[R] Problem in performing goodness of fit test in R.
I am trying to perform goodness of fit test using R. I am using this website http://wiener.math.csi.cuny.edu/Statistics/R/simpleR/stat013.html for help. However, I am unable to carry out the test successfully. My code follows. It is taken from the website just mentioned. freq=c(22,21,22,27,22,36) # frequencies obtained after rolling the dice 150 times. prob=c(1,1,1,1,1,1)/6 # specify expected frequency for each category. chisq.test(freq,p=prob) # I do not know what this line means. I just followed instructions on the website. The erorr I receive is "erorr in chisq.test(freq,p=prob)/6 probabilities must sum to 1" I am very new to R, so any help would be appreciated. Faiz. [[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] SAS and RODBC
> -Original Message-
> From: Frank E Harrell Jr [mailto:f.harr...@vanderbilt.edu]
> Sent: Saturday, February 13, 2010 5:49 AM
> To: Daniel Nordlund
> Cc: r-help@r-project.org
> Subject: Re: [R] SAS and RODBC
>
> Daniel Nordlund wrote:
> . . .
>
> >
> > This is just a quick follow-up to my previous post. Based on Prof.
> Ripley's response I went back and looked at the SAS log file and reread
> the RODBC help pages. The problem of writing a SAS dataset was solved by
> setting colQuote=NULL in the call to the odbcConnect() function.
> >
> > ch <- odbcConnect('sasodbc', believeNRows=FALSE, colQuote=NULL)
> >
> > I hope this will be useful to others who may have the SAS BASE product
> and want to do graphics or statistical analyses with their SAS data, but
> can't afford the high licensing fees for the SAS STAT and GRAPH modules.
> Thanks to Prof. Ripley for the fine RODBC package.
> >
> > Dan
> >
> > Daniel Nordlund
> > Bothell, WA USA
> >
> >
>
> Daniel since you have SAS BASE installed why not use sas.get in the
> Hmisc package and also get access to metadata such as variable labels
> that ODBC does not handle? Besides providing better documentation, the
> labels are very useful as axis labels in plotting, etc.
>
> Frank
>
Frank,
I have used sas.get from Hmisc before, and I will continue to use it. I
appreciate the work that you and your colleagues have done with Hmisc and the
Design and rms packages. However, the sas.get function still appears to be
broken on Windows platforms (or Windows is broken :-). I know how to fix the
problem, but I am always looking for approaches where I don't have to fix
things. It may well be that the better documentation provided by sas.get will
prove to out weigh the inconvenience of having to source an edited version of
sas.get for my regular use.
As for moving data from R to SAS, I don't know of any methods other than the
RODBC package with the SAS ODBC driver for writing SAS datasets. Yes, I can
write to csv or other file types that SAS can import, but if I can eliminate
extra steps when going from R to SAS then that is a plus for me.
Thanks again for the great tools,
Dan
Daniel Nordlund
Bothell, WA USA
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Re: [R] NextMethod() example from S Programming by Vena bles and Ripley (page 78)
blue sky gmail.com> writes:
> S Programming by Venables and Ripley (page 78) has the example listed
> at the end of this email. However, I get the following error when I
> try the example. I don't understand the descriptions of NextMethod on
> its help page. Could somebody let me know how to fix the error of this
> example?
> > test(x)
> c1
> c2
> Error in NextMethod() : no method to invoke
> Calls: test -> test.c1 -> NextMethod -> test.c2 -> NextMethod
> Execution halted
>
>
> test=function(x) UseMethod('test')
>
> test.c1=function(x) {
> cat('c1\n')
> NextMethod()
> x
> }
>
> test.c2=function(x) {
> cat('c2\n')
> NextMethod()
> x
> }
>
> test.c3=function(x) {
> cat('c3\n')
> x
> }
>
> x=1
> class(x)=c('c1','c2')
> test(x)
It works fine for me if you add a default method,
which I think is what it is looking for.
test.default <- function(x){
cat("default")
x
}
test(x)
c1
c2
default[1] 1
attr(,"class")
[1] "c1" "c2"
--
Ken Knoblauch
Inserm U846
Stem-cell and Brain Research Institute
Department of Integrative Neurosciences
18 avenue du Doyen Lépine
69500 Bron
France
tel: +33 (0)4 72 91 34 77
fax: +33 (0)4 72 91 34 61
portable: +33 (0)6 84 10 64 10
http://www.sbri.fr/members/kenneth-knoblauch.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.
[R] NextMethod() example from S Programming by Venables and Ripley (page 78)
S Programming by Venables and Ripley (page 78) has the example listed
at the end of this email. However, I get the following error when I
try the example. I don't understand the descriptions of NextMethod on
its help page. Could somebody let me know how to fix the error of this
example?
> test(x)
c1
c2
Error in NextMethod() : no method to invoke
Calls: test -> test.c1 -> NextMethod -> test.c2 -> NextMethod
Execution halted
test=function(x) UseMethod('test')
test.c1=function(x) {
cat('c1\n')
NextMethod()
x
}
test.c2=function(x) {
cat('c2\n')
NextMethod()
x
}
test.c3=function(x) {
cat('c3\n')
x
}
x=1
class(x)=c('c1','c2')
test(x)
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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] Dimensional reduction package
Is there any R package which implements non-linear dimensionality reduction (LLE, ISOMAP, GTM, and so on) and/or intrinsic dimensionality estimation ? Thank you, Maura tutti i telefonini TIM! [[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] lm function in R
Dear Something,
I think you have no hope of completing this project without going back
to an introductory regression text. The two models you listed (Y-Hat =
b0 + b1X1 + b2X2 + . . . bkXk + · · · + bkXk) and (Y-Hat = b0 + b1X1 +
b2x2 + ... + bkXk + b12 X12+ b13 X13 + + c) are identical
except for the typos. Cohen and Cohen [1] might be a good place to
start.
Best,
Ista
[1]
http://www.amazon.com/Multiple-Regression-Correlation-Analysis-Behavioral/dp/0805822232
On Sat, Feb 13, 2010 at 8:03 PM, Something Something
wrote:
>>>From your question it is difficult to determine what sort of tutoring you
> are expecting.
> The kind of tutoring you are providing is definitely helpful :)
>
> I guess I should quickly explain what I am trying to accomplish. I have
> been a computer scientist for quite a few years, but I studied Statistics
> long time ago. Now trying to get back into the Statistics field.
>
> For my latest project I have been asked to implement Multiple Regression
> Analysis (using interactions - I guess) in Java & Distributed Computing
> (more specifically Hadoop). I looked at the possibility of using rJava, but
> it's 'single threaded' model plus complex deployment on 1000s of machines
> does not make it very appealing. I looked at using 3rd party libraries such
> as 'Apache Commons Math' & Flanagan's JAVA statistics library. Both of them
> use this formula:
> Y-Hat = b0 + b1X1 + b2X2 + . . . bkXk + · · · + bkXk
>
> and NOT the one I have been asked to implement - which is -
> Y-Hat = b0 + b1X1 + b2x2 + ... + bkXk + b12 X12+ b13 X13 + + c
>
> (Not to mention these tools do not use BigDecimal so the answers given by
> them are not as precise as those from R).
>
> So now the deadline is approaching... and I can't find any Java library that
> matches results from R.. so I have to roll up my sleeves and start coding in
> Java - which means I need to get up to speed on Y-Hat calculations very
> quickly. Ideally, looking for a paper (or a text book) that gives step by
> step instructions on calculating coefficients such as (b0, b1... b13). I
> have been searching but haven't found anything. My last resort is to look
> at R source code and start converting it to Java.
>
> I am new to R and a little rusty on Statistics, so I apologize for all the
> dumb questions, and GREATLY appreciate your patience and help. Thanks.
>
>
> On Sat, Feb 13, 2010 at 3:20 PM, David Winsemius
> wrote:
>
>>
>> On Feb 13, 2010, at 5:03 PM, Something Something wrote:
>>
>> I tried..
>>>
>>> mod = lm(Y ~ X1*X2*X3, na.action = na.exclude)
>>> formula(mod)
>>>
>>> This produced
>>> Y ~ X1 * X2 * X3
>>>
>>>
>>> When I typed just mod I got:
>>>
>>> Call:
>>> lm(formula = Y ~ X1 * X2 * X3, na.action = na.exclude)
>>>
>>> Coefficients:
>>> (Intercept) X11 X21 X31 X11:X21
>>> X11:X31
>>> X21:X31 X11:X21:X31
>>> 177.9245 0.2005 2.4482 3.1216 0.8127 -26.6166
>>> -3.0398 29.6049
>>>
>>>
>>> I am trying to figure out how R computed all these coefficients.
>>>
>>
>> From your question it is difficult to determine what sort of tutoring you
>> are expecting. To get the code of an R formula, you just type its name:
>>
>> lm
>>
>> Leads to lm.fit:
>>
>> lm.fit
>>
>> Reading further it appears the lm and lm.fit functions are really front
>> ends for this call:
>>
>> .Fortran("dqrls", qr = x, n = n, p = p, y = y, ny = ny,
>> tol = as.double(tol), coefficients = mat.or.vec(p, ny),
>> residuals = y, effects = y, rank = integer(1L), pivot = 1L:p,
>> qraux = double(p), work = double(2 * p), PACKAGE = "base")
>>
>> Seems pretty likely that is a QR decomposition-based method that i
>> implemented in compiled code.
>>
>> So if you want to go deeper, at least you know what to search for. Or if
>> you want to know how regression works on a matrix level, you should consult
>> a good reference text or Wikipedia, which is surprisingly good for that sort
>> of question these days.
>>
>> --
>> David.
>>
>>>
>>>
>>>
>>>
>>>
>>> On Sat, Feb 13, 2010 at 1:30 PM, Bert Gunter
>>> wrote:
>>>
>>> ?formula
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of Something Something
Sent: Saturday, February 13, 2010 1:24 PM
To: Daniel Nordlund
Cc: r-help@r-project.org
Subject: Re: [R] lm function in R
Thanks Dan. Yes that was very helpful. I didn't see the change from '*'
to
'+'.
Seems like when I put * it means - interaction & when I put + it's not an
interaction.
Is it correct to assume then that...
When I put + R evaluates the following equation:
Y-Hat = b0 + b1X1 + b2X2 + . . . bkXk + 7 7 7 + bkXk
But when I put * R evaluates the following equation;
Y-Hat = b0 + b1X1 + b2x2 + ... + bkXk + b12 X12
Re: [R] Hierarchical data sets: which software to use?
Dear Anton, 4Mb is not a lot of data. A Gb still wouldn't be that troublesome in a flat file. Your data can be migrated to a relational database at a future point. Sincerely, KeithC. -Original Message- From: Anton du Toit [mailto:atdutoitrh...@gmail.com] Sent: Saturday, February 13, 2010 3:54 AM To: r-help Subject: Re: [R] Hierarchical data sets: which software to use? Hi Douglas, Thanks for your helpful response. I've commented on some of the points you raised below: > Although this may not be helpful for your immediate goal, storing and manipulating data of this size and complexity (and, I expect, cost for collection) really calls for tools like relational databases. A single flat file of 2500 variables by 1500 cases is almost never the best way to organize such data. A normalized representation as a collection of interlinked tables in a relational data base is much more effective and less error prone. The widespread use of spreadsheets or SPSS data sets or SAS data sets which encourage the "single table with a gargantuan number of columns, most of which are missing data in most cases" approach to organization of longitudinal data is regrettable. I'm both relieved and daunted by this. Daunted because it means I'll need to learn another package (probably postGreSQL or MySQL?), but relieved because constructing a 2500 by 1500 file seemed intuitively wrong, as well as introducing the possibility of errors unnecessarily--surely it makes more sense to leave the data as is. As far as immediate goals go--I am at the beginning of a thesis, and I have more research planned after that, so I want to get things right from the start. > For later analysis in R it is better to start with "long" form of the data, as opposed to the "wide" form, even if it means repeating demographic information over several occasions. Using a relational database allows for a long view to be generated without the possibility of inconsistency in the demographics. I am using the descriptions "long" and "wide" in the sense that they are used in the reshape help page. See ?reshape > in R. The long view is also called the subject/occasion view in the sense that each row corresponds to one subject on one occasion. > Robert Gentleman's book "R Programming for Bioinformatics" provides background on linking R to relational databases. Thanks--I'll look this one up. > As I said at the beginning, you may not want to undertake the necessary study and effort to reorganize your data for this specific project but if you do this a lot you may want to consider it. As above: a stitch in time, I suppose. Thanks again. Anton On Sat, Feb 6, 2010 at 3:22 AM, Douglas Bates wrote: > On Sun, Jan 31, 2010 at 10:24 PM, Anton du Toit wrote: >> Dear R-helpers, >> >> Im writing for advice on whether I should use R or a different >> package or language. Ive looked through the R-help archives, some >> manuals, and some other sites as well, and I havent done too well >> finding relevant info, hence my question here. >> >> Im working with hierarchical data (in SPSS lingo). That is, for each >> case >> (person) I read in three types of (medical) record: >> >> 1. demographic data: name, age, sex, address, etc >> >> 2. admissions data: this generally repeats, so I will have 20 or so >> variables relating to their first hospital admission, then the same >> 20 again for their second admission, and so on >> >> 3. collections data, about 100 variables containing the results of >> a battery of standard tests. These are administered at intervals and >> so this is repeating data as well. >> >> The number of repetitions varies between cases, so in its one case >> per line format the data is non-rectangular. >> >> At present I have shoehorned all of this into SPSS, with each case on >> one line. My test database has 2,500 variables and 1,500 cases (or >> persons), and in SPSSs *.SAV format is ~4MB. The one I finally work >> with will be larger again, though likely within one order of >> magnitude. Down the track, funding permitting, I hope to be working with tens of thousands of cases. > > Although this may not be helpful for your immediate goal, storing and > manipulating data of this size and complexity (and, I expect, cost for > collection) really calls for tools like relational databases. A > single flat file of 2500 variables by 1500 cases is almost never the > best way to organize such data. A normalized representation as a > collection of interlinked tables in a relational data base is much > more effective and less error prone. The widespread use of > spreadsheets or SPSS data sets or SAS data sets which encourage the > "single table with a gargantuan number of columns, most of which are > missing data in most cases" approach to organization of longitudinal > data is regrettable. > > For later analysis in R it is better to start with "long" form of the > data, as opposed to the "wide" form, even if it me
Re: [R] lm function in R
It seems to me that your question is more about the econometrics than about
R. Any introductory econometric textbook or compendium on econometrics will
cover this as it is a basic. See, for example, Greene 2006 or Wooldridge
2002.
