Re: [R] defining color sequence in image()
one solution is: img1-matrix(1:5) img2-matrix(2:5) col-1:5 # col-c(green,yellow,...) image(img1,col=col[sort(unique(img1))]) image(img2,col=col[sort(unique(img2))]) On 12/26/06, Milton Cezar Ribeiro [EMAIL PROTECTED] wrote: Dear All, How can I define a color sequence for each image value? I have several images with values varying from 1 to 5, and I would like to assing a fixed color for each value (green for 1, yellow for 2...). I used somethink like: image(img,col=c(green,yellow,...)) but unfortunately whem I have absent values, the color that I defined for each values change. Thanks a lot, Miltinho Brazil PS : Merry Christmas!!! __ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] extend summary.lm for hccm?
On Sun, 24 Dec 2006, John Fox wrote: If I remember Freedman's recent paper correctly, he argues that sandwich variance estimator, though problematic in general, is not problematic in the case that White described -- an otherwise correctly specified linear model with heteroscedasticity estimated by least-squares. More generally, sandwich-type estimators are valid (i.e., estimate the right quantity, although not necessarily precisely, as Frank pointed out) in situations where the estimating functions are correctly specified but remaining aspets of the likelihood (not captured in the estimating functions) are potentially not. In linear models, it is easy to see what this means: the mean function has to be correctly specified (i.e., the errors have zero mean) but the correlation structure of the errors (i.e., their (co-)variances) might differ from the usual assumptions. In GLMs, in particular logistic regression, it is much more difficult to see against which types of misspecification sandwich-based inference is robust. Freedman's paper stresses the point that many model misspecifications also imply misspecified estimating functions (and in his example this is rather obvious) so that consequently the sandwich-type estimators estimate the wrong quantity. Best wishes, Z __ R-help@stat.math.ethz.ch 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: vectorize an iterative process.
SGkgRXZlcnlvbmUsDQogDQpJIGFtIHN0dWNrIHdpdGggYSBzaW1wbGUgcHJvYmxlbS4gU3Vw cG9zZSBJIGhhdmUgYSB2ZWN0b3IgeCwgYW5kIEkgd2FudCB0byBjYWxjdWxhdGUgeVtpXT14 W2krMV0teFtpXSwgaXQgaXMgdmVyeSBlYXN5LiBJIGp1c3QgbmVlZCB0byB3cml0ZSB5PC14 WzI6bGVuZ3RoKHgpXS14WzE6bGVuZ3RoKHgpLTFdLiANCiANCk5vdyBpZiBJIGtub3cgeSwg YW5kIHdhbnQgdG8ga25vdyB0aGUgdmVjdG9yIHggZGVmaW5lZCBieSB4W2ldPXhbaS0xXSt5 W2ktMV0gZm9yIGFsbCBpLCBob3cgY2FuIEkgZG8gdGhpcyB3aXRob3V0IGEgbG9vcD8gDQog DQpUaGFua3MsDQpHZW9mZnJleQ0KIA0KDQpfX19fX19fX19fX19fX19fX19fX19fX19fX19f X19fX19fX19fX19fX19fX19fX19fX19fX19fCgoKVGhlIGluZm9ybWF0aW9uIGluIHRoaXMg ZW1haWwgb3IgaW4gYW55IGZpbGUgYXR0YWNoZWQgaGVyZXRvIGlzCmludGVuZGVkIG9ubHkg Zm9yIHRoZSBwZXJzb25hbCBhbmQgY29uZmlkZW50aWFsIHVzZSBvZiB0aGUgaW5kaXZpZHVh bApvciBlbnRpdHkgdG8gd2hpY2ggaXQgaXMgYWRkcmVzc2VkIGFuZCBtYXkgY29udGFpbiBp bmZvcm1hdGlvbiB0aGF0IGlzCnByb3ByaWV0YXJ5IGFuZCBjb25maWRlbnRpYWwuIElmIHlv dSBhcmUgbm90IHRoZSBpbnRlbmRlZCByZWNpcGllbnQgb2YKdGhpcyBtZXNzYWdlIHlvdSBh cmUgaGVyZWJ5IG5vdGlmaWVkIHRoYXQgYW55IHJldmlldywgZGlzc2VtaW5hdGlvbiwKZGlz dHJpYnV0aW9uIG9yIGNvcHlpbmcgb2YgdGhpcyBtZXNzYWdlIGlzIHN0cmljdGx5IHByb2hp Yml0ZWQuIFRoaXMgY29tbXVuaWNhdGlvbiBpcyBmb3IgaW5mb3JtYXRpb24gcHVycG9zZXMg b25seSBhbmQgc2hvdWxkIG5vdCBiZSByZWdhcmRlZCBhcyBhbiBvZmZlciB0byBzZWxsIG9y IGFzIGEgc29saWNpdGF0aW9uIG9mIGFuIG9mZmVyIHRvIGJ1eSBhbnkgZmluYW5jaWFsIHBy b2R1Y3QuIEVtYWlsIHRyYW5zbWlzc2lvbiBjYW5ub3QgYmUgZ3VhcmFudGVlZCB0byBiZSBz ZWN1cmUgb3IgZXJyb3ItZnJlZS4NCg== __ R-help@stat.math.ethz.ch 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] writing to S-PLUS .dat file
Dear Users, I am new to R. I use write() to write my data in .txt format. I'd like to write to a disc any kind of data in a .dat S-PLUS format. Please help. SM __ R-help@stat.math.ethz.ch 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] sequential row selection in dataframe
On Dec 26, 2006, at 12:07 AM, Pedro Mardones wrote: I'm wondering if there is any 'efficient' approach for selecting a sample of 'every nth rows' from a dataframe. For example, let's use the dataframe GAGurine in MASS library: length(GAGurine[,1]) [1] 314 # select an 75% of the dataset, i.e. = 236 rows, every 2 rows starting from row 1 test-GAGurine[seq(1,314,2),] length(test[,1]) [1] 157 # so, I still need another 79 rows, one way could be: test2-GAGurine[-seq(1,314,2),] length(test2[,1]) [1] 157 test3-test2[seq(1,157,2),] # and then final-rbind(test2,test3) length(final[,1]) [1] 236 Does anyone have a better idea to get the same results but without creating different datasets like test2 and test3? A probabilistic approach: len - length(GAGurine[,1]) GAGu - GAGurine[sample(1:len, round(.75 * len)), ] # 236 rows A deterministic one: nr - 1 #or 2 GAGu2 - GAGurine[-seq(nr, len, 4),] # drop every 4th, giving 235 rows nr - 3 # or 4 will give 236 rows. _ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400Charlottesville, VA 22904-4400 Parcels:Room 102Gilmer Hall McCormick RoadCharlottesville, VA 22903 Office:B011+1-434-982-4729 Lab:B019+1-434-982-4751 Fax:+1-434-982-4766 WWW:http://www.people.virginia.edu/~mk9y/ __ R-help@stat.math.ethz.ch 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] vectorizing an iterative process.
