[R] problem in finding sizes of objects using a for loop
Dear All,
I wanted to extract the sizes of all created objects. For E.g when I
created 2 objects(x and y), I got their sizes using the following
code:
x-rnorm(1)
y-runif(100,min=40,max=1000)
ls()
[1] x y
object.size(x)
80024 bytes
object.size(y)
824 bytes
However, I was unable to get their sizes when I used a for loop in the
following way:
objects-ls()
for (i in seq_along(objects)){
+ print(c(objects[i],object.size(objects[i])))
+
+ }
[1] x 64
[1] y 64
The result obtained by me is wrong in second case.
I understood that variables x and y are treated as characters. But to
rectify this problem.
Regards,
Purna
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Re: [R] problem in finding sizes of objects using a for loop
Dear Purna,
You need the get() function around object[i] in order to accomplish the
same results:
# data
x-rnorm(1)
y-runif(100,min=40,max=1000)
# sizes
objects-ls()
for (i in seq_along(objects)){
print(c(objects[i],object.size(get(objects[i] # get() is added here
}
[1] x 80040
[1] y 840
get() is needed because each element of objects is a character and the
object.size() function does not operates on characters, but on objects (not
your variable, the definition in R). See ?get and ?object.size for more
information.
HTH,
Jorge.-
On Thu, Oct 25, 2012 at 5:24 PM, Purna chander wrote:
Dear All,
I wanted to extract the sizes of all created objects. For E.g when I
created 2 objects(x and y), I got their sizes using the following
code:
x-rnorm(1)
y-runif(100,min=40,max=1000)
ls()
[1] x y
object.size(x)
80024 bytes
object.size(y)
824 bytes
However, I was unable to get their sizes when I used a for loop in the
following way:
objects-ls()
for (i in seq_along(objects)){
+ print(c(objects[i],object.size(objects[i])))
+
+ }
[1] x 64
[1] y 64
The result obtained by me is wrong in second case.
I understood that variables x and y are treated as characters. But to
rectify this problem.
Regards,
Purna
__
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http://www.R-project.org/posting-guide.html
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[R] Instrumental variables for the competitive price
Dear R user, I have been setting up the models for predicting the volume based on the price information of own product and competitive products. one option is to use instrumental variable to break price into two parts: one part that might be correlated with error term, and the another part that is not. But now I met the problem of choosing instumental variables. I have searched many papers. it has been mentioned that cost can be used as one instrumental variable for the price. yes. I can use cost for own price. but how about the competitive price's instrumental variables? I don't know their cost information at all except their price information and market share. I will be really appreciated if someone can suggest me the possible choices of instrumental variables for the competitive price? Thanks in advance. Kind regards, Tammy [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] correlated events in time series
Apologies in advance for the basic nature of my question.
I've never worked with time series, but I am, at present, dealing
with evolution in time of certain scalar quantities.
By looking at the plots, scalar quantity vs time, for several of these
quantities, I am observing a correlation of events happening at specific,
non-regularly spaced instants of time. The fact of observing them in all
plots, to me means that they are due to some kind of objective physical
cause.
My question is: are there well known and effective methods in time series
analysis
to extract such occurrences from multiple series? Is there any package in R
that deals
with this?
Kind regards,
J
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Re: [R] Defining categories
Hi Maybe also findInterval can be used dat1$cat-findInterval(dat1$V1, 1:6, rightmost.closed = T, all.inside = T) It is said to be more efficient than cut. Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of arun Sent: Thursday, October 25, 2012 3:32 AM To: Jorge I Velez; bibek sharma Cc: R help Subject: Re: [R] Defining categories Hi, (Jorge: Thanks for the suggestion.) cut? will be much easier. dat1-read.table(text= 2.880556 0.616667 5.08 0.858333 0.47 2.936111 4.258333 0.258333 2.03 2.58 1.09 0.447222 1.87 0.080556 4.03 4.116667 1.63 2.147222 ,sep=,header=FALSE) dat1$Categ-cut(dat1$V1,breaks=c(0,1,2,3,4,5,6)) #Either library(car) dat1$Categ- recode(dat1$Categ,'(0,1]'=1;'(1,2]'=2;'(2,3]'=3;'(3,4]'=4;'(4,5]'=5;'( 5,6]'=6) #or dat1$Categ-as.numeric(gsub(.*\\,(\\d+).*,\\1,dat1$Categ)) #formats the Categ column. head(dat1) # V1 Categ #1 2.880556 3 #2 0.616667 1 #3 5.08 6 #4 0.858333 1 #5 0.47 1 #6 2.936111 3 A.K. From: Jorge I Velez jorgeivanve...@gmail.com To: arun smartpink...@yahoo.com Cc: bibek sharma mbhpat...@gmail.com; R help r-help@r-project.org Sent: Wednesday, October 24, 2012 7:27 PM Subject: Re: [R] Defining categories See ?cut for a simpler way of doing this. HTH, Jorge.- On Thu, Oct 25, 2012 at 10:02 AM, arun wrote: Hi, May be this: dat1-read.table(text= 2.880556 0.616667 5.08 0.858333 0.47 2.936111 4.258333 0.258333 2.03 2.58 1.09 0.447222 1.87 0.080556 4.03 4.116667 1.63 2.147222 ,sep=,header=FALSE) dat1$category-ifelse(dat1$V1=1 dat1$V10,1,ifelse(dat1$V11 dat1$V1=2,2,ifelse(dat1$V12dat1$V1=3,3,ifelse(dat1$V13dat1$V1=4, 4,ifelse(dat1$V14dat1$V1=5,5,6) head(dat1) # V1 category #1 2.880556 3 #2 0.616667 1 #3 5.08 6 #4 0.858333 1 #5 0.47 1 #6 2.936111 3 A.K. - Original Message - From: bibek sharma To: r-help@r-project.org Cc: Sent: Wednesday, October 24, 2012 6:52 PM Subject: [R] Defining categories Hello R user, Data below represent year in decimal. I would like to catagorize it in such a way that any valye [0,1] goes to catagory 1 , (1,2] goes to catagory 2 and so on.. Any suggestion how it can be done with if else statement or any other way? 2.880556 0.616667 5.08 0.858333 0.47 2.936111 4.258333 0.258333 2.03 2.58 1.09 0.447222 1.87 0.080556 4.03 4.116667 1.63 2.147222 Thank you for your help. Bibek __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Regarding the memory allocation problem
Dear All, My main objective was to compute the distance of 10 vectors from a set having 900 other vectors. I've a file named seq_vec containing 10 records and 256 columns. While computing, the memory was not sufficient and resulted in error cannot allocate vector of size 152.1Mb So I've approached the problem in the following: Rather than reading the data completely at a time, I read the data in chunks of 2 records using scan() function. After reading each chunk, I've computed distance of each of these vectors with a set of another vectors. Even though I was successful in computing the distances for first 3 chunks, I obtained similar error (cannot allocate vector of size 102.3Mb). Q) Here what I could not understand is, how come memory become insufficient when dealing with 4th chunk? Q) Suppose if i computed a matrix 'm' during calculation associated with chunk1, then is this matrix not replaced when I again compute 'm' when dealing with chunk 2? Regards, Purna __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Defining categories
Thank you Dr. Pikal for this alternative. Best, Jorge.- Sent from my phone. Please excuse my brevity and misspelling. On Oct 25, 2012, at 8:29 PM, PIKAL Petr petr.pi...@precheza.cz wrote: Hi Maybe also findInterval can be used dat1$cat-findInterval(dat1$V1, 1:6, rightmost.closed = T, all.inside = T) It is said to be more efficient than cut. Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of arun Sent: Thursday, October 25, 2012 3:32 AM To: Jorge I Velez; bibek sharma Cc: R help Subject: Re: [R] Defining categories Hi, (Jorge: Thanks for the suggestion.) cut? will be much easier. dat1-read.table(text= 2.880556 0.616667 5.08 0.858333 0.47 2.936111 4.258333 0.258333 2.03 2.58 1.09 0.447222 1.87 0.080556 4.03 4.116667 1.63 2.147222 ,sep=,header=FALSE) dat1$Categ-cut(dat1$V1,breaks=c(0,1,2,3,4,5,6)) #Either library(car) dat1$Categ- recode(dat1$Categ,'(0,1]'=1;'(1,2]'=2;'(2,3]'=3;'(3,4]'=4;'(4,5]'=5;'( 5,6]'=6) #or dat1$Categ-as.numeric(gsub(.*\\,(\\d+).*,\\1,dat1$Categ)) #formats the Categ column. head(dat1) #V1 Categ #1 2.880556 3 #2 0.616667 1 #3 5.08 6 #4 0.858333 1 #5 0.47 1 #6 2.936111 3 A.K. From: Jorge I Velez jorgeivanve...@gmail.com To: arun smartpink...@yahoo.com Cc: bibek sharma mbhpat...@gmail.com; R help r-help@r-project.org Sent: Wednesday, October 24, 2012 7:27 PM Subject: Re: [R] Defining categories See ?cut for a simpler way of doing this. HTH, Jorge.- On Thu, Oct 25, 2012 at 10:02 AM, arun wrote: Hi, May be this: dat1-read.table(text= 2.880556 0.616667 5.08 0.858333 0.47 2.936111 4.258333 0.258333 2.03 2.58 1.09 0.447222 1.87 0.080556 4.03 4.116667 1.63 2.147222 ,sep=,header=FALSE) dat1$category-ifelse(dat1$V1=1 dat1$V10,1,ifelse(dat1$V11 dat1$V1=2,2,ifelse(dat1$V12dat1$V1=3,3,ifelse(dat1$V13dat1$V1=4, 4,ifelse(dat1$V14dat1$V1=5,5,6) head(dat1) #V1 category #1 2.8805563 #2 0.6166671 #3 5.086 #4 0.8583331 #5 0.471 #6 2.9361113 A.K. - Original Message - From: bibek sharma To: r-help@r-project.org Cc: Sent: Wednesday, October 24, 2012 6:52 PM Subject: [R] Defining categories Hello R user, Data below represent year in decimal. I would like to catagorize it in such a way that any valye [0,1] goes to catagory 1 , (1,2] goes to catagory 2 and so on.. Any suggestion how it can be done with if else statement or any other way? 2.880556 0.616667 5.08 0.858333 0.47 2.936111 4.258333 0.258333 2.03 2.58 1.09 0.447222 1.87 0.080556 4.03 4.116667 1.63 2.147222 Thank you for your help. Bibek __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] correlated events in time series
Yes, but you'll need to learn vector (multivariate) time series methods. See,
perhaps firstly, B Pfaff's book and corresponding R packages. It's dense, but
not too long and will get you going the right way. Terms like VAR and VECM will
help guide your googling
Michael
On Oct 25, 2012, at 8:34 AM, james.fo...@diamond.ac.uk wrote:
Apologies in advance for the basic nature of my question.
I've never worked with time series, but I am, at present, dealing
with evolution in time of certain scalar quantities.
By looking at the plots, scalar quantity vs time, for several of these
quantities, I am observing a correlation of events happening at specific,
non-regularly spaced instants of time. The fact of observing them in all
plots, to me means that they are due to some kind of objective physical
cause.
My question is: are there well known and effective methods in time series
analysis
to extract such occurrences from multiple series? Is there any package in R
that deals
with this?
Kind regards,
J
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[R] cut point in ROC
var1 var2 var3 var4 var5 var6 var7 var8 var9 var10 gold 2 3 1 2 4 0 1 4 4 3 2 2 4 2 4 3 4 2 4 4 4 2 3 3 0 0 4 1 0 2 4 4 2 1 4 0 3 2 0 0 2 4 4 2 3 4 0 2 2 0 0 0 3 4 2 2 2 3 2 2 0 0 0 2 4 2 2 4 1 1 2 0 0 3 3 3 2 3 4 1 4 0 0 0 0 3 4 2 3 1 0 2 2 1 0 2 3 3 2 0 3 1 1 1 1 2 1 2 3 2 1 1 0 1 0 0 0 0 1 2 1 1 1 0 0 0 0 0 0 2 2 1 1 2 0 0 0 0 0 0 2 1 1 1 1 0 0 0 0 0 0 2 2 1 0 1 1 0 1 0 0 0 0 3 1 1 2 0 1 0 0 0 0 1 1 1 1 2 1 0 0 0 0 0 1 1 1 1 2 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 2 2 1 0 2 0 1 0 0 0 0 1 2 1 Dear All I try to find a diagnosis test instead of the gold standard. The gold standard is based on 10 vars. Now, I try to find a new test based on several vars. For example, based on 3 or 4 vars. There are a lot of choices to test, I am confused. I will be happy if you guide me. Best Regards, Soheila [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Changing radii line type in radial plots
On 10/25/2012 05:06 AM, bwone wrote: I am using the package plotrix radial.plot(). Yes, radial.plot() has a line type argument, lty, but that is for the polygons or the radial lines, not the radii or axes of the radial plot.unless I am doing something wrong. Hi bwone, No, there is no way to change the line type of the grid in radial.plot at the moment. If you need to have different line types, you could change two lines in the function: polygon(xpos,ypos,border=grid.col,col=grid.bg) to polygon(xpos,ypos,border=grid.col,col=grid.bg,lty=my.lty) if(show.radial.grid) segments(0,0,xpos,ypos,col=grid.col) to if(show.radial.grid) segments(0,0,xpos,ypos,col=grid.col,lty=my.lty) I might be persuaded to add yet another argument to the function if anyone else wants different line types. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [r] How to pick colums from a ragged array?
Sorry, forgot to cc to rhelp
Petr
-Original Message-
From: PIKAL Petr
Sent: Thursday, October 25, 2012 11:19 AM
To: 'Stuart Leask'; arun (smartpink...@yahoo.com)
Subject: RE: [r] How to pick colums from a ragged array?
Hi
If I understand correctly you now want only to identify rows for which
for a given ID, two or more first or last DATEs are same but DG is
different and put TRUE/FALSE to new column
fff-function(data) {
data$Identify - FALSE
testfirst - function(x) (x[1,DATE]==x[2,DATE])
(x[1,DG]!=x[2,DG]) testlast - function(x) {
(x[nrow(x),DATE]==x[nrow(x)-1,DATE]) (x[nrow(x),DG]!=x[nrow(x)-
1,DG])
}
sel - as.numeric(names(which(unlist(sapply(split(data,data[,1]),
testfirst)
sel - c(sel, as.numeric(names(which(unlist(sapply(split(data,
data[,1]), testlast))
data[data[,1] %in% sel,Identify] - TRUE data }
I slightly modified my code to get rid of necessary user selection of
first or last variant and put both together, add a new column and
extended testing functions to evaluate DG and look if they are the same
or different.
Does it suit to your purpose?
Regards
Petr
-Original Message-
From: Stuart Leask [mailto:stuart.le...@nottingham.ac.uk]
Sent: Wednesday, October 24, 2012 5:25 PM
To: arun (smartpink...@yahoo.com); PIKAL Petr; Rui Barradas
(ruipbarra...@sapo.pt)
Subject: RE: [r] How to pick colums from a ragged array?
Arun, Petr, Rui, many thanks for your help, and the functions you
have
written.
You'll recall I wanted to remove these first (or last) duplicates,
because they represented instances where two different diagnoses (in
this case, variable DG, value 1, 2, 3, 4 or 5) had been recorded on
the same day - so I can't say which was 'first' (or 'last').
Your functions have revealed something I wasn't expecting: In some
cases, the diagnoses recorded on the duplicated DATEs are the same!
This is a surprise to me, but probably reflects someone going to two
different departments in a clinic, and both departments submit data.
I
have to say that crazy things like this are often a feature of real
data, which I'm sure you've come across yourselves.
Of course, I don't want to remove records in which I can determine an
unambiguous 'first diagnosis'.
You have all put in so much effort on my behalf, I'm ashamed to ask,
but I wonder if any of the functions you've written could do this
with
a little more Indexing and the 'duplicate' function So the function
should only exclude an ID, having identified a first (or last) DATE
duplicate, the DGs for these two dates are different.
Test dataset:
ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
,547,794,814,814,814,814,814,814,841,841,841,841,841
,841,841,841,841,910,910,910,910,910,910,999,1019,1019
,1019)
DATE -
c(20060821,20061207,20080102,20090904,20040205,20040205,2005
,20060111,20071119,20080107,20080407,20080521,20080521,20041005
,20070905,20020814,20021125,20040429,20040429,20071205,20071205
,20050421,20050421,20060428,20060602,20060816,20061025,20061129
,20070112,20070514, 19870508,20040205,20040205, 20080521,20080521
,20091224,20050503,19870508,19870508,19880330)
DG-
c(1,2,1,1,4,4,3,2,3,2,1,2,3,2,1,2,2,2,2,2,2,1,2,1,1,1,1,1,1,4,3,3,3,4,
3
,2,2,2,1,1)
id.d-data.frame(ID,DATE,DG)
id.d
# Considering Ruis getRepeat function:
g.r-getRepeat(id.d)# defaults to first = TRUE getRepeat(id.d,
first = FALSE) to get the last ones
g.rr-do.call(rbind, g.r) # put the data into a matrix
# I can remove the date duplicates with:
g.rr[rep(!duplicated(g.rr)[(1:(dim(g.rr)[1]/2))*2],each=2),]
I'm not sure how to add this to your suggestions, Arun Petr...
Stuart
-Original Message-
From: PIKAL Petr [mailto:petr.pi...@precheza.cz]
Sent: 23 October 2012 15:24
To: Stuart Leask
Subject: RE: [r] How to pick colums from a ragged array?
Hi
I assumed that id.d is data frame
id.d - data.frame (ID,DATE )
and
fff(id.d)
works for me
Petr
-Original Message-
From: Stuart Leask [mailto:stuart.le...@nottingham.ac.uk]
Sent: Tuesday, October 23, 2012 3:13 PM
To: PIKAL Petr
Subject: RE: [r] How to pick colums from a ragged array?
Hi Petr.
I see what you mean it should do, but when I run it I get an error
(see below).