Say X is your data matrix, that contains columns for each of the individual
variables (x), columns with their interactions, and one column of 1s for the
intercept. Let y be your dependent variable. Then, OLS estimates are
computed by
X'X inverse X'y
Or in R
solve(t(X)%*%X)%*%t(X)%*%y
Best,
Daniel
-
cuncta stricte discussurus
-
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Something Something
Sent: Saturday, February 13, 2010 5:04 PM
To: Bert Gunter
Cc: r-help@r-project.org
Subject: Re: [R] lm function in R
I tried..
mod = lm(Y ~ X1*X2*X3, na.action = na.exclude)
formula(mod)
This produced
Y ~ X1 * X2 * X3
When I typed just mod I got:
Call:
lm(formula = Y ~ X1 * X2 * X3, na.action = na.exclude)
Coefficients:
(Intercept) X11 X21 X31 X11:X21 X11:X31
X21:X31 X11:X21:X31
177.9245 0.2005 2.4482 3.1216 0.8127 -26.6166
-3.0398 29.6049
I am trying to figure out how R computed all these coefficients.
On Sat, Feb 13, 2010 at 1:30 PM, Bert Gunter wrote:
> ?formula
>
>
> Bert Gunter
> Genentech Nonclinical Statistics
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On
> Behalf Of Something Something
> Sent: Saturday, February 13, 2010 1:24 PM
> To: Daniel Nordlund
> Cc: r-help@r-project.org
> Subject: Re: [R] lm function in R
>
> Thanks Dan. Yes that was very helpful. I didn't see the change from '*'
> to
> '+'.
>
> Seems like when I put * it means - interaction & when I put + it's not an
> interaction.
>
> Is it correct to assume then that...
>
> When I put + R evaluates the following equation:
> Y-Hat = b0 + b1X1 + b2X2 + . . . bkXk + 7 7 7 + bkXk
>
>
> But when I put * R evaluates the following equation;
> Y-Hat = b0 + b1X1 + b2x2 + ... + bkXk + b12 X12+ b13 X13 + + c
>
> Is this correct? If it is then can someone point me to any sources that
> will explain how the coefficients (such as b0... bk, b12.. , b123..) are
> calculated. I guess, one source is the R source code :) but is there any
> other documentation anywhere?
>
> Please let me know. Thanks.
>
>
>
> On Fri, Feb 12, 2010 at 5:54 PM, Daniel Nordlund
> wrote:
>
> > > -Original Message-
> > > From: r-help-boun...@r-project.org [mailto:
> r-help-boun...@r-project.org]
> > > On Behalf Of Something Something
> > > Sent: Friday, February 12, 2010 5:28 PM
> > > To: Phil Spector; r-help@r-project.org
> > > Subject: Re: [R] lm function in R
> > >
> > > Thanks for the replies everyone. Greatly appreciate it. Some
> progress,
> > > but
> > > now I am getting the following values when I don't use "as.factor"
> > >
> > > 13.14167 25.11667 28.34167 49.14167 40.39167 66.86667
> > >
> > > Is that what you guys get?
> >
> >
> > If you look at Phil's response below, no, that is not what he got. The
> > difference is that you are specifying an interaction, whereas Phil did
> not
> > (because the equation you initially specified did not include an
> > interaction. Use Y ~ X1 + X2 instead of Y ~ X1*X2 for your formula.
> >
> > >
> > >
> > > On Fri, Feb 12, 2010 at 5:00 PM, Phil Spector
> > > wrote:
> > >
> > > > By converting the two variables to factors, you are fitting
> > > > an entirely different model. Leave out the as.factor stuff
> > > > and it will work exactly as you want it to.
> > > >
> > > > dat
> > > >>
> > > > V1 V2 V3 V4
> > > > 1 s1 14 4 1
> > > > 2 s2 23 4 2
> > > > 3 s3 30 7 2
> > > > 4 s4 50 7 4
> > > > 5 s5 39 10 3
> > > > 6 s6 67 10 6
> > > >
> > > >> names(dat) = c('id','y','x1','x2')
> > > >> z = lm(y~x1+x2,dat)
> > > >> predict(z)
> > > >>
> > > > 123456 15.16667
> 24.7
> > > > 27.7 46.7 40.16667 68.7
> > > >
> > > >
> > > >- Phil Spector
> > > > Statistical Computing
> Facility
> > > > Department of Statistics
> > > > UC Berkeley
> > > > spec...@stat.berkeley.edu
> > > >
> > > >
> > > >
> > > > On Fri, 12 Feb 2010, Something Something wrote:
> > > >
> > > > Hello,
> > > >>
> > > >> I am trying to learn how to perform Multiple Regression Analysis in
> R.
> > > I
> > > >> decided to take a simple example given in this PDF:
> > > >> http://www.utdallas.edu/~herve/abdi-prc-pretty.pdf
> > > >>
> > > >> I created a small CSV called, students.csv that contains the
> following
> > > >> data:
> > > >>
> > > >> s1 14 4 1
> > > >> s2 23 4 2
> > > >> s3 30 7 2
Re: [R] lm function in R
>>From your question it is difficult to determine what sort of tutoring you
are expecting.
The kind of tutoring you are providing is definitely helpful :)
I guess I should quickly explain what I am trying to accomplish. I have
been a computer scientist for quite a few years, but I studied Statistics
long time ago. Now trying to get back into the Statistics field.
For my latest project I have been asked to implement Multiple Regression
Analysis (using interactions - I guess) in Java & Distributed Computing
(more specifically Hadoop). I looked at the possibility of using rJava, but
it's 'single threaded' model plus complex deployment on 1000s of machines
does not make it very appealing. I looked at using 3rd party libraries such
as 'Apache Commons Math' & Flanagan's JAVA statistics library. Both of them
use this formula:
Y-Hat = b0 + b1X1 + b2X2 + . . . bkXk + · · · + bkXk
and NOT the one I have been asked to implement - which is -
Y-Hat = b0 + b1X1 + b2x2 + ... + bkXk + b12 X12+ b13 X13 + + c
(Not to mention these tools do not use BigDecimal so the answers given by
them are not as precise as those from R).
So now the deadline is approaching... and I can't find any Java library that
matches results from R.. so I have to roll up my sleeves and start coding in
Java - which means I need to get up to speed on Y-Hat calculations very
quickly. Ideally, looking for a paper (or a text book) that gives step by
step instructions on calculating coefficients such as (b0, b1... b13). I
have been searching but haven't found anything. My last resort is to look
at R source code and start converting it to Java.
I am new to R and a little rusty on Statistics, so I apologize for all the
dumb questions, and GREATLY appreciate your patience and help. Thanks.
On Sat, Feb 13, 2010 at 3:20 PM, David Winsemius wrote:
>
> On Feb 13, 2010, at 5:03 PM, Something Something wrote:
>
> I tried..
>>
>> mod = lm(Y ~ X1*X2*X3, na.action = na.exclude)
>> formula(mod)
>>
>> This produced
>> Y ~ X1 * X2 * X3
>>
>>
>> When I typed just mod I got:
>>
>> Call:
>> lm(formula = Y ~ X1 * X2 * X3, na.action = na.exclude)
>>
>> Coefficients:
>> (Intercept) X11 X21 X31 X11:X21
>> X11:X31
>>X21:X31 X11:X21:X31
>> 177.9245 0.2005 2.4482 3.1216 0.8127 -26.6166
>>-3.0398 29.6049
>>
>>
>> I am trying to figure out how R computed all these coefficients.
>>
>
> From your question it is difficult to determine what sort of tutoring you
> are expecting. To get the code of an R formula, you just type its name:
>
> lm
>
> Leads to lm.fit:
>
> lm.fit
>
> Reading further it appears the lm and lm.fit functions are really front
> ends for this call:
>
> .Fortran("dqrls", qr = x, n = n, p = p, y = y, ny = ny,
>tol = as.double(tol), coefficients = mat.or.vec(p, ny),
>residuals = y, effects = y, rank = integer(1L), pivot = 1L:p,
>qraux = double(p), work = double(2 * p), PACKAGE = "base")
>
> Seems pretty likely that is a QR decomposition-based method that i
> implemented in compiled code.
>
> So if you want to go deeper, at least you know what to search for. Or if
> you want to know how regression works on a matrix level, you should consult
> a good reference text or Wikipedia, which is surprisingly good for that sort
> of question these days.
>
> --
> David.
>
>>
>>
>>
>>
>>
>> On Sat, Feb 13, 2010 at 1:30 PM, Bert Gunter
>> wrote:
>>
>> ?formula
>>>
>>>
>>> Bert Gunter
>>> Genentech Nonclinical Statistics
>>>
>>> -Original Message-
>>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
>>> On
>>> Behalf Of Something Something
>>> Sent: Saturday, February 13, 2010 1:24 PM
>>> To: Daniel Nordlund
>>> Cc: r-help@r-project.org
>>> Subject: Re: [R] lm function in R
>>>
>>> Thanks Dan. Yes that was very helpful. I didn't see the change from '*'
>>> to
>>> '+'.
>>>
>>> Seems like when I put * it means - interaction & when I put + it's not an
>>> interaction.
>>>
>>> Is it correct to assume then that...
>>>
>>> When I put + R evaluates the following equation:
>>> Y-Hat = b0 + b1X1 + b2X2 + . . . bkXk + 7 7 7 + bkXk
>>>
>>>
>>> But when I put * R evaluates the following equation;
>>> Y-Hat = b0 + b1X1 + b2x2 + ... + bkXk + b12 X12+ b13 X13 + + c
>>>
>>> Is this correct? If it is then can someone point me to any sources that
>>> will explain how the coefficients (such as b0... bk, b12.. , b123..) are
>>> calculated. I guess, one source is the R source code :) but is there any
>>> other documentation anywhere?
>>>
>>> Please let me know. Thanks.
>>>
>>>
>>>
>>> On Fri, Feb 12, 2010 at 5:54 PM, Daniel Nordlund
>>> wrote:
>>>
>>> -Original Message-
> From: r-help-boun...@r-project.org [mailto:
>
r-help-boun...@r-project.org]
>>>
On Behalf Of Something Something
> Sent: Friday, February 12, 2010 5:28 PM
> To: Phil Spector; r-help@r-project.org
> Subject: Re: [R] lm funct
Re: [R] lm function in R
You can find out the model matrix like this where we use the builtin
data frame CO2 to illustrate:
fo <- uptake ~ Treatment * Type
mod <- lm(fo, CO2); mod
mm <- model.matrix(fo, CO2)
View(mm)
On Sat, Feb 13, 2010 at 5:03 PM, Something Something
wrote:
> I tried..
>
> mod = lm(Y ~ X1*X2*X3, na.action = na.exclude)
> formula(mod)
>
> This produced
> Y ~ X1 * X2 * X3
>
>
> When I typed just mod I got:
>
> Call:
> lm(formula = Y ~ X1 * X2 * X3, na.action = na.exclude)
>
> Coefficients:
> (Intercept) X11 X21 X31 X11:X21 X11:X31
> X21:X31 X11:X21:X31
> 177.9245 0.2005 2.4482 3.1216 0.8127 -26.6166
> -3.0398 29.6049
>
>
> I am trying to figure out how R computed all these coefficients.
>
>
>
>
>
> On Sat, Feb 13, 2010 at 1:30 PM, Bert Gunter wrote:
>
>> ?formula
>>
>>
>> Bert Gunter
>> Genentech Nonclinical Statistics
>>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
>> On
>> Behalf Of Something Something
>> Sent: Saturday, February 13, 2010 1:24 PM
>> To: Daniel Nordlund
>> Cc: r-help@r-project.org
>> Subject: Re: [R] lm function in R
>>
>> Thanks Dan. Yes that was very helpful. I didn't see the change from '*'
>> to
>> '+'.
>>
>> Seems like when I put * it means - interaction & when I put + it's not an
>> interaction.
>>
>> Is it correct to assume then that...
>>
>> When I put + R evaluates the following equation:
>> Y-Hat = b0 + b1X1 + b2X2 + . . . bkXk + 7 7 7 + bkXk
>>
>>
>> But when I put * R evaluates the following equation;
>> Y-Hat = b0 + b1X1 + b2x2 + ... + bkXk + b12 X12+ b13 X13 + + c
>>
>> Is this correct? If it is then can someone point me to any sources that
>> will explain how the coefficients (such as b0... bk, b12.. , b123..) are
>> calculated. I guess, one source is the R source code :) but is there any
>> other documentation anywhere?
>>
>> Please let me know. Thanks.
>>
>>
>>
>> On Fri, Feb 12, 2010 at 5:54 PM, Daniel Nordlund
>> wrote:
>>
>> > > -Original Message-
>> > > From: r-help-boun...@r-project.org [mailto:
>> r-help-boun...@r-project.org]
>> > > On Behalf Of Something Something
>> > > Sent: Friday, February 12, 2010 5:28 PM
>> > > To: Phil Spector; r-help@r-project.org
>> > > Subject: Re: [R] lm function in R
>> > >
>> > > Thanks for the replies everyone. Greatly appreciate it. Some
>> progress,
>> > > but
>> > > now I am getting the following values when I don't use "as.factor"
>> > >
>> > > 13.14167 25.11667 28.34167 49.14167 40.39167 66.86667
>> > >
>> > > Is that what you guys get?
>> >
>> >
>> > If you look at Phil's response below, no, that is not what he got. The
>> > difference is that you are specifying an interaction, whereas Phil did
>> not
>> > (because the equation you initially specified did not include an
>> > interaction. Use Y ~ X1 + X2 instead of Y ~ X1*X2 for your formula.
>> >
>> > >
>> > >
>> > > On Fri, Feb 12, 2010 at 5:00 PM, Phil Spector
>> > > wrote:
>> > >
>> > > > By converting the two variables to factors, you are fitting
>> > > > an entirely different model. Leave out the as.factor stuff
>> > > > and it will work exactly as you want it to.
>> > > >
>> > > > dat
>> > > >>
>> > > > V1 V2 V3 V4
>> > > > 1 s1 14 4 1
>> > > > 2 s2 23 4 2
>> > > > 3 s3 30 7 2
>> > > > 4 s4 50 7 4
>> > > > 5 s5 39 10 3
>> > > > 6 s6 67 10 6
>> > > >
>> > > >> names(dat) = c('id','y','x1','x2')
>> > > >> z = lm(y~x1+x2,dat)
>> > > >> predict(z)
>> > > >>
>> > > > 1 2 3 4 5 6 15.16667
>> 24.7
>> > > > 27.7 46.7 40.16667 68.7
>> > > >
>> > > >
>> > > > - Phil Spector
>> > > > Statistical Computing
>> Facility
>> > > > Department of Statistics
>> > > > UC Berkeley
>> > > > spec...@stat.berkeley.edu
>> > > >
>> > > >
>> > > >
>> > > > On Fri, 12 Feb 2010, Something Something wrote:
>> > > >
>> > > > Hello,
>> > > >>
>> > > >> I am trying to learn how to perform Multiple Regression Analysis in
>> R.