Hi Everyone, I am stuck with a simple problem. Suppose I have a vector x, and I want to calculate y[i]=3Dx[i+1]-x[i], it is very easy. I just need to write y-x[2:length(x)]-x[1:length(x)-1]. Now if I know y, and want to know the vector x defined by x[i]=3Dx[i-1]+y[i-1] for all i, how can I do this without a loop? Thanks, Geoffrey PS. Sorry if you see a duplicate message. The previous one was in a weird format that most people would not be able to read. ___=0A= =0A= =0A= The information in this email or in any file attached hereto is=0A= intended only for the personal and confidential use of the individual=0A= or entity to which it is addressed and may contain information that is=0A= proprietary and confidential. If you are not the intended recipient of=0A= this message you are hereby notified that any review, dissemination,=0A= distribution or copying of this message is strictly prohibited. This communi= cation is for information purposes only and should not be regarded as an off= er to sell or as a solicitation of an offer to buy any financial product. Em= ail transmission cannot be guaranteed to be secure or error-free. __ R-help@stat.math.ethz.ch 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] vectorizing an iterative process.
x[i]=3Dx[i-1]+y[i-1] for all i, how can I do this without a loop? It looks like x - cumsum(y) What does 3D mean? __ R-help@stat.math.ethz.ch 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] vectorizing an iterative process.
Hi Richard, 3D is automatically generated by the mailing list software, probably because I had ] followed by =3D without a space in the original post. What I meant was to calculate x[i] =3D x[i-1] + y[i-1] For example, if X - 1:10 Then I want the vector Y to be 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, or in other words Y[i] =3D y[i-1] + x[i]. Yes, cumsum does the trick for this. This is what I need. Thanks. Just curious, do you know how to calculate the more generic x[i] - f( x[i-1], y[i-1] )? Thanks, Geoffrey -Original Message- From: Richard M. Heiberger [mailto:[EMAIL PROTECTED] Sent: Tuesday, December 26, 2006 9:56 AM To: Geoffrey Zhu; r-help@stat.math.ethz.ch Subject: Re: [R] vectorizing an iterative process. x[i]=3D3Dx[i-1]+y[i-1] for all i, how can I do this without a loop? It looks like x - cumsum(y) What does 3D mean? ___=0A= =0A= =0A= The information in this email or in any file attached hereto is=0A= intended only for the personal and confidential use of the individual=0A= or entity to which it is addressed and may contain information that is=0A= proprietary and confidential. If you are not the intended recipient of=0A= this message you are hereby notified that any review, dissemination,=0A= distribution or copying of this message is strictly prohibited. This communi= cation is for information purposes only and should not be regarded as an off= er to sell or as a solicitation of an offer to buy any financial product. Em= ail transmission cannot be guaranteed to be secure or error-free. __ R-help@stat.math.ethz.ch 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] vectorizing an iterative process.
x[i]=3Dx[i-1]+y[i-1] for all i, how can I do this without a loop? It looks like x - cumsum(y) What does 3D mean? The =3D is probably an encoding error - it should just be =. In general to vectorise an iterative problem, you will need to solve the recurrence relation (http://en.wikipedia.org/wiki/Recurrence_relation), although I think it's pretty tricky in general, and there's no guarantee that the vectorised/non-recursive form will be more efficient. Hadley __ R-help@stat.math.ethz.ch 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] vectorizing an iterative process.
I meant x[i] - x[i-1] + y[i-1] and Y[i] - y[i-1] + x[i] below. The mailing list software just keep adding 3D's. Sorry. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Geoffrey Zhu Sent: Tuesday, December 26, 2006 10:03 AM To: Richard M. Heiberger; r-help@stat.math.ethz.ch Subject: Re: [R] vectorizing an iterative process. Hi Richard, 3D is automatically generated by the mailing list software, probably because I had ] followed by =3D3D without a space in the original post. What I meant was to calculate x[i] =3D3D x[i-1] + y[i-1] For example, if X - 1:10 Then I want the vector Y to be 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, or in other words Y[i] =3D3D y[i-1] + x[i]. Yes, cumsum does the trick for this. This is what I need. Thanks. Just curious, do you know how to calculate the more generic x[i] - f( x[i-1], y[i-1] )? Thanks, Geoffrey -Original Message- From: Richard M. Heiberger [mailto:[EMAIL PROTECTED] Sent: Tuesday, December 26, 2006 9:56 AM To: Geoffrey Zhu; r-help@stat.math.ethz.ch Subject: Re: [R] vectorizing an iterative process. x[i]=3D3D3Dx[i-1]+y[i-1] for all i, how can I do this without a loop? It looks like x - cumsum(y) What does 3D mean? ___=3D0A=3D =3D0A=3D =3D0A=3D The information in this email or in any file attached hereto is=3D0A=3D intended only for the personal and confidential use of the individual=3D0A=3D or entity to which it is addressed and may contain information that is=3D0A=3D proprietary and confidential. If you are not the intended recipient of=3D0A=3D this message you are hereby notified that any review, dissemination,=3D0A=3D distribution or copying of this message is strictly prohibited. This communi=3D cation is for information purposes only and should not be regarded as an off=3D er to sell or as a solicitation of an offer to buy any financial product. Em=3D ail transmission cannot be guaranteed to be secure or error-free. __ R-help@stat.math.ethz.ch 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. ___=0A= =0A= =0A= The information in this email or in any file attached hereto is=0A= intended only for the personal and confidential use of the individual=0A= or entity to which it is addressed and may contain information that is=0A= proprietary and confidential. If you are not the intended recipient of=0A= this message you are hereby notified that any review, dissemination,=0A= distribution or copying of this message is strictly prohibited. This communi= cation is for information purposes only and should not be regarded as an off= er to sell or as a solicitation of an offer to buy any financial product. Em= ail transmission cannot be guaranteed to be secure or error-free. __ R-help@stat.math.ethz.ch 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] vectorizing an iterative process.