Stuart
ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
+ ,547,794,814,814,814,814,814,814,841,841,841,841,841
+ ,841,841,841,841,910,910,910,910,910,910,999,1019,1019
+ ,1019)
DATE -
+ c(20060821,20061207,20080102,20090904,20040205,20040205,2005
+ ,20060111,20071119,20080107,20080407,20080521,20080711,20041005
+ ,20070905,20020814,20021125,20040429,20040429,20071205,20080227
+ ,20050421,20050421,20060428,20060602,20060816,20061025,20061129
+ ,20070112,20070514, 19870508,20040205,20040205,
Re: [R] How to quit R script return to R prompt
you can see it goes wrong at: barplot(xtab(`profits,data=Forbes2000)) You typed a ` without closing it: barplot(xtab(`profits`,data=Forbes2000)) anyway: pushing the escape button should also return you to the R-prompt (at least on a Windows platform) -- View this message in context: http://r.789695.n4.nabble.com/How-to-quit-R-script-return-to-R-prompt-tp4647376p4647382.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Expression in legend of plot
I would like to have an expression f^(-1)(x) in a legend of plot. For this I
used expression(f^{-1}(x)), but variable is always in exponent. How could I
change it in order to (x) be in a line, not in exponent?
Thank you for your responses!
--
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[R] How to quit R script return to R prompt
Hi can you please tell me how to quit R script return to R prompt. As i tried following but still cannot able to return on to R prompt.. R barplot(Forbes2000$profits) R barplot(xtab(`profits,data=Forbes2000)) + barplot(xtab(~profits,data=Forbes2000)) + ) + Q() + ?barplots + + + + + barplot? + ? + ?? + + stop() + exit() + + + ctrl+c + q() + I'm beginner in R trying to learn Thanks Regards Amitesh -- View this message in context: http://r.789695.n4.nabble.com/How-to-quit-R-script-return-to-R-prompt-tp4647376.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] parallel processing with foreach
Hi,
I am trying to parallel computing with foreach function, but not able to
get the result. I know that in parallel processing, all result is collected
in list format, but I am not able to get input there.
Any help is really appreciated.
esf.m -foreach (i = 1:n.s, .combine=rbind) %dopar% {
EV - as.data.frame(eig$vectors[,1:n.candid[i]])
colnames(EV) - paste(EV, 1:NCOL(EV), sep=)
r25.esf.f - lm(y ~ x1 + x2 +., data = EV)
assign(paste(r25.esf., i, sep=), stepwise.forward(r25.esf.f, lm(y ~ x1
+ x2, data = EV), 0.1, verbose = F))}
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Re: [R] equation solver
Dear Berend, Many thanks for the advice. I'll try to work with these packages and I hope to find a solution. again, many thanks Pina. -- View this message in context: http://r.789695.n4.nabble.com/equation-solver-tp4647287p4647388.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [r] How to pick colums from a ragged array?
Hi Stuart, So, I guess my result (below) serves the purpose! A.K. - Original Message - From: Stuart Leask stuart.le...@nottingham.ac.uk To: arun smartpink...@yahoo.com Cc: Sent: Thursday, October 25, 2012 3:13 AM Subject: RE: [r] How to pick colums from a ragged array? Even confusing myself now, serves me right for replying late at night! ** If DGs are the same, then the first (or last) diagnosis is unambiguous even if date is duplicated - so I can use the data.** Consider we want INCLUDE.FIRST to look at first dates. Duplicate dates: 167, 323,814, 841, 910 1019 AND This dup is the first date: 167, 841, 1019 AND This dup has different DGs: 841 1019 = give all rows of 841 and 1019 FALSE. (All other rows TRUE) Now consider we want INCLUDE.LAST to look at last dates. Duplicate dates: 167, 323,814, 841, 910 1019 AND This dup is the last date: 167, 323, 814 AND This dup has different DGs: 323 = give all rows of 323 FALSE. (All others TRUE) Of course, I'm happy to run a function twice, either one with a 'first/last' switch, or one that assumes initial order of sort by DATE determines whether you end up with first or last date duplicates. -Original Message- From: arun [mailto:smartpink...@yahoo.com] Sent: 24 October 2012 22:59 To: Stuart Leask Subject: Re: [r] How to pick colums from a ragged array? Hi Stuart, So, 167 should be FALSE eventhough DG is same because it comes under earliest/first date, but TRUE for 814 because it comes under latest/last date. 167 comes under both cases. Let me try to make sense of that: I am just pasting my earlier solution and its results again to see whether we are on the same page: res1- data.frame(flag=tapply(id.d[,2],id.d[,1],FUN=function(x) head(duplicated(x)|duplicated(x,fromLast=TRUE),1)|tail(duplicated(x)|duplicated(x,fromLast=TRUE),1))) res2-id.d[id.d[,1]%in%names(res1[res1$flag==TRUE,])(duplicated(id.d[,1:2])|duplicated(id.d[,1:2],fromLast=TRUE)),] res3-res2[!res2$ID%in% res2[duplicated(res2)|duplicated(res2,fromLast=TRUE),]$ID,] id.d1-id.d bad-id.d1[id.d1$ID%in%res3$ID,] bad$INCLUDE-FALSE res4-merge(id.d1,bad,all=TRUE) res4$INCLUDE[is.na(res4$INCLUDE)]-TRUE res4 ID DATE DG INCLUDE 1 58 20060821 1 TRUE 2 58 20061207 2 TRUE 3 58 20080102 1 TRUE 4 58 20090904 1 TRUE 5 167 20040205 4 TRUE 6 167 20040205 4 TRUE 7 323 2005 3 FALSE 8 323 20060111 2 FALSE 9 323 20071119 3 FALSE 10 323 20080107 2 FALSE 11 323 20080407 1 FALSE 12 323 20080521 2 FALSE 13 323 20080521 3 FALSE 14 547 20041005 2 TRUE 15 794 20070905 1 TRUE 16 814 20020814 2 TRUE 17 814 20021125 2 TRUE 18 814 20040429 2 TRUE 19 814 20040429 2 TRUE 20 814 20071205 2 TRUE 21 814 20071205 2 TRUE 22 841 20050421 1 FALSE 23 841 20050421 2 FALSE 24 841 20060428 1 FALSE 25 841 20060602 1 FALSE 26 841 20060816 1 FALSE 27 841 20061025 1 FALSE 28 841 20061129 1 FALSE 29 841 20070112 1 FALSE 30 841 20070514 4 FALSE 31 910 19870508 3 TRUE 32 910 20040205 3 TRUE 33 910 20040205 3 TRUE 34 910 20080521 3 TRUE 35 910 20080521 4 TRUE 36 910 20091224 2 TRUE 37 999 20050503 2 TRUE 38 1019 19870508 1 FALSE 39 1019 19870508 2 FALSE 40 1019 19880330 1 FALSE A.K. - Original Message - From: Stuart Leask stuart.le...@nottingham.ac.uk To: arun smartpink...@yahoo.com; Rui Barradas ruipbarra...@sapo.pt Cc: R help r-help@r-project.org Sent: Wednesday, October 24, 2012 5:40 PM Subject: RE: [r] How to pick colums from a ragged array? I mis-typed, missing an if. I think you've got it, but let me try again: The function should: - put FALSE in a column for every instance of an ID IF ( that ID has a first (or last) DATE duplicated ) AND IF (the DGs for the duplicated dates are different). So for the earliest/first date function, INCLUDE should be TRUE, apart from FALSE for _all_ the instances of IDs 167, 841 and 1019 For the latest/last date function, INCLUDE should be TRUE, apart from FALSE for all the instances of ID 323. Stuart -Original Message- From: arun [mailto:smartpink...@yahoo.com] Sent: 24 October 2012 21:30 To: Rui Barradas Cc: R help; Stuart Leask Subject: Re: [r] How to pick colums from a ragged array? Hi, According to the OP So the function should only exclude an ID, having identified a first (or last) DATE duplicate, the DGs for these two dates are different. Rui: By running your modified function (using dte - tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x, 2),fromLast = TRUE))), id.d$INCLUDE - !(rm1 | rm2) head(id.d) # ID DATE DG INCLUDE #1 58 20060821 1 TRUE #2 58 20061207 2 TRUE #3 58 20080102 1 TRUE #4 58 20090904 1 TRUE #5 167 20040205 4 FALSE #6 167 20040205 4 FALSE For #167, DGs are same. Not sure whether to exclude it or not. My modified solution is similar but I am excluding 167 and 814.
Re: [R] Expression in legend of plot
Hello,
Use an asterisk to put a space between the exponent an (x).
plot(1, main = expression(f^{-1}*(x)))
Hope this helps,
Rui Barradas
Em 25-10-2012 12:21, stat.kk escreveu:
I would like to have an expression f^(-1)(x) in a legend of plot. For this I
used expression(f^{-1}(x)), but variable is always in exponent. How could I
change it in order to (x) be in a line, not in exponent?
Thank you for your responses!
--
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Re: [R] Defining categories
HI Petr, Thanks for sharing the function. True, very efficient than cut. dat1$cat-findInterval(dat1$V1, 1:6, rightmost.closed = T, all.inside = T) ^^^ # May be the interval is 0:6. dat1$cat1-findInterval(dat1$V1, 1:6, rightmost.closed = T, all.inside = T) dat1$cat2-findInterval(dat1$V1, 0:6, rightmost.closed = T, all.inside = T) head(dat1) # V1 cat1 cat2 #1 2.880556 2 3 #2 0.616667 1 1 #3 5.08 5 6 #4 0.858333 1 1 #5 0.47 1 1 #6 2.936111 2 3 A.K. - Original Message - From: PIKAL Petr petr.pi...@precheza.cz To: arun smartpink...@yahoo.com; Jorge I Velez jorgeivanve...@gmail.com; bibek sharma mbhpat...@gmail.com Cc: R help r-help@r-project.org Sent: Thursday, October 25, 2012 5:29 AM Subject: RE: [R] Defining categories Hi Maybe also findInterval can be used dat1$cat-findInterval(dat1$V1, 1:6, rightmost.closed = T, all.inside = T) It is said to be more efficient than cut. Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of arun Sent: Thursday, October 25, 2012 3:32 AM To: Jorge I Velez; bibek sharma Cc: R help Subject: Re: [R] Defining categories Hi, (Jorge: Thanks for the suggestion.) cut? will be much easier. dat1-read.table(text= 2.880556 0.616667 5.08 0.858333 0.47 2.936111 4.258333 0.258333 2.03 2.58 1.09 0.447222 1.87 0.080556 4.03 4.116667 1.63 2.147222 ,sep=,header=FALSE) dat1$Categ-cut(dat1$V1,breaks=c(0,1,2,3,4,5,6)) #Either library(car) dat1$Categ- recode(dat1$Categ,'(0,1]'=1;'(1,2]'=2;'(2,3]'=3;'(3,4]'=4;'(4,5]'=5;'( 5,6]'=6) #or dat1$Categ-as.numeric(gsub(.*\\,(\\d+).*,\\1,dat1$Categ)) #formats the Categ column. head(dat1) # V1 Categ #1 2.880556 3 #2 0.616667 1 #3 5.08 6 #4 0.858333 1 #5 0.47 1 #6 2.936111 3 A.K. From: Jorge I Velez jorgeivanve...@gmail.com To: arun smartpink...@yahoo.com Cc: bibek sharma mbhpat...@gmail.com; R help r-help@r-project.org Sent: Wednesday, October 24, 2012 7:27 PM Subject: Re: [R] Defining categories See ?cut for a simpler way of doing this. HTH, Jorge.- On Thu, Oct 25, 2012 at 10:02 AM, arun wrote: Hi, May be this: dat1-read.table(text= 2.880556 0.616667 5.08 0.858333 0.47 2.936111 4.258333 0.258333 2.03 2.58 1.09 0.447222 1.87 0.080556 4.03 4.116667 1.63 2.147222 ,sep=,header=FALSE) dat1$category-ifelse(dat1$V1=1 dat1$V10,1,ifelse(dat1$V11 dat1$V1=2,2,ifelse(dat1$V12dat1$V1=3,3,ifelse(dat1$V13dat1$V1=4, 4,ifelse(dat1$V14dat1$V1=5,5,6) head(dat1) # V1 category #1 2.880556 3 #2 0.616667 1 #3 5.08 6 #4 0.858333 1 #5 0.47 1 #6 2.936111 3 A.K. - Original Message - From: bibek sharma To: r-help@r-project.org Cc: Sent: Wednesday, October 24, 2012 6:52 PM Subject: [R] Defining categories Hello R user, Data below represent year in decimal. I would like to catagorize it in such a way that any valye [0,1] goes to catagory 1 , (1,2] goes to catagory 2 and so on.. Any suggestion how it can be done with if else statement or any other way? 2.880556 0.616667 5.08 0.858333 0.47 2.936111 4.258333 0.258333 2.03 2.58 1.09 0.447222 1.87 0.080556 4.03 4.116667 1.63 2.147222 Thank you for your help. Bibek __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] List of multidimensional arrays
x and y are the frequencies of missing participants in the intervention and the control treatment respectively! vector t contains the code of the interventions (we have 11 interventions). I re-attach the PDF with some small modifications. I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. On Wed, Oct 24, 2012 at 5:17 PM, Richard M. Heiberger r...@temple.eduwrote: I am looking at your pdf file. it doesn't match your description. There are no missing values in the vectors x, y, t. Each vector has length 55 which is not a multiple of 4, so we don't know where the 2x2 matrices come from. The code doesn't run. There are are too many }. PDF files are formatted and are not ascii. Please send your code in ascii text in the body of the email. Please pick it up from your email and paste it into a fresh R session to be sure that it works. Perhaps consrtruct manually an example of what you want the answer to look like. Rich On Wed, Oct 24, 2012 at 3:51 AM, Loukia Spineli spinelilouki...@gmail.com wrote: Dear all, I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. Any suggestion would be really helpful Thank you in advance, Loukia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. List of multidimensional arrays.pdf Description: Adobe PDF document __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Kaplan Meier Post Hoc?
No, you didn't miss anything in the survival package. I've never found post-hoc tests
interesting so have little motivation to add such (and a very long to do list of things
I would like to add).
If you simply must have them, why not do all pairwise tests?
chisq - matrix(0., 4,4)
for (i in 1:4) {
for (j in (1:4)[-i]) {
temp - survdiff(Surv(time, status) ~ group, data=mydata,
subset=(group %in% (unique(group))[c(i,j)]))
chisq[i,j] - temp$chisq
}
}
Terry Therneau
On 10/25/2012 05:00 AM, r-help-requ...@r-project.org wrote:
This is more of a general question without data. After doing 'survdiff',
from the 'survival' package, on strata including four groups (so 4 curves
on a Kaplan Meier curve) you get a chi squared p-value whether to reject
the null hypothesis or not. Is there a method to followup with pairwise
testing on the respective groups? I have searched the library but have
come up with nothing. Perhaps I am mistaken in something here.
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Re: [R] problem in finding sizes of objects using a for loop
Hi,
You can also use lapply() or sapply() without the get() function.
list1-list(x=x,y=y)
sapply(list1,object.size)
# x y
#80040 840
do.call(rbind,lapply(list1,object.size))
# [,1]
#x 80040
#y 840
A.K.
- Original Message -
From: Jorge I Velez jorgeivanve...@gmail.com
To: Purna chander chander...@gmail.com
Cc: r-help r-help@r-project.org
Sent: Thursday, October 25, 2012 2:40 AM
Subject: Re: [R] problem in finding sizes of objects using a for loop
Dear Purna,
You need the get() function around object[i] in order to accomplish the
same results:
# data
x-rnorm(1)
y-runif(100,min=40,max=1000)
# sizes
objects-ls()
for (i in seq_along(objects)){
print(c(objects[i],object.size(get(objects[i] # get() is added here
}
[1] x 80040
[1] y 840
get() is needed because each element of objects is a character and the
object.size() function does not operates on characters, but on objects (not
your variable, the definition in R). See ?get and ?object.size for more
information.
HTH,
Jorge.-
On Thu, Oct 25, 2012 at 5:24 PM, Purna chander wrote:
Dear All,
I wanted to extract the sizes of all created objects. For E.g when I
created 2 objects(x and y), I got their sizes using the following
code:
x-rnorm(1)
y-runif(100,min=40,max=1000)
ls()
[1] x y
object.size(x)
80024 bytes
object.size(y)
824 bytes
However, I was unable to get their sizes when I used a for loop in the
following way:
objects-ls()
for (i in seq_along(objects)){
+ print(c(objects[i],object.size(objects[i])))
+
+ }
[1] x 64
[1] y 64
The result obtained by me is wrong in second case.
I understood that variables x and y are treated as characters. But to
rectify this problem.
Regards,
Purna
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[[alternative HTML version deleted]]
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Re: [R] Expression in legend of plot
... and note that the { ,} are unnecessary:
plot(1, main = bquote(f^-1*(x)))
-- Bert
On Thu, Oct 25, 2012 at 4:32 AM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Use an asterisk to put a space between the exponent an (x).
plot(1, main = expression(f^{-1}*(x)))
Hope this helps,
Rui Barradas
Em 25-10-2012 12:21, stat.kk escreveu:
I would like to have an expression f^(-1)(x) in a legend of plot. For this
I
used expression(f^{-1}(x)), but variable is always in exponent. How could
I
change it in order to (x) be in a line, not in exponent?
Thank you for your responses!
--
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http://r.789695.n4.nabble.com/Expression-in-legend-of-plot-tp4647393.html
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--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
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Re: [R] bugs and misfeatures in polr(MASS).... fixed!