>> > > I
>> > > >> decided to take a simple example given in this PDF:
>> > > >> http://www.utdallas.edu/~herve/abdi-prc-pretty.pdf
>> > > >>
>> > > >> I created a small CSV called, students.csv that contains the
>> following
>> > > >> data:
>> > > >>
>> > > >> s1 14 4 1
>> > > >> s2 23 4 2
>> > > >> s3 30 7 2
>> > > >> s4 50 7 4
>> > > >> s5 39 10 3
>> > > >> s6 67 10 6
>> > > >>
>> > > >> Col headers: Student id, Memory span(Y), age(X1), speech rate(X2)
>> > > >>
>> > > >> Now the expected results are:
>> > > >>
>> > > >> yHat[0]:15.168
>> > > >> yHat[1]:24.668
>> > > >> yHat[2]:27.664
>> > > >> yHat[3]:46.664
>> > > >> yHat[4]:40.164
>> > > >> yHat[5]:68.67
>> > > >>
>> > > >>
Re: [R] lm function in R
On Feb 13, 2010, at 5:03 PM, Something Something wrote:
I tried..
mod = lm(Y ~ X1*X2*X3, na.action = na.exclude)
formula(mod)
This produced
Y ~ X1 * X2 * X3
When I typed just mod I got:
Call:
lm(formula = Y ~ X1 * X2 * X3, na.action = na.exclude)
Coefficients:
(Intercept) X11 X21 X31 X11:X21
X11:X31
X21:X31 X11:X21:X31
177.9245 0.2005 2.4482 3.1216 0.8127
-26.6166
-3.0398 29.6049
I am trying to figure out how R computed all these coefficients.
From your question it is difficult to determine what sort of tutoring
you are expecting. To get the code of an R formula, you just type its
name:
lm
Leads to lm.fit:
lm.fit
Reading further it appears the lm and lm.fit functions are really
front ends for this call:
.Fortran("dqrls", qr = x, n = n, p = p, y = y, ny = ny,
tol = as.double(tol), coefficients = mat.or.vec(p, ny),
residuals = y, effects = y, rank = integer(1L), pivot = 1L:p,
qraux = double(p), work = double(2 * p), PACKAGE = "base")
Seems pretty likely that is a QR decomposition-based method that i
implemented in compiled code.
So if you want to go deeper, at least you know what to search for. Or
if you want to know how regression works on a matrix level, you should
consult a good reference text or Wikipedia, which is surprisingly good
for that sort of question these days.
--
David.
On Sat, Feb 13, 2010 at 1:30 PM, Bert Gunter
wrote:
?formula
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
]
On
Behalf Of Something Something
Sent: Saturday, February 13, 2010 1:24 PM
To: Daniel Nordlund
Cc: r-help@r-project.org
Subject: Re: [R] lm function in R
Thanks Dan. Yes that was very helpful. I didn't see the change
from '*'
to
'+'.
Seems like when I put * it means - interaction & when I put + it's
not an
interaction.
Is it correct to assume then that...
When I put + R evaluates the following equation:
Y-Hat = b0 + b1X1 + b2X2 + . . . bkXk + 7 7 7 + bkXk
But when I put * R evaluates the following equation;
Y-Hat = b0 + b1X1 + b2x2 + ... + bkXk + b12 X12+ b13 X13 +
+ c
Is this correct? If it is then can someone point me to any sources
that
will explain how the coefficients (such as b0... bk, b12.. ,
b123..) are
calculated. I guess, one source is the R source code :) but is
there any
other documentation anywhere?
Please let me know. Thanks.
On Fri, Feb 12, 2010 at 5:54 PM, Daniel Nordlund
wrote:
-Original Message-
From: r-help-boun...@r-project.org [mailto:
r-help-boun...@r-project.org]
On Behalf Of Something Something
Sent: Friday, February 12, 2010 5:28 PM
To: Phil Spector; r-help@r-project.org
Subject: Re: [R] lm function in R
Thanks for the replies everyone. Greatly appreciate it. Some
progress,
but
now I am getting the following values when I don't use "as.factor"
13.14167 25.11667 28.34167 49.14167 40.39167 66.86667
Is that what you guys get?
If you look at Phil's response below, no, that is not what he
got. The
difference is that you are specifying an interaction, whereas Phil
did
not
(because the equation you initially specified did not include an
interaction. Use Y ~ X1 + X2 instead of Y ~ X1*X2 for your formula.
On Fri, Feb 12, 2010 at 5:00 PM, Phil Spector
wrote:
By converting the two variables to factors, you are fitting
an entirely different model. Leave out the as.factor stuff
and it will work exactly as you want it to.
dat
V1 V2 V3 V4
1 s1 14 4 1
2 s2 23 4 2
3 s3 30 7 2
4 s4 50 7 4
5 s5 39 10 3
6 s6 67 10 6
names(dat) = c('id','y','x1','x2')
z = lm(y~x1+x2,dat)
predict(z)
123456 15.16667
24.7
27.7 46.7 40.16667 68.7
- Phil Spector
Statistical Computing
Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Fri, 12 Feb 2010, Something Something wrote:
Hello,
I am trying to learn how to perform Multiple Regression
Analysis in
R.
I
decided to take a simple example given in this PDF:
http://www.utdallas.edu/~herve/abdi-prc-pretty.pdf
I created a small CSV called, students.csv that contains the
following
data:
s1 14 4 1
s2 23 4 2
s3 30 7 2
s4 50 7 4
s5 39 10 3
s6 67 10 6
Col headers: Student id, Memory span(Y), age(X1), speech
rate(X2)
Now the expected results are:
yHat[0]:15.168
yHat[1]:24.668
yHat[2]:27.664
yHat[3]:46.664
yHat[4]:40.164
yHat[5]:68.67
This is based on the following equation (given in the PDF): Y =
1.67
+
X1
+
9.50 X2
I ran the following commands in R:
data = re
Re: [R] NMDS ordination
Aisyah ioz.ac.uk> writes: > > > Hi > > Im currently trying to plot my NMDS data together with fitted variables > (envfit funct) on an ordination plot. The plot function shows two > displays="sites" and "sp". I was wondering how to plot it so that the sites > come up as different points for different sites but the species come up as > actual names? It looks a little busy at the moment with everything in. Please provide an example. I.e. working code creating the plot as you have it at the moment. Include an example data set and be explicit about what packages are needed. Don't post a large data set -- just create a minimal example demonstrating the problem you are having. > > Sya __ 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] lm function in R
I tried..
mod = lm(Y ~ X1*X2*X3, na.action = na.exclude)
formula(mod)
This produced
Y ~ X1 * X2 * X3
When I typed just mod I got:
Call:
lm(formula = Y ~ X1 * X2 * X3, na.action = na.exclude)
Coefficients:
(Intercept) X11 X21 X31 X11:X21 X11:X31
X21:X31 X11:X21:X31
177.9245 0.2005 2.4482 3.1216 0.8127 -26.6166
-3.0398 29.6049
I am trying to figure out how R computed all these coefficients.
On Sat, Feb 13, 2010 at 1:30 PM, Bert Gunter wrote:
> ?formula
>
>
> Bert Gunter
> Genentech Nonclinical Statistics
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On
> Behalf Of Something Something
> Sent: Saturday, February 13, 2010 1:24 PM
> To: Daniel Nordlund
> Cc: r-help@r-project.org
> Subject: Re: [R] lm function in R
>
> Thanks Dan. Yes that was very helpful. I didn't see the change from '*'
> to
> '+'.
>
> Seems like when I put * it means - interaction & when I put + it's not an
> interaction.
>
> Is it correct to assume then that...
>
> When I put + R evaluates the following equation:
> Y-Hat = b0 + b1X1 + b2X2 + . . . bkXk + 7 7 7 + bkXk
>
>
> But when I put * R evaluates the following equation;
> Y-Hat = b0 + b1X1 + b2x2 + ... + bkXk + b12 X12+ b13 X13 + + c
>
> Is this correct? If it is then can someone point me to any sources that
> will explain how the coefficients (such as b0... bk, b12.. , b123..) are
> calculated. I guess, one source is the R source code :) but is there any
> other documentation anywhere?
>
> Please let me know. Thanks.
>
>
>
> On Fri, Feb 12, 2010 at 5:54 PM, Daniel Nordlund
> wrote:
>
> > > -Original Message-
> > > From: r-help-boun...@r-project.org [mailto:
> r-help-boun...@r-project.org]
> > > On Behalf Of Something Something
> > > Sent: Friday, February 12, 2010 5:28 PM
> > > To: Phil Spector; r-help@r-project.org
> > > Subject: Re: [R] lm function in R
> > >
> > > Thanks for the replies everyone. Greatly appreciate it. Some
> progress,
> > > but
> > > now I am getting the following values when I don't use "as.factor"
> > >
> > > 13.14167 25.11667 28.34167 49.14167 40.39167 66.86667
> > >
> > > Is that what you guys get?
> >
> >
> > If you look at Phil's response below, no, that is not what he got. The
> > difference is that you are specifying an interaction, whereas Phil did
> not
> > (because the equation you initially specified did not include an
> > interaction. Use Y ~ X1 + X2 instead of Y ~ X1*X2 for your formula.
> >
> > >
> > >
> > > On Fri, Feb 12, 2010 at 5:00 PM, Phil Spector
> > > wrote:
> > >
> > > > By converting the two variables to factors, you are fitting
> > > > an entirely different model. Leave out the as.factor stuff
> > > > and it will work exactly as you want it to.
> > > >
> > > > dat
> > > >>
> > > > V1 V2 V3 V4
> > > > 1 s1 14 4 1
> > > > 2 s2 23 4 2
> > > > 3 s3 30 7 2
> > > > 4 s4 50 7 4
> > > > 5 s5 39 10 3
> > > > 6 s6 67 10 6
> > > >
> > > >> names(dat) = c('id','y','x1','x2')
> > > >> z = lm(y~x1+x2,dat)
> > > >> predict(z)
> > > >>
> > > > 123456 15.16667
> 24.7
> > > > 27.7 46.7 40.16667 68.7
> > > >
> > > >
> > > >- Phil Spector
> > > > Statistical Computing
> Facility
> > > > Department of Statistics
> > > > UC Berkeley
> > > > spec...@stat.berkeley.edu
> > > >
> > > >
> > > >
> > > > On Fri, 12 Feb 2010, Something Something wrote:
> > > >
> > > > Hello,
> > > >>
> > > >> I am trying to learn how to perform Multiple Regression Analysis in
> R.
> > > I
> > > >> decided to take a simple example given in this PDF:
> > > >> http://www.utdallas.edu/~herve/abdi-prc-pretty.pdf
> > > >>
> > > >> I created a small CSV called, students.csv that contains the
> following
> > > >> data:
> > > >>
> > > >> s1 14 4 1
> > > >> s2 23 4 2
> > > >> s3 30 7 2
> > > >> s4 50 7 4
> > > >> s5 39 10 3
> > > >> s6 67 10 6
> > > >>
> > > >> Col headers: Student id, Memory span(Y), age(X1), speech rate(X2)
> > > >>
> > > >> Now the expected results are:
> > > >>
> > > >> yHat[0]:15.168
> > > >> yHat[1]:24.668
> > > >> yHat[2]:27.664
> > > >> yHat[3]:46.664
> > > >> yHat[4]:40.164
> > > >> yHat[5]:68.67
> > > >>
> > > >> This is based on the following equation (given in the PDF): Y =
> 1.67
> > +
> > > X1
> > > >> +
> > > >> 9.50 X2
> > > >>
> > > >> I ran the following commands in R:
> > > >>
> > > >> data = read.table("students.csv", head=F, as.is=T, na.string=".",
> > > >> row.nam=NULL)
> > > >> X1 = as.factor(data[[3]])
> > > >> X2 = as.factor(data[[4]])
> > > >> Y = data[[2]]
> > > >> mod = lm(Y ~ X1*X2, na.action = na.exclude)
> > > >> Y.ha
Re: [R] (no subject)
[cc'ing back to r-help]
You need to get the separator right: you specify slashes below where
your date is space-separated
as.Date("02 03 2008", format="%m %d %Y")
note that the solution I gave you below assumed that your dates were
in d/m/Y order (since you didn't specify). You will have to modify
things slightly (i.e. specifying format explicitly) if that's not true.
Hussain Abusaaq wrote:
> I hope that will work. just to make my problem clear
> when i used as.date for e "02 03 2008" it gives me NA
>
> for example
>
>> as.Date("02 03 2008",format="%m/%d/%Y")
> [1] NA
> I hope you can help me here
> - Original Message -
> From: Ben Bolker
> Date: Saturday, February 13, 2010 1:03 pm
> Subject: Re: [R] (no subject)
> To: r-h...@stat.math.ethz.ch
>
>> Hussain Abusaaq fsu.edu> writes:
>>
>>> I have this vector and I want to change it to date.
>>>
>>> for example G=[05 12 2008]
>>>
>>> or g=[2]
>>> f=[3]
>>> y=[2208]
>>>
>> Something like as.Date(paste(rev(G),collapse="-"))
>>
>> __
>> 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.htmland provide commented, minimal, self-contained,
>> reproducible code.
>>
--
Ben Bolker
Associate professor, Biology Dep't, Univ. of Florida
bol...@ufl.edu / people.biology.ufl.edu/bolker
GPG key: people.biology.ufl.edu/bolker/benbolker-publickey.asc
signature.asc
Description: OpenPGP digital signature
__
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] Highlighting points in a quantile plot for different values of a second column
Saptarshi Guha wrote: Hello, I have the following data frame, using lattice and qqmath, how do I display the quantile plot (e.g distribution = qunif), but highlight the f==TRUE and f==FALSE differently? library(lattice) pp <- function(n) ((1:n)-0.5)/n x=data.frame(x=runif(20),f=sapply(runif(20),function(r) r>0.5)) I would like a single curve, the following example displays two curves. qqmath(~x,groups=f,data=x,distribution=qunif,f.value=pp) Here's one way: qqmath(~x, data=x, type='b', col=ifelse(x$f, 2, 4), pch=ifelse(x$f, 17, 19),cex=2) -Peter Ehlers Thank you Saptarshi __ 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. -- Peter Ehlers University of Calgary __ 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] lm function in R
?formula
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Something Something
Sent: Saturday, February 13, 2010 1:24 PM
To: Daniel Nordlund
Cc: r-help@r-project.org
Subject: Re: [R] lm function in R
Thanks Dan. Yes that was very helpful. I didn't see the change from '*' to
'+'.
Seems like when I put * it means - interaction & when I put + it's not an
interaction.
Is it correct to assume then that...
When I put + R evaluates the following equation:
Y-Hat = b0 + b1X1 + b2X2 + . . . bkXk + 7 7 7 + bkXk
But when I put * R evaluates the following equation;
Y-Hat = b0 + b1X1 + b2x2 + ... + bkXk + b12 X12+ b13 X13 + + c
Is this correct? If it is then can someone point me to any sources that
will explain how the coefficients (such as b0... bk, b12.. , b123..) are
calculated. I guess, one source is the R source code :) but is there any
other documentation anywhere?
Please let me know. Thanks.