In your case, the recurrence relationship for x can be solved easily:
Notice that
sum{i=1,n}(x[i]-x[i-1]) = x[n] - x[0]
and therefore
x[n] = x[0] + sum{i=1,n}(y[i-1] for n=1, N, with the appropriate initial
condition for i=0, (x[0],y[0]).
Thus cumsum on y will give you a direct answer.
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Geoffrey Zhu
Sent: Tuesday, December 26, 2006 11:06 AM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] vectorizing an iterative process.
I meant x[i] - x[i-1] + y[i-1] and Y[i] - y[i-1] + x[i] below.
The mailing list software just keep adding 3D's. Sorry.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Geoffrey Zhu
Sent: Tuesday, December 26, 2006 10:03 AM
To: Richard M. Heiberger; r-help@stat.math.ethz.ch
Subject: Re: [R] vectorizing an iterative process.
Hi Richard,
3D is automatically generated by the mailing list software, probably because
I had ] followed by =3D3D without a space in the original post.
What I meant was to calculate x[i] =3D3D x[i-1] + y[i-1]
For example, if X - 1:10
Then I want the vector Y to be 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, or in
other words Y[i] =3D3D y[i-1] + x[i].
Yes, cumsum does the trick for this. This is what I need. Thanks.
Just curious, do you know how to calculate the more generic x[i] - f(
x[i-1], y[i-1] )?
Thanks,
Geoffrey
-Original Message-
From: Richard M. Heiberger [mailto:[EMAIL PROTECTED]
Sent: Tuesday, December 26, 2006 9:56 AM
To: Geoffrey Zhu; r-help@stat.math.ethz.ch
Subject: Re: [R] vectorizing an iterative process.
x[i]=3D3D3Dx[i-1]+y[i-1] for all i, how can I do this without a loop?
It looks like
x - cumsum(y)
What does 3D mean?
___=3D0A=3D
=3D0A=3D
=3D0A=3D
The information in this email or in any file attached hereto is=3D0A=3D
intended only for the personal and confidential use of the
individual=3D0A=3D or entity to which it is addressed and may contain
information that is=3D0A=3D proprietary and confidential. If you are not the
intended recipient of=3D0A=3D this message you are hereby notified that any
review, dissemination,=3D0A=3D distribution or copying of this message is
strictly prohibited. This communi=3D cation is for information purposes only
and should not be regarded as an off=3D er to sell or as a solicitation of
an offer to buy any financial product. Em=3D ail transmission cannot be
guaranteed to be secure or error-free.
__
R-help@stat.math.ethz.ch 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.
___=0A=
=0A=
=0A=
The information in this email or in any file attached hereto is=0A= intended
only for the personal and confidential use of the individual=0A= or entity
to which it is addressed and may contain information that is=0A= proprietary
and confidential. If you are not the intended recipient of=0A= this message
you are hereby notified that any review, dissemination,=0A= distribution or
copying of this message is strictly prohibited. This communi= cation is for
information purposes only and should not be regarded as an off= er to sell
or as a solicitation of an offer to buy any financial product. Em= ail
transmission cannot be guaranteed to be secure or error-free.
__
R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Problem to generate training data set and test data set
What you describe is called stratified sampling. It was discusssed last month (and other times) on this list: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/90220.html Using RSiteSearch(stratified sampling) will produce many hits to relevant articles and packages. On Mon, 25 Dec 2006, Aimin Yan wrote: I have a full data set like this: aa basaas bms ams bcuacu omega y 1 ALA 0 127.71 0 69.99 0 -0.2498560 79.91470 outward 2 PRO 0 68.55 0 55.44 0 -0.0949008 76.60380 outward 3 ALA 0 52.72 0 47.82 0 -0.0396550 52.19970 outward 4 PHE 0 22.62 0 31.21 0 0.1270330 169.52500 inward 5 SER 0 71.32 0 52.84 0 -0.1312380 7.47528 outward 6 VAL 0 12.92 0 22.40 0 0.1728390 149.09400 inward .. aa have 19 levels, and there are different number of observation for each levels. I want to pick 75% of observations of each levels randomly to generate a training set, and 25% of observation of each levels to generate a testing set. Does anyone know to do this? Thanks Aimin Yan __ R-help@stat.math.ethz.ch 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. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0717 __ R-help@stat.math.ethz.ch 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] vectorizing an iterative process.
Yes, this solves my problem. Thanks for your help.
-Original Message-
From: Christos Hatzis [mailto:[EMAIL PROTECTED]
Sent: Tuesday, December 26, 2006 10:58 AM
To: Geoffrey Zhu; r-help@stat.math.ethz.ch
Subject: RE: [R] vectorizing an iterative process.
In your case, the recurrence relationship for x can be solved easily:
Notice that
sum{i=3D1,n}(x[i]-x[i-1]) =3D x[n] - x[0]
and therefore
x[n] =3D x[0] + sum{i=3D1,n}(y[i-1] for n=3D1, N, with the appropriate initi=
al
condition for i=3D0, (x[0],y[0]).
Thus cumsum on y will give you a direct answer.