This could well be out of date because I have not paid any attention to the official POLR code in a year or more. Attached is fixed-polr.R. cheers, Tim On Wed, Oct 24, 2012 at 10:38 PM, ahs [via R] ml-node+s789695n4647311...@n4.nabble.com wrote: Great! You can skip my question about s0 though, I found where it is being used, but I still struggle with the code and convergence problem. I also found something here http://biostat.mc.vanderbilt.edu/wiki/pub/Main/CharlesDupontStuff/newPolr.R that seems like someone tried to fix it, but with this code I get mu starting values out of range, nor do I get it right when I try to force starting values. If you got your code to work fine, you might have the answer to my problem. If you want, maybe you could post your whole *fixed-polr.R*? :-) I also can't see where the new object s0 is used in the old code? -- If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/bugs-and-misfeatures-in-polr-MASS-fixed-tp3024677p4647311.html To unsubscribe from bugs and misfeatures in polr(MASS) fixed!, click herehttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=3024677code=dGltb3RoeS5iZW5oYW1AdXFjb25uZWN0LmVkdS5hdXwzMDI0Njc3fDE5NTE2NDMxMjk= . NAMLhttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=macro_viewerid=instant_html%21nabble%3Aemail.namlbase=nabble.naml.namespaces.BasicNamespace-nabble.view.web.template.NabbleNamespace-nabble.view.web.template.NodeNamespacebreadcrumbs=notify_subscribers%21nabble%3Aemail.naml-instant_emails%21nabble%3Aemail.naml-send_instant_email%21nabble%3Aemail.naml fixed-polr.R (28K) http://r.789695.n4.nabble.com/attachment/4647403/0/fixed-polr.R - Tim J. Benham -- View this message in context: http://r.789695.n4.nabble.com/bugs-and-misfeatures-in-polr-MASS-fixed-tp3024677p4647403.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cut point in ROC
Please use dput while supplying data. Best Regards, Bhupendrasinh Thakre Sent from my iPhone On Oct 25, 2012, at 5:31 AM, Soheila Khodakarim lkhodaka...@gmail.com wrote: var1 var2 var3 var4 var5 var6 var7 var8 var9 var10 gold 2 3 1 2 4 0 1 4 4 3 2 2 4 2 4 3 4 2 4 4 4 2 3 3 0 0 4 1 0 2 4 4 2 1 4 0 3 2 0 0 2 4 4 2 3 4 0 2 2 0 0 0 3 4 2 2 2 3 2 2 0 0 0 2 4 2 2 4 1 1 2 0 0 3 3 3 2 3 4 1 4 0 0 0 0 3 4 2 3 1 0 2 2 1 0 2 3 3 2 0 3 1 1 1 1 2 1 2 3 2 1 1 0 1 0 0 0 0 1 2 1 1 1 0 0 0 0 0 0 2 2 1 1 2 0 0 0 0 0 0 2 1 1 1 1 0 0 0 0 0 0 2 2 1 0 1 1 0 1 0 0 0 0 3 1 1 2 0 1 0 0 0 0 1 1 1 1 2 1 0 0 0 0 0 1 1 1 1 2 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 2 2 1 0 2 0 1 0 0 0 0 1 2 1 Dear All I try to find a diagnosis test instead of the gold standard. The gold standard is based on 10 vars. Now, I try to find a new test based on several vars. For example, based on 3 or 4 vars. There are a lot of choices to test, I am confused. I will be happy if you guide me. Best Regards, Soheila [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to quit R script return to R prompt
You typed a ` without closing it:
barplot(xtab(`profits`,data=Forbes2000))
anyway: pushing the escape button should also return you to
the R-prompt (at least on a Windows platform)
But why is the OP using a backtick at all? It's not necessary in this instance
(and in fact it's very rarely necessary at all unless you have used a
syntactically invalid name with something silly like spaces in it). See ?Quotes
for a little more detail on what ` does.
Also, xtab should possibly be xtabs, and the first argument to that is a
one-sided formula, not a vector:
library(HSAUR)
barplot(xtabs( ~ profits,data=Forbes2000))
should work fine.
***
This email and any attachments are confidential. Any use...{{dropped:8}}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] List of multidimensional arrays
You were asked to provide R code in ascii text. Also use dput to get the objects into a mail. Nobody is going to make the effort to copy from your pdf. I certainly will not. Berend On 25-10-2012, at 13:45, Loukia Spineli wrote: x and y are the frequencies of missing participants in the intervention and the control treatment respectively! vector t contains the code of the interventions (we have 11 interventions). I re-attach the PDF with some small modifications. I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. On Wed, Oct 24, 2012 at 5:17 PM, Richard M. Heiberger r...@temple.eduwrote: I am looking at your pdf file. it doesn't match your description. There are no missing values in the vectors x, y, t. Each vector has length 55 which is not a multiple of 4, so we don't know where the 2x2 matrices come from. The code doesn't run. There are are too many }. PDF files are formatted and are not ascii. Please send your code in ascii text in the body of the email. Please pick it up from your email and paste it into a fresh R session to be sure that it works. Perhaps consrtruct manually an example of what you want the answer to look like. Rich On Wed, Oct 24, 2012 at 3:51 AM, Loukia Spineli spinelilouki...@gmail.com wrote: Dear all, I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. Any suggestion would be really helpful Thank you in advance, Loukia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. List of multidimensional arrays.pdf__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] List of multidimensional arrays
Loukia, Please send your replies to the R-help list. I am CC'ing this to the R-help list. On 25-10-2012, at 15:30, Loukia Spineli wrote: Dear Berend, I am not familiar at all with ascii text. If you could suggest me a good link to learn about it, then it would be really helpful! If that is so, then how are you making your R code or R script? You must be using a plain text editor but we don't know your version of R, your OS, or anything else. If you are only using an R console you should look in the documentation of your system to find out how to work with plain text. Berend Best wishes, Loukia On Thu, Oct 25, 2012 at 4:23 PM, Berend Hasselman b...@xs4all.nl wrote: You were asked to provide R code in ascii text. Also use dput to get the objects into a mail. Nobody is going to make the effort to copy from your pdf. I certainly will not. Berend On 25-10-2012, at 13:45, Loukia Spineli wrote: x and y are the frequencies of missing participants in the intervention and the control treatment respectively! vector t contains the code of the interventions (we have 11 interventions). I re-attach the PDF with some small modifications. I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. On Wed, Oct 24, 2012 at 5:17 PM, Richard M. Heiberger r...@temple.eduwrote: I am looking at your pdf file. it doesn't match your description. There are no missing values in the vectors x, y, t. Each vector has length 55 which is not a multiple of 4, so we don't know where the 2x2 matrices come from. The code doesn't run. There are are too many }. PDF files are formatted and are not ascii. Please send your code in ascii text in the body of the email. Please pick it up from your email and paste it into a fresh R session to be sure that it works. Perhaps consrtruct manually an example of what you want the answer to look like. Rich On Wed, Oct 24, 2012 at 3:51 AM, Loukia Spineli spinelilouki...@gmail.com wrote: Dear all, I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. Any suggestion would be really helpful Thank you in advance, Loukia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. List of multidimensional arrays.pdf__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained,
Re: [R] List of multidimensional arrays
Thank you very much Berend!:) On Thu, Oct 25, 2012 at 4:37 PM, Berend Hasselman b...@xs4all.nl wrote: Loukia, Please send your replies to the R-help list. I am CC'ing this to the R-help list. On 25-10-2012, at 15:30, Loukia Spineli wrote: Dear Berend, I am not familiar at all with ascii text. If you could suggest me a good link to learn about it, then it would be really helpful! If that is so, then how are you making your R code or R script? You must be using a plain text editor but we don't know your version of R, your OS, or anything else. If you are only using an R console you should look in the documentation of your system to find out how to work with plain text. Berend Best wishes, Loukia On Thu, Oct 25, 2012 at 4:23 PM, Berend Hasselman b...@xs4all.nl wrote: You were asked to provide R code in ascii text. Also use dput to get the objects into a mail. Nobody is going to make the effort to copy from your pdf. I certainly will not. Berend On 25-10-2012, at 13:45, Loukia Spineli wrote: x and y are the frequencies of missing participants in the intervention and the control treatment respectively! vector t contains the code of the interventions (we have 11 interventions). I re-attach the PDF with some small modifications. I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. On Wed, Oct 24, 2012 at 5:17 PM, Richard M. Heiberger r...@temple.edu wrote: I am looking at your pdf file. it doesn't match your description. There are no missing values in the vectors x, y, t. Each vector has length 55 which is not a multiple of 4, so we don't know where the 2x2 matrices come from. The code doesn't run. There are are too many }. PDF files are formatted and are not ascii. Please send your code in ascii text in the body of the email. Please pick it up from your email and paste it into a fresh R session to be sure that it works. Perhaps consrtruct manually an example of what you want the answer to look like. Rich On Wed, Oct 24, 2012 at 3:51 AM, Loukia Spineli spinelilouki...@gmail.com wrote: Dear all, I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. Any suggestion would be really helpful Thank you in advance, Loukia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. List of multidimensional arrays.pdf__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative
Re: [R] List of multidimensional arrays
With effort I re-inserted the line feeds in your code that the pdf file deleted. There is an error in your code Error in 0:(x[t == i][l]) (from #9) : NA/NaN argument x[t == i] [1] 8 2 9 l [1] 4 On Thu, Oct 25, 2012 at 7:45 AM, Loukia Spineli spinelilouki...@gmail.comwrote: x and y are the frequencies of missing participants in the intervention and the control treatment respectively! vector t contains the code of the interventions (we have 11 interventions). I re-attach the PDF with some small modifications. I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. On Wed, Oct 24, 2012 at 5:17 PM, Richard M. Heiberger r...@temple.eduwrote: I am looking at your pdf file. it doesn't match your description. There are no missing values in the vectors x, y, t. Each vector has length 55 which is not a multiple of 4, so we don't know where the 2x2 matrices come from. The code doesn't run. There are are too many }. PDF files are formatted and are not ascii. Please send your code in ascii text in the body of the email. Please pick it up from your email and paste it into a fresh R session to be sure that it works. Perhaps consrtruct manually an example of what you want the answer to look like. Rich On Wed, Oct 24, 2012 at 3:51 AM, Loukia Spineli spinelilouki...@gmail.com wrote: Dear all, I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. Any suggestion would be really helpful Thank you in advance, Loukia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] List of multidimensional arrays
Please use this model to send your data and code mydata - data.frame(x, y, fx, fy) head(mydata) ## just the first 6 rows are displayed on screen x y fx fy 1 4 3 37 40 2 4 3 21 20 3 0 0 65 50 4 12 0 91 49 5 1 5 22 20 6 2 3 6 10 dump(mydata, ) mydata - structure(list(x = c(4, 4, 0, 12, 1, 2, 0, 2, 6, 1, 0, 20, 6, 5, 20, 0, 0, 0, 4, 1, 0, 16, 0, 9, 7, 11, 0, 2, 3, 0, 10, 2, 1, 0, 3, 6, 4, 0, 20, 11, 1, 8, 0, 6, 15, 0, 2, 5, 2, 6, 24, 1, 11, 6, 9), y = c(3, 3, 0, 0, 5, 3, 0, 5, 4, 2, 0, 20, 6, 1, 18, 0, 0, 3, 3, 1, 0, 4, 0, 13, 7, 12, 0, 2, 2, 0, 2, 4, 0, 4, 2, 5, 0, 0, 23, 8, 0, 14, 0, 9, 20, 0, 2, 0, 2, 2, 14, 4, 1, 4, 7), fx = c(37, 21, 65, 91, 22, 6, 31, 41, 27, 5, 77, 33, 90, 26, 35, 35, 34, 30, 13, 24, 107, 122, 33, 18, 34, 23, 27, 8, 81, 6, 9, 38, 44, 18, 30, 37, 96, 8, 53, 56, 10, 71, 26, 36, 24, 73, 27, 46, 85, 46, 117, 17, 45, 51, 67), fy = c(40, 20, 50, 49, 20, 10, 36, 48, 39, 7, 72, 38, 71, 2, 69, 7, 7, 32, 32, 32, 34, 17, 26, 25, 140, 132, 36, 15, 35, 23, 19, 35, 35, 38, 30, 38, 103, 13, 61, 48, 17, 74, 33, 44, 28, 94, 24, 52, 70, 47, 118, 21, 52, 59, 84)), .Names = c(x, y, fx, fy), row.names = c(NA, -55L), class = data.frame) Now the reader can pick up the mydata - lines and paste them into the R session and reproduce what you want us to see. On Thu, Oct 25, 2012 at 9:53 AM, Richard M. Heiberger r...@temple.eduwrote: With effort I re-inserted the line feeds in your code that the pdf file deleted. There is an error in your code Error in 0:(x[t == i][l]) (from #9) : NA/NaN argument x[t == i] [1] 8 2 9 l [1] 4 On Thu, Oct 25, 2012 at 7:45 AM, Loukia Spineli spinelilouki...@gmail.com wrote: x and y are the frequencies of missing participants in the intervention and the control treatment respectively! vector t contains the code of the interventions (we have 11 interventions). I re-attach the PDF with some small modifications. I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. On Wed, Oct 24, 2012 at 5:17 PM, Richard M. Heiberger r...@temple.eduwrote: I am looking at your pdf file. it doesn't match your description. There are no missing values in the vectors x, y, t. Each vector has length 55 which is not a multiple of 4, so we don't know where the 2x2 matrices come from. The code doesn't run. There are are too many }. PDF files are formatted and are not ascii. Please send your code in ascii text in the body of the email. Please pick it up from your email and paste it into a fresh R session to be sure that it works. Perhaps consrtruct manually an example of what you want the answer to look like. Rich On Wed, Oct 24, 2012 at 3:51 AM, Loukia Spineli spinelilouki...@gmail.com wrote: Dear all, I am trying to create a list, where each list element is a vector of different length arrays that contain 2by2 matrices. To be more specific there are 11 treatments that are compared with placebo (we have 11 comparisons) and each comparison is studied by a different number of trials and each trial has a different number of missing participants in both arms. The length of the list is equal to the number of comparisons. In each comparison the number of arrays is equal to the number of trials that study this comparison. For instance 4 trials compare PAR with placebo. So, for this comparison we have 4 arrays and each array has a length equal to the producy of the number of participants in each arm. These arrays contain 2x2 matrices. I have attached a document with the data and the code. I cannot create the list results the way I have described above. It creates only the matrices for the first array (that has length equal to 135) of the first comparison leaving the rest 10 comparisons NULL. Any suggestion would be really helpful Thank you in advance, Loukia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
[R] Plot lmer model with Effects package
Hi everyone! I have a simple model that i would like to plot with 95% CIs. It is like follows: m1-lmer(Richness~Grazing+I(Grazing^2)+(1|Plot),family=poisson) By using the effects package I get two plots, one for the linear term and one for the squared term. Q1: Can I get all in one? I.e. with one line for the whole model? Q2: Can I also visualize the random effects? I would be very happy for your answers! -- Mvh Marte S. Lilleeng __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] mean of a value of the last 2 hours
Hello, I have a data frame somewhat like that: myframe - data.frame (ID=c(Ernie, Ernie, Ernie, Bert, Bert, Bert), Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00, 24.09.2012 11:00), Hunger=c(1,1,1,2,2,1) ) myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp), %d.%m.%Y %H:%M), tz=GMT) myframe2 - cbind (myframe,myframestime) myframe2$Timestamp - NULL myframe2 I want to add an additional column at the right and get in each row a value which shows the mean of hunger of the last two hours. Does anyone know how that works? That would be very helpful. -- View this message in context: http://r.789695.n4.nabble.com/mean-of-a-value-of-the-last-2-hours-tp4647415.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Egarch (1,1) with Student t distribution in RExcel
Hi I want to implement Egarch (1,1) with t distribution model using RExcel and VBA. May I know the syntax. Following is the code that I 'm using. rinterface.RRun spec=ugarchspec(variance.model=list(model=(eGARCH),garchOrder=c(1,1)), mean.model=list(armaOrder=c(1,1), arfima=FALSE), distribution.model=(std)) rinterface.RRun fit = ugarchfit(Data = b, spec = spec) rinterface.RRun output=sigma(fit) Please let me know the error and it's solution. Thanks Dheeraj [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Expression in legend of plot
So easy solution..., thank you very much :) -- View this message in context: http://r.789695.n4.nabble.com/Expression-in-legend-of-plot-tp4647393p4647397.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] After sorting a dataframe by date
As David said, you sorted by Date. But sorting by rownames is not really the point. The point is that rownames are not line numbers. The rownames were assigned when the data frame was created, and then preserved when you sorted. Sometimes, rownames contain meaningful information that is associated with the data in that row, and therefore must be sorted along with the rows. If you want to change the row names to be analogous to line numbers after sorting you can do, for example, rownames(sorted.df) - seq(nrow(sorted.df)) -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 10/23/12 9:34 AM, martiny my2...@outlook.com wrote: HI, I have created a dataframe df and try to sort it by its date using the order() as below: df-read.csv(constr,header=T) sorted.df-df[order(as.Date(df$Date), decreasing = F),] print(sorted.df) The dataframe was sorted, but the output from the command console shows reserved line order..rather than starting at 1it does not really affect my result, but I want to understand why is that... Date Open HighLow Close Volume Adj.Close 252 2011-01-03 325.64 330.26 324.84 329.57 15897800328.16 251 2011-01-04 332.44 332.50 328.15 331.29 11038600329.87 250 2011-01-05 329.55 334.34 329.50 334.00 9125700332.57 249 2011-01-06 334.72 335.25 332.90 333.73 10729600332.30 248 2011-01-07 333.99 336.35 331.90 336.12 11140400334.68 247 2011-01-10 338.83 343.23 337.17 342.45 1602340.99 246 2011-01-11 344.88 344.96 339.47 341.64 15861000340.18 245 2011-01-12 343.25 344.43 342.00 344.42 10806800342.95 244 2011-01-13 345.16 346.64 343.85 345.68 10599300344.20 ... ... 3 2011-12-28 406.89 408.25 401.34 402.64 8166500400.92 2 2011-12-29 403.40 405.65 400.51 405.12 7713500403.39 1 2011-12-30 403.51 406.28 403.49 405.00 6416500403.27 Any advice would be highly appreciated. -- View this message in context: http://r.789695.n4.nabble.com/After-sorting-a-dataframe-by-date-tp4647173. html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] trying ti use a function in aggregate
Hi -I am using R v 2.13.0. I am trying to use the aggregate function to calculate the percent at length for each Trip_id and CommonName. Here is a small subset of the data. Trip_id Vessel CommonName Length Count 1 230SunlightShad,American 19 1 2 230SunlightShad,American 20 1 3 230SunlightShad,American 21 1 4 230SunlightShad,American 23 1 5 230SunlightShad,American 26 1 6 230SunlightShad,American 27 1 7 230SunlightShad,American 30 2 8 230SunlightShad,American 33 1 9 230SunlightShad,American 34 1 10 230SunlightShad,American 37 1 11 230Sunlight Herring,Blueback 20 1 12 230Sunlight Herring,Blueback 21 2 13 230Sunlight Herring,Blueback 22 5 14 230Sunlight Herring,Blueback 26 1 15 230Sunlight Alewife 17 1 16 230Sunlight Alewife 18 1 17 230Sunlight Alewife 20 2 18 230Sunlight Alewife 21 4 19 230Sunlight Alewife 2216 20 230Sunlight Alewife 2322 21 230Sunlight Alewife 2416 22 230Sunlight Alewife 25 4 23 230Sunlight Alewife 26 1 24 230Sunlight Alewife 27 2 25 230Sunlight Alewife 28 2 26 231 Western VentureShad,American 23 1 27 231 Western VentureShad,American 24 1 28 231 Western VentureShad,American 25 1 29 231 Western VentureShad,American 28 2 30 231 Western VentureShad,American 29 2 My code is: myfun-function (x) x/sum(x) b-with(data,aggregate(x=list(Percent=Count),by=list(Trip_id=Trip_id,Length=Length,Species=CommonName), FUN=myfun)) My issue is that the percent is not be calculated by Trip_id and CommonName. The result is that each row has a percent of 1 indicating that myfun is not dividing by the sum of counts with a Trip_id/CommonName group. Any help would be appreciated. Thank you -- View this message in context: http://r.789695.n4.nabble.com/trying-ti-use-a-function-in-aggregate-tp4647414.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Egarch (1,1) with Student t distribution in RExcel
How Can I run all these codes in VBA using RExcel library(rugarch) spec=ugarchspec(variance.model=list(model=sGARCH,garchOrder=c(1,1)), mean.model=list(armaOrder=c(1,1), arfima=FALSE), distribution.model=std) fit=ugarchfit(data=d, spec=spec) z=sigma(fit) Dheeraj [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help appending randomly selected observation based on criteria
Hello,
I wrote a request yesterday on how to select a random observation that met
certain criteria. I think mostly I have it figured out, but I can't figure
out how to append the observations. The output just has one observation in
it.