On Fri, Feb 12, 2010 at 5:54 PM, Daniel Nordlund
wrote:
> > -Original Message-
> > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> > On Behalf Of Something Something
> > Sent: Friday, February 12, 2010 5:28 PM
> > To: Phil Spector; r-help@r-project.org
> > Subject: Re: [R] lm function in R
> >
> > Thanks for the replies everyone. Greatly appreciate it. Some progress,
> > but
> > now I am getting the following values when I don't use "as.factor"
> >
> > 13.14167 25.11667 28.34167 49.14167 40.39167 66.86667
> >
> > Is that what you guys get?
>
>
> If you look at Phil's response below, no, that is not what he got. The
> difference is that you are specifying an interaction, whereas Phil did not
> (because the equation you initially specified did not include an
> interaction. Use Y ~ X1 + X2 instead of Y ~ X1*X2 for your formula.
>
> >
> >
> > On Fri, Feb 12, 2010 at 5:00 PM, Phil Spector
> > wrote:
> >
> > > By converting the two variables to factors, you are fitting
> > > an entirely different model. Leave out the as.factor stuff
> > > and it will work exactly as you want it to.
> > >
> > > dat
> > >>
> > > V1 V2 V3 V4
> > > 1 s1 14 4 1
> > > 2 s2 23 4 2
> > > 3 s3 30 7 2
> > > 4 s4 50 7 4
> > > 5 s5 39 10 3
> > > 6 s6 67 10 6
> > >
> > >> names(dat) = c('id','y','x1','x2')
> > >> z = lm(y~x1+x2,dat)
> > >> predict(z)
> > >>
> > > 123456 15.16667 24.7
> > > 27.7 46.7 40.16667 68.7
> > >
> > >
> > >- Phil Spector
> > > Statistical Computing Facility
> > > Department of Statistics
> > > UC Berkeley
> > > spec...@stat.berkeley.edu
> > >
> > >
> > >
> > > On Fri, 12 Feb 2010, Something Something wrote:
> > >
> > > Hello,
> > >>
> > >> I am trying to learn how to perform Multiple Regression Analysis in
R.
> > I
> > >> decided to take a simple example given in this PDF:
> > >> http://www.utdallas.edu/~herve/abdi-prc-pretty.pdf
> > >>
> > >> I created a small CSV called, students.csv that contains the
following
> > >> data:
> > >>
> > >> s1 14 4 1
> > >> s2 23 4 2
> > >> s3 30 7 2
> > >> s4 50 7 4
> > >> s5 39 10 3
> > >> s6 67 10 6
> > >>
> > >> Col headers: Student id, Memory span(Y), age(X1), speech rate(X2)
> > >>
> > >> Now the expected results are:
> > >>
> > >> yHat[0]:15.168
> > >> yHat[1]:24.668
> > >> yHat[2]:27.664
> > >> yHat[3]:46.664
> > >> yHat[4]:40.164
> > >> yHat[5]:68.67
> > >>
> > >> This is based on the following equation (given in the PDF): Y = 1.67
> +
> > X1
> > >> +
> > >> 9.50 X2
> > >>
> > >> I ran the following commands in R:
> > >>
> > >> data = read.table("students.csv", head=F, as.is=T, na.string=".",
> > >> row.nam=NULL)
> > >> X1 = as.factor(data[[3]])
> > >> X2 = as.factor(data[[4]])
> > >> Y = data[[2]]
> > >> mod = lm(Y ~ X1*X2, na.action = na.exclude)
> > >> Y.hat = fitted(mod)
> > >> Y.hat
> > >>
> > >> This gives me the following output:
> > >>
> > >> Y.hat
> > >>>
> > >> 1 2 3 4 5 6
> > >> 14 23 30 50 39 67
> > >>
> > >> Obviously I am doing something wrong. Please help. Thanks.
> > >>
>
> Hope this is helpful,
>
> Dan
>
> Daniel Nordlund
> Bothell, WA 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.
>
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Re: [R] lm function in R
Thanks Dan. Yes that was very helpful. I didn't see the change from '*' to
'+'.
Seems like when I put * it means - interaction & when I put + it's not an
interaction.
Is it correct to assume then that...
When I put + R evaluates the following equation:
Y-Hat = b0 + b1X1 + b2X2 + . . . bkXk + · · · + bkXk
But when I put * R evaluates the following equation;
Y-Hat = b0 + b1X1 + b2x2 + ... + bkXk + b12 X12+ b13 X13 + + c
Is this correct? If it is then can someone point me to any sources that
will explain how the coefficients (such as b0... bk, b12.. , b123..) are
calculated. I guess, one source is the R source code :) but is there any
other documentation anywhere?
Please let me know. Thanks.
On Fri, Feb 12, 2010 at 5:54 PM, Daniel Nordlund wrote:
> > -Original Message-
> > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> > On Behalf Of Something Something
> > Sent: Friday, February 12, 2010 5:28 PM
> > To: Phil Spector; r-help@r-project.org
> > Subject: Re: [R] lm function in R
> >
> > Thanks for the replies everyone. Greatly appreciate it. Some progress,
> > but
> > now I am getting the following values when I don't use "as.factor"
> >
> > 13.14167 25.11667 28.34167 49.14167 40.39167 66.86667
> >
> > Is that what you guys get?
>
>
> If you look at Phil's response below, no, that is not what he got. The
> difference is that you are specifying an interaction, whereas Phil did not
> (because the equation you initially specified did not include an
> interaction. Use Y ~ X1 + X2 instead of Y ~ X1*X2 for your formula.
>
> >
> >
> > On Fri, Feb 12, 2010 at 5:00 PM, Phil Spector
> > wrote:
> >
> > > By converting the two variables to factors, you are fitting
> > > an entirely different model. Leave out the as.factor stuff
> > > and it will work exactly as you want it to.
> > >
> > > dat
> > >>
> > > V1 V2 V3 V4
> > > 1 s1 14 4 1
> > > 2 s2 23 4 2
> > > 3 s3 30 7 2
> > > 4 s4 50 7 4
> > > 5 s5 39 10 3
> > > 6 s6 67 10 6
> > >
> > >> names(dat) = c('id','y','x1','x2')
> > >> z = lm(y~x1+x2,dat)
> > >> predict(z)
> > >>
> > > 123456 15.16667 24.7
> > > 27.7 46.7 40.16667 68.7
> > >
> > >
> > >- Phil Spector
> > > Statistical Computing Facility
> > > Department of Statistics
> > > UC Berkeley
> > > spec...@stat.berkeley.edu
> > >
> > >
> > >
> > > On Fri, 12 Feb 2010, Something Something wrote:
> > >
> > > Hello,
> > >>
> > >> I am trying to learn how to perform Multiple Regression Analysis in R.
> > I
> > >> decided to take a simple example given in this PDF:
> > >> http://www.utdallas.edu/~herve/abdi-prc-pretty.pdf
> > >>
> > >> I created a small CSV called, students.csv that contains the following
> > >> data:
> > >>
> > >> s1 14 4 1
> > >> s2 23 4 2
> > >> s3 30 7 2
> > >> s4 50 7 4
> > >> s5 39 10 3
> > >> s6 67 10 6
> > >>
> > >> Col headers: Student id, Memory span(Y), age(X1), speech rate(X2)
> > >>
> > >> Now the expected results are:
> > >>
> > >> yHat[0]:15.168
> > >> yHat[1]:24.668
> > >> yHat[2]:27.664
> > >> yHat[3]:46.664
> > >> yHat[4]:40.164
> > >> yHat[5]:68.67
> > >>
> > >> This is based on the following equation (given in the PDF): Y = 1.67
> +
> > X1
> > >> +
> > >> 9.50 X2
> > >>
> > >> I ran the following commands in R:
> > >>
> > >> data = read.table("students.csv", head=F, as.is=T, na.string=".",
> > >> row.nam=NULL)
> > >> X1 = as.factor(data[[3]])
> > >> X2 = as.factor(data[[4]])
> > >> Y = data[[2]]
> > >> mod = lm(Y ~ X1*X2, na.action = na.exclude)
> > >> Y.hat = fitted(mod)
> > >> Y.hat
> > >>
> > >> This gives me the following output:
> > >>
> > >> Y.hat
> > >>>
> > >> 1 2 3 4 5 6
> > >> 14 23 30 50 39 67
> > >>
> > >> Obviously I am doing something wrong. Please help. Thanks.
> > >>
>
> Hope this is helpful,
>
> Dan
>
> Daniel Nordlund
> Bothell, WA 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.
>
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Re: [R] Plot different regression models on one graph
Brian Joiner, my teacher and insightful guide to the world of applied statistics when I was a student, succinctly summarized this discussion by: "Even the data aren't sufficient." (This is an inside joke for statisticians). Cheers, Bert Bert Gunter Genentech Nonclinical Statistics -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Peter Ehlers Sent: Saturday, February 13, 2010 12:35 PM To: David Winsemius Cc: Rhonda Reidy; r-help@r-project.org Subject: Re: [R] Plot different regression models on one graph Rhonda: As David points out, a cubic fit is rather speculative. I think that one needs to distinguish two situations: 1) theoretical justification for a cubic model is available, or 2) you're dredging the data for the "best" fit. Your case is the second. That puts you in the realm of EDA (exploratory data analysis). You're free to fit any model you wish, but you should assess the value of the model. One convenient way is to use the influence.measures() function, which will show that points 9 and 10 in your data have a strong influence on your cubic fit (as, of course, your eyes would tell you). A good thing to do at this point is to fit 3 more cubic models: 1) omitting point 9, 2) omitting point 10, 3) omitting both. Try it and plot the resulting fits. You may be surprised. Conclusion: Without more data, you can conclude nothing about a non-straightline fit. (And this hasn't even addressed the relative abundance of x=0 data.) -Peter Ehlers David Winsemius wrote: > > On Feb 13, 2010, at 1:35 PM, Rhonda Reidy wrote: > >> The following variables have the following significant relationships >> (x is the explanatory variable): linear, cubic, exponential, logistic. >> The linear relationship plots without any trouble. >> >> Cubic is the 'best' model, but it is not plotting as a smooth curve >> using the following code: >> >> cubic.lm<- lm(y~poly(x,3)) > > Try: > > lines(0:80, predict(cubic.lm, data.frame(x=0:80)),lwd=2) > > But I really must say the superiority of that relationship over a linear > one is far from convincing to my eyes. Seems to be caused by one data > point. I hope the quotes around "best" mean tha tyou have the same qualms. > > >> lines(x,predict(cubic.lm),lwd=2) >> >> How do I plot the data and the estimated curves for all of these >> regression models in the same plot? >> >> x <- c(62.5,68.5,0,52,0,52,0,52,23.5,86,0,0,0,0,0,0,0,0,0,0) >> >> y <- >> c(0.054,0.055,0.017,0.021,0.020,0.028,0.032,0.073,0.076,0.087,0.042,0.042,0. 041,0.045,0.021,0.018,0.017,0.018,0.028,0.022) >> >> >> Thanks in advance. >> >> Rhonda Reidy >> -- Peter Ehlers University of Calgary __ 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] NMDS ordination
Hi Im currently trying to plot my NMDS data together with fitted variables (envfit funct) on an ordination plot. The plot function shows two displays="sites" and "sp". I was wondering how to plot it so that the sites come up as different points for different sites but the species come up as actual names? It looks a little busy at the moment with everything in. Sya -- View this message in context: http://n4.nabble.com/NMDS-ordination-tp1554632p1554632.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] max and in the zoo package
Try this: > library(zoo) > set.seed(1) > z <- zoo(rnorm(100), as.Date(0) + 1:100) > mx <- z[which.max(z)]; mx 1970-03-03 2.401618 > coredata(mx) [1] 2.401618 > time(mx) [1] "1970-03-03" On Sat, Feb 13, 2010 at 3:39 PM, Edouard Tallent wrote: > hi everyone. > i am dealing on a zoo-class object.i am aware of the 'min' and 'max' function > to get the minimum and the maximum values.getting these maximum and minimum > values is ont thing.but, how to get the (zoo) dates these values occur ? > in fact, in analysing a time series i am interested in the particular > (extreme) values AND in the moments they happen. > regards.edouard. > > [[alternative(swapped) HTML version deleted]] > > This is MIME Epilogue > > __ > 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] Highlighting points in a quantile plot for different values of a second column
Hello, I have the following data frame, using lattice and qqmath, how do I display the quantile plot (e.g distribution = qunif), but highlight the f==TRUE and f==FALSE differently? library(lattice) pp <- function(n) ((1:n)-0.5)/n x=data.frame(x=runif(20),f=sapply(runif(20),function(r) r>0.5)) I would like a single curve, the following example displays two curves. qqmath(~x,groups=f,data=x,distribution=qunif,f.value=pp) Thank you Saptarshi __ 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] max and in the zoo package
hi everyone. i am dealing on a zoo-class object.i am aware of the 'min' and 'max' function to get the minimum and the maximum values.getting these maximum and minimum values is ont thing.but, how to get the (zoo) dates these values occur ? in fact, in analysing a time series i am interested in the particular (extreme) values AND in the moments they happen. regards.edouard. [[alternative(swapped) HTML version deleted]] This is MIME Epilogue __ 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 different regression models on one graph
Rhonda: As David points out, a cubic fit is rather speculative. I think that one needs to distinguish two situations: 1) theoretical justification for a cubic model is available, or 2) you're dredging the data for the "best" fit. Your case is the second. That puts you in the realm of EDA (exploratory data analysis). You're free to fit any model you wish, but you should assess the value of the model. One convenient way is to use the influence.measures() function, which will show that points 9 and 10 in your data have a strong influence on your cubic fit (as, of course, your eyes would tell you). A good thing to do at this point is to fit 3 more cubic models: 1) omitting point 9, 2) omitting point 10, 3) omitting both. Try it and plot the resulting fits. You may be surprised. Conclusion: Without more data, you can conclude nothing about a non-straightline fit. (And this hasn't even addressed the relative abundance of x=0 data.) -Peter Ehlers David Winsemius wrote: On Feb 13, 2010, at 1:35 PM, Rhonda Reidy wrote: The following variables have the following significant relationships (x is the explanatory variable): linear, cubic, exponential, logistic. The linear relationship plots without any trouble. Cubic is the 'best' model, but it is not plotting as a smooth curve using the following code: cubic.lm<- lm(y~poly(x,3)) Try: lines(0:80, predict(cubic.lm, data.frame(x=0:80)),lwd=2) But I really must say the superiority of that relationship over a linear one is far from convincing to my eyes. Seems to be caused by one data point. I hope the quotes around "best" mean tha tyou have the same qualms. lines(x,predict(cubic.lm),lwd=2) How do I plot the data and the estimated curves for all of these regression models in the same plot? x <- c(62.5,68.5,0,52,0,52,0,52,23.5,86,0,0,0,0,0,0,0,0,0,0) y <- c(0.054,0.055,0.017,0.021,0.020,0.028,0.032,0.073,0.076,0.087,0.042,0.042,0.041,0.045,0.021,0.018,0.017,0.018,0.028,0.022) Thanks in advance. Rhonda Reidy -- Peter Ehlers University of Calgary __ 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 do calculations in data matrices?