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Geoffrey Zhu
Sent: Tuesday, December 26, 2006 11:06 AM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] vectorizing an iterative process.
I meant x[i] - x[i-1] + y[i-1] and Y[i] - y[i-1] + x[i] below.
The mailing list software just keep adding 3D's. Sorry.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Geoffrey Zhu
Sent: Tuesday, December 26, 2006 10:03 AM
To: Richard M. Heiberger; r-help@stat.math.ethz.ch
Subject: Re: [R] vectorizing an iterative process.
Hi Richard,
3D is automatically generated by the mailing list software, probably
because I had ] followed by =3D3D3D without a space in the original post.
What I meant was to calculate x[i] =3D3D3D x[i-1] + y[i-1]
For example, if X - 1:10
Then I want the vector Y to be 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, or
in other words Y[i] =3D3D3D y[i-1] + x[i].
Yes, cumsum does the trick for this. This is what I need. Thanks.
Just curious, do you know how to calculate the more generic x[i] - f(
x[i-1], y[i-1] )?
Thanks,
Geoffrey
-Original Message-
From: Richard M. Heiberger [mailto:[EMAIL PROTECTED]
Sent: Tuesday, December 26, 2006 9:56 AM
To: Geoffrey Zhu; r-help@stat.math.ethz.ch
Subject: Re: [R] vectorizing an iterative process.
x[i]=3D3D3D3Dx[i-1]+y[i-1] for all i, how can I do this without a loop?
It looks like
x - cumsum(y)
What does 3D mean?
___=3D3D0A=3D3D
=3D3D0A=3D3D
=3D3D0A=3D3D
The information in this email or in any file attached hereto is=3D3D0A=3D3D
intended only for the personal and confidential use of the
individual=3D3D0A=3D3D or entity to which it is addressed and may contain
information that is=3D3D0A=3D3D proprietary and confidential. If you are not
the intended recipient of=3D3D0A=3D3D this message you are hereby notified
that any review, dissemination,=3D3D0A=3D3D distribution or copying of this
message is strictly prohibited. This communi=3D3D cation is for
information purposes only and should not be regarded as an off=3D3D er to
sell or as a solicitation of an offer to buy any financial product.
Em=3D3D ail transmission cannot be guaranteed to be secure or error-free.
__
R-help@stat.math.ethz.ch 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.
___=3D0A=3D
=3D0A=3D
=3D0A=3D
The information in this email or in any file attached hereto is=3D0A=3D
intended only for the personal and confidential use of the
individual=3D0A=3D or entity to which it is addressed and may contain
information that is=3D0A=3D proprietary and confidential. If you are not the
intended recipient of=3D0A=3D this message you are hereby notified that any
review, dissemination,=3D0A=3D distribution or copying of this message is
strictly prohibited. This communi=3D cation is for information purposes
only and should not be regarded as an off=3D er to sell or as a
solicitation of an offer to buy any financial product. Em=3D ail
transmission cannot be guaranteed to be secure or error-free.
__
R-help@stat.math.ethz.ch 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.
___=0A=
=0A=
=0A=
The information in this email or in any file attached hereto is=0A=
intended only for the personal and confidential use of the individual=0A=
or entity to which it is addressed and may contain information that is=0A=
proprietary and confidential. If you are not the intended recipient of=0A=
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[R] Rating competitors
I am looking for hints on how to estimate ratings for competitors in an ongoing pairwise competition using R... my particular area of interest being the game of Go, but the idea of identifying ratings (on a continuous scale) rather than relative rankings seems easily generalized to other competitions so I thought someone might be studying something related already. I presume the rating of a competitor would be best modeled as a random variate on the rating scale, and an encounter between two competitors would be represented by a binary result. Logistic regression seems promising, but I am at a loss how to represent the model since the pairings are arbitrary and not necessarily repeated often. I have read about some approaches to estimating ratings for Go, but they seem to involve optimization using assumed distributions rather than model fitting which characterizes analysis in R. Does any of this sound familiar? Suggestions for reading, anyone? -- --- Jeff NewmillerThe . . Go Live... DCN:[EMAIL PROTECTED]Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k __ R-help@stat.math.ethz.ch 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] vectorizing an iterative process.
Geoffrey Zhu wrote: I meant x[i] - x[i-1] + y[i-1] and Y[i] - y[i-1] + x[i] below. The mailing list software just keep adding 3D's. Sorry. Rather, I suspect that *your* mailer is sending in Quoted-Printable, without setting the appropriate headers. Take a look at http://mail.python.org/pipermail/mailman-users/2003-December/033425.html -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Geoffrey Zhu Sent: Tuesday, December 26, 2006 10:03 AM To: Richard M. Heiberger; r-help@stat.math.ethz.ch Subject: Re: [R] vectorizing an iterative process. Hi Richard, 3D is automatically generated by the mailing list software, probably because I had ] followed by =3D3D without a space in the original post. What I meant was to calculate x[i] =3D3D x[i-1] + y[i-1] For example, if X - 1:10 Then I want the vector Y to be 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, or in other words Y[i] =3D3D y[i-1] + x[i]. Yes, cumsum does the trick for this. This is what I need. Thanks. Just curious, do you know how to calculate the more generic x[i] - f( x[i-1], y[i-1] )? Thanks, Geoffrey -Original Message- From: Richard M. Heiberger [mailto:[EMAIL PROTECTED] Sent: Tuesday, December 26, 2006 9:56 AM To: Geoffrey Zhu; r-help@stat.math.ethz.ch Subject: Re: [R] vectorizing an iterative process. x[i]=3D3D3Dx[i-1]+y[i-1] for all i, how can I do this without a loop? It looks like x - cumsum(y) What does 3D mean? ___=3D0A=3D =3D0A=3D =3D0A=3D The information in this email or in any file attached hereto is=3D0A=3D intended only for the personal and confidential use of the individual=3D0A=3D or entity to which it is addressed and may contain information that is=3D0A=3D proprietary and confidential. If you are not the intended recipient of=3D0A=3D this message you are hereby notified that any review, dissemination,=3D0A=3D distribution or copying of this message is strictly prohibited. This communi=3D