Any help would be appreciated,
thanks.
Christy
setwd (C:/christy/roads/)
roads=read.csv(streamland23.csv)
for (i in 1:nrow (roads)){
Sitetype= roads$Sitetype[i]
yr=roads$REACH_Yr[i]
initRchid=roads$RchID[i]
huc1=roads$HUC[i]
sample.df - function(df, n) df[sample(nrow(df), n), , drop = FALSE]
selected=sample.df(roads[roads$HUC == huc1, ], 1)
selectedb=selected
selectedb$oldrch=initRchid
write.csv(selectedb,file=outhuc3.csv,append=TRUE)
}
--
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Re: [R] mean of a value of the last 2 hours
Hi, May be this helps: new1-with(myframe2,aggregate(cbind(Hunger,myframestime),by=list(ID=ID), function(x) mean(tail(x,2[,1:2] merge(myframe2,new1,by=ID,all=TRUE) # ID Hunger.x myframestime Hunger.y #1 Bert 2 2012-09-24 09:00:00 1.5 #2 Bert 2 2012-09-24 10:00:00 1.5 #3 Bert 1 2012-09-24 11:00:00 1.5 #4 Ernie 1 2012-09-24 09:00:00 1.0 #5 Ernie 1 2012-09-24 10:00:00 1.0 #6 Ernie 1 2012-09-24 11:00:00 1.0 A.K. - Original Message - From: Tagmarie ramga...@gmx.net To: r-help@r-project.org Cc: Sent: Thursday, October 25, 2012 10:35 AM Subject: [R] mean of a value of the last 2 hours Hello, I have a data frame somewhat like that: myframe - data.frame (ID=c(Ernie, Ernie, Ernie, Bert, Bert, Bert), Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00, 24.09.2012 11:00), Hunger=c(1,1,1,2,2,1) ) myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp), %d.%m.%Y %H:%M), tz=GMT) myframe2 - cbind (myframe,myframestime) myframe2$Timestamp - NULL myframe2 I want to add an additional column at the right and get in each row a value which shows the mean of hunger of the last two hours. Does anyone know how that works? That would be very helpful. -- View this message in context: http://r.789695.n4.nabble.com/mean-of-a-value-of-the-last-2-hours-tp4647415.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] parallel processing with foreach
It seems you don't quite understand how foreach works. foreach (..)
%dopar% { ... } takes the last value from each of the second {...}
evaluations and feeds them to the .combine function (in your case
rbind()). Since your last call in the %dopar% {...} block is assign(),
you are not getting anything meaningful.
Make the last value a vector that you want to be rbind-ed to the result.
HTH,
Peter
On Thu, Oct 25, 2012 at 1:47 AM, pxs101220 pxs101...@utdallas.edu wrote:
Hi,
I am trying to parallel computing with foreach function, but not able to
get the result. I know that in parallel processing, all result is collected
in list format, but I am not able to get input there.
Any help is really appreciated.
esf.m -foreach (i = 1:n.s, .combine=rbind) %dopar% {
EV - as.data.frame(eig$vectors[,1:n.candid[i]])
colnames(EV) - paste(EV, 1:NCOL(EV), sep=)
r25.esf.f - lm(y ~ x1 + x2 +., data = EV)
assign(paste(r25.esf., i, sep=), stepwise.forward(r25.esf.f, lm(y ~ x1
+ x2, data = EV), 0.1, verbose = F))}
--
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Re: [R] Egarch (1,1) with Student t distribution in RExcel
On Oct 25, 2012, at 8:00 AM, Dheeraj Pandey wrote: How Can I run all these codes in VBA using RExcel library(rugarch) I predict there are a very small number (possibly zero) of persons reading this list that are both Excel coders and users of that package. You might consider posting it on the mailing list where RExcel is supported and so more cross-functional eyes and brains would be seeing it. (And when you do post again, please avoid double posting.) -- David. spec=ugarchspec(variance.model=list(model=sGARCH,garchOrder=c(1,1)), mean.model=list(armaOrder=c(1,1), arfima=FALSE), distribution.model=std) fit=ugarchfit(data=d, spec=spec) z=sigma(fit) Dheeraj [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to quit R script return to R prompt
On Oct 25, 2012, at 6:00 AM, S Ellison wrote: You typed a ` without closing it: barplot(xtab(`profits`,data=Forbes2000)) anyway: pushing the escape button should also return you to the R-prompt (at least on a Windows platform) Also on Macs. But why is the OP using a backtick at all? There was suggestive evidence in the rest of his post that his shift key is wonky and he did mean to type a tilde. It's not necessary in this instance (and in fact it's very rarely necessary at all unless you have used a syntactically invalid name with something silly like spaces in it). See ?Quotes for a little more detail on what ` does. Also, xtab should possibly be xtabs, and the first argument to that is a one-sided formula, not a vector: library(HSAUR) barplot(xtabs( ~ profits,data=Forbes2000)) Actually any unmatched single or double quote would have the same effect. Similar conundrums occur when Turing machines of any sort are reading input data where unmatched quotes result in huge missing gaps as line after line of perfectly useful data gets gathered up into one ungodly large single character element. (The human brain is thus demonstrated not to be a Turing machine.) should work fine. David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mean of a value of the last 2 hours
Hello, Or ct - cut(myframe2$myframestime, breaks = 2 hour) ave(myframe2$Hunger, ct) And assign the output of 'ave' to a new column. Hope this helps, Rui Barradas Em 25-10-2012 15:50, arun escreveu: Hi, May be this helps: new1-with(myframe2,aggregate(cbind(Hunger,myframestime),by=list(ID=ID), function(x) mean(tail(x,2[,1:2] merge(myframe2,new1,by=ID,all=TRUE) # ID Hunger.xmyframestime Hunger.y #1 Bert2 2012-09-24 09:00:00 1.5 #2 Bert2 2012-09-24 10:00:00 1.5 #3 Bert1 2012-09-24 11:00:00 1.5 #4 Ernie1 2012-09-24 09:00:00 1.0 #5 Ernie1 2012-09-24 10:00:00 1.0 #6 Ernie1 2012-09-24 11:00:00 1.0 A.K. - Original Message - From: Tagmarie ramga...@gmx.net To: r-help@r-project.org Cc: Sent: Thursday, October 25, 2012 10:35 AM Subject: [R] mean of a value of the last 2 hours Hello, I have a data frame somewhat like that: myframe - data.frame (ID=c(Ernie, Ernie, Ernie, Bert, Bert, Bert), Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00, 24.09.2012 11:00), Hunger=c(1,1,1,2,2,1) ) myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp), %d.%m.%Y %H:%M), tz=GMT) myframe2 - cbind (myframe,myframestime) myframe2$Timestamp - NULL myframe2 I want to add an additional column at the right and get in each row a value which shows the mean of hunger of the last two hours. Does anyone know how that works? That would be very helpful. -- View this message in context: http://r.789695.n4.nabble.com/mean-of-a-value-of-the-last-2-hours-tp4647415.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] arima.sim
Hi, For the sake of completeness, I thought I might follow up my email from earlier in the month with what I found out by digging into the code. The answer to my earlier question perhaps should have been obvious to me from the Usage description from the documentation. In the three examples I originally included, the difference between a1 and the other two AR(1) models (i.e. a2 and a3) is the sd parameter for start.innov. In a1, the sd for innov is set to 0.1, but the sd for start.innov remains the default parameter for rnorm (i.e. sd = 1). In a2 and a3, by passing in sd as an additional argument via ..., this sd is used by both innov and start.innov. So a1 is equivalent to: set.seed(2012) a4 - arima.sim(list(order = c(1, 0, 0), ar = 0.5), n = 250, innov = rnorm(n = 250, mean = 0, sd = 0.1), n.start = 10, start.innov = rnorm(n = 10, mean = 0, sd = 1)) Whereas, a2 and a3 are equivalent to: set.seed(2012) a5 - arima.sim(list(order = c(1, 0, 0), ar = 0.5), n = 250, innov = rnorm(n = 250, mean = 0, sd = 0.1), n.start = 10, start.innov = rnorm(n = 10, mean = 0, sd = 0.1)) The only difference being the sd used by start.innov. Of course this leads to an additional question, which is, why might someone choose to use differing values for sd in innov and start.innov? I'd love to hear any insight into this as I'm still learning to use this tool. Thanks. James On Mon, Oct 8, 2012 at 9:23 AM, J Toll jct...@gmail.com wrote: Hi, I have been using arima.sim from the stats package recently, and I'm wondering why I get different results when using what seem to be the same parameters. For example, I've given examples of three different ways to run arima.sim with what I believe are the same parameters. It's my understanding from the R documentation that rnorm is the default function for rand.gen if not provided in innov. So it would seem that there's quite a bit of redundancy in the assignment to a1. My expectation is that a1, a2, and a3 should be identical. It turns out that a1 is only partially identical to a2 and a3. The first 56 values of a1 are different from a2 and a3, but the rest are the same. What is it that makes an explicit declaration of the innov parameters different from using the defaults? At this point, my best guess is that this might have something to do with n.start and the length of the burn-in period. Any suggestions as to why all three of these aren't the same? Thanks. James set.seed(2012) a1 - arima.sim(list(order = c(1, 0, 0), ar = 0.5), n = 250, innov = rnorm(n = 250, mean = 0, sd = 0.1)) summary(a1) Min. 1st Qu. Median Mean 3rd Qu. Max. -0.45630 -0.11060 -0.03225 -0.02525 0.05484 0.27120 set.seed(2012) a2 - arima.sim(list(order = c(1, 0, 0), ar = 0.5), n = 250, sd = 0.1) summary(a2) Min. 1st Qu. Median Mean 3rd Qu. Max. -0.31830 -0.10200 -0.03016 -0.02174 0.05593 0.27120 set.seed(2012) a3 - arima.sim(list(ar = 0.5), n = 250, sd = 0.1) summary(a3) Min. 1st Qu. Median Mean 3rd Qu. Max. -0.31830 -0.10200 -0.03016 -0.02174 0.05593 0.27120 a1 == a2 Time Series: Start = 1 End = 250 Frequency = 1 [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE [22] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE [43] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [64] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [85] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot lmer model with Effects package
Marte Lilleeng mlilleeng at gmail.com writes: [snip] I have a simple model that i would like to plot with 95% CIs. It is like follows: m1-lmer(Richness~Grazing+I(Grazing^2)+(1|Plot),family=poisson) By using the effects package I get two plots, one for the linear term and one for the squared term. Q1: Can I get all in one? I.e. with one line for the whole model? Q2: Can I also visualize the random effects? It would probably be best to send this to the r-sig-mixed-models list. You may want to look at the code at http://glmm.wikidot.com/faq for generating predictions and confidence intervals from lmer fits. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mean of a value of the last 2 hours
I want to add an additional column at the right and get in
each row a value which shows the mean of hunger of the last two hours.
Isn't that just a moving average?
if so, see
http://stackoverflow.com/questions/743812/calculating-moving-average-in-r,
which mentions zoo for RollingMeans, TTR for MovingAverages, forecast for ma
and a possible solution using filter. One of those might help.
S Ellison
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Re: [R] problem in finding sizes of objects using a for loop
Here is a function I use to get the size of objects:
Here is an example output:
my.ls()
Size Mode
allStores 7,303,224 list
convertedStores 0 NULL
f.createCluster40,508 function
x 41,672 list
**Total 7,385,404 ---
my.ls - function (pos = 1, sorted = FALSE, envir = as.environment(pos))
{
.result - sapply(ls(envir = envir, all.names = TRUE),
function(..x) object.size(eval(as.symbol(..x),
envir = envir)))
if (sorted) {
.result - rev(sort(.result))
}
.ls - as.data.frame(rbind(as.matrix(.result), `**Total` = sum(.result)))
names(.ls) - Size
.ls$Size - formatC(.ls$Size, big.mark = ,, digits = 0,
format = f)
.ls$Mode - c(unlist(lapply(rownames(.ls)[-nrow(.ls)], function(x)
mode(eval(as.symbol(x),
envir = envir, ---)
.ls
}
On Thu, Oct 25, 2012 at 2:24 AM, Purna chander chander...@gmail.com wrote:
Dear All,
I wanted to extract the sizes of all created objects. For E.g when I
created 2 objects(x and y), I got their sizes using the following
code:
x-rnorm(1)
y-runif(100,min=40,max=1000)
ls()
[1] x y
object.size(x)
80024 bytes
object.size(y)
824 bytes
However, I was unable to get their sizes when I used a for loop in the
following way:
objects-ls()
for (i in seq_along(objects)){
+ print(c(objects[i],object.size(objects[i])))
+
+ }
[1] x 64
[1] y 64
The result obtained by me is wrong in second case.
I understood that variables x and y are treated as characters. But to
rectify this problem.
Regards,
Purna
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and provide commented, minimal, self-contained, reproducible code.
--
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Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
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Re: [R] problem in finding sizes of objects using a for loop
On Oct 25, 2012, at 10:56 AM, jim holtman wrote:
Here is a function I use to get the size of objects:
Here is an example output:
my.ls()
Size Mode
allStores 7,303,224 list
convertedStores 0 NULL
f.createCluster40,508 function
x 41,672 list
**Total 7,385,404 ---
That's far more elegant that the one I use;
getsizes - function() {z - sapply(ls(envir=globalenv()),
function(x) object.size(get(x)))
(tmp - as.matrix(rev(sort(z))[1:10]))}
getsizes()
Only returns the sorted-by-size matrix of the largest ten objects, but
modifying it to return all of them should be trivial.
--
david.
my.ls - function (pos = 1, sorted = FALSE, envir = as.environment(pos))
{
.result - sapply(ls(envir = envir, all.names = TRUE),
function(..x) object.size(eval(as.symbol(..x),
envir = envir)))
if (sorted) {
.result - rev(sort(.result))
}
.ls - as.data.frame(rbind(as.matrix(.result), `**Total` = sum(.result)))
names(.ls) - Size
.ls$Size - formatC(.ls$Size, big.mark = ,, digits = 0,
format = f)
.ls$Mode - c(unlist(lapply(rownames(.ls)[-nrow(.ls)], function(x)
mode(eval(as.symbol(x),
envir = envir, ---)
.ls
}
On Thu, Oct 25, 2012 at 2:24 AM, Purna chander chander...@gmail.com wrote:
Dear All,
I wanted to extract the sizes of all created objects. For E.g when I
created 2 objects(x and y), I got their sizes using the following
code:
x-rnorm(1)
y-runif(100,min=40,max=1000)
ls()
[1] x y
object.size(x)
80024 bytes
object.size(y)
824 bytes
However, I was unable to get their sizes when I used a for loop in the
following way:
objects-ls()
for (i in seq_along(objects)){
+ print(c(objects[i],object.size(objects[i])))
+
+ }
[1] x 64
[1] y 64
The result obtained by me is wrong in second case.
I understood that variables x and y are treated as characters. But to
rectify this problem.