Zoppoli, Gabriele (NIH/NCI) [G] mail.nih.gov> writes: > > Please give me just a reference where I can find something useful. The others are right that rather than randomly googling, you should bite the bullet and sit down for a couple of hours with some introductory material on R (a book, or one of the freely available pdfs). Unless you are never going to use R again, it will be worth it. But seeing as you asked your question clearly, here's one way to do the steps you specify. Hopefully this will help as well. First, make a matrix to work with: > mat1 <- matrix(sample(1:10, size=12, replace=TRUE), ncol=4) > mat1 [,1] [,2] [,3] [,4] [1,]57 109 [2,]4 1082 [3,]8 1095 > > > In summary, I need to : > > - find the median of each row of a matrix You can use apply for that: > row.medians <- apply(mat1, 1, median) > row.medians [1] 8.0 6.0 8.5 > - create a new matrix with each value in the first matrix divided by > the median of its row That's easy to *do*: > mat2 <- mat1 / row.medians > mat2 [,1] [,2] [,3] [,4] [1,] 0.625 0.875000 1.25 1.125 [2,] 0.667 1.67 1.33 0.333 [3,] 0.9411765 1.176471 1.058824 0.5882353 but it may take more time to understand why that worked. How come it knew that we wanted to divide each row by the median of the row? (Hint: understand the "byrow" argument in ?matrix and the mentions of the word "recyling" in ?Arithmetic). > - if a value "a" in the second matrix is < 1, I need to substitute it > with 1/a First make a logical vector which identifies the elements of the matrix you want to operate on: > is.small <- mat2 < 1 Then perform the operation on those elements: > mat2[is.small] <- 1 / mat2[is.small] > mat2 [,1] [,2] [,3] [,4] [1,] 1.6000 1.142857 1.25 1.125 [2,] 1.5000 1.67 1.33 3.000 [3,] 1.0625 1.176471 1.058824 1.700 # you could also use ifelse(mat2 < 1, 1/mat2, mat2) dan > > I know that for some of you it must be overeasy, but I swear I googled > for two hours with keywords "operations", "calculations", "data > matrices", "data tables", and "CRAN", and I didn't find anything > useful. > > Thank you all > > Gabriele Zoppoli, MD > Ph.D. Fellow, Experimental and Clinical Oncology and Hematology, > University of Genova, Genova, Italy > Guest Researcher, LMP, NCI, NIH, Bethesda MD > > Work: 301-451-8575 > Mobile: 301-204-5642 > Email: zoppolig mail.nih.gov > __ 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 different regression models on one graph
On Feb 13, 2010, at 1:35 PM, Rhonda Reidy wrote: The following variables have the following significant relationships (x is the explanatory variable): linear, cubic, exponential, logistic. The linear relationship plots without any trouble. Cubic is the 'best' model, but it is not plotting as a smooth curve using the following code: cubic.lm<- lm(y~poly(x,3)) Try: lines(0:80, predict(cubic.lm, data.frame(x=0:80)),lwd=2) But I really must say the superiority of that relationship over a linear one is far from convincing to my eyes. Seems to be caused by one data point. I hope the quotes around "best" mean tha tyou have the same qualms. lines(x,predict(cubic.lm),lwd=2) How do I plot the data and the estimated curves for all of these regression models in the same plot? x <- c(62.5,68.5,0,52,0,52,0,52,23.5,86,0,0,0,0,0,0,0,0,0,0) y <- c (0.054,0.055,0.017,0.021,0.020,0.028,0.032,0.073,0.076,0.087,0.042,0.042,0.041,0.045,0.021,0.018,0.017,0.018,0.028,0.022 ) Thanks in advance. Rhonda Reidy -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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 different regression models on one graph
Rhonda Reidy wrote: > > The following variables have the following significant relationships (x is > the explanatory variable): linear, cubic, exponential, logistic. The > linear relationship plots without any trouble. > > Cubic is the 'best' model, but it is not plotting as a smooth curve using > the following code: > > cubic.lm<- lm(y~poly(x,3)) > lines(x,predict(cubic.lm),lwd=2) > > How do I plot the data and the estimated curves for all of these > regression models in the same plot? > > x <- c(62.5,68.5,0,52,0,52,0,52,23.5,86,0,0,0,0,0,0,0,0,0,0) > > y <- > c(0.054,0.055,0.017,0.021,0.020,0.028,0.032,0.073,0.076,0.087,0.042,0.042,0.041,0.045,0.021,0.018,0.017,0.018,0.028,0.022) > Hi Rhonda, The problem is that the default behavior of predict() only produces output corresponding to the original values of x used in the model. To get a "smooth" curve, you will want to predict at a large number of x values that fall within the range of the data. This can be done using the pretty() function, or using a composition of the seq() and range() functions if you desire more control: # Create a new vector of X values at which to generate predictions: predX <- pretty( x, n = 100 ) plot( y ~ x ) # Add the regression model, but generate predictions using the # points in predX-- the default is to predict using points in x. lines( predX, predict( cubic.lm, newdata = list( x = predX )), lwd = 2 ) The above should produce the output you want. However, I suggest Hadley Wickham's excellent ggplot2 package-- I feel the following is more expressive than using base graphics functions: require( ggplot2 ) Plot <- qplot( x, y ) + stat_smooth( method = 'lm', formula = 'y ~ poly(x,3)' ) print( Plot ) The amazing power of ggplot2 is that when you want to alter or extend your plot, you simply "add" more ggplot2 commands to your plot object: # Add a linear regression model and change the color theme used in the output Plot <- Plot + stat_smooth( method = 'lm', color = 'red' ) + theme_bw() print( Plot ) Hope this helps! -Charlie -- View this message in context: http://n4.nabble.com/Plot-different-regression-models-on-one-graph-tp1554606p1554622.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] how to do calculations in data matrices?
On Sat, Feb 13, 2010 at 5:39 PM, Zoppoli, Gabriele (NIH/NCI) [G] wrote: > Please give me just a reference where I can find something useful. > To complement Sarah's suggestions, a good place to start is Quick-R [1]. Liviu [1] http://www.statmethods.net/stats/descriptives.html __ 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] legend mathematical annotation problem
On Feb 13, 2010, at 1:45 PM, David Winsemius wrote:
On Feb 13, 2010, at 1:13 PM, Mark Heckmann wrote:
1) I need to plot a legend containing the mathematical symbol
greater-than-or-equal sign.
And I want the text to start with that symbol.
plot(110, 0.8)
categories <- expression(blank >= 85, 84.9 - 80, 79.9 - 75, 74.9 -
70, 69.9 - 65, 64.9 - 60, blank< 60)
legend(110, 0.8, categories, lty=1:7, lwd=3, col=1, merge=TRUE)
What I want is just ">=85" to be printed, but without something
(here "blank") in front it is no correct expression.
2) Also I want German type decimals, that is a comma instead of a
point.
But the problem is, that the comma is used as argument separator in
expression.
plot(110, 0.8)
categories <- expression(blank >= 85, 84,9 - 80, 79,9 - 75, 74,9 -
70, 69,9 - 65, 64,9 - 60, blank< 60)
legend(110, 0.8, categories, lty=1:7, lwd=3, col=1, merge=TRUE)
Try:
> plot(110, 0.8)
> categories <-legend(110, 0.8, c( expression(" ">= " 85"),
"84,9 - 80", "79,9 - 75", "74,9 - 70", "69,9 - 65", "64,9 - 60",
expression(""<" 60"))
+ , lty=1:7, lwd=3, col=1, merge=TRUE)
That worked because of side-effects. Better might be:
categories <- c( expression(" " >= " 85"), "84,9 - 80", "79,9 -
75", "74,9 - 70", "69,9 - 65", "64,9 - 60", expression("" < " 60"))
plot(110, 0.8)
legend(110, 0.8, categories, lty=1:7, lwd=3, col=1, merge=TRUE)
The commas inside quotes do not cause problems.
This does obviously not work. Any ideas?
Thanks,
Mark
–––
Mark Heckmann
Dipl. Wirt.-Ing. cand. Psych.
Vorstraße 93 B01
28359 Bremen
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.com
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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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
Heritage Laboratories
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] legend mathematical annotation problem
On Feb 13, 2010, at 1:13 PM, Mark Heckmann wrote:
1) I need to plot a legend containing the mathematical symbol
greater-than-or-equal sign.
And I want the text to start with that symbol.
plot(110, 0.8)
categories <- expression(blank >= 85, 84.9 - 80, 79.9 - 75, 74.9 -
70, 69.9 - 65, 64.9 - 60, blank< 60)
legend(110, 0.8, categories, lty=1:7, lwd=3, col=1, merge=TRUE)
What I want is just ">=85" to be printed, but without something
(here "blank") in front it is no correct expression.
2) Also I want German type decimals, that is a comma instead of a
point.
But the problem is, that the comma is used as argument separator in
expression.
plot(110, 0.8)
categories <- expression(blank >= 85, 84,9 - 80, 79,9 - 75, 74,9 -
70, 69,9 - 65, 64,9 - 60, blank< 60)
legend(110, 0.8, categories, lty=1:7, lwd=3, col=1, merge=TRUE)
Try:
> plot(110, 0.8)
> categories <-legend(110, 0.8, c( expression(" ">= " 85"), "84,9
- 80", "79,9 - 75", "74,9 - 70", "69,9 - 65", "64,9 - 60",
expression(""<" 60"))
+ , lty=1:7, lwd=3, col=1, merge=TRUE)
The commas inside quotes do not cause problems.
This does obviously not work. Any ideas?
Thanks,
Mark
–––
Mark Heckmann
Dipl. Wirt.-Ing. cand. Psych.
Vorstraße 93 B01
28359 Bremen
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.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.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Plot different regression models on one graph
The following variables have the following significant relationships (x is the explanatory variable): linear, cubic, exponential, logistic. The linear relationship plots without any trouble. Cubic is the 'best' model, but it is not plotting as a smooth curve using the following code: cubic.lm<- lm(y~poly(x,3)) lines(x,predict(cubic.lm),lwd=2) How do I plot the data and the estimated curves for all of these regression models in the same plot? x <- c(62.5,68.5,0,52,0,52,0,52,23.5,86,0,0,0,0,0,0,0,0,0,0) y <- c(0.054,0.055,0.017,0.021,0.020,0.028,0.032,0.073,0.076,0.087,0.042,0.042,0.041,0.045,0.021,0.018,0.017,0.018,0.028,0.022) Thanks in advance. Rhonda Reidy [[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] legend mathematical annotation problem
1) I need to plot a legend containing the mathematical symbol greater- than-or-equal sign. And I want the text to start with that symbol. plot(110, 0.8) categories <- expression(blank >= 85, 84.9 - 80, 79.9 - 75, 74.9 - 70, 69.9 - 65, 64.9 - 60, blank< 60) legend(110, 0.8, categories, lty=1:7, lwd=3, col=1, merge=TRUE) What I want is just ">=85" to be printed, but without something (here "blank") in front it is no correct expression. 2) Also I want German type decimals, that is a comma instead of a point. But the problem is, that the comma is used as argument separator in expression. plot(110, 0.8) categories <- expression(blank >= 85, 84,9 - 80, 79,9 - 75, 74,9 - 70, 69,9 - 65, 64,9 - 60, blank< 60) legend(110, 0.8, categories, lty=1:7, lwd=3, col=1, merge=TRUE) This does obviously not work. Any ideas? Thanks, Mark ––– Mark Heckmann Dipl. Wirt.-Ing. cand. Psych. Vorstraße 93 B01 28359 Bremen Blog: www.markheckmann.de R-Blog: http://ryouready.wordpress.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] (no subject)
Hussain Abusaaq fsu.edu> writes: > I have this vector and I want to change it to date. > > for example G=[05 12 2008] > > or g=[2] > f=[3] > y=[2208] > Something like as.Date(paste(rev(G),collapse="-")) __ 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] (no subject)
Hi, I havw some porblem in R and i need your help. I have this vector and I want to change it to date. for example G=[05 12 2008] or g=[2] f=[3] y=[2208] __ 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 do calculations in data matrices?
Some places to start apply() sweep() and any of the good introductory guides to R available online. For finding R material, it's vastly easier to use the custom search engine at www.rseek.org than to wander aimlessly around google trying to find things with "R" in them. We'd be happy to give more specific answers if you need them, especially if you do as the posting guide asks and provide a reproducible example. (Not that you need one for this question, since you did ask just for references.) Sarah On Sat, Feb 13, 2010 at 12:39 PM, Zoppoli, Gabriele (NIH/NCI) [G] wrote: > Please give me just a reference where I can find something useful. > > In summary, I need to : > > - find the median of each row of a matrix > - create a new matrix with each value in the first matrix divided by the > median of its row > - if a value "a" in the second matrix is < 1, I need to substitute it with 1/a > > I know that for some of you it must be overeasy, but I swear I googled for > two hours with keywords "operations", "calculations", "data matrices", "data > tables", and "CRAN", and I didn't find anything useful. > > Thank you all > > > Gabriele Zoppoli, MD > Ph.D. Fellow, Experimental and Clinical Oncology and Hematology, University > of Genova, Genova, Italy > Guest Researcher, LMP, NCI, NIH, Bethesda MD > > Work: 301-451-8575 > Mobile: 301-204-5642 > Email: zoppo...@mail.nih.gov > __ > 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. > -- Sarah Goslee http://www.stringpage.com http://www.astronomicum.com http://www.functionaldiversity.org __ 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] logical operations with lists
The fact that A and B have dimensions indicates that they are more than simple
vectors. It appears that when you create D you are subsetting columns rather
than rows, try:
> D <- C[which(C %in% A ==FALSE),] # note the ","
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Zoppoli, Gabriele (NIH/NCI) [G]
> Sent: Friday, February 12, 2010 3:58 PM
> To: r-help@r-project.org
> Subject: Re: [R] logical operations with lists
>
> I'm sorry but here's what I get:
>
> > A[1:10,]
> [1] UQCRC1 IDH3B PDHA1 SUCLA2 COX5B SDHB SDHA MDH2 DLD
> COQ7
>
> > dim(A)
> [1] 10131
>
> > B[1:10,]
> [1] 3.8-1.2 3.8-1.3 3.8-1.4 3.8-1.5 5-HT3c2 A1BGA1CFA2BP1
> A2LD1 A2M
>
> > dim(B)
> [1] 55546 1
>
> > C<-rbind(A,B)
> > dim(C)
> [1] 56559 1
>
> > D <- C[which(C %in% A ==FALSE)]
> > dim(D)
> [1] 56559 0
>
> and so with any other proposed method.
>
> I imported the list A and B this way:
>
> > A<-as.vector(read.delim("E:/A.txt",sep="\t",header=FALSE))
>
> and then removed the redundant rows with:
>
> > A<-unique(A)
>
> Guess I'm doing something really wrong here... Sorry for the
> inexperience, I'm trying to improve...