cation is for information purposes only and should not be regarded as an off=3D er to sell or as a solicitation of an offer to buy any financial product. Em=3D ail transmission cannot be guaranteed to be secure or error-free. __ R-help@stat.math.ethz.ch 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. ___=0A= =0A= =0A= The information in this email or in any file attached hereto is=0A= intended only for the personal and confidential use of the individual=0A= or entity to which it is addressed and may contain information that is=0A= proprietary and confidential. If you are not the intended recipient of=0A= this message you are hereby notified that any review, dissemination,=0A= distribution or copying of this message is strictly prohibited. This communi= cation is for information purposes only and should not be regarded as an off= er to sell or as a solicitation of an offer to buy any financial product. Em= ail transmission cannot be guaranteed to be secure or error-free. __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Rating competitors
Have you considered Bradley-Terry models? RSiteSearch(bradley, functions) just returned 31 hits for me. Hope this helps. Spencer Graves Jeff Newmiller wrote: I am looking for hints on how to estimate ratings for competitors in an ongoing pairwise competition using R... my particular area of interest being the game of Go, but the idea of identifying ratings (on a continuous scale) rather than relative rankings seems easily generalized to other competitions so I thought someone might be studying something related already. I presume the rating of a competitor would be best modeled as a random variate on the rating scale, and an encounter between two competitors would be represented by a binary result. Logistic regression seems promising, but I am at a loss how to represent the model since the pairings are arbitrary and not necessarily repeated often. I have read about some approaches to estimating ratings for Go, but they seem to involve optimization using assumed distributions rather than model fitting which characterizes analysis in R. Does any of this sound familiar? Suggestions for reading, anyone? __ R-help@stat.math.ethz.ch 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] Rating competitors
There is a substantial literature on 'statistics in sports' and pairwise comparisons are of obvious interest. Here is a starting point: http://www.amstat.org/sections/sis/ You might browse the newsletters posted there. You might enjoy: Bridging Different Eras in Sports by Scott M. Berry, Patrick D. Larkey, C. Shane Reese; Journal of the American Statistical Association, Vol. 94, 1999 or Baseball's All-Time Best Hitters: How Statistics Can Level the Playing Field by Michael J. Schell http://press.princeton.edu/titles/6550.html On Tue, 26 Dec 2006, Jeff Newmiller wrote: I am looking for hints on how to estimate ratings for competitors in an ongoing pairwise competition using R... my particular area of interest being the game of Go, but the idea of identifying ratings (on a continuous scale) rather than relative rankings seems easily generalized to other competitions so I thought someone might be studying something related already. I presume the rating of a competitor would be best modeled as a random variate on the rating scale, and an encounter between two competitors would be represented by a binary result. Logistic regression seems promising, but I am at a loss how to represent the model since the pairings are arbitrary and not necessarily repeated often. I have read about some approaches to estimating ratings for Go, but they seem to involve optimization using assumed distributions rather than model fitting which characterizes analysis in R. Does any of this sound familiar? Suggestions for reading, anyone? -- --- Jeff NewmillerThe . . Go Live... DCN:[EMAIL PROTECTED]Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k __ R-help@stat.math.ethz.ch 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. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0717 __ R-help@stat.math.ethz.ch 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] slightly inconsistent behavior
dear R experts: This is just a minor, minor nuisance, but I thought I would point it out: dataset - read.table(file=pipe(cmdline), header =T, + na.strings=c(NaN, C,I,M, E), sep=,, as.is=T, nrows=); Error: cannot allocate vector of size 781249 Kb If I extend nrows by a few more 9's, the error goes away. Similarly, if I use much fewer observations, the error goes away. regards, /iaw __ R-help@stat.math.ethz.ch 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] Rating competitors
I would start with elimination-by-aspects models: ?eba I would read the Tversky 1972 paper (cited on the help page for the eba() function), which is brilliant. Jeff Newmiller wrote: I am looking for hints on how to estimate ratings for competitors in an ongoing pairwise competition using R... my particular area of interest being the game of Go, but the idea of identifying ratings (on a continuous scale) rather than relative rankings seems easily generalized to other competitions so I thought someone might be studying something related already. I presume the rating of a competitor would be best modeled as a random variate on the rating scale, and an encounter between two competitors would be represented by a binary result. Logistic regression seems promising, but I am at a loss how to represent the model since the pairings are arbitrary and not necessarily repeated often. _ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400Charlottesville, VA 22904-4400 Parcels:Room 102Gilmer Hall McCormick RoadCharlottesville, VA 22903 Office:B011+1-434-982-4729 Lab:B019+1-434-982-4751 Fax:+1-434-982-4766 WWW:http://www.people.virginia.edu/~mk9y/ __ R-help@stat.math.ethz.ch 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] Rating competitors