Regards,
Purna
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
Alameda, CA, USA
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Change to daily digest
Hello folks, I am currently receiving a lot of emails from the list which proves that this is a very important place to get good feedbacks and tips and that the community is here to help .. Excellent thing. I am not though able to login to my subscriber space to change the email reception into daily digest, or I am not looking to the right place, if someone can point me to the right URL that would be very appreciated. Cheers Rad __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] trying ti use a function in aggregate
Hello, Try the following. (I've changed your function a bit. And named the data.frame 'dat', not 'data', which is an R function.) myfun - function (x) ifelse(sum(x) == 0, 0, x/sum(x)) aggregate(Count ~ Trip_id + Length + CommonName, data = dat, myfun) The output shows that each and every group corresponds to a single row of the original df. The 1's represent 100%, myfun _is_ dividing by sum(Count) Trip_id/Length/CommonName group. If you want just Trip_id/CommonName, use aggregate(Count ~ Trip_id + CommonName, data = dat, myfun) Or use your instruction without 'Length' in the by list: b - with(dat, aggregate(x=list(Percent=Count), by=list(Trip_id=Trip_id, Species=CommonName), FUN = myfun)) b Trip_id SpeciesPercent 1 230 Alewife 0.01408451 2 230 Herring,Blueback 0. 3 230Shad,American 0.09090909 4 231Shad,American 0.14285714 As you can see, the results are the same, with different output colnames. Hope this helps, Rui Barradas Em 25-10-2012 15:19, Sally_roman escreveu: Hi -I am using R v 2.13.0. I am trying to use the aggregate function to calculate the percent at length for each Trip_id and CommonName. Here is a small subset of the data. Trip_id Vessel CommonName Length Count 1 230SunlightShad,American 19 1 2 230SunlightShad,American 20 1 3 230SunlightShad,American 21 1 4 230SunlightShad,American 23 1 5 230SunlightShad,American 26 1 6 230SunlightShad,American 27 1 7 230SunlightShad,American 30 2 8 230SunlightShad,American 33 1 9 230SunlightShad,American 34 1 10 230SunlightShad,American 37 1 11 230Sunlight Herring,Blueback 20 1 12 230Sunlight Herring,Blueback 21 2 13 230Sunlight Herring,Blueback 22 5 14 230Sunlight Herring,Blueback 26 1 15 230Sunlight Alewife 17 1 16 230Sunlight Alewife 18 1 17 230Sunlight Alewife 20 2 18 230Sunlight Alewife 21 4 19 230Sunlight Alewife 2216 20 230Sunlight Alewife 2322 21 230Sunlight Alewife 2416 22 230Sunlight Alewife 25 4 23 230Sunlight Alewife 26 1 24 230Sunlight Alewife 27 2 25 230Sunlight Alewife 28 2 26 231 Western VentureShad,American 23 1 27 231 Western VentureShad,American 24 1 28 231 Western VentureShad,American 25 1 29 231 Western VentureShad,American 28 2 30 231 Western VentureShad,American 29 2 My code is: myfun-function (x) x/sum(x) b-with(data,aggregate(x=list(Percent=Count),by=list(Trip_id=Trip_id,Length=Length,Species=CommonName), FUN=myfun)) My issue is that the percent is not be calculated by Trip_id and CommonName. The result is that each row has a percent of 1 indicating that myfun is not dividing by the sum of counts with a Trip_id/CommonName group. Any help would be appreciated. Thank you -- View this message in context: http://r.789695.n4.nabble.com/trying-ti-use-a-function-in-aggregate-tp4647414.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] trying ti use a function in aggregate
Hi, May be this helps: dat1-read.table(text= Trip_id Vessel CommonName Length Count 1 230 Sunlight ShadAmerican 19 1 2 230 Sunlight ShadAmerican 20 1 3 230 Sunlight ShadAmerican 21 1 4 230 Sunlight ShadAmerican 23 1 5 230 Sunlight ShadAmerican 26 1 6 230 Sunlight ShadAmerican 27 1 7 230 Sunlight ShadAmerican 30 2 8 230 Sunlight ShadAmerican 33 1 9 230 Sunlight ShadAmerican 34 1 10 230 Sunlight ShadAmerican 37 1 11 230 Sunlight HerringBlueback 20 1 12 230 Sunlight HerringBlueback 21 2 13 230 Sunlight HerringBlueback 22 5 14 230 Sunlight HerringBlueback 26 1 15 230 Sunlight Alewife 17 1 16 230 Sunlight Alewife 18 1 17 230 Sunlight Alewife 20 2 18 230 Sunlight Alewife 21 4 19 230 Sunlight Alewife 22 16 20 230 Sunlight Alewife 23 22 21 230 Sunlight Alewife 24 16 22 230 Sunlight Alewife 25 4 23 230 Sunlight Alewife 26 1 24 230 Sunlight Alewife 27 2 25 230 Sunlight Alewife 28 2 26 231 Western_Venture ShadAmerican 23 1 27 231 Western_Venture ShadAmerican 24 1 28 231 Western_Venture ShadAmerican 25 1 29 231 Western_Venture ShadAmerican 28 2 30 231 Western_Venture ShadAmerican 29 2 ,sep=,header=TRUE,stringsAsFactors=FALSE) with(dat1,aggregate(Count,by=list(Trip_id=Trip_id,Species=CommonName),function(x) x/sum(x))) Trip_id Species 1 230 Alewife 2 230 HerringBlueback 3 230 ShadAmerican 4 231 ShadAmerican x #1 0.01408451, 0.01408451, 0.02816901, 0.05633803, 0.22535211, 0.30985915, 0.22535211, 0.05633803, 0.01408451, 0.02816901, 0.02816901 #2 0.111, 0.222, 0.556, 0.111 #3 0.09090909, 0.09090909, 0.09090909, 0.09090909, 0.09090909, 0.09090909, 0.18181818, 0.09090909, 0.09090909, 0.09090909 #4 0.1428571, 0.1428571, 0.1428571, 0.2857143, 0.2857143 #or library(plyr) res-ddply(dat1,.(Trip_id=Trip_id,Vessel=Vessel,CommonName=CommonName), summarize, Count/sum(Count)) colnames(res)[4]-value head(res) # Trip_id Vessel CommonName value #1 230 Sunlight Alewife 0.01408451 #2 230 Sunlight Alewife 0.01408451 #3 230 Sunlight Alewife 0.02816901 #4 230 Sunlight Alewife 0.05633803 #5 230 Sunlight Alewife 0.22535211 #6 230 Sunlight Alewife 0.30985915 A.K. - Original Message - From: Sally_roman sro...@umassd.edu To: r-help@r-project.org Cc: Sent: Thursday, October 25, 2012 10:19 AM Subject: [R] trying ti use a function in aggregate Hi -I am using R v 2.13.0. I am trying to use the aggregate function to calculate the percent at length for each Trip_id and CommonName. Here is a small subset of the data. Trip_id Vessel CommonName Length Count 1 230 Sunlight Shad,American 19 1 2 230 Sunlight Shad,American 20 1 3 230 Sunlight Shad,American 21 1 4 230 Sunlight Shad,American 23 1 5 230 Sunlight Shad,American 26 1 6 230 Sunlight Shad,American 27 1 7 230 Sunlight Shad,American 30 2 8 230 Sunlight Shad,American 33 1 9 230 Sunlight Shad,American 34 1 10 230 Sunlight Shad,American 37 1 11 230 Sunlight Herring,Blueback 20 1 12 230 Sunlight Herring,Blueback 21 2 13 230 Sunlight Herring,Blueback 22 5 14 230 Sunlight Herring,Blueback 26 1 15 230 Sunlight Alewife 17 1 16 230 Sunlight Alewife 18 1 17 230 Sunlight Alewife 20 2 18 230 Sunlight Alewife 21 4 19 230 Sunlight Alewife 22 16 20 230 Sunlight Alewife 23 22 21 230 Sunlight Alewife 24 16 22 230 Sunlight Alewife 25 4 23 230 Sunlight Alewife 26 1 24 230 Sunlight Alewife 27 2 25 230 Sunlight Alewife 28 2 26 231 Western Venture
Re: [R] Change to daily digest
Which one ? :) On Oct 25, 2012, at 2:34 PM, R. Michael Weylandt michael.weyla...@gmail.com michael.weyla...@gmail.com wrote: On Oct 25, 2012, at 7:30 PM, Mohamed Radhouane Aniba arad...@gmail.com wrote: Hello folks, I am currently receiving a lot of emails from the list which proves that this is a very important place to get good feedbacks and tips and that the community is here to help .. Excellent thing. I am not though able to login to my subscriber space to change the email reception into daily digest, or I am not looking to the right place, if someone can point me to the right URL that would be very appreciated. Cheers Rad __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help follow this link RMW PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mean of a value of the last 2 hours
Or using ddply from plyr, library(plyr) myframe - data.frame (ID=c(Ernie, Ernie, Ernie, Bert, Bert, Bert), Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00, 24.09.2012 11:00), Hunger=c(1,1,1,2,2,1) ) myframe myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp), %d.%m.%Y %H:%M), tz=GMT) myframestime myframe2 - cbind (myframe,myframestime) myframe2 ddply(myframe2,.(ID),summarise,Last2=mean(tail(Hunger,2))) Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA http://www.fws.gov/redbluff/rbdd_jsmp.aspx From: Tagmarie ramga...@gmx.net To: r-help@r-project.org Sent: Thursday, October 25, 2012 7:35 AM Subject: [R] mean of a value of the last 2 hours Hello, I have a data frame somewhat like that: myframe - data.frame (ID=c(Ernie, Ernie, Ernie, Bert, Bert, Bert), Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00, 24.09.2012 11:00), Hunger=c(1,1,1,2,2,1) ) myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp), %d.%m.%Y %H:%M), tz=GMT) myframe2 - cbind (myframe,myframestime) myframe2$Timestamp - NULL myframe2 I want to add an additional column at the right and get in each row a value which shows the mean of hunger of the last two hours. Does anyone know how that works? That would be very helpful. -- View this message in context: http://r.789695.n4.nabble.com/mean-of-a-value-of-the-last-2-hours-tp4647415.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change to daily digest
On Oct 25, 2012, at 7:30 PM, Mohamed Radhouane Aniba arad...@gmail.com wrote: Hello folks, I am currently receiving a lot of emails from the list which proves that this is a very important place to get good feedbacks and tips and that the community is here to help .. Excellent thing. I am not though able to login to my subscriber space to change the email reception into daily digest, or I am not looking to the right place, if someone can point me to the right URL that would be very appreciated. Cheers Rad __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help follow this link RMW PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] trying ti use a function in aggregate
Sally, It's great that you provided data and code. To make it even more user-friendly for R-help readers, supply your data as Rcode, using (for example) the dput() function. The reason you were getting all 1s with your code, is that you had told it to aggregate by trip, LENGTH, and species. But the data are already summarized by trip, LENGTH, and species, so your myfun() function is calculating the count/count=1 for each row. You could get rid of LENGTH to use your myfun() function, but the results aren't pretty ... with(data, aggregate(data.frame(Total=Count), data.frame(Trip_id, CommonName), myfun)) Instead, I suggest you can use the aggregate function to calculate the total counts, then merge these totals with your original data to calculate the proportions. # small subset of data data - structure(list(Trip_id = c(230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 231L, 231L, 231L, 231L, 231L), Vessel = c(Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Western Venture, Western Venture, Western Venture, Western Venture, Western Venture), CommonName = c(Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Herring,Blueback, Herring,Blueback, Herring,Blueback, Herring,Blueback, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American), Length = c(19L, 20L, 21L, 23L, 26L, 27L, 30L, 33L, 34L, 37L, 20L, 21L, 22L, 26L, 17L, 18L, 20L, 21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 23L, 24L, 25L, 28L, 29L), Count = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 5L, 1L, 1L, 1L, 2L, 4L, 16L, 22L, 16L, 4L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L)), .Names = c(Trip_id, Vessel, CommonName, Length, Count), row.names = c(NA, -30L), class = data.frame) # calculate the total count for each trip and Species agg - with(data, aggregate(data.frame(Total=Count), data.frame(Trip_id, CommonName), sum)) # combine the totals with the full data frame data2 - merge(data, agg) # then calculate proportions data2$Prop - data2$Count/data2$Total data2 Jean Sally_roman sro...@umassd.edu wrote on 10/25/2012 09:19:57 AM: Hi -I am using R v 2.13.0. I am trying to use the aggregate function to calculate the percent at length for each Trip_id and CommonName. Here is a small subset of the data. Trip_id Vessel CommonName Length Count 1 230SunlightShad,American 19 1 2 230SunlightShad,American 20 1 3 230SunlightShad,American 21 1 4 230SunlightShad,American 23 1 5 230SunlightShad,American 26 1 6 230SunlightShad,American 27 1 7 230SunlightShad,American 30 2 8 230SunlightShad,American 33 1 9 230SunlightShad,American 34 1 10 230SunlightShad,American 37 1 11 230Sunlight Herring,Blueback 20 1 12 230Sunlight Herring,Blueback 21 2 13 230Sunlight Herring,Blueback 22 5 14 230Sunlight Herring,Blueback 26 1 15 230Sunlight Alewife 17 1 16 230Sunlight Alewife 18 1 17 230Sunlight Alewife 20 2 18 230Sunlight Alewife 21 4 19 230Sunlight Alewife 2216 20 230Sunlight Alewife 2322 21 230Sunlight Alewife 2416 22 230Sunlight Alewife 25 4 23 230Sunlight Alewife 26 1 24 230Sunlight Alewife 27 2 25 230Sunlight Alewife 28 2 26 231 Western VentureShad,American 23 1 27 231 Western VentureShad,American 24 1 28 231 Western VentureShad,American 25 1 29 231 Western VentureShad,American 28 2 30 231 Western VentureShad,American 29 2 My code is: myfun-function (x) x/sum(x) b-with(data,aggregate(x=list(Percent=Count),by=list (Trip_id=Trip_id,Length=Length,Species=CommonName), FUN=myfun)) My issue is that the percent is not be calculated by Trip_id and CommonName. The result is that each row has a percent of 1 indicating that myfun is not dividing by the sum of counts with a Trip_id/CommonName group. Any help would be appreciated. Thank you
Re: [R] trying ti use a function in aggregate
Hi, Try this: dat1$Percent-unlist(tapply(dat1$Count,list(dat1$Trip_id,dat1$CommonName),function(x) x/sum(x))) head(dat1) # Trip_id Vessel CommonName Length Count Percent #1 230 Sunlight ShadAmerican 19 1 0.01408451 #2 230 Sunlight ShadAmerican 20 1 0.01408451 #3 230 Sunlight ShadAmerican 21 1 0.02816901 #4 230 Sunlight ShadAmerican 23 1 0.05633803 #5 230 Sunlight ShadAmerican 26 1 0.22535211 #6 230 Sunlight ShadAmerican 27 1 0.30985915 I also have a doubt from your first email. I am trying to use the aggregate function to calculate the percent at length for each Trip_id and CommonName. So, I am thinking you need the percentage of length. If that is the case, res1-with(dat1,aggregate(Count,by=list(Trip_id=Trip_id,CommonName=CommonName),function(x) length(x))) library(plyr) res1$Percent-ddply(res1,.(Trip_id),summarize,x/sum(x))[,2] res1 # Trip_id CommonName x Percent #1 230 Alewife 11 0.44 #2 230 HerringBlueback 4 0.16 #3 230 ShadAmerican 10 0.40 #4 231 ShadAmerican 5 1.00 A.K. From: Sally A Roman sro...@umassd.edu To: arun smartpink...@yahoo.com Sent: Thursday, October 25, 2012 12:44 PM Subject: Re: [R] trying ti use a function in aggregate Thank you for your help. Your code does what I need it to, but the output that I need is Trip_id, Vessel, CommonName, Length, Percent. When I run your functions there is no length output. Do you have any suggestions on how to get the Length variable into the output? Thanks again- Sally From: arun smartpink...@yahoo.com To: Sally_roman sro...@umassd.edu Cc: R help r-help@r-project.org Sent: Thursday, October 25, 2012 12:04:23 PM Subject: Re: [R] trying ti use a function in aggregate Hi, May be this helps: dat1-read.table(text= Trip_id Vessel CommonName Length Count 1 230 Sunlight ShadAmerican 19 1 2 230 Sunlight ShadAmerican 20 1 3 230 Sunlight ShadAmerican 21 1 4 230 Sunlight ShadAmerican 23 1 5 230 Sunlight ShadAmerican 26 1 6 230 Sunlight ShadAmerican 27 1 7 230 Sunlight ShadAmerican 30 2 8 230 Sunlight ShadAmerican 33 1 9 230 Sunlight ShadAmerican 34 1 10 230 Sunlight ShadAmerican 37 1 11 230 Sunlight HerringBlueback 20 1 12 230 Sunlight HerringBlueback 21 2 13 230 Sunlight HerringBlueback 22 5 14 230 Sunlight HerringBlueback 26 1 15 230 Sunlight Alewife 17 1 16 230 Sunlight Alewife 18 1 17 230 Sunlight Alewife 20 2 18 230 Sunlight Alewife 21 4 19 230 Sunlight Alewife 22 16 20 230 Sunlight Alewife 23 22 21 230 Sunlight Alewife 24 16 22 230 Sunlight Alewife 25 4 23 230 Sunlight Alewife 26 1 24 230 Sunlight Alewife 27 2 25 230 Sunlight Alewife 28 2 26 231 Western_Venture ShadAmerican 23 1 27 231 Western_Venture ShadAmerican 24 1 28 231 Western_Venture ShadAmerican 25 1 29 231 Western_Venture ShadAmerican 28 2 30 231 Western_Venture ShadAmerican 29 2 ,sep=,header=TRUE,stringsAsFactors=FALSE) with(dat1,aggregate(Count,by=list(Trip_id=Trip_id,Species=CommonName),function(x) x/sum(x))) Trip_id Species 1 230 Alewife 2 230 HerringBlueback 3 230 ShadAmerican 4 231 ShadAmerican x #1 0.01408451, 0.01408451, 0.02816901, 0.05633803, 0.22535211, 0.30985915, 0.22535211, 0.05633803, 0.01408451, 0.02816901, 0.02816901 #2 0.111, 0.222, 0.556, 0.111 #3 0.09090909, 0.09090909, 0.09090909, 0.09090909, 0.09090909, 0.09090909, 0.18181818, 0.09090909, 0.09090909, 0.09090909 #4 0.1428571, 0.1428571, 0.1428571, 0.2857143, 0.2857143 #or library(plyr) res-ddply(dat1,.(Trip_id=Trip_id,Vessel=Vessel,CommonName=CommonName), summarize, Count/sum(Count)) colnames(res)[4]-value head(res) # Trip_id Vessel CommonName value #1 230 Sunlight Alewife 0.01408451 #2 230 Sunlight Alewife 0.01408451 #3 230 Sunlight Alewife 0.02816901 #4 230 Sunlight Alewife 0.05633803 #5 230 Sunlight Alewife 0.22535211 #6 230 Sunlight Alewife 0.30985915 A.K. -
[R] How to extract auc, specificity and sensitivity
I am running my code in a loop and it does not work but when I run it
outside the loop I get the values I want.
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
sentiv-performance(rocpreds,sens)@y.values
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
specs-performance(rocpreds,spec)@y.values
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
return(output)
}
y-fitglm(.05,1)
y
The code runs without the sentiv and specs but when I remove the loop i can
get the sensitivity and spec. values ???
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to extract auc, specificity and sensitivity
On 25-10-2012, at 21:28, Adel Powell wrote:
I am running my code in a loop and it does not work but when I run it
outside the loop I get the values I want.
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
sentiv-performance(rocpreds,sens)@y.values
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
specs-performance(rocpreds,spec)@y.values
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
A missing before Specificity?
return(output)
}
y-fitglm(.05,1)
y
Running this after correction of the missing one gets en error
Error in fitglm(0.05, 1) : could not find function prediction
How are you using a loop?
Your example is not reproducible.
Berend
The code runs without the sentiv and specs but when I remove the loop i can
get the sensitivity and spec. values ???
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to extract auc, specificity and sensitivity
Your code is still not runnable.
It gives the error message
Error in fitglm(0.05, 1) : could not find function prediction
Berend
On 25-10-2012, at 21:55, Adel Powell wrote:
I think I have corrected it. Can you tell me are my spec and sens values
correct
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
fp-performance(rocpreds,sens)
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
fp2-performance(rocpreds,spec)
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
return(output)
}
y-fitglm(.05,1)
y
On Thu, Oct 25, 2012 at 3:43 PM, Berend Hasselman b...@xs4all.nl wrote:
On 25-10-2012, at 21:28, Adel Powell wrote:
I am running my code in a loop and it does not work but when I run it
outside the loop I get the values I want.
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
sentiv-performance(rocpreds,sens)@y.values
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
specs-performance(rocpreds,spec)@y.values
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
A missing before Specificity?
return(output)
}
y-fitglm(.05,1)
y
Running this after correction of the missing one gets en error
Error in fitglm(0.05, 1) : could not find function prediction
How are you using a loop?