>
>
>
>
>
> Gabriele Zoppoli, MD
> Ph.D. Fellow, Experimental and Clinical Oncology and Hematology,
> University of Genova, Genova, Italy
> Guest Researcher, LMP, NCI, NIH, Bethesda MD
>
> Work: 301-451-8575
> Mobile: 301-204-5642
> Email: zoppo...@mail.nih.gov
>
> From: jbreic...@gmail.com [jbreic...@gmail.com] On Behalf Of Jonathan
> [jonsle...@gmail.com]
> Sent: Friday, February 12, 2010 5:21 PM
> To: Zoppoli, Gabriele (NIH/NCI) [G]
> Cc: r-help@r-project.org
> Subject: Re: [R] logical operations with lists
>
> This is probably not the best way, but (assuming you had vectors and
> not lists, since I'm not sure what your list looks like):
>
> C <- B[which(B %in% A ==FALSE)]
>
> Regards,
> Jonathan
>
>
> On Fri, Feb 12, 2010 at 5:06 PM, Zoppoli, Gabriele (NIH/NCI) [G]
> wrote:
> > Sorry, maybe it's easy but I haven't found anything useful:
> >
> > how can I obtain a list C that contains all the members in the list B
> that are not in list A? This are lists of nanes, not numbers!
> >
> > Thank you
> >
> >
> > Gabriele Zoppoli, MD
> > Ph.D. Fellow, Experimental and Clinical Oncology and Hematology,
> University of Genova, Genova, Italy
> > Guest Researcher, LMP, NCI, NIH, Bethesda MD
> >
> > Work: 301-451-8575
> > Mobile: 301-204-5642
> > Email: zoppo...@mail.nih.gov
> > __
> > 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] how to do calculations in data matrices?
Please give me just a reference where I can find something useful. In summary, I need to : - find the median of each row of a matrix - create a new matrix with each value in the first matrix divided by the median of its row - if a value "a" in the second matrix is < 1, I need to substitute it with 1/a I know that for some of you it must be overeasy, but I swear I googled for two hours with keywords "operations", "calculations", "data matrices", "data tables", and "CRAN", and I didn't find anything useful. Thank you all Gabriele Zoppoli, MD Ph.D. Fellow, Experimental and Clinical Oncology and Hematology, University of Genova, Genova, Italy Guest Researcher, LMP, NCI, NIH, Bethesda MD Work: 301-451-8575 Mobile: 301-204-5642 Email: zoppo...@mail.nih.gov __ 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] Multiple figures margin problem
I still have not received a copy of the esp package, so I don't know what exactly you tried, or what the results are. The following works for me and seems to fit your description, but without code and details of the results and how they differ from what you expect, we can only guess. x1 <- runif(100, 1, 25) x2 <- runif(100, 1, 25) y1 <- rnorm(100, x1) y2 <- rnorm(100, x2) y3 <- rnorm(100, x1+x2) par(mfcol=c(3,2), mar=c(0,0,0,0), oma=c(5,4,1,1)+0.1, xpd=NA) plot(x1,y1, xaxt='n') plot(x1,y2, xaxt='n') plot(x1,y3) plot(x2,y1, xaxt='n', yaxt='n') plot(x2,y2, xaxt='n', yaxt='n') plot(x2,y3, yaxt='n') you may also want to look at the pairs function or the pairs2 function (TeachingDemos package) for examples of axes in the margins. In fact you can get similar to the above by pairs2( cbind(x1,x2), cbind(y1,y2,y3), gap=0 ) -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- > project.org] On Behalf Of mnstn > Sent: Friday, February 12, 2010 3:27 PM > To: r-help@r-project.org > Subject: Re: [R] Multiple figures margin problem > > > Hello Greg, > I tried that and got a similar result. The axes are still hidden. I am > studying R Intro and Fig2A and 3B in > http://research.stowers-institute.org/efg/R/Graphics/Basics/mar- > oma/index.htm > . They seem to indicate that the "mar" parameter alone controls the > visibility of axis labels. Am I missing something obvious? > MoonStone > -- > View this message in context: http://n4.nabble.com/Multiple-figures- > margin-problem-tp1490455p1490493.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] Labels on a pyramide
This is good!
Yes I must learn about dput.
Thank you
Caveman
On Sat, Feb 13, 2010 at 6:57 PM, David Winsemius wrote:
> I took a shot at getting "assymetric lableling but it's by no means perfect.
> You really, really ,really, should learn to offer data with dput. See below:
>
>
> On Feb 13, 2010, at 9:27 AM, Orvalho Augusto wrote:
>
>> I am using pyramid.plot() from the plotrix package.
>>
>> I have something like this
>>
>>
>> xy.pop<-dados$masfr
>> xx.pop<-dados$femfr
>> #agelabels<-dados$femlab
>> xycol<-color.gradient(c(0,0,0.5,1),c(0,0,0.5,1),c(1,1,0.5,1),11)
>> xxcol<-color.gradient(c(1,1,0.5,1),c(0.5,0.5,0.5,1),c(0.5,0.5,0.5,1),11)
>> xylab<-dados$maslab
>> xxlab<-dados$femlab
>> agelabels<-xylab
>>
>> png("piramide9808.png")
>>
>> par(mar=pyramid.plot(xy.pop,xx.pop,labels=agelabels,top.labels=c("Masculino","","Feminino"),
>> main="Primeiras 10 cancros mais frequentes por
>> sexo...",xycol=xycol,xxcol=xxcol,gap=0, labelcex=0))
>>
>> dev.off()
>>
>> #
>>
>> the dados a dataframe fabircated by someother program and looks like:
>>
>> ##
>> ordem femlab femfa femfr maslab masfa masfr
>>
>> 1 Colo do utero 258 26.76348548 Prostata 613
>> 43.81701215
>>
>> 2 Mama 110 11.41078838 Figado 84 6.004288778
>>
>> 3 Esofago 62 6.43153527 Pele 70 5.003573981
>>
>> 4 Figado 60 6.22406639 Sarcoma de Kaposi 65
>> 4.64617584
>>
>> 5 Pele 48 4.979253112 Esofago 63 4.503216583
>>
>> 6 Bexiga 37 3.838174274 Pulmao 46 3.288062902
>>
>> 7 Corpo do utero 34 3.526970954 Bexiga 43
>> 3.073624017
>>
>> 8 "Utero, SOE" 28 2.904564315 Penis 33
>> 2.358827734
>>
>> 9 Sarcoma de Kaposi 28 2.904564315 Laringe 27
>> 1.929949964
>>
>> 10 Vulva e Vagina 24 2.489626556 Colon 24
>> 1.715511079
>>
>> 11 Outras localizacoes 275 28.52697095 Outras
>> localizacoes 331 23.65975697
>
>
> In case anyone wants to take a crack at this without the hassle of
> recreating htis dataset...
>
> dput(dados) can be used to recreate the complete dataframe
>
> dados <- structure(list(ordem = 1:11, femlab = structure(c(2L, 6L, 4L,
> 5L, 8L, 1L, 3L, 10L, 9L, 11L, 7L), .Label = c("Bexiga", "Colo do utero",
> "Corpo do utero", "Esofago", "Figado", "Mama", "Outras localizacoes",
> "Pele", "Sarcoma de Kaposi", "Utero SOE", "Vulva e Vagina"), class =
> "factor"),
> femfa = c(258L, 110L, 62L, 60L, 48L, 37L, 34L, 28L, 28L,
> 24L, 275L), femfr = c(26.76348548, 11.41078838, 6.43153527,
> 6.22406639, 4.979253112, 3.838174274, 3.526970954, 2.904564315,
> 2.904564315, 2.489626556, 28.52697095), maslab = structure(c(9L,
> 4L, 7L, 11L, 3L, 10L, 1L, 8L, 5L, 2L, 6L), .Label = c("Bexiga",
> "Colon", "Esofago", "Figado", "Laringe", "Outras localizacoes",
> "Pele", "Penis", "Prostata", "Pulmao", "Sarcoma de Kaposi"
> ), class = "factor"), masfa = c(613L, 84L, 70L, 65L, 63L,
> 46L, 43L, 33L, 27L, 24L, 331L), masfr = c(43.81701215, 6.004288778,
> 5.003573981, 4.64617584, 4.503216583, 3.288062902, 3.073624017,
> 2.358827734, 1.929949964, 1.715511079, 23.65975697)), .Names = c("ordem",
> "femlab", "femfa", "femfr", "maslab", "masfa", "masfr"), class =
> "data.frame", row.names = c(NA,
> -11L))
>
>
>
> xy.pop<-dados$masfr
> xx.pop<-dados$femfr
> #agelabels<-paste(dados$femlab, "\t\t\t", dados$maslab, sep='')
> xycol<-color.gradient(c(0,0,0.5,1),c(0,0,0.5,1),c(1,1,0.5,1),11)
> xxcol<-color.gradient(c(1,1,0.5,1),c(0.5,0.5,0.5,1),c(0.5,0.5,0.5,1),11)
> #xylab<-dados$maslab
> #xxlab<-dados$femlab
> #agelabels<-xylab
>
>
> par(mar=pyramid.plot(xy.pop, xx.pop, labels= rep("",length(xy.pop)),
> top.labels=c("Masculino", "", "Feminino"),
> main="Primeiras 10 cancros mais frequentes por
> sexo...", xycol=xycol,xxcol=xxcol, gap=0))
> text(-xy.pop-5*nchar(dados$maslab), 1:length(xy.pop), dados$maslab
> );text(xx.pop+7*nchar(dados$femlab), 1:length(xx.pop), dados$femlab )
>
> # Could not get the right ratio of xx.pop to nchar to get completely correct
> placement and given the small number you might just want to put in a
> "hand-crafted" vector,
>
>>
>>
>> #
>>
>> The problem is (1) I do not want plot agelabels on the center and (2)
>> I want plot different labels for each pair of the bars (one label for
>> masculine and the other feminine).
>>
>> The data represent the 10 most frequent cancer in a group of individuals.
>>
>> Can some one help please?
>>
>> Caveman
>>
>>
>>
>> --
>> OpenSource Software Consultant
>> CENFOSS (www.cenfoss.co.mz)
>> SP Tech (www.sptech.co.mz)
>> email: orvaq...@cenfoss.co.mz
>> cell: +258828810980
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>>
Re: [R] Labels on a pyramide
I took a shot at getting "assymetric lableling but it's by no means
perfect. You really, really ,really, should learn to offer data with
dput. See below:
On Feb 13, 2010, at 9:27 AM, Orvalho Augusto wrote:
I am using pyramid.plot() from the plotrix package.
I have something like this
xy.pop<-dados$masfr
xx.pop<-dados$femfr
#agelabels<-dados$femlab
xycol<-color.gradient(c(0,0,0.5,1),c(0,0,0.5,1),c(1,1,0.5,1),11)
xxcol<-color.gradient(c(1,1,0.5,1),c(0.5,0.5,0.5,1),c(0.5,0.5,0.5,1),
11)
xylab<-dados$maslab
xxlab<-dados$femlab
agelabels<-xylab
png("piramide9808.png")
par
(mar
=
pyramid
.plot
(xy
.pop,xx.pop,labels=agelabels,top.labels=c("Masculino","","Feminino"),
main="Primeiras 10 cancros mais frequentes por
sexo...",xycol=xycol,xxcol=xxcol,gap=0, labelcex=0))
dev.off()
#
the dados a dataframe fabircated by someother program and looks like:
##
ordem femlab femfa femfr maslab masfa masfr
1 Colo do utero 258 26.76348548 Prostata613
43.81701215
2 Mama110 11.41078838 Figado 84 6.004288778
3 Esofago 62 6.43153527 Pele70 5.003573981
4 Figado 60 6.22406639 Sarcoma de Kaposi 65
4.64617584
5 Pele48 4.979253112 Esofago 63 4.503216583
6 Bexiga 37 3.838174274 Pulmao 46 3.288062902
7 Corpo do utero 34 3.526970954 Bexiga 43 3.073624017
8 "Utero, SOE" 28 2.904564315 Penis 33 2.358827734
9 Sarcoma de Kaposi 28 2.904564315 Laringe 27
1.929949964
10 Vulva e Vagina 24 2.489626556 Colon 24 1.715511079
11 Outras localizacoes 275 28.52697095 Outras localizacoes 331
23.65975697
In case anyone wants to take a crack at this without the hassle of
recreating htis dataset...
dput(dados) can be used to recreate the complete dataframe
dados <- structure(list(ordem = 1:11, femlab = structure(c(2L, 6L, 4L,
5L, 8L, 1L, 3L, 10L, 9L, 11L, 7L), .Label = c("Bexiga", "Colo do utero",
"Corpo do utero", "Esofago", "Figado", "Mama", "Outras localizacoes",
"Pele", "Sarcoma de Kaposi", "Utero SOE", "Vulva e Vagina"), class =
"factor"),
femfa = c(258L, 110L, 62L, 60L, 48L, 37L, 34L, 28L, 28L,
24L, 275L), femfr = c(26.76348548, 11.41078838, 6.43153527,
6.22406639, 4.979253112, 3.838174274, 3.526970954, 2.904564315,
2.904564315, 2.489626556, 28.52697095), maslab = structure(c(9L,
4L, 7L, 11L, 3L, 10L, 1L, 8L, 5L, 2L, 6L), .Label = c("Bexiga",
"Colon", "Esofago", "Figado", "Laringe", "Outras localizacoes",
"Pele", "Penis", "Prostata", "Pulmao", "Sarcoma de Kaposi"
), class = "factor"), masfa = c(613L, 84L, 70L, 65L, 63L,
46L, 43L, 33L, 27L, 24L, 331L), masfr = c(43.81701215, 6.004288778,
5.003573981, 4.64617584, 4.503216583, 3.288062902, 3.073624017,
2.358827734, 1.929949964, 1.715511079, 23.65975697)), .Names =
c("ordem",
"femlab", "femfa", "femfr", "maslab", "masfa", "masfr"), class =
"data.frame", row.names = c(NA,
-11L))
xy.pop<-dados$masfr
xx.pop<-dados$femfr
#agelabels<-paste(dados$femlab, "\t\t\t", dados$maslab, sep='')
xycol<-color.gradient(c(0,0,0.5,1),c(0,0,0.5,1),c(1,1,0.5,1),11)
xxcol<-color.gradient(c(1,1,0.5,1),c(0.5,0.5,0.5,1),c(0.5,0.5,0.5,1),11)
#xylab<-dados$maslab
#xxlab<-dados$femlab
#agelabels<-xylab
par(mar=pyramid.plot(xy.pop, xx.pop, labels= rep("",length(xy.pop)),
top.labels=c("Masculino", "", "Feminino"),
main="Primeiras 10 cancros mais frequentes por
sexo...", xycol=xycol,xxcol=xxcol, gap=0))
text(-xy.pop-5*nchar(dados$maslab), 1:length(xy.pop), dados
$maslab );text(xx.pop+7*nchar(dados$femlab), 1:length(xx.pop), dados
$femlab )
# Could not get the right ratio of xx.pop to nchar to get completely
correct placement and given the small number you might just want to
put in a "hand-crafted" vector,
#
The problem is (1) I do not want plot agelabels on the center and (2)
I want plot different labels for each pair of the bars (one label for
masculine and the other feminine).