One approach that is already coded in R is the Bradley-Terry model (found in the BradleyTerry package of all places). This could be a good place to start if you want something quick, others have given you references if you want more detail and/or control. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jeff Newmiller Sent: Tuesday, December 26, 2006 10:54 AM To: r-help@stat.math.ethz.ch Subject: [R] Rating competitors I am looking for hints on how to estimate ratings for competitors in an ongoing pairwise competition using R... my particular area of interest being the game of Go, but the idea of identifying ratings (on a continuous scale) rather than relative rankings seems easily generalized to other competitions so I thought someone might be studying something related already. I presume the rating of a competitor would be best modeled as a random variate on the rating scale, and an encounter between two competitors would be represented by a binary result. Logistic regression seems promising, but I am at a loss how to represent the model since the pairings are arbitrary and not necessarily repeated often. I have read about some approaches to estimating ratings for Go, but they seem to involve optimization using assumed distributions rather than model fitting which characterizes analysis in R. Does any of this sound familiar? Suggestions for reading, anyone? -- --- Jeff NewmillerThe . . Go Live... DCN:[EMAIL PROTECTED]Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] McNemar test in R SPSS
Hello, I am hoping someone can clarify why I might obtain a quite different value in R SPSS for a McNemar test I ran. Firstly, here is the R syntax output R OUTPUT mctest - as.table(matrix(c(128,29,331,430), + ncol =2, dimnames = list(group=c(preMHT,postMHT), + assault=c(yes,no mctest assault group yes no preMHT 128 331 postMHT 29 430 mcnemar.test(mctest,correct=F) McNemar's Chi-squared test data: mctest McNemar's chi-squared = 253.3444, df = 1, p-value 2.2e-16 SPSS OUTPUT The same data was inputted in SPSS. Regarding the first table SPSS generated - it lists the number of cases in each combination of categories. The diagonal contains the number of cases with the same response on both variables, while the off diagonal contains cases that have different responses on the 2 variables. The overall chisquare value is much lower than the value obtained using R, though still significant. pre02 post02 pre02 post02 0 1 0 311 20 1 119 9 Test Statistics(b) pre02 post02 N 459 Chi-Square(a) 69.094 Asymp. Sig. .000 a Continuity Corrected b McNemar Test Any assistance is much appreciated, Bob Green __ R-help@stat.math.ethz.ch 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] McNemar test in R SPSS
Bob Green wrote: Hello, I am hoping someone can clarify why I might obtain a quite different value in R SPSS for a McNemar test I ran. Firstly, here is the R syntax output R OUTPUT mctest - as.table(matrix(c(128,29,331,430), + ncol =2, dimnames = list(group=c(preMHT,postMHT), + assault=c(yes,no mctest assault group yes no preMHT 128 331 postMHT 29 430 mcnemar.test(mctest,correct=F) McNemar's Chi-squared test data: mctest McNemar's chi-squared = 253.3444, df = 1, p-value 2.2e-16 SPSS OUTPUT The same data was inputted in SPSS. Regarding the first table SPSS generated - it lists the number of cases in each combination of categories. The diagonal contains the number of cases with the same response on both variables, while the off diagonal contains cases that have different responses on the 2 variables. The overall chisquare value is much lower than the value obtained using R, though still significant. pre02 post02 pre02 post02 0 1 0 311 20 1 119 9 Test Statistics(b) pre02 post02 N 459 Chi-Square(a) 69.094 Asymp. Sig. .000 a Continuity Corrected b McNemar Test Any assistance is much appreciated, Well, you can't expect R to give the correct result if you feed it the wrong matrix, can you? d - read.table(stdin()) 0: 0 1 1: 0 311 20 2: 1 119 9 3: d X0 X1 0 311 20 1 119 9 mcnemar.test(as.matrix(d),correct=F) McNemar's Chi-squared test data: as.matrix(d) McNemar's chi-squared = 70.5108, df = 1, p-value 2.2e-16 mcnemar.test(as.matrix(d),correct=T) McNemar's Chi-squared test with continuity correction data: as.matrix(d) McNemar's chi-squared = 69.0935, df = 1, p-value 2.2e-16 Bob Green __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] extend summary.lm for hccm?
On Mon, 25 Dec 2006, Achim Zeileis wrote: On Sun, 24 Dec 2006, John Fox wrote: If I remember Freedman's recent paper correctly, he argues that sandwich variance estimator, though problematic in general, is not problematic in the case that White described -- an otherwise correctly specified linear model with heteroscedasticity estimated by least-squares. More generally, sandwich-type estimators are valid (i.e., estimate the right quantity, although not necessarily precisely, as Frank pointed out) in situations where the estimating functions are correctly specified but remaining aspets of the likelihood (not captured in the estimating functions) are potentially not. Not exactly. The asymptotic properites are good, but in samples of moderate size the properties (including both biasedness and variance) can be surprisingly bad. And if you are trying to calculate a p-value, getting a too-small variance estimate gives you spuriously small p-values. FWIW, I've seen a case in which the nominal size of a test based on the sandwich estimator was several orders of magnitude smaller than a test with correct nominal size. There is a modest literature on this. Some refs: Background Papers: Drum M, McCullagh P. Comment. Statistical Science 1993; 8:300-301. Freedman DA. On the So-Called Huber Sandwich Estimator and Robust Standard Errors. The American Statistician, Volume 60, Number 4, November 2006, pp. 299-302(4) - Some Proposed Corrections: Fay MP and Graubard BI. Small-Sample Adjustments for Walt-Type Tests Using Sandwich Estimators. Biometrics 2001; 57: 1198-1206. Guo X, Pan W, Connett JE, Hannan PJ and French SA. Small-sample performance of the robust score test and its modifications in generalized estimating equations Statistics in Medicine 2005; 24:3479-3495 Mancl LA and DeRouen TA, A Covariance Estimator for GEE with Improved Small-Sample Properties. Biometrics 2001; 57:126-134. Morel JG, Bokossa MC, and Neerchal NK. Small Sample Correction for the Variance of GEE Estimators Biometrical Journal 2003; 45(4): 395-409. Pan W and Wall MM. Small-sample adjustments in using the sandwich variance estimator in generalized estimating equations. Statistics in Medicine 2002; 21:1429-1441. HTH, Chuck In linear models, it is easy to see what this means: the mean function has to be correctly specified (i.e., the errors have zero mean) but the correlation structure of the errors (i.e., their (co-)variances) might differ from the usual assumptions. In GLMs, in particular logistic regression, it is much more difficult to see against which types of misspecification sandwich-based inference is robust. Freedman's paper stresses the point that many model misspecifications also imply misspecified estimating functions (and in his example this is rather obvious) so that consequently the sandwich-type estimators estimate the wrong quantity. Best wishes, Z __ R-help@stat.math.ethz.ch 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. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0717 __ R-help@stat.math.ethz.ch 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] McNemar test in R SPSS