Your example is not reproducible.
Berend
The code runs without the sentiv and specs but when I remove the loop i can
get the sensitivity and spec. values ???
[[alternative HTML version deleted]]
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[R] Estimating credit card default probabilities.
Folks, I am working on a credit card defaults and transition probabilities. For example a single credit card account could be in a number of states: up-to-date, 30, 60, 90 days in arrears or in default. * Are there packages in R that do estimation of the transition probabilities given historical cohort defaults? * Any pointers to papers specific to this type of estimation? * Simulation of future paths of defaults. I know this is not strictly an R question so please feel free to slam me. Afterwards I would appreciate any pointers you might have. Thanks much for your time, KW -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change to daily digest
Mohamed Radhouane Aniba aradwen at gmail.com writes: Which one ? :) https://stat.ethz.ch/mailman/listinfo/r-help __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to extract auc, specificity and sensitivity
n - 10; # Sample size
fitglm - function(sigma,tau){
+ x - rnorm(n,0,sigma)
+ intercept - 0
+ beta - 0
+ ystar - intercept+beta*x
+ z - rbinom(n,1,plogis(ystar))
+ xerr - x + rnorm(n,0,tau)
+ model-glm(z ~ xerr, family=binomial(logit))
+ int-coef(model)[1]
+ slope-coef(model)[2]
+ pred-predict(model)
+
+ result-ifelse(pred.5,1,0)
+
+ accuracy-length(which(result==z))/length(z)
+ accuracy
+
+ rocpreds-prediction(result,z)
+ auc-performance(rocpreds,auc)@y.values
+ fp-performance(rocpreds,sens)
+ sentiv-slot(fp,y.values)[[1]]
+ sentiv-sentiv[2]
+ sentiv
+ fp2-performance(rocpreds,spec)
+ specs-slot(fp2,y.values)[[1]]
+ specs-specs[2]
+ specs
+ output-c(int,slope,.5,accuracy,auc,sentiv,specs)
+
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
+ return(output)
+
+ }
y-fitglm(2,1)
y
$Intercept
[1] 1.335284
$Slope
[1] 0.1562984
$CutPoint
[1] 0.5
$Accuracy
[1] 0.8
$AUC
[1] 0.5
$Sentivity
[1] 1
$Specificity
[1] 0
Don't get error message but wrong values
On Thu, Oct 25, 2012 at 4:05 PM, Berend Hasselman b...@xs4all.nl wrote:
Your code is still not runnable.
It gives the error message
Error in fitglm(0.05, 1) : could not find function prediction
Berend
On 25-10-2012, at 21:55, Adel Powell wrote:
I think I have corrected it. Can you tell me are my spec and sens values
correct
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
fp-performance(rocpreds,sens)
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
fp2-performance(rocpreds,spec)
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
return(output)
}
y-fitglm(.05,1)
y
On Thu, Oct 25, 2012 at 3:43 PM, Berend Hasselman b...@xs4all.nl wrote:
On 25-10-2012, at 21:28, Adel Powell wrote:
I am running my code in a loop and it does not work but when I run it
outside the loop I get the values I want.
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
sentiv-performance(rocpreds,sens)@y.values
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
specs-performance(rocpreds,spec)@y.values
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
A missing before Specificity?
return(output)
}
y-fitglm(.05,1)
y
Running this after correction of the missing one gets en error
Error in fitglm(0.05, 1) : could not find function prediction
How are you using a loop?
Your example is not reproducible.
Berend
The code runs without the sentiv and specs but when I remove the loop
i can
get the sensitivity and spec. values ???
[[alternative HTML version deleted]]
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Estimating credit card default probabilities.
You might have better luck posting on stats.stackexchange.com, a statistical help list. -- Bert On Thu, Oct 25, 2012 at 1:08 PM, Keith Weintraub kw1...@gmail.com wrote: Folks, I am working on a credit card defaults and transition probabilities. For example a single credit card account could be in a number of states: up-to-date, 30, 60, 90 days in arrears or in default. * Are there packages in R that do estimation of the transition probabilities given historical cohort defaults? * Any pointers to papers specific to this type of estimation? * Simulation of future paths of defaults. I know this is not strictly an R question so please feel free to slam me. Afterwards I would appreciate any pointers you might have. Thanks much for your time, KW -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R5: Lock a class field from within a method?
David Winsemius wrote:
On Oct 24, 2012, at 2:14 AM, Johannes Graumann wrote:
Hello,
testclass - setRefClass(
testclass,
fields = list(testfield = logical),
methods = list(validate=function(){testfield-TRUE}))
test - testclass$new()
test$testfield
logical(0)
test$validate()
test$testfield
[1] TRUE
Works just fine for me.
I would love to be able to do something like
testclass - setRefClass(
testclass,
fields = list(testfield = logical),
methods = list(validate=function(){
testfield-TRUE
.self$lock(testfield)
}))
but am unabel to achieve that. Can anyone point out how to go about
rendering a field immutable after execution of a specific method?
The fact that you used only lock in your code and I am unable to
find such a function makes me wonder whether that was an implicit
psuedo-code effort and that you do not know about:
?lockBinding
lockbinding does in fact what I want, but your statement about pseudo-code
is not entirely correct ... try the following (?setRefClass will light the
way).
testclass - setRefClass(
testclass,
fields = list(testfield = logical))
testclass$lock(testfield)
test - testclass$new()
test$testfield - TRUE
test$testfield - FALSE
I was assuming to be able to have access to the lock method from a class
instance as well (rather than just in the class definition) ... Is that
indeed impossible?
Cheers, Joh
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[R] How generate random numbers from given vector???
I wanna generate random numbers from a vector...
for example number-c(0,1,3,4,5,6,8)
so
rsidp-function(x){
i=0
for (i in seq(1:x))
{y-sample(number,x, replace=T)}
return(y)
}
so all random numbers have to be from vector number;
so if I type rsidp(5). it has to give me 5 random numbers except 2,7,9
(because they are not in the vector numbers). help me plz with it (((
--
View this message in context:
http://r.789695.n4.nabble.com/How-generate-random-numbers-from-given-vector-tp4647447.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] ifelse reformulation
Hi, Did you mean this? group1-c(40,50,60,70) #or group2-c(50,var1,var2,60) In either of the above cases, when you check str(group1) # all are converted to character. # chr [1:4] 40 50 60 70 str(group2) # chr [1:4] 50 var1 var2 60 Suppose, I am comparing the test1 dataset (x4: x6 columns) with group2 apply(test1,1,function(x) ifelse(x[5:7]%in%group2,0,1)) # 2nd row is 50 for x4 column # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 #[1,] 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 #[2,] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 #[3,] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 # final result as.vector(apply(test1,1,function(x) ifelse(any(x[5:7]%in%group2),0,1))) # [1] 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 I guess it works. BTW, in my previous reply, I made a mistake as.vector(apply(test1,1,function(x) ifelse(any(x[4:6]%in%group),0,1))) ^^^ It should be 5:7. group-c(40,50,60,70) as.vector(apply(test1,1,function(x) ifelse(any(x[5:7]%in%group),0,1))) #[1] 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1 ifelse(test1$x4==40|test1$x4==50|test1$x4==60|test1$x4==70|test1$x5==40|test1$x5==50|test1$x5==60|test1$x5==70|test1$x6==40|test1$x6==50|test1$x6==60|test1$x7==70,0,1) # [1] 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1 as.vector(apply(test1,1,function(x) ifelse(any(x[5:7]==40|x[5:7]==50|x[5:7]==60|x[5:7]==70),0,1))) #[1] 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1 A.K. - Original Message - From: brunosm brunos...@gmail.com To: r-help@r-project.org Cc: Sent: Thursday, October 25, 2012 12:55 PM Subject: Re: [R] ifelse reformulation Arun, thank you very much... but a new problem arose... What if... in the variable group that i want to compare, there is numeric and non numeric types? Or, if you think its better, can i have two variables, one numeric and one non numeric, and made the comparision? Best regards, Bruno 2012/10/12 arun kirshna [via R] ml-node+s789695n4645988...@n4.nabble.com HI, Try this: test1-read.table(text= id x1 x2 x3 x4 x5 x6 x7 1 1 36 26 21 32 31 27 31 2 2 45 21 46 50 22 36 29 3 3 49 47 35 44 33 31 46 4 4 42 32 38 28 39 45 32 5 5 29 42 39 48 25 35 34 6 6 39 31 30 37 46 43 44 7 7 41 40 25 23 42 40 24 8 8 27 29 47 34 26 38 28 9 9 25 35 29 36 43 34 23 10 10 24 44 37 26 27 46 22 11 11 38 50 32 49 37 24 40 12 12 20 34 48 25 30 41 36 13 13 26 46 20 40 29 20 43 14 14 33 37 49 31 47 30 30 15 15 43 39 27 35 48 47 27 ,sep=,header=TRUE) count40- ifelse(test1$x1==40|test1$x2==40|test1$x3==40|test1$x4==40|test1$x5==40|test1$x6==40|test1$x7==40,0,1) count40 #[1] 1 1 1 1 1 1 0 1 1 1 0 1 0 1 1 #if you want to get the same result, apply(test1,1,function(x) ifelse(any(x==40),0,1)) # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 #1 1 1 1 1 1 0 1 1 1 0 1 0 1 1 A.K. -- If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/ifelse-reformulation-tp4645981p4645988.html To unsubscribe from ifelse reformulation, click herehttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=4645981code=YnJ1bm9zbTg3QGdtYWlsLmNvbXw0NjQ1OTgxfDIwMjc4MjE3MDg= . NAMLhttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=macro_viewerid=instant_html%21nabble%3Aemail.namlbase=nabble.naml.namespaces.BasicNamespace-nabble.view.web.template.NabbleNamespace-nabble.view.web.template.NodeNamespacebreadcrumbs=notify_subscribers%21nabble%3Aemail.naml-instant_emails%21nabble%3Aemail.naml-send_instant_email%21nabble%3Aemail.naml -- View this message in context: http://r.789695.n4.nabble.com/ifelse-reformulation-tp4645981p4647431.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] trying ti use a function in aggregate
Hi Jean - Thank you for your help. The code worked great. - Original Message - From: Jean V Adams [via R] ml-node+s789695n4647444...@n4.nabble.com To: Sally_roman sro...@umassd.edu Sent: Thursday, October 25, 2012 2:48:45 PM Subject: Re: trying ti use a function in aggregate Sally, It's great that you provided data and code. To make it even more user-friendly for R-help readers, supply your data as Rcode, using (for example) the dput() function. The reason you were getting all 1s with your code, is that you had told it to aggregate by trip, LENGTH, and species. But the data are already summarized by trip, LENGTH, and species, so your myfun() function is calculating the count/count=1 for each row. You could get rid of LENGTH to use your myfun() function, but the results aren't pretty ... with(data, aggregate(data.frame(Total=Count), data.frame(Trip_id, CommonName), myfun)) Instead, I suggest you can use the aggregate function to calculate the total counts, then merge these totals with your original data to calculate the proportions. # small subset of data data - structure(list(Trip_id = c(230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 231L, 231L, 231L, 231L, 231L), Vessel = c(Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Western Venture, Western Venture, Western Venture, Western Venture, Western Venture), CommonName = c(Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Herring,Blueback, Herring,Blueback, Herring,Blueback, Herring,Blueback, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Alewife, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American), Length = c(19L, 20L, 21L, 23L, 26L, 27L, 30L, 33L, 34L, 37L, 20L, 21L, 22L, 26L, 17L, 18L, 20L, 21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 23L, 24L, 25L, 28L, 29L), Count = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 5L, 1L, 1L, 1L, 2L, 4L, 16L, 22L, 16L, 4L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L)), .Names = c(Trip_id, Vessel, CommonName, Length, Count), row.names = c(NA, -30L), class = data.frame) # calculate the total count for each trip and Species agg - with(data, aggregate(data.frame(Total=Count), data.frame(Trip_id, CommonName), sum)) # combine the totals with the full data frame data2 - merge(data, agg) # then calculate proportions data2$Prop - data2$Count/data2$Total data2 Jean Sally_roman [hidden email] wrote on 10/25/2012 09:19:57 AM: Hi -I am using R v 2.13.0. I am trying to use the aggregate function to calculate the percent at length for each Trip_id and CommonName. Here is a small subset of the data. Trip_id Vessel CommonName Length Count 1 230 Sunlight Shad,American 19 1 2 230 Sunlight Shad,American 20 1 3 230 Sunlight Shad,American 21 1 4 230 Sunlight Shad,American 23 1 5 230 Sunlight Shad,American 26 1 6 230 Sunlight Shad,American 27 1 7 230 Sunlight Shad,American 30 2 8 230 Sunlight Shad,American 33 1 9 230 Sunlight Shad,American 34 1 10 230 Sunlight Shad,American 37 1 11 230 Sunlight Herring,Blueback 20 1 12 230 Sunlight Herring,Blueback 21 2 13 230 Sunlight Herring,Blueback 22 5 14 230 Sunlight Herring,Blueback 26 1 15 230 Sunlight Alewife 17 1 16 230 Sunlight Alewife 18 1 17 230 Sunlight Alewife 20 2 18 230 Sunlight Alewife 21 4 19 230 Sunlight Alewife 22 16 20 230 Sunlight Alewife 23 22 21 230 Sunlight Alewife 24 16 22 230 Sunlight Alewife 25 4 23 230 Sunlight Alewife 26 1 24 230 Sunlight Alewife 27 2 25 230 Sunlight Alewife 28 2 26 231 Western Venture Shad,American 23 1 27 231 Western Venture Shad,American 24 1 28 231 Western Venture Shad,American 25 1 29 231 Western Venture Shad,American 28 2 30 231 Western Venture Shad,American 29 2 My code is: myfun-function (x) x/sum(x) b-with(data,aggregate(x=list(Percent=Count),by=list (Trip_id=Trip_id,Length=Length,Species=CommonName), FUN=myfun)) My issue is that the percent is not be calculated by Trip_id and CommonName. The result is that each row has a percent of 1 indicating that myfun is not dividing by the sum of counts with a Trip_id/CommonName group. Any help would be appreciated. Thank you [[alternative HTML version deleted]] __ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. If you reply to this email, your message will be added to
Re: [R] trying ti use a function in aggregate
HI, In my previous solution, the order got messed up. I should have ordered the columns. Try this: dat1-read.table(text= Trip_id Vessel CommonName Length Count 1 230 Sunlight ShadAmerican 19 1 2 230 Sunlight ShadAmerican 20 1 3 230 Sunlight ShadAmerican 21 1 4 230 Sunlight ShadAmerican 23 1 5 230 Sunlight ShadAmerican 26 1 6 230 Sunlight ShadAmerican 27 1 7 230 Sunlight ShadAmerican 30 2 8 230 Sunlight ShadAmerican 33 1 9 230 Sunlight ShadAmerican 34 1 10 230 Sunlight ShadAmerican 37 1 11 230 Sunlight HerringBlueback 20 1 12 230 Sunlight HerringBlueback 21 2 13 230 Sunlight HerringBlueback 22 5 14 230 Sunlight HerringBlueback 26 1 15 230 Sunlight Alewife 17 1 16 230 Sunlight Alewife 18 1 17 230 Sunlight Alewife 20 2 18 230 Sunlight Alewife 21 4 19 230 Sunlight Alewife 22 16 20 230 Sunlight Alewife 23 22 21 230 Sunlight Alewife 24 16 22 230 Sunlight Alewife 25 4 23 230 Sunlight Alewife 26 1 24 230 Sunlight Alewife 27 2 25 230 Sunlight Alewife 28 2 26 231 Western_Venture ShadAmerican 23 1 27 231 Western_Venture ShadAmerican 24 1 28 231 Western_Venture ShadAmerican 25 1 29 231 Western_Venture ShadAmerican 28 2 30 231 Western_Venture ShadAmerican 29 2 ,sep=,header=TRUE,stringsAsFactors=FALSE) dat2-dat1[order(dat1$Trip_id,dat1$Vessel,dat1$CommonName,dat1$Length,dat1$Count),] dat3-dat2 dat3$Prop-unlist(tapply(dat3$Count,list(dat3$Trip_id,dat3$CommonName),function(x) x/sum(x))) #Jean's method: agg - with(dat2, aggregate(data.frame(Total=Count), data.frame(Trip_id, CommonName), sum)) # combine the totals with the full data frame data2 - merge(dat2, agg) # then calculate proportions data2$Prop - data2$Count/data2$Total data3-data2[,-6] data4-data3[,c(1,3,2,4:6)] rownames(dat3)-1:nrow(dat3) identical(dat3,data4) #[1] TRUE head(dat3) # Trip_id Vessel CommonName Length Count Prop #1 230 Sunlight Alewife 17 1 0.01408451 #2 230 Sunlight Alewife 18 1 0.01408451 #3 230 Sunlight Alewife 20 2 0.02816901 #4 230 Sunlight Alewife 21 4 0.05633803 #5 230 Sunlight Alewife 22 16 0.22535211 #6 230 Sunlight Alewife 23 22 0.30985915 head(data4) # Trip_id Vessel CommonName Length Count Prop #1 230 Sunlight Alewife 17 1 0.01408451 #2 230 Sunlight Alewife 18 1 0.01408451 #3 230 Sunlight Alewife 20 2 0.02816901 #4 230 Sunlight Alewife 21 4 0.05633803 #5 230 Sunlight Alewife 22 16 0.22535211 #6 230 Sunlight Alewife 23 22 0.30985915 A.K. - Original Message - From: Jean V Adams jvad...@usgs.gov To: Sally_roman sro...@umassd.edu Cc: r-help@r-project.org Sent: Thursday, October 25, 2012 2:45 PM Subject: Re: [R] trying ti use a function in aggregate Sally, It's great that you provided data and code. To make it even more user-friendly for R-help readers, supply your data as Rcode, using (for example) the dput() function. The reason you were getting all 1s with your code, is that you had told it to aggregate by trip, LENGTH, and species. But the data are already summarized by trip, LENGTH, and species, so your myfun() function is calculating the count/count=1 for each row. You could get rid of LENGTH to use your myfun() function, but the results aren't pretty ... with(data, aggregate(data.frame(Total=Count), data.frame(Trip_id, CommonName), myfun)) Instead, I suggest you can use the aggregate function to calculate the total counts, then merge these totals with your original data to calculate the proportions. # small subset of data data - structure(list(Trip_id = c(230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 230L, 231L, 231L, 231L, 231L, 231L), Vessel = c(Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Sunlight, Western Venture, Western Venture, Western Venture, Western Venture, Western Venture), CommonName = c(Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Shad,American, Herring,Blueback, Herring,Blueback,
Re: [R] How to extract auc, specificity and sensitivity
n - 10; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
fp-performance(rocpreds,sens)
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
fp2-performance(rocpreds,spec)
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
return(output)
}
y-fitglm(2,1)
y
I corrected the code. I am put everything in a loop because I am running
monte carlo reps outside of this later.