The data represent the 10 most frequent cancer in a group of
individuals.
Can some one help please?
Caveman
--
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CENFOSS (www.cenfoss.co.mz)
SP Tech (www.sptech.co.mz)
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Re: [R] Help for numbers close to 1 in R
Check out ?expm1 On Sat, Feb 13, 2010 at 9:17 AM, cjmr wrote: > > Hi all. > > I have a problem with adding 1 and very small numbers > > ex: >>p=1e-49 >>1+p > [1] 1 > > My problem comes from some values of x for the value: > 1-exp(-sum((x-1/sqrt(3))^2)) > the return values are always equivalent. > > Please help > Thanks > -- > View this message in context: > http://n4.nabble.com/Help-for-numbers-close-to-1-in-R-tp1554416p1554416.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] Why double quote is preferred?
On Fri, Feb 12, 2010 at 4:25 PM, blue sky wrote: > ?'`' shows the following: > > "Single and double quotes delimit character constants. They can be > used interchangeably but double quotes are *preferred* (and character > constants are printed using double quotes), so single quotes are > normally only used to delimit character constants containing double > quotes." > > It is not clear to me why double quote is preferred (I don't think > that "character constants are printed using double quotes" should be > the reason, in the sense that single quote can be used for printing if > we want to do so). It seems that double quote and single quote can be > used interchangeably, except that "Single quotes need to be escaped by > backslash in single-quoted strings, and double quotes in double-quoted > strings." > > Could somebody why double quote is preferred? To avoid confusion for those who are accustomed to programming in the C family of languages (C, C++, Java), where there is a difference in the meaning of single quotes and double quotes. A C programmer reads 'a' as a single character and "a" as a character string consisting of the letter 'a' followed by a null character to terminate the string. In R there is no character data type, there are only character strings. For consistency with other languages it helps if character strings are delimited by double quotes. The single quote version in R is for convenience. On most keyboards you don't need to use the shift key to type a single quote but you do need the shift for a double quote. P.S. I do not plan to get into an extended debate on this issue, Peng. (As others have pointed out, it is considered bad form to disguise your identity on this list.) Your opinions on the design of the language have been noted but if you want the language to be redesigned to your specifications, you will need to fork your own version. __ 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 for numbers close to 1 in R
On Sat, Feb 13, 2010 at 2:17 PM, cjmr wrote:
>>p=1e-49
>>1+p
> [1] 1
>
Try this:
> source("http://r-bc.googlecode.com/svn/trunk/R/bc.R";)
> p <- bc(1e-49)
> p
[1] ".1"
> 1+p
[1] "1.1"
You might want to check this [1] and search the archives for "precision".
Liviu
[1] http://code.google.com/p/r-bc/
> My problem comes from some values of x for the value:
> 1-exp(-sum((x-1/sqrt(3))^2))
> the return values are always equivalent.
>
> Please help
> Thanks
> --
> View this message in context:
> http://n4.nabble.com/Help-for-numbers-close-to-1-in-R-tp1554416p1554416.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.
>
--
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Re: [R] How to create probeAnno object?
On 02/11/2010 08:03 PM, AMBUJ wrote:
> Hi,
> Thank you Viviana for the description to create probeAnno object.
> The below link was very helpful:
> http://svitsrv25.epfl.ch/R-doc/library/Ringo/html/probeAnnoClass.html
This is just the help page for one of Ringo's classes. It's available in
R with
> library(Ringo)
> ?probeAnno
Have you looked at
> browseVignettes("Ringo")
? Ringo is a Bioconductor package so ask on the Bioconductor mailing list
http://bioconductor.org
Might as well include the maintainer in your email, since they're likely
to be the expert
> packageDescription("Ringo")
and always useful to include the output of
> sessionInfo()
to unambiguously report the version of R and Ringo you're using. A first
action is usually to update to the latest released R and corresponding
Bioconductor packages,
http://bioconductor.org/install
>
> I tried creating the object in the following ways where: startProbe &
> endProbe are the vectors which has the genomic start co-ordinates and end
> co-ordinates and index is a vector to store identifier of the probes.
> intensityData is the vector that stores data to be segmented
>
> Method 1.
>> map<-new("environment",startProbe,endProbe,index)
calling 'new' without naming the second and subsequent arguments is not
likely to create the object you expect. Maybe
map = new("environment",
startProbe=startProbe, endProbe=endProbe,
index=index)
?
>> arrayName<-"2009_01_18_veg1028_w"
>> genome<-"genome"
>> probeAnnotation<-new("probeAnno",map,arrayName,genome)
Here you're probably aiming for something like
new("probeAnno", map=map, arrayName=arrrayName, genome=genome)
Martin
>> segEnv<-segChrom(intensityData,probeAnno=probeAnnotation,chr="1",strands="+",nrBasesPerSegment=750)
>
> Running 'segment' on chromosome 1.+Error in mget(paste(chrstrd[j],
what, sep = "."), probeAnno) : second argument must be an environment
> Method 2.
>> pa3<-posToProbeAnno("~/523/POSFormat_tab.csv");Creating probeAnno mapping
>> for chromosome 1 Done.
>> arrayName(pa3)<-"S.Pombe"
>> genome(pa3)<-"genome"
>> show(pa3)
> A 'probeAnno' object holding the mapping between reporters and genomic
> positions.
> Chromosomes: 1
> Microarray platform: S.Pombe
> Genome: genome
>> segEnv<-segChrom(intensityData,probeAnno=pa3,chr="1",strands="+",nrBasesPerSegment=750)
>
> Running 'segment' on chromosome 1.+Error in mget(paste(chrstrd[j], what, sep
> = "."), probeAnno) : second argument must be an environment
>
>
> Both the methods gave the same error "second argument must be an
> environment". I am unable to execute segChrom() of tilingArray package. Any
> sugetions would be helpful. Thanks in advance.
>
>
> Regards,
>
> Ambuj A Thacker
>
>
> The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.
> http://in.yahoo.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.
--
Martin Morgan
Computational Biology / Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N.
PO Box 19024 Seattle, WA 98109
Location: Arnold Building M1 B861
Phone: (206) 667-2793
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Re: [R] Using getSYMBOL, annotate package on a list with empty elements.
On 02/13/2010 07:17 AM, David Winsemius wrote: > > On Feb 13, 2010, at 10:09 AM, Martin Morgan wrote: > >> On 02/13/2010 06:21 AM, David Winsemius wrote: >>> >>> On Feb 13, 2010, at 3:18 AM, Sahil Seth wrote: >>> Hi, I have been trying to find a solution to this issue, but have not been able to so ! I am trying to use sapply on the function getSYMBOL, >>> >>> The annotate package is from BioConductor. >>> an extract from the list is: > test.goP[13:14] $`GO:050` IEA IEA IEA IEA TAS TAS TAS IEA "5270753" "5720725" "1690128" "4850681" "110433" "2640544" "4900370" "1430280" IEA NAS TAS IEA "6110044" "1170615" "6590546" "1690632" $`GO:052` [1] NA goG=sapply(test.goP,getSYMBOL,data="hgu95av2") >>> >>> I was a bit surprised to see a data= argument to sapply. That didn't >>> seem typical but perhaps I am not aware of all the tricks available. >> >>> args(sapply) >> function (X, FUN, ..., simplify = TRUE, USE.NAMES = TRUE) >> >> The ... arguments are passed to FUN. Here FUN is getSYMBOL >> >>> args(getSYMBOL) >> function (x, data) >> >> so data="hgu95av2" is used in each application of getSYMBOL. This could >> have been written as >> >> sapply(test.goP, getSYMBOL, "hgu95av2") >> >> using positional matching to getSYMBOL's arguments. A 'trick' that is >> sometimes useful is to use named arguments to force the apply to vary >> the second (or other) argument, rather than the first. >> >> error: "Error in .checkKeysAreWellFormed(keys) : keys must be supplied in a character vector with no NAs " In this the 14th element has missing values, thus getSYMBOL raises issues. >>> >>> The way this is being displayed makes me think you are processing a list >>> with named elements. You should use str to determine what test.goP >>> really is. Then you will have a better idea what functions would be >>> appropriate. you may want to compose a function to go with the sapply >>> loop that omots the NA's: >>> >>> ?na.omit >>> >>> You should also post the results of dput or dump on the test objects you >>> are working with. That way the list readers can also get access to that >>> information. GetSYMBOL has to be given a char array, so a simple solution is infact to delete the missing elements from the list. I have been trying to find a solution for it, but in vain: tried: completecases(goP), na.omit(goP) and several other things. Any suggestions please ? >> >> > > Would that be: > > sapply(!is.na(test.goP), getSYMBOL, "hgu95av2") actually, sapply(test.goP[!is.na(test.goP)], getSYMBOL, "hgu95av2") ;). Martin > > ? > >> >>> >>> In addition to the above suggestions ... post on the correct list? >> >> http://bioconductor.org/ >> >> Martin >> >>> -- > > >> Phone: (206) 667-2793 > > David Winsemius, MD > Heritage Laboratories > West Hartford, CT > -- Martin Morgan Computational Biology / Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: Arnold Building M1 B861 Phone: (206) 667-2793 __ 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] Using getSYMBOL, annotate package on a list with empty elements.
On Feb 13, 2010, at 10:09 AM, Martin Morgan wrote: On 02/13/2010 06:21 AM, David Winsemius wrote: On Feb 13, 2010, at 3:18 AM, Sahil Seth wrote: Hi, I have been trying to find a solution to this issue, but have not been able to so ! I am trying to use sapply on the function getSYMBOL, The annotate package is from BioConductor. an extract from the list is: test.goP[13:14] $`GO:050` IEA IEA IEA IEA TAS TAS TAS IEA "5270753" "5720725" "1690128" "4850681" "110433" "2640544" "4900370" "1430280" IEA NAS TAS IEA "6110044" "1170615" "6590546" "1690632" $`GO:052` [1] NA goG=sapply(test.goP,getSYMBOL,data="hgu95av2") I was a bit surprised to see a data= argument to sapply. That didn't seem typical but perhaps I am not aware of all the tricks available. args(sapply) function (X, FUN, ..., simplify = TRUE, USE.NAMES = TRUE) The ... arguments are passed to FUN. Here FUN is getSYMBOL args(getSYMBOL) function (x, data) so data="hgu95av2" is used in each application of getSYMBOL. This could have been written as sapply(test.goP, getSYMBOL, "hgu95av2") using positional matching to getSYMBOL's arguments. A 'trick' that is sometimes useful is to use named arguments to force the apply to vary the second (or other) argument, rather than the first. error: "Error in .checkKeysAreWellFormed(keys) : keys must be supplied in a character vector with no NAs " In this the 14th element has missing values, thus getSYMBOL raises issues. The way this is being displayed makes me think you are processing a list with named elements. You should use str to determine what test.goP really is. Then you will have a better idea what functions would be appropriate. you may want to compose a function to go with the sapply loop that omots the NA's: ?na.omit You should also post the results of dput or dump on the test objects you are working with. That way the list readers can also get access to that information. GetSYMBOL has to be given a char array, so a simple solution is infact to delete the missing elements from the list. I have been trying to find a solution for it, but in vain: tried: completecases(goP), na.omit(goP) and several other things. Any suggestions please ? Would that be: sapply(!is.na(test.goP), getSYMBOL, "hgu95av2") ? In addition to the above suggestions ... post on the correct list? http://bioconductor.org/ Martin -- Phone: (206) 667-2793 David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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 for numbers close to 1 in R
Hi all. I have a problem with adding 1 and very small numbers ex: >p=1e-49 >1+p [1] 1 My problem comes from some values of x for the value: 1-exp(-sum((x-1/sqrt(3))^2)) the return values are always equivalent. Please help Thanks -- View this message in context: http://n4.nabble.com/Help-for-numbers-close-to-1-in-R-tp1554416p1554416.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] Using getSYMBOL, annotate package on a list with empty elements.
On 02/13/2010 06:21 AM, David Winsemius wrote: > > On Feb 13, 2010, at 3:18 AM, Sahil Seth wrote: > >> Hi, >> I have been trying to find a solution to this issue, but have not been >> able >> to so ! >> I am trying to use sapply on the function getSYMBOL, > > The annotate package is from BioConductor. > >> an extract from the list is: >>> test.goP[13:14] >> $`GO:050` >> IEA IEA IEA IEA TAS TAS TAS >> IEA >> "5270753" "5720725" "1690128" "4850681" "110433" "2640544" "4900370" >> "1430280" >> IEA NAS TAS IEA >> "6110044" "1170615" "6590546" "1690632" >> >> $`GO:052` >> [1] NA >> >> goG=sapply(test.goP,getSYMBOL,data="hgu95av2") > > I was a bit surprised to see a data= argument to sapply. That didn't > seem typical but perhaps I am not aware of all the tricks available. > args(sapply) function (X, FUN, ..., simplify = TRUE, USE.NAMES = TRUE) The ... arguments are passed to FUN. Here FUN is getSYMBOL > args(getSYMBOL) function (x, data) so data="hgu95av2" is used in each application of getSYMBOL. This could have been written as sapply(test.goP, getSYMBOL, "hgu95av2") using positional matching to getSYMBOL's arguments. A 'trick' that is sometimes useful is to use named arguments to force the apply to vary the second (or other) argument, rather than the first. >> error: "Error in .checkKeysAreWellFormed(keys) : >> keys must be supplied in a character vector with no NAs " >> In this the 14th element has missing values, thus getSYMBOL raises >> issues. > > The way this is being displayed makes me think you are processing a list > with named elements. You should use str to determine what test.goP > really is. Then you will have a better idea what functions would be > appropriate. you may want to compose a function to go with the sapply > loop that omots the NA's: > > ?na.omit > > You should also post the results of dput or dump on the test objects you > are working with. That way the list readers can also get access to that > information. >> >> GetSYMBOL has to be given a char array, so a simple solution is infact to >> delete the missing elements from the list. >> >> I have been trying to find a solution for it, but in vain: >> tried: completecases(goP), na.omit(goP) and several other things. >> >> Any suggestions please ? sapply(is.na(test.goP), getSYMBOL, "hgu95av2") > > In addition to the above suggestions ... post on the correct list? http://bioconductor.org/ Martin > >> -- > > David Winsemius, MD > Heritage Laboratories > West Hartford, CT > > __ > 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. -- Martin Morgan Computational Biology / Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: Arnold Building M1 B861 Phone: (206) 667-2793 __ 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] Failed install of package xlsReadWrite
2010/2/11 : > Does anyone have a work-around for a failed installation of this package? >> library(xlsReadWrite) ... > xls.getshlib() > > However, the xls.getshlib() command fails with this error: (For the record) It probably was a corporate firewall problem. Apart from info on www.swissr.org (thanks for pointing this out) you can always download the file directly. The full listing is here: http://dl.dropbox.com/u/2602516/swissrpkg/listing.html Direct links to the current (R2.10.x/1.5.1 pkg) version, either: - download the zipped dll: http://dl.dropbox.com/u/2602516/swissrpkg/bin/win32/shlib/xlsReadWrite_1.5.1_dll.zip and manually replace xlsReadWrite.dll (in e.g. 'C:\Programme\R\R-2.10.1\library\xlsReadWrite\libs') or - download the full package: http://dl.dropbox.com/u/2602516/swissrpkg/bin/win32/2.10/xlsReadWrite_1.5.1.zip and re-install. Note: 0.0.0-versions are development versions. They will be updated when there is something 'fixworthy' while a new official version is 'in wait mode'. Currently 0.0.0 fixes a bug with non-absolute file paths in 1.5.1. Cheers, Hans-Peter __ 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] Labels on a pyramide
I am using pyramid.plot() from the plotrix package.
