch=rep(17,2),
panel =
Hello R users,
This is more of a convenience question that I hope others might find useful
if there is a better answer. I work with large datasets that requires
multiple parsing stages for different analysis. For example, compare group
3 vs. group 4. A more complicated comparison would be time
On Sep 28, 2012, at 12:44 PM, Jonathan Greenberg wrote:
Rui:
Quick follow-up -- it looks like seek does do what I want (I see Simon
suggested it some time ago) -- what do mean by trash your disk?
I can't speak for Rui, but the difference between seeking and explicit write is
that the FS
I would like to select a all the duplicate rows of a data frame including
the original. Any help would be much appreciated. This is where I'm at so
far. Thanks.
#Sample data frame:
df - read.table(header=T, con - textConnection('
label value
A 4
B 3
C 6
B 3
Hello,
Try the following.
idx - duplicated(df) | duplicated(df, fromLast = TRUE)
df[idx, ]
Note that they are returned in their original order in the df.
Hope this helps,
Rui Barradas
Em 28-09-2012 21:11, Adam Gabbert escreveu:
I would like to select a all the duplicate rows of a data
You have not specified the objective function you are trying to optimize with
your term efficient, or what you do with all of these subsets once you have
them.
For notational simplification and completeness of coverage (not necessarily
computational speedup) you might want to look at tapply
Good Evening-
I have a dataframe that has 10 columns that has a header and 7306 rows in
each column, I want to combine these columns into one. I utilized the stack
function but it only returned 3/4 of the data...my code is:
where nfcuy_bw is the dataframe with 7305 obs. and 10 variables
Once I
On Sep 28, 2012, at 11:59 AM, Charles Determan Jr wrote:
Hello R users,
This is more of a convenience question that I hope others might find useful
if there is a better answer. I work with large datasets that requires
multiple parsing stages for different analysis. For example, compare
On Sep 28, 2012, at 2:51 PM, Meredith Ballard LaBeau wrote:
Good Evening-
I have a dataframe that has 10 columns that has a header and 7306 rows in
each column, I want to combine these columns into one. I utilized the stack
function but it only returned 3/4 of the data...my code is:
where
Hello R-Users!
I'm using a heatmap to visualize a matrix of values between -1 and 3.
How can I set the colors so that white is zero, below zero is blue of
increasing intensity towards -1 and above zero is red of increasing
intensity towards red?
I tried like this (using the marray and gplots
?unlist
(A data frame is a list, as ?data.frame explains. Also the Intro to R
tutorial, which should be read by everyone beginning with R).
-- Bert
On Fri, Sep 28, 2012 at 2:51 PM, Meredith Ballard LaBeau
mmbal...@mtu.edu wrote:
Good Evening-
I have a dataframe that has 10 columns that has a
Hello Ilai,
Thank you for the response.
It did help a lot.
However, a beginner to lattice has three questions.
Q1
Please kindly explain why in this case OP is using it with no at
argument,
so it is possible to display the median and the outliers with different pch?
Q2.
what is the
On Sep 28, 2012, at 3:16 PM, Nick Fankhauser wrote:
Hello R-Users!
I'm using a heatmap to visualize a matrix of values between -1 and 3.
How can I set the colors so that white is zero, below zero is blue of
increasing intensity towards -1 and above zero is red of increasing intensity
On Sep 28, 2012, at 4:52 PM, David Winsemius wrote:
On Sep 28, 2012, at 3:16 PM, Nick Fankhauser wrote:
Hello R-Users!
I'm using a heatmap to visualize a matrix of values between -1 and 3.
How can I set the colors so that white is zero, below zero is blue of
increasing intensity
Hi guys, I have many rows (1000) and columns (30) of geno matrix. I use the
following loop and condition statement (adapted from someone else code). I
always have an error below. I was wondering if anyone knows what's the problem
how to fix it.
Thanks,Zhengyu ### geno matrix P1
Hi,
You can also use grep() to subset:
LD-paste0(rep(rep(c(3,4),each=4),2),c(rep(L,8),rep(D,8)))
set.seed(1)
dat1-data.frame(LD=LD,value=sample(1:15,16,replace=TRUE))
dat2-within(dat1,{LD-as.character(LD)})
dat2[grepl(.*L,dat2$LD),] # subset all L values
dat2[grepl(.*D,dat2$LD),] # subset all D
Happy Friday Everyone,
Hope Friday afternoon doesn't turn out to be a terrible time to post a
question. I've been doing a little data mining of patient text medical records
as of late. I started out trying to predict whether or not cancer patients had
received KRAS mutation testing and did
That works. Thank you!
On Fri, Sep 28, 2012 at 4:22 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Try the following.
idx - duplicated(df) | duplicated(df, fromLast = TRUE)
df[idx, ]
Note that they are returned in their original order in the df.
Hope this helps,
Rui Barradas
HI,
You can also try:
idx-data.frame(t(sapply(df,function(x) !is.na(match(x,x[duplicated(x)])
df1-df[sapply(idx,function(x) all(x==TRUE)),]
df1
# label value
#1 A 4
#2 B 3
#4 B 3
#7 A 4
#8 A 4
A.K.
- Original Message -
From: Rui Barradas
Hi,
I have a 3d array as below, I want to make this array to a matrix of p=50(rows)
and n=20(columns) with the coverage values .
The code before the array is:
library(binom)
Loading required package: lattice
pi.seq-seq(from = 0.01, to = 0.5, by = 0.01)
no.seq-seq(from = 5, to = 100, by = 5)
On Sep 28, 2012, at 3:59 PM, farnoosh sheikhi wrote:
Hi,
I have a 3d array as below, I want to make this array to a matrix of
p=50(rows) and n=20(columns) with the coverage values .
The code before the array is:
?matrix
mat - matrix(datfrm$coverage, 50, 20)
filled.contour(mat) #
On Sep 28, 2012, at 1:16 PM, JiangZhengyu wrote:
Hi guys, I have many rows (1000) and columns (30) of geno matrix. I use
the following loop and condition statement (adapted from someone else code).
I always have an error below. I was wondering if anyone knows what's the
problem how to
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