It works but the value returned is wrong.
Here is the manual manipulation:
* x - rnorm(10,0,2)*
* *
* intercept - 0*
* *
* beta - 5*
* *
* ystar - intercept+beta*x*
* *
* ystar*
[1] 16.5436337 7.7740329 -10.1653928 -2.8338118 -21.5410780 2.6902171
5.1156558 5.0729933 -10.8556430 0.9208434
* test - plogis(ystar)*
* *
* test*
[1] 9.99e-01 9.995797e-01 3.847772e-05 5.552417e-02 4.413963e-10
9.364469e-01 9.940338e-01 9.937753e-01 1.929504e-05 7.152139e-01
* z - rbinom(10,1,plogis(ystar))*
* *
* z*
[1] 1 1 0 0 0 1 1 1 0 1
* xerr - x + rnorm(10,0,1) *
* *
* xerr*
[1] 0.5610573 3.1741687 -2.3915066 -0.2546224 -4.1790037 -1.4387786
1.4211227
-1.1141176 -1.6230087 0.7595021
* model-glm(z ~ xerr, family=binomial(logit))*
* *
* model*
Call: glm(formula = z ~ xerr, family = binomial(logit))
Coefficients:
(Intercept) xerr
1.5001.309
* int-coef(model)[1]*
* *
* slope-coef(model)[2]*
* *
* pred1-predict(model)*
* *
* pred2-predict(model,type=response)*
* *
* pred1*
1 2 3 4 5
6
7 8 9 10
2.23478077 5.65499178 -1.62972799 1.16716581 -3.96932102 -0.38273530
3.36049056 0.04220225 -0.62386760 2.49451832
* pred2*
1 2 3 4 5 6
7
8 9 10
0.90332965 0.99651221 0.16386763 0.76263234 0.01853617 0.40546735
0.96644669 0.51054900 0.34890234 0.92375664
* result-ifelse(pred2.5,1,0) *
* *
* result*
1 2 3 4 5 6 7 8 9 10
1 1 0 1 0 0 1 1 0 1
* accuracy-length(which(result==z))/length(z)*
* *
* accuracy*
[1] 0.8
* rocpreds-prediction(result,z)*
* *
* rocpreds*
* auc-performance(rocpreds,auc)@y.values*
* *
* auc*
[[1]]
[1] 0.7916667
* fp-performance(rocpreds,sens)*
* *
* sentiv-slot(fp,y.values)[[1]]*
* *
* sentiv-sentiv[2]*
* *
* sentiv*
[1] 0.833
* *
* fp2-performance(rocpreds,spec)*
* *
* specs-slot(fp2,y.values)[[1]]*
* *
* specs*
[1] 1.00 0.75 0.00
* specs-specs[2]*
* *
* specs*
[1] 0.75
On Thu, Oct 25, 2012 at 3:43 PM, Berend Hasselman b...@xs4all.nl wrote:
On 25-10-2012, at 21:28, Adel Powell wrote:
I am running my code in a loop and it does not work but when I run it
outside the loop I get the values I want.
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
sentiv-performance(rocpreds,sens)@y.values
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
specs-performance(rocpreds,spec)@y.values
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
A missing before Specificity?
return(output)
}
y-fitglm(.05,1)
y
Running this after correction of the missing one gets en error
Error in fitglm(0.05, 1) : could not find function prediction
How are you using a loop?
Your example is not reproducible.
Berend
The code runs without the sentiv and specs but when I remove the loop i
can
get the sensitivity and spec. values ???
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Re: [R] trying ti use a function in aggregate
Hi Rui, I thought the OP was looking for something like this: May be I am wrong. dat2-dat1[order(dat1$Trip_id,dat1$Vessel,dat1$CommonName,dat1$Length,dat1$Count),] dat3-dat2 dat3$Prop-unlist(tapply(dat3$Count,list(dat3$Trip_id,dat3$CommonName),function(x) x/sum(x))) head(dat3) # Trip_id Vessel CommonName Length Count Prop #1 230 Sunlight Alewife 17 1 0.01408451 #2 230 Sunlight Alewife 18 1 0.01408451 #3 230 Sunlight Alewife 20 2 0.02816901 #4 230 Sunlight Alewife 21 4 0.05633803 #5 230 Sunlight Alewife 22 16 0.22535211 #6 230 Sunlight Alewife 23 22 0.30985915 A.K. - Original Message - From: Rui Barradas ruipbarra...@sapo.pt To: Sally_roman sro...@umassd.edu Cc: r-help@r-project.org Sent: Thursday, October 25, 2012 11:59 AM Subject: Re: [R] trying ti use a function in aggregate Hello, Try the following. (I've changed your function a bit. And named the data.frame 'dat', not 'data', which is an R function.) myfun - function (x) ifelse(sum(x) == 0, 0, x/sum(x)) aggregate(Count ~ Trip_id + Length + CommonName, data = dat, myfun) The output shows that each and every group corresponds to a single row of the original df. The 1's represent 100%, myfun _is_ dividing by sum(Count) Trip_id/Length/CommonName group. If you want just Trip_id/CommonName, use aggregate(Count ~ Trip_id + CommonName, data = dat, myfun) Or use your instruction without 'Length' in the by list: b - with(dat, aggregate(x=list(Percent=Count), by=list(Trip_id=Trip_id, Species=CommonName), FUN = myfun)) b Trip_id Species Percent 1 230 Alewife 0.01408451 2 230 Herring,Blueback 0. 3 230 Shad,American 0.09090909 4 231 Shad,American 0.14285714 As you can see, the results are the same, with different output colnames. Hope this helps, Rui Barradas Em 25-10-2012 15:19, Sally_roman escreveu: Hi -I am using R v 2.13.0. I am trying to use the aggregate function to calculate the percent at length for each Trip_id and CommonName. Here is a small subset of the data. Trip_id Vessel CommonName Length Count 1 230 Sunlight Shad,American 19 1 2 230 Sunlight Shad,American 20 1 3 230 Sunlight Shad,American 21 1 4 230 Sunlight Shad,American 23 1 5 230 Sunlight Shad,American 26 1 6 230 Sunlight Shad,American 27 1 7 230 Sunlight Shad,American 30 2 8 230 Sunlight Shad,American 33 1 9 230 Sunlight Shad,American 34 1 10 230 Sunlight Shad,American 37 1 11 230 Sunlight Herring,Blueback 20 1 12 230 Sunlight Herring,Blueback 21 2 13 230 Sunlight Herring,Blueback 22 5 14 230 Sunlight Herring,Blueback 26 1 15 230 Sunlight Alewife 17 1 16 230 Sunlight Alewife 18 1 17 230 Sunlight Alewife 20 2 18 230 Sunlight Alewife 21 4 19 230 Sunlight Alewife 22 16 20 230 Sunlight Alewife 23 22 21 230 Sunlight Alewife 24 16 22 230 Sunlight Alewife 25 4 23 230 Sunlight Alewife 26 1 24 230 Sunlight Alewife 27 2 25 230 Sunlight Alewife 28 2 26 231 Western Venture Shad,American 23 1 27 231 Western Venture Shad,American 24 1 28 231 Western Venture Shad,American 25 1 29 231 Western Venture Shad,American 28 2 30 231 Western Venture Shad,American 29 2 My code is: myfun-function (x) x/sum(x) b-with(data,aggregate(x=list(Percent=Count),by=list(Trip_id=Trip_id,Length=Length,Species=CommonName), FUN=myfun)) My issue is that the percent is not be calculated by Trip_id and CommonName. The result is that each row has a percent of 1 indicating that myfun is not dividing by the sum of counts with a Trip_id/CommonName group. Any help would be appreciated. Thank you -- View this message in context: http://r.789695.n4.nabble.com/trying-ti-use-a-function-in-aggregate-tp4647414.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
Re: [R] List of multidimensional arrays
hello, I want to make a two dimensional matrix from the data table.That data table has three column 1st is userid,2nditemid and 3rd is rating corresponding to the itemid given by userid.I want to make a m*n matrix in which 'm' is userid and 'n' to be itemid .That 2D matrix should b filled with that corresponding rating for that itemid for that userid. please help me out. -- View this message in context: http://r.789695.n4.nabble.com/List-of-multidimensional-arrays-tp4647286p4647458.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] List of multidimensional arrays
You are very lucky. This question was answered last week in https://stat.ethz.ch/pipermail/r-help/2012-October/326597.html For the future, please read the posting guide referenced below and include an example of your data and the result of the calculation you would like to see. Please use R-help directly and not the nabble interface. On Thu, Oct 25, 2012 at 4:21 PM, pinki751 raj4technol...@gmail.com wrote: hello, I want to make a two dimensional matrix from the data table.That data table has three column 1st is userid,2nditemid and 3rd is rating corresponding to the itemid given by userid.I want to make a m*n matrix in which 'm' is userid and 'n' to be itemid .That 2D matrix should b filled with that corresponding rating for that itemid for that userid. please help me out. -- View this message in context: http://r.789695.n4.nabble.com/List-of-multidimensional-arrays-tp4647286p4647458.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How generate random numbers from given vector???
Hi,
Try this:
number1-c(0,1,3,4,5,6,8)
rsidp-function(x){
y-sample(x,5,replace=TRUE)
y
}
rsidp(number1)
#[1] 3 0 6 8 4
rsidp(number1)
#[1] 1 8 8 6 4
rsidp(number1)
#[1] 8 3 6 6 6
A.K.
- Original Message -
From: Rlotus yerl...@hotmail.com
To: r-help@r-project.org
Cc:
Sent: Thursday, October 25, 2012 3:24 PM
Subject: [R] How generate random numbers from given vector???
I wanna generate random numbers from a vector...
for example number-c(0,1,3,4,5,6,8)
so
rsidp-function(x){
i=0
for (i in seq(1:x))
{y-sample(number,x, replace=T)}
return(y)
}
so all random numbers have to be from vector number;
so if I type rsidp(5). it has to give me 5 random numbers except 2,7,9
(because they are not in the vector numbers). help me plz with it (((
--
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http://r.789695.n4.nabble.com/How-generate-random-numbers-from-given-vector-tp4647447.html
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] How generate random numbers from given vector???
thank u so much! i got it. -- View this message in context: http://r.789695.n4.nabble.com/How-generate-random-numbers-from-given-vector-tp4647447p4647469.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to draw the graph???
I have his code. I wanna draw the graph according to x and y. But at the end
when I run the code it gives me only one dot...it is strange cuz from
the loop it has to give 20 dots. Help me plz to draw a graph(((
for (i in seq(1:20))
{rsidpVector= rsidp(1)
x-mean(rsidpVector)
y-length(number)
plot(y,x,ylab=sample mean,xlab=sample size)
print (rsidpVector)
print(x)
}
rsidp-function(x){
i=0
{y-sample(number,x), replace=TRUE}
return(y)
}
--
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Demo code
I have code in R. I need also demo code of my output and demo code of code. I dont know what is demo. An d how to create it...can you tell how to do that plz. thank you. -- View this message in context: http://r.789695.n4.nabble.com/Demo-code-tp4647471.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] error bars
Hello R-help, I am using R version 2.15.1. I upgraded from R version 2.13 a few months back. Previously, I was able to plot error bars on an xy scatter plot using the errbar function: errbar(RAEthylene$TIME,RAEthylene$AVE,RAEthylene$AVE+RAEthylene$STD,RAEthylene$AVE-RAEthylene$STD,add = TRUE,lty=2,pch=17); Today, I went to update my plot. However, in R version 2.15.1 I get error code saying that this function cannot be found: Error: could not find function errbar I would perfer to avoid using the xy.error.bars-function(x, y, xbar, ybar) coding for these error bars, as I have many data to put on one plot. I've searched the .pdf file for R 2.15.1 version and cannot find any updates for this function. Do I have to reinstall version 2.13 again?? I need to generate these plots today!! Please advise. Regards, Franklin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] system is computationally singular: reciprocal condition number
Hi folks,
I know, this is a fairly common question and I am really disappointed that I
could not find a solution.
I am trying to calculate Mahanalobis distances in a data frame, where I have
several hundreds groups and several hundreds of variables.
Whatever I do, however I subset it I get the system is computationally
singular: reciprocal condition number error.
I know what it means and I know what should be the problem, but there is no
way this is a singular matrix.
I have uploaded the input file to my ftp:
http://mkk.szie.hu/dep/talt/lv/CentInpDuplNoHeader.txt
It is a tab delimited txt file with no headers.
I tried the StatMatch Mahanalobis function and also this function:
mahal_dist -function (data, nclass, nvariable) {
dist - matrix(0, nclass, nclass)
n=0
w - cov(data)
print(w)
for(i in 1:nclass) {
for(c in 1:nclass){
diffl - vector(length = nvariable)
for(l in 1:nvariable){
diffl[l]=abs(data[i,l]-data[c,l])
}
### matrixes
print(diffl)
dist[i,c]= (t(diffl))%*%(solve(w))%*%(diffl)
}
n=n+1
print(n)
}
return(dist)
sqrt_dist - sqrt(dist)
print(sqrt_dist) }
I have a deadline for this project (not a homework:)), and I could always
use this codes, so I thought I will be able to quit the calculations short,
but now I am just lost.
I would really appreciate any help.
Thanks for any help
--
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How generate random numbers from given vector???
Hello,
You don't need the loop, the sample() argument 'size' is there for that.
See 'sample.
number - c(0,1,3,4,5,6,8)
rsidp - function(n) sample(number, n, replace = TRUE)
rsidp(5)
Hope this helps,
Rui Barradas
Em 25-10-2012 20:24, Rlotus escreveu:
I wanna generate random numbers from a vector...
for example number-c(0,1,3,4,5,6,8)
so
rsidp-function(x){
i=0
for (i in seq(1:x))
{y-sample(number,x, replace=T)}
return(y)
}
so all random numbers have to be from vector number;
so if I type rsidp(5). it has to give me 5 random numbers except 2,7,9
(because they are not in the vector numbers). help me plz with it (((
--
View this message in context:
http://r.789695.n4.nabble.com/How-generate-random-numbers-from-given-vector-tp4647447.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] system is computationally singular: reciprocal condition number
1. I don't know what StatMatch is. Try using stats::mahalanobis.
2. It's the covariance matrix that is **numerically** singular and
can't be inverted. Why do you claim that there's no way this could
be true when there are hundreds of variables (= dimensions).
3. Try calculating the svd of your matrix and see what you get if you
haven't already done so.
Cheers,
Bert
On Thu, Oct 25, 2012 at 4:14 PM, langvince la...@purdue.edu wrote:
Hi folks,
I know, this is a fairly common question and I am really disappointed that I
could not find a solution.
I am trying to calculate Mahanalobis distances in a data frame, where I have
several hundreds groups and several hundreds of variables.
Whatever I do, however I subset it I get the system is computationally
singular: reciprocal condition number error.
I know what it means and I know what should be the problem, but there is no
way this is a singular matrix.
I have uploaded the input file to my ftp:
http://mkk.szie.hu/dep/talt/lv/CentInpDuplNoHeader.txt
It is a tab delimited txt file with no headers.
I tried the StatMatch Mahanalobis function and also this function:
mahal_dist -function (data, nclass, nvariable) {
dist - matrix(0, nclass, nclass)
n=0
w - cov(data)
print(w)
for(i in 1:nclass) {
for(c in 1:nclass){
diffl - vector(length = nvariable)
for(l in 1:nvariable){
diffl[l]=abs(data[i,l]-data[c,l])
}
### matrixes
print(diffl)
dist[i,c]= (t(diffl))%*%(solve(w))%*%(diffl)
}
n=n+1
print(n)
}
return(dist)
sqrt_dist - sqrt(dist)
print(sqrt_dist) }
I have a deadline for this project (not a homework:)), and I could always
use this codes, so I thought I will be able to quit the calculations short,
but now I am just lost.
I would really appreciate any help.
Thanks for any help
--
View this message in context:
http://r.789695.n4.nabble.com/system-is-computationally-singular-reciprocal-condition-number-tp4647472.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to draw the graph???
On Oct 25, 2012, at 2:24 PM, Rlotus wrote:
I have his code. I wanna draw the graph according to x and y. But at the end
when I run the code it gives me only one dot...it is strange cuz from
the loop it has to give 20 dots. Help me plz to draw a graph(((
for (i in seq(1:20))
{rsidpVector= rsidp(1)
x-mean(rsidpVector)
y-length(number)
plot(y,x,ylab=sample mean,xlab=sample size)
print (rsidpVector)
print(x)
}
rsidp-function(x){
i=0
{y-sample(number,x), replace=TRUE}
return(y)
}
Set up the plot before the loo and then use the points function inside the
loop.