I have something like this
xy.pop<-dados$masfr
xx.pop<-dados$femfr
#agelabels<-dados$femlab
xycol<-color.gradient(c(0,0,0.5,1),c(0,0,0.5,1),c(1,1,0.5,1),11)
xxcol<-color.gradient(c(1,1,0.5,1),c(0.5,0.5,0.5,1),c(0.5,0.5,0.5,1),11)
xylab<-dados$maslab
xxlab<-dados$femlab
agelabels<-xylab
png("piramide9808.png")
par(mar=pyramid.plot(xy.pop,xx.pop,labels=agelabels,top.labels=c("Masculino","","Feminino"),
main="Primeiras 10 cancros mais frequentes por
sexo...",xycol=xycol,xxcol=xxcol,gap=0, labelcex=0))
dev.off()
#
the dados a dataframe fabircated by someother program and looks like:
##
ordem femlab femfa femfr maslab masfa masfr
1 Colo do utero 258 26.76348548 Prostata613
43.81701215
2 Mama110 11.41078838 Figado 84 6.004288778
3 Esofago 62 6.43153527 Pele70 5.003573981
4 Figado 60 6.22406639 Sarcoma de Kaposi 65
4.64617584
5 Pele48 4.979253112 Esofago 63 4.503216583
6 Bexiga 37 3.838174274 Pulmao 46 3.288062902
7 Corpo do utero 34 3.526970954 Bexiga 43 3.073624017
8 "Utero, SOE"28 2.904564315 Penis 33 2.358827734
9 Sarcoma de Kaposi 28 2.904564315 Laringe 27
1.929949964
10 Vulva e Vagina 24 2.489626556 Colon 24 1.715511079
11 Outras localizacoes 275 28.52697095 Outras localizacoes
331 23.65975697
#
The problem is (1) I do not want plot agelabels on the center and (2)
I want plot different labels for each pair of the bars (one label for
masculine and the other feminine).
The data represent the 10 most frequent cancer in a group of individuals.
Can some one help please?
Caveman
--
OpenSource Software Consultant
CENFOSS (www.cenfoss.co.mz)
SP Tech (www.sptech.co.mz)
email: orvaq...@cenfoss.co.mz
cell: +258828810980
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using getSYMBOL, annotate package on a list with empty elements.
On Feb 13, 2010, at 3:18 AM, Sahil Seth wrote: Hi, I have been trying to find a solution to this issue, but have not been able to so ! I am trying to use sapply on the function getSYMBOL, The annotate package is from BioConductor. an extract from the list is: test.goP[13:14] $`GO:050` IEA IEA IEA IEA TAS TAS TAS IEA "5270753" "5720725" "1690128" "4850681" "110433" "2640544" "4900370" "1430280" IEA NAS TAS IEA "6110044" "1170615" "6590546" "1690632" $`GO:052` [1] NA goG=sapply(test.goP,getSYMBOL,data="hgu95av2") I was a bit surprised to see a data= argument to sapply. That didn't seem typical but perhaps I am not aware of all the tricks available. error: "Error in .checkKeysAreWellFormed(keys) : keys must be supplied in a character vector with no NAs " In this the 14th element has missing values, thus getSYMBOL raises issues. The way this is being displayed makes me think you are processing a list with named elements. You should use str to determine what test.goP really is. Then you will have a better idea what functions would be appropriate. you may want to compose a function to go with the sapply loop that omots the NA's: ?na.omit You should also post the results of dput or dump on the test objects you are working with. That way the list readers can also get access to that information. GetSYMBOL has to be given a char array, so a simple solution is infact to delete the missing elements from the list. I have been trying to find a solution for it, but in vain: tried: completecases(goP), na.omit(goP) and several other things. Any suggestions please ? In addition to the above suggestions ... post on the correct list? -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] "drop if missing" command?
On Feb 12, 2010, at 6:48 PM, Nora Kozloff wrote:
This will probably seem very simple to experienced R programmers:
I am doing a snp association analysis and am at the model-fitting
stage. I
am using the Stats package's "drop1" with the following code:
##geno is the dataset
## the dependent variable (casectrln) is dichotomous and coded 0,1
## rs743572_2 is one of the snps (which is coded 0,1,2 for the 3
genotypes)
library(stats)
modadd = glm(geno$casectrln ~rs743572_2 + factor(racegrp)+
factor(smokgp)+
factor(alcgp)+ factor(bmigp) + factor(ipssgp)
+ agebase, family="binomial")
drop1(mod,scale=0,test=c("Chisq"), x=NULL, k=2)
There is a great deal of missing data in this dataset for both snps
and for
covariates--so I have been instructed not to simply drop all cases
with
missing genotype or covariate data .
It sounds as though you have been asked to do some sort of imputation
of missing data.
How can I drop the observations which
are missing only for the snp I am modeling at the time, then
reinstate
those observations to model the next snp?
Thanks for any help,
Nora
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] SAS and RODBC
Daniel Nordlund wrote:
. . .
This is just a quick follow-up to my previous post. Based on Prof. Ripley's
response I went back and looked at the SAS log file and reread the RODBC help
pages. The problem of writing a SAS dataset was solved by setting
colQuote=NULL in the call to the odbcConnect() function.
ch <- odbcConnect('sasodbc', believeNRows=FALSE, colQuote=NULL)
I hope this will be useful to others who may have the SAS BASE product and want
to do graphics or statistical analyses with their SAS data, but can't afford
the high licensing fees for the SAS STAT and GRAPH modules. Thanks to Prof.
Ripley for the fine RODBC package.
Dan
Daniel Nordlund
Bothell, WA USA
Daniel since you have SAS BASE installed why not use sas.get in the
Hmisc package and also get access to metadata such as variable labels
that ODBC does not handle? Besides providing better documentation, the
labels are very useful as axis labels in plotting, etc.
Frank
--
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
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Re: [R] help with EXPASY HTML form submission in RCurl package
Sunando Roy wrote:
>
> Hi Duncan,
>
> Thanks for your help. I changed the P but the output that I get is not
> what I expect. The form gets aborted without any actual output. I get
> the same result with
>
> postForm("http://www.expasy.ch/tools/protscale.html";)
That URL (...protscale.html) is the HTML page that contains the form.
It is not the URL to which you are supposed to submit the form request.
That information is in the attributes of the .
That is
http://www.expasy.ch/cgi-bin/protscale.pl?1
So you have to know a little about HTML forms in order to
figure out how to map the HTML description to a request.
That is what your browser does ( including hidden fields in
the form, etc.)
It is also the purpose of the R package RHTMLForms (www.omegahat.org/RHTMLForms
and
install.packages("RHTMLForms", repos = "http://www.omegahat.org/R";, type =
"source")
)
e.g.
library(RHTMLForms)
f = getHTMLFormDescription("http://www.expasy.ch/tools/protscale.html";)
fun = createFunction(f[[2]])
o = fun(prot_id = "P05130", weight_var = "exponential", style = "POST")
(The protscale.html is malformed with an < that messes up parsing the
linear option.)
Now, of course, you have to parse the resulting HTML (in the string given in
the variable o)
to get the information the form submission generated.
D.
>
> just with an added message that there was no input passed on. But with
> the input like I presented I get the same output. I could make some of
> the examples work like for e.g
>
> postForm("http://www.omegahat.org/RCurl/testPassword/form_validation.cgi";,
> your_name = "Duncan",
> your_age = "35-55",
> your_sex = "m",
> submit = "submit",
> .opts = list(userpwd = "bob:welcome"))
>
> which would suggest atleast the setup is correct.
> I parsed the expasy protscale source code to identify the variables but
> the form does not seem to go through. I can post the html body code if
> needed.
>
> Regards
>
> Sunando
> On Fri, Feb 12, 2010 at 3:54 PM, Duncan Temple Lang
> mailto:dun...@wald.ucdavis.edu>> wrote:
>
>
>
> Sunando Roy wrote:
> > Hi,
> >
> > I am trying to submit a form to the EXPASY protscale server (
> > http://www.expasy.ch/tools/protscale.html). I am using the RCurl
> package and
> > the postForm function available in it. I have extracted the
> variables for
> > the form from the HTML source page. According to the syntax of
> postForm, I
> > just need to mention the url and assign values to the input
> mentioned in the
> > HTML code.
> > The code that I am using is:
> > postForm("http://www.expasy.ch/tools/protscale.html";,
> > "sequence" = "
> > ",
> > scale = "Molecular weight",
> > window = "5",
> > weight_edges = "100",
> > weight_var = "linear",
> > norm = "no",
> > submit = "Submit"), .checkparams = TRUE)
>
>
> I don't think that is what you actually submitted to R.
> It is a syntax error as you end the cal to postForm) after "Submit"
> and then have an extra ", .checkparams = TRUE)" afterwards.
>
> But, when you remove the ')' after "Submit",
> the problem you get is that .checkparams is not a parameter
> of postForm(), but .checkParams is. R is case-sensitive
> so the problem is that .checkparams is being treated as a parameter
> of your form.
>
> So change the p to P in .checkparams, and it works.
>
> D.
>
> > the constant error that I get is:
> > Error in postForm("http://www.expasy.ch/tools/protscale.html";,
> .params =
> > list(sequence = "not", : STRING_ELT() can only be applied to a
> 'character
> > vector', not a 'logical'
> >
> > Is there any other way to submit an HTML form in R ?
> >
> > Thanks for the help
> >
> > Regards
> >
> > Sunando
> >
> > [[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.
>
>
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Re: [R] Hierarchical data sets: which software to use?
Hi Douglas, Thanks for your helpful response. I've commented on some of the points you raised below: > Although this may not be helpful for your immediate goal, storing and manipulating data of this size and complexity (and, I expect, cost for collection) really calls for tools like relational databases. A single flat file of 2500 variables by 1500 cases is almost never the best way to organize such data. A normalized representation as a collection of interlinked tables in a relational data base is much more effective and less error prone. The widespread use of spreadsheets or SPSS data sets or SAS data sets which encourage the "single table with a gargantuan number of columns, most of which are missing data in most cases" approach to organization of longitudinal data is regrettable. I'm both relieved and daunted by this. Daunted because it means I'll need to learn another package (probably postGreSQL or MySQL?), but relieved because constructing a 2500 by 1500 file seemed intuitively wrong, as well as introducing the possibility of errors unnecessarily--surely it makes more sense to leave the data as is. As far as immediate goals go--I am at the beginning of a thesis, and I have more research planned after that, so I want to get things right from the start. > For later analysis in R it is better to start with "long" form of the data, as opposed to the "wide" form, even if it means repeating demographic information over several occasions. Using a relational database allows for a long view to be generated without the possibility of inconsistency in the demographics. I am using the descriptions "long" and "wide" in the sense that they are used in the reshape help page. See ?reshape > in R. The long view is also called the subject/occasion view in the sense that each row corresponds to one subject on one occasion. > Robert Gentleman's book "R Programming for Bioinformatics" provides background on linking R to relational databases. Thanks--I'll look this one up. > As I said at the beginning, you may not want to undertake the necessary study and effort to reorganize your data for this specific project but if you do this a lot you may want to consider it. As above: a stitch in time, I suppose. Thanks again. Anton On Sat, Feb 6, 2010 at 3:22 AM, Douglas Bates wrote: > On Sun, Jan 31, 2010 at 10:24 PM, Anton du Toit > wrote: >> Dear R-helpers, >> >> I’m writing for advice on whether I should use R or a different package or >> language. I’ve looked through the R-help archives, some manuals, and some >> other sites as well, and I haven’t done too well finding relevant info, >> hence my question here. >> >> I’m working with hierarchical data (in SPSS lingo). That is, for each case >> (person) I read in three types of (medical) record: >> >> 1. demographic data: name, age, sex, address, etc >> >> 2. ‘admissions’ data: this generally repeats, so I will have 20 or so >> variables relating to their first hospital admission, then the same 20 again >> for their second admission, and so on >> >> 3. ‘collections’ data, about 100 variables containing the results of a >> battery of standard tests. These are administered at intervals and so this >> is repeating data as well. >> >> The number of repetitions varies between cases, so in its one case per line >> format the data is non-rectangular. >> >> At present I have shoehorned all of this into SPSS, with each case on one >> line. My test database has 2,500 variables and 1,500 cases (or persons), and >> in SPSS’s *.SAV format is ~4MB. The one I finally work with will be larger >> again, though likely within one order of magnitude. Down the track, funding >> permitting, I hope to be working with tens of thousands of cases. > > Although this may not be helpful for your immediate goal, storing and > manipulating data of this size and complexity (and, I expect, cost for > collection) really calls for tools like relational databases. A > single flat file of 2500 variables by 1500 cases is almost never the > best way to organize such data. A normalized representation as a > collection of interlinked tables in a relational data base is much > more effective and less error prone. The widespread use of > spreadsheets or SPSS data sets or SAS data sets which encourage the > "single table with a gargantuan number of columns, most of which are > missing data in most cases" approach to organization of longitudinal > data is regrettable. > > For later analysis in R it is better to start with "long" form of the > data, as opposed to the "wide" form, even if it means repeating > demographic information over several occasions. Using a relational > database allows for a long view to be generated without the > possibility of inconsistency in the demographics. I am using the > descriptions "long" and "wide" in the sense that they are used in the > reshape help page. See > > ?reshape > > in R. The long view is also called the subject/occasion view in the > sense th
[R] Using getSYMBOL, annotate package on a list with empty elements.
Hi, I have been trying to find a solution to this issue, but have not been able to so ! I am trying to use sapply on the function getSYMBOL, an extract from the list is: > test.goP[13:14] $`GO:050` IEA IEA IEA IEA TAS TAS TAS IEA "5270753" "5720725" "1690128" "4850681" "110433" "2640544" "4900370" "1430280" IEA NAS TAS IEA "6110044" "1170615" "6590546" "1690632" $`GO:052` [1] NA goG=sapply(test.goP,getSYMBOL,data="hgu95av2") error: "Error in .checkKeysAreWellFormed(keys) : keys must be supplied in a character vector with no NAs " In this the 14th element has missing values, thus getSYMBOL raises issues. GetSYMBOL has to be given a char array, so a simple solution is infact to delete the missing elements from the list. I have been trying to find a solution for it, but in vain: tried: completecases(goP), na.omit(goP) and several other things. Any suggestions please ? Thanks a lot ! [[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.