Peter, Thanks for your reply. To perform the analysis in R, I used the table format suggested in the book by Everitt Hothorn, whereas in SPSS the analysis was performed directly from the 2 variables, rather than using count data. There is still something I don't understand - I tried to replicate the syntax from your e-mail but my computer just kept waiting - what am I still doing wrong? mctest - as.table(matrix(c(128,29,331,430), ncol =2, dimnames = list(group=c(preMHT,postMHT), assault=c(yes,no d - read.table(stdin()) 0: mcnemar.test(as.matrix(d),correct=F) 1: Thanks again Bob At 10:52 PM 26/12/2006 +0100, Peter Dalgaard wrote: Bob Green wrote: Hello, I am hoping someone can clarify why I might obtain a quite different value in R SPSS for a McNemar test I ran. Firstly, here is the R syntax output R OUTPUT mctest - as.table(matrix(c(128,29,331,430), + ncol =2, dimnames = list(group=c(preMHT,postMHT), + assault=c(yes,no mctest assault group yes no preMHT 128 331 postMHT 29 430 mcnemar.test(mctest,correct=F) McNemar's Chi-squared test data: mctest McNemar's chi-squared = 253.3444, df = 1, p-value 2.2e-16 SPSS OUTPUT The same data was inputted in SPSS. Regarding the first table SPSS generated - it lists the number of cases in each combination of categories. The diagonal contains the number of cases with the same response on both variables, while the off diagonal contains cases that have different responses on the 2 variables. The overall chisquare value is much lower than the value obtained using R, though still significant. pre02 post02 pre02 post02 0 1 0 311 20 1 119 9 Test Statistics(b) pre02 post02 N 459 Chi-Square(a) 69.094 Asymp. Sig. .000 a Continuity Corrected b McNemar Test Any assistance is much appreciated, Well, you can't expect R to give the correct result if you feed it the wrong matrix, can you? d - read.table(stdin()) 0: 0 1 1: 0 311 20 2: 1 119 9 3: d X0 X1 0 311 20 1 119 9 mcnemar.test(as.matrix(d),correct=F) McNemar's Chi-squared test data: as.matrix(d) McNemar's chi-squared = 70.5108, df = 1, p-value 2.2e-16 mcnemar.test(as.matrix(d),correct=T) McNemar's Chi-squared test with continuity correction data: as.matrix(d) McNemar's chi-squared = 69.0935, df = 1, p-value 2.2e-16 Bob Green __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] McNemar test in R SPSS
Bob Green wrote: Peter, Thanks for your reply. To perform the analysis in R, I used the table format suggested in the book by Everitt Hothorn, whereas in SPSS the analysis was performed directly from the 2 variables, rather than using count data. You still need the right table. matrix(c(128,29,331,430), ncol =2,) consists of the two marginal tables, which has strictly less information than the crosstabulation of pre and post values. I expect every text on the McNemar test makes this point, and I'd be highly surprised if EH really suggested that you should use the table that you did. There is still something I don't understand - I tried to replicate the syntax from your e-mail but my computer just kept waiting - what am I still doing wrong? You need to give it the data when it starts with the 0:-style prompt, end with a blank line, as shown. It's just a device for cutting and pasting your table. I might as well have used d - matrix(c(311,119,20,9), ncol=2) Or, with the raw data to hand d - table(preMHT, postMHT) mctest - as.table(matrix(c(128,29,331,430), ncol =2, dimnames = list(group=c(preMHT,postMHT), assault=c(yes,no d - read.table(stdin()) 0: mcnemar.test(as.matrix(d),correct=F) 1: Thanks again Bob At 10:52 PM 26/12/2006 +0100, Peter Dalgaard wrote: Bob Green wrote: Hello, I am hoping someone can clarify why I might obtain a quite different value in R SPSS for a McNemar test I ran. Firstly, here is the R syntax output R OUTPUT mctest - as.table(matrix(c(128,29,331,430), + ncol =2, dimnames = list(group=c(preMHT,postMHT), + assault=c(yes,no mctest assault group yes no preMHT 128 331 postMHT 29 430 mcnemar.test(mctest,correct=F) McNemar's Chi-squared test data: mctest McNemar's chi-squared = 253.3444, df = 1, p-value 2.2e-16 SPSS OUTPUT The same data was inputted in SPSS. Regarding the first table SPSS generated - it lists the number of cases in each combination of categories. The diagonal contains the number of cases with the same response on both variables, while the off diagonal contains cases that have different responses on the 2 variables. The overall chisquare value is much lower than the value obtained using R, though still significant. pre02 post02 pre02 post02 0 1 0 311 20 1 119 9 Test Statistics(b) pre02 post02 N 459 Chi-Square(a) 69.094 Asymp. Sig. .000 a Continuity Corrected b McNemar Test Any assistance is much appreciated, Well, you can't expect R to give the correct result if you feed it the wrong matrix, can you? d - read.table(stdin()) 0: 0 1 1: 0 311 20 2: 1 119 9 3: d X0 X1 0 311 20 1 119 9 mcnemar.test(as.matrix(d),correct=F) McNemar's Chi-squared test data: as.matrix(d) McNemar's chi-squared = 70.5108, df = 1, p-value 2.2e-16 mcnemar.test(as.matrix(d),correct=T) McNemar's Chi-squared test with continuity correction data: as.matrix(d) McNemar's chi-squared = 69.0935, df = 1, p-value 2.2e-16 Bob Green __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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@stat.math.ethz.ch 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] xyplot line colors
Hello, I have a longitudinal data with about 30 subjects. I used xyplot() to plot the longitudinal data. One problem is that xyplot() recycles the color of auto.key so that every 7th subject has the same color (symbol if setps() was used). Is there a way so that every subject will have a unique color or symbol? Thanks Osman -- Osman O. Al-Radi, MD, MSc, FRCSC Fellow, Cardiovascular Surgery The Hospital for Sick Children University of Toronto, Canada [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] xyplot line colors