--
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
Alameda, CA, USA
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Z score
Hi Arun, Thank you ! Very much appreciated. [Only added t() at the end to preserve the original orientation.] Also, many thanks to Rui Barradas. Cheers, Ved On Thu, Oct 25, 2012 at 12:06 PM, arun smartpink...@yahoo.com wrote: Hi Ved, Sorry, I didn't test it well enough at that time. In your example file, #there were NAs MyFile1 - read.csv( text= Names,'Sample_1','Sample_2','Sample_3' Gene_1,87,77,88 Gene_2,98,22,34 Gene_3,33,43,33 Gene_4,78,,81 , header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= ) #Here, the apply() function outputs a list when I remove the NA from the last row. apply(MyFile1,1,function(x) x[!is.na(x)]) #outputs a list #$Gene_1 #Sample_1 Sample_2 Sample_3 # 87 77 88 #$Gene_2 #Sample_1 Sample_2 Sample_3 # 98 22 34 #$Gene_3 #Sample_1 Sample_2 Sample_3 # 33 43 33 #$Gene_4 #Sample_1 Sample_3 # 78 81 # Without NAs MyFile2 - read.csv( text= Names,'Sample_1','Sample_2','Sample_3' Gene_1,87,77,88 Gene_2,98,22,34 Gene_3,33,43,33 Gene_4,78,48,81 , header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= ) apply(dat3,1,function(x) x[!is.na(x)]) # the output is a matrix # Gene_1 Gene_2 Gene_3 Gene_4 #Sample_1 87 98 33 78 #Sample_2 77 22 43 48 #Sample_3 88 34 33 81 is.matrix(apply(dat3,1,function(x) x[!is.na(x)]) ) #[1] TRUE #Consider another case MyFile3 - read.csv( text= Names,'Sample_1','Sample_2','Sample_3' Gene_1,87,77,88 Gene_2,,22,34 Gene_3,33,43,33 Gene_4,78,,81 , header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= ) t(sapply(lapply(apply(MyFile3,1,function(x) x[!is.na(x)]),function(x) (x-mean(x))/sd(x)),function(x) x[colnames(MyFile3)] )) #works because the apply() output is a list #Sample_1 Sample_2 Sample_3 #Gene_1 0.4931970 -1.1507929 0.6575959 #Gene_2 NA -0.7071068 0.7071068 #Gene_3 -0.5773503 1.1547005 -0.5773503 #Gene_4 -0.7071068 NA 0.7071068 #Yet another case: MyFile4 - read.csv( text= Names,'Sample_1','Sample_2','Sample_3' Gene_1,87,77 Gene_2,,22,34 Gene_3,33,,33 Gene_4,78,,81 , header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= ) apply(MyFile4,1,function(x) x[!is.na(x)]) #output is a matrix because equal number of NAs were present in each row # Gene_1 Gene_2 Gene_3 Gene_4 #[1,] 87 22 33 78 #[2,] 77 34 33 81 t(sapply(lapply(apply(MyFile4,1,function(x) x[!is.na(x)]),function(x) (x-mean(x))/sd(x)),function(x) x[colnames(MyFile4)] )) #doesn't work #In your dataset, there were no NAs dat1-read.csv(Bcl2_With_expressions.csv,sep=\t,row.names=1) MyFile-dat1[,-1] str(apply(MyFile,1,function(x) x[!is.na(x)])) # a matrix # num [1:29, 1:18] 10.48 10.96 9.28 11.1 10.95 ... #- attr(*, dimnames)=List of 2 # ..$ : chr [1:29] ALL2 MLL8 ALL42 MLL5 ... # ..$ : chr [1:18] BAX BCL2L15 BCL2 BMF ... #In this case, either res2-apply(MyFile,1,function(x) (x-mean(x))/sd(x)) #or res1-apply(apply(MyFile,1,function(x) x[!is.na(x)]),2,function(x) (x-mean(x))/sd(x)) #works identical(res1,res2) #[1] TRUE head(res1,2) # BAX BCL2L15 BCL2BMFBAD MCL1 BCL2L1 #ALL2 0.1216373 -0.215256 1.040758 -0.4078606 -0.2427741 0.6967070 -0.1054749 #MLL8 0.6565878 -1.446252 1.052566 -0.1825442 -0.2312166 0.9882503 -0.9687260 # BOK BCL2A1BCL2L14 BAK1 BBC3BCL2L11 #ALL2 -0.1465807 0.5353133 -0.1772439 -0.3751981 0.6341806 -1.2432273 #MLL8 0.2918296 -0.8466821 0.3088331 -1.4025846 0.7056799 0.9944288 # BID NOXA1BIK HRKBCL2L2 #ALL2 -2.2961643 0.2105960 -0.9195998 -0.001731806 1.6691590 #MLL8 -0.5103087 0.3433778 1.2352986 -0.568548518 0.3674839 Hope it helps A.K. From: Vedant Sharma vedantg...@gmail.com To: arun smartpink...@yahoo.com Sent: Wednesday, October 24, 2012 7:56 PM Subject: Re: [R] Z score Hello Arun, Thank you. I could manage to get the answer. However, this particular code, however, doesn't seem to work when I try to read from a .csv file (as attached). And, I am inquisitive to find out the reason ! MyFile - read.csv (file.choose(), header=T, row.names=1) MyFile - MyFile [,-1] res2-t(sapply(lapply(apply(MyFile,1,function(x) x[!is.na(x)]),function(x) (x-mean(x))/sd(x)),function(x) x[colnames(MyFile)] )) Thanks again !! Cheers, Ved = On Wed, Oct 24, 2012 at 9:53 PM, arun smartpink...@yahoo.com wrote: Hi, In cases, with more sample columns, you could also use this: res2-t(sapply(lapply(apply(MyFile,1,function(x) x[!is.na(x)]),function(x) (x-mean(x))/sd(x)),function(x) x[colnames(MyFile)] )) res2 #Sample_1 Sample_2 Sample_3 #Gene_1 0.4931970 -1.1507929 0.6575959
Re: [R] error bars
On Oct 25, 2012, at 2:45 PM, Johnson, Franklin Theodore wrote:
Hello R-help,
I am using R version 2.15.1.
I upgraded from R version 2.13 a few months back.
Previously, I was able to plot error bars on an xy scatter plot using the
errbar function:
errbar(RAEthylene$TIME,RAEthylene$AVE,RAEthylene$AVE+RAEthylene$STD,RAEthylene$AVE-RAEthylene$STD,add
= TRUE,lty=2,pch=17);
Today, I went to update my plot.
However, in R version 2.15.1 I get error code saying that this function
cannot be found:
Error: could not find function errbar
I would perfer to avoid using the xy.error.bars-function(x, y, xbar, ybar)
coding for these error bars, as I have many data to put on one plot.
I've searched the .pdf file for R 2.15.1 version and cannot find any updates
for this function.
Do I have to reinstall version 2.13 again??
There is no base::errbar function. You probably were using a package and have
not yet reinstalled that package under the new version of R. When I run
sos::findFn('errbar') I see two packages (Hmisc and sfsmisc) with a function of
that name.
--
David Winsemius, MD
Alameda, CA, USA
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Re: [R] system is computationally singular: reciprocal condition number
On Oct 25, 2012, at 4:41 PM, Bert Gunter wrote:
1. I don't know what StatMatch is. Try using stats::mahalanobis.
2. It's the covariance matrix that is **numerically** singular and
can't be inverted. Why do you claim that there's no way this could
be true when there are hundreds of variables (= dimensions).
3. Try calculating the svd of your matrix and see what you get if you
haven't already done so.
This was crossposted to StackOverflow where Josh O'Brien has responded that his
code using svd() shows the matrix to be highly collinear. This is the upper
left corner of the correlation matrix:
V1 V2 V3 V4 V5
V11. 0.97250825 0.93390424 0.918813118 0.89705917
V20.97250825 1. 0.97118079 0.954020724 0.93992361
V30.93390424 0.97118079 1. 0.991508026 0.97602188
V40.91881312 0.95402072 0.99150803 1.0 0.98837387
V50.89705917 0.93992361 0.97602188 0.988373865 1.
length( which(cor(mat)==1) )
[1] 374
Just looking at it should give a good idea why. I can see bands of columns that
are identically zero.
--
david.
Cheers,
Bert
On Thu, Oct 25, 2012 at 4:14 PM, langvince la...@purdue.edu wrote:
Hi folks,
I know, this is a fairly common question and I am really disappointed that I
could not find a solution.
I am trying to calculate Mahanalobis distances in a data frame, where I have
several hundreds groups and several hundreds of variables.
Whatever I do, however I subset it I get the system is computationally
singular: reciprocal condition number error.
I know what it means and I know what should be the problem, but there is no
way this is a singular matrix.
I have uploaded the input file to my ftp:
http://mkk.szie.hu/dep/talt/lv/CentInpDuplNoHeader.txt
It is a tab delimited txt file with no headers.
I tried the StatMatch Mahanalobis function and also this function:
mahal_dist -function (data, nclass, nvariable) {
dist - matrix(0, nclass, nclass)
n=0
w - cov(data)
print(w)
for(i in 1:nclass) {
for(c in 1:nclass){
diffl - vector(length = nvariable)
for(l in 1:nvariable){
diffl[l]=abs(data[i,l]-data[c,l])
}
### matrixes
print(diffl)
dist[i,c]= (t(diffl))%*%(solve(w))%*%(diffl)
}
n=n+1
print(n)
}
return(dist)
sqrt_dist - sqrt(dist)
print(sqrt_dist) }
I have a deadline for this project (not a homework:)), and I could always
use this codes, so I thought I will be able to quit the calculations short,
but now I am just lost.
I would really appreciate any help.
Thanks for any help
--
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and provide commented, minimal, self-contained, reproducible code.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
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David Winsemius, MD
Alameda, CA, USA
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Re: [R] How generate random numbers from given vector???
Hello,
The other options is to use the sample() function.
test2 - matrix (rep(sample(number1, size = 5), times=3), nrow=3)
Pradip Muhuri
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Rui Barradas [ruipbarra...@sapo.pt]
Sent: Thursday, October 25, 2012 7:19 PM
To: Rlotus
Cc: r-help@r-project.org
Subject: Re: [R] How generate random numbers from given vector???
Hello,
You don't need the loop, the sample() argument 'size' is there for that.
See 'sample.
number - c(0,1,3,4,5,6,8)
rsidp - function(n) sample(number, n, replace = TRUE)
rsidp(5)
Hope this helps,
Rui Barradas
Em 25-10-2012 20:24, Rlotus escreveu:
I wanna generate random numbers from a vector...
for example number-c(0,1,3,4,5,6,8)
so
rsidp-function(x){
i=0
for (i in seq(1:x))
{y-sample(number,x, replace=T)}
return(y)
}
so all random numbers have to be from vector number;
so if I type rsidp(5). it has to give me 5 random numbers except 2,7,9
(because they are not in the vector numbers). help me plz with it (((
--
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Re: [R] How generate random numbers from given vector???
You can also just replicate() as follows:
# x is the vector
# s is the size of the sample
# B is the number of samples
# ... arguments passed to sample()
f - function(x, s, B, ...) replicate(B, sample(x, s, ...))
f(x, 3, 10, TRUE)
f(x, 3, 10, FALSE)
HTH,
Jorge.-
On Fri, Oct 26, 2012 at 12:05 PM, Muhuri, Pradip (SAMHSA/CBHSQ)
pradip.muh...@samhsa.hhs.gov wrote:
Hello,
The other options is to use the sample() function.
test2 - matrix (rep(sample(number1, size = 5), times=3), nrow=3)
Pradip Muhuri
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On
Behalf Of Rui Barradas [ruipbarra...@sapo.pt]
Sent: Thursday, October 25, 2012 7:19 PM
To: Rlotus
Cc: r-help@r-project.org
Subject: Re: [R] How generate random numbers from given vector???
Hello,
You don't need the loop, the sample() argument 'size' is there for that.
See 'sample.
number - c(0,1,3,4,5,6,8)
rsidp - function(n) sample(number, n, replace = TRUE)
rsidp(5)
Hope this helps,
Rui Barradas
Em 25-10-2012 20:24, Rlotus escreveu:
I wanna generate random numbers from a vector...
for example number-c(0,1,3,4,5,6,8)
so
rsidp-function(x){
i=0
for (i in seq(1:x))
{y-sample(number,x, replace=T)}
return(y)
}
so all random numbers have to be from vector number;
so if I type rsidp(5). it has to give me 5 random numbers except
2,7,9
(because they are not in the vector numbers). help me plz with it (((
--
View this message in context:
http://r.789695.n4.nabble.com/How-generate-random-numbers-from-given-vector-tp4647447.html
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__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
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[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] system is computationally singular: reciprocal condition number
On Fri, Oct 26, 2012 at 12:14 PM, langvince la...@purdue.edu wrote: Whatever I do, however I subset it I get the system is computationally singular: reciprocal condition number error. I know what it means and I know what should be the problem, but there is no way this is a singular matrix. I have uploaded the input file to my ftp: http://mkk.szie.hu/dep/talt/lv/CentInpDuplNoHeader.txt It is a tab delimited txt file with no headers. It's a singular matrix. The data matrix has rank 300 according to either qr() or svd(). The 301st singular value is about ten orders of magnitude smaller than the 300th one. The problem is the rounding of the values -- if you take 372 vectors in 380-dimensional space they should be linearly independent, but if you force them to lie on a relatively coarse grid there are quite likely to be linear dependencies. When I add random noise in the fourth decimal place, the matrix stops being singular. -thomas -thomas -- Thomas Lumley Professor of Biostatistics University of Auckland __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Merge matrices with different column names
A general question that I have been pursuing for some time but have set aside. When finishing some analysis, I can have multiple matrices that have specific column names. Ideally, I would like to combine these separate matrices for a final output as a csv file. A generic example: Matrix 1 var1A var1B var1C x x x x x x Matrix 2 var2A var2B var2C x x x x x x I would like a final exportable matrix or dataframe or whichever format is most workable. Matrix 3 var1A var1B var1C x x x x x x var2A var2B var2C x x x x x x However, no matter which function I try reports an error that the column names are not the same. Any insights would be appreciated. Thanks as always, Charles [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to draw the graph???
plot(y,x, ylab=sample mean,xlab=sample size)
for (i in seq(1:20))
{rsidpVector= rsidp(1)
x-mean(rsidpVector)
y-length(number)
print (rsidpVector)
print(x)
sorry, but how i can use points function in the loop?
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Re: [R] Merge matrices with different column names
If they have the same number of rows, you can use cbind() to create one object to write out. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Charles Determan Jr deter...@umn.edu wrote: A general question that I have been pursuing for some time but have set aside. When finishing some analysis, I can have multiple matrices that have specific column names. Ideally, I would like to combine these separate matrices for a final output as a csv file. A generic example: Matrix 1 var1A var1B var1C x x x x x x Matrix 2 var2A var2B var2C x x x x x x I would like a final exportable matrix or dataframe or whichever format is most workable. Matrix 3 var1A var1B var1C x x x x x x var2A var2B var2C x x x x x x However, no matter which function I try reports an error that the column names are not the same. Any insights would be appreciated. Thanks as always, Charles [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] system is computationally singular: reciprocal condition number
Hey Bert, thanks for your fast reply. Yes, based on svd it is singular. The no way statement was because of the source of the dataset. I would not expect that. I never used the stats Maha dist calc, but after giving it a shot, not a surprise still singular. Any idea how to manipulate the data to have it run, or other idea to solve the problem? thanks -- View this message in context: http://r.789695.n4.nabble.com/system-is-computationally-singular-reciprocal-condition-number-tp4647472p4647483.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] system is computationally singular: reciprocal condition number
Hey David, my answers are delayed here, although I am not using my gmail email address:) Yep thats right, those bands of zeros are one of the most important values to define one group, and have a nice distance from the rest of the groups :). I cannot really get rid of those, I bet it would not help if I would change all of them to a really small (but same) value. Thanks -- View this message in context: http://r.789695.n4.nabble.com/system-is-computationally-singular-reciprocal-condition-number-tp4647472p4647486.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Generate two bell curves
How do I generate two bell curves, both with the same mean but different standard deviations? I used to have a script for it but I lost it. Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Generate-two-bell-curves-tp4647488.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] wanna create all points.....
Plz help me ;(( I lost 7 hours for thinking and doing this code ;( I need to
create points.20 points(dots). All these dots have to be in one graph...
when i run this codeit gives me only one dote.
number- (0,2,3,4,5,6,8)
for (i in seq(1:20))
{rsidpVector=rsidp(i)
mean(rsidpVector)
plot (length(rsidpVector), mean(rsidpVector), ylab=sample mean, xlab=
sample size)
}
rsidp- function(x){
i+0
{y-sample(number,x,replace = TRUE)}
return(y)
}
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[R] how to make simulation faster
Dear R users, I need to run 1000 simulations to find maximum likelihood estimates. I print my output as a vector. However, it is taking too long. I am running 50 simulations at a time and it is taking me 30 minutes. Once I tried to run 200 simulations at once, after 2 hours I stopped it and saw that only about 40 of them are simulated in those 2 hours. Is there any way to make my simulations faster? (I can post my code if needed, I'm just looking for general ideas here). Thank you in advance. -- View this message in context: http://r.789695.n4.nabble.com/how-to-make-simulation-faster-tp4647492.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] wanna create all points.....
One dote..and its a last dote of the loop ;( -- View this message in context: http://r.789695.n4.nabble.com/wanna-create-all-points-tp4647494p4647495.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Interpreting and visualising lme results
Dear R users, I have used the following function (in blue) aiming to find the linear regression between MOE and XLA and nesting my data by Species. I have obtained the following results (in green). model4-lme(MOE~XLA, random = ~ XLA|Species, method=ML)summary(model4) Linear mixed-effects model fit by maximum likelihood Data: NULL AIC BIC logLik -1.040187 8.78533 6.520094 Random effects: Formula: ~XLA | Species Structure: General positive-definite, Log-Cholesky parametrization StdDev Corr (Intercept) 1.944574e-01 (Intr)XLA 6.134158e-06 -0.884Residual 1.636428e-01 Fixed effects: MOE ~ XLA Value Std.Error DF t-value p-value(Intercept) 3.0558697 0.15075939 32 20.269847 0.XLA 0.005 0.0335 32 0.150811 0.8811 Correlation: (Intr)XLA -0.861 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -1.8354171 -0.4704322 0.1414749 0.5500273 1.5950338 Number of Observations: 38Number of Groups: 5 I have read that large correlation values such as,Correlation: (Intr)XLA -0.861reflect an ill-conditioned model, in addition XLA does not have an effect on the model p=0.88. These results are not logic when I look at my data and therefore I think I am missing something in the model? It would be very helpful if someone has some tips on this? In addition, I was wondering if somebody knows what is the best way to visualise this kind of data (nested data)? Thank you very much for any help and time. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to draw the graph???
On Oct 25, 2012, at 5:14 PM, Rlotus wrote:
plot(y,x, ylab=sample mean,xlab=sample size)
for (i in seq(1:20))
{rsidpVector= rsidp(1)
x-mean(rsidpVector)
y-length(number)
print (rsidpVector)
print(x)
sorry, but how i can use points function in the loop?
Tested code provided in response to reproducible examples. The points function
will not redraw a new plot with each iteration. It might be the case that you
do not even need a loop, but you have not described the problem yet.
--
David Winsemius, MD
Alameda, CA, USA
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Re: [R] Generate two bell curves
On Oct 26, 2012, at 03:17 , phantastic wrote: How do I generate two bell curves, both with the same mean but different standard deviations? I used to have a script for it but I lost it. Thanks. curve(, from=..., to=...) curve(, add=TRUE) If you plot the tallest one first, you won't need to diddle the ylim setting. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