Set the colors yourself using the par.settings= argument of xyplot or trellis.par.set. See ?xyplot and ?trellis.par.set for more info. If you follow the instructions on the last line of every post to r-help you will get more detailed answers. On 12/26/06, Osman Al-Radi [EMAIL PROTECTED] wrote: Hello, I have a longitudinal data with about 30 subjects. I used xyplot() to plot the longitudinal data. One problem is that xyplot() recycles the color of auto.key so that every 7th subject has the same color (symbol if setps() was used). Is there a way so that every subject will have a unique color or symbol? Thanks Osman -- Osman O. Al-Radi, MD, MSc, FRCSC Fellow, Cardiovascular Surgery The Hospital for Sick Children University of Toronto, Canada [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Bayesian data mining
Frank, You might want to use a package that is associated with some textbook that can give you guidance and examples. A lot of people would do this with WinBUGS, which is usually run separately although it has some kind of R interface. I'm sure there are lots of nice texts about that package. You might have a look at the bayesm package. For that, you can find lots of examples in a text on Bayes in marketing (Rossi et al.) David Farrar New River Analytic 540-818-7373 Frank Grex [EMAIL PROTECTED] wrote: Hi, I need a help to know whether I can perform the following in R: I have a set of observations (Ns) and each observation is drawn from a poisson distribution with an unkown mean, lambda. The set of lambdas in their turn are drawn from a common prior distribution which is supposed to be a a mixture of two gamma distributions. Is there a way to determine the poisson means in R, given the Ns and the probabilities? And how can one determine the two alphas and the two betas of the gamma mixtures? I am assuming there will be an MLE somewhere. This is to help me understand and apply William DuMouchel concept of datamining especially in his article: Bayesian data mining in large frequency tables. Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] plotting time series with zoo pckg
Hi all, I am using the zoo package to plot time series. I have a problem with formatting the axes. my zoo object (z) looks like the following. c1 1992-01-10 21 1992-01-17 34 1992-01-24 33 1992-01-31 41 1992-02-07 39 1992-02-14 38 1992-02-21 37 1992-02-28 28 1992-03-06 33 1992-03-13 40 plot.zoo(z) produces a plot with the labels on the x-axis that I cannot control. I want a an xtick every 10 data points with corresponding date labels. I have tried different combination of axis command without success any idea? Thanks __ R-help@stat.math.ethz.ch 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] plotting time series with zoo pckg
Try this: # test data library(zoo) z - structure(c(21, 34, 33, 41, 39, 38, 37, 28, 33, 40), index = structure(c(8044, 8051, 8058, 8065, 8072, 8079, 8086, 8093, 8100, 8107), class = Date), class = zoo) z # plot without X axis plot(z, xaxt = n) # unlabelled tick at each point axis(1, time(z), lab = FALSE) # labelled tick every third point dd - time(z)[seq(1, length(z), 3)] axis(1, dd, as.character(dd), cex.axis = 0.7, tcl = -0.7) On 12/26/06, ahmad ajakh [EMAIL PROTECTED] wrote: Hi all, I am using the zoo package to plot time series. I have a problem with formatting the axes. my zoo object (z) looks like the following. c1 1992-01-10 21 1992-01-17 34 1992-01-24 33 1992-01-31 41 1992-02-07 39 1992-02-14 38 1992-02-21 37 1992-02-28 28 1992-03-06 33 1992-03-13 40 plot.zoo(z) produces a plot with the labels on the x-axis that I cannot control. I want a an xtick every 10 data points with corresponding date labels. I have tried different combination of axis command without success any idea? Thanks __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] plotting time series with zoo pckg
Dear Gabor, Thank you for your quick reply. This solution works for my univariate zoo class time series. I first tried it for a timeseries with 4 columns of data, it did not plot the labels nor the ticks, I tried it on a one dim timeseries (one column zoo class data as the example in the question) and it worked! is there something that I am missing? Thanks again. AA. - Original Message From: Gabor Grothendieck [EMAIL PROTECTED] To: ahmad ajakh [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Tuesday, December 26, 2006 8:31:07 PM Subject: Re: [R] plotting time series with zoo pckg Try this: # test data library(zoo) z - structure(c(21, 34, 33, 41, 39, 38, 37, 28, 33, 40), index = structure(c(8044, 8051, 8058, 8065, 8072, 8079, 8086, 8093, 8100, 8107), class = Date), class = zoo) z # plot without X axis plot(z, xaxt = n) # unlabelled tick at each point axis(1, time(z), lab = FALSE) # labelled tick every third point dd - time(z)[seq(1, length(z), 3)] axis(1, dd, as.character(dd), cex.axis = 0.7, tcl = -0.7) On 12/26/06, ahmad ajakh [EMAIL PROTECTED] wrote: Hi all, I am using the zoo package to plot time series. I have a problem with formatting the axes. my zoo object (z) looks like the following. c1 1992-01-10 21 1992-01-17 34 1992-01-24 33 1992-01-31 41 1992-02-07 39 1992-02-14 38 1992-02-21 37 1992-02-28 28 1992-03-06 33 1992-03-13 40 plot.zoo(z) produces a plot with the labels on the x-axis that I cannot control. I want a an xtick every 10 data points with corresponding date labels. I have tried different combination of axis command without success any idea? Thanks __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] writing to S-PLUS .dat file
Hi Sebastian try save(x,y, file = filename.txt, ascii = TRUE) where x, y are your R objects. You should read ?save good luck AA. - Original Message From: Sebastian Michalski [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Monday, December 25, 2006 7:36:50 AM Subject: [R] writing to S-PLUS .dat file Dear Users, I am new to R. I use write() to write my data in .txt format. I'd like to write to a disc any kind of data in a .dat S-PLUS format. Please help. SM __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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 transform string to variable name in a fuction?
there is a data frame, like this:
df
aa bb
1 a 20.27802
2 b 22.10664
3 c 21.33470
4 a 22.32898
5 b 19.73760
6 c 20.38979
.(suppressed)
what I want to do is to copy the data frame's rows into different data frames
according to the levels of 'aa' column,
df.a - df[df[,1]=='a',] ; df.b - df[df[,1]=='b',] ;
df.a
aa bb
1 a 20.27802
4 a 22.32898
...
So, when completed, there should be df.a, df.b,df.c, etc.
If we could do this by hand, it is pretty fine. But could I write a loop to do
this ?
when I tried this using a funciton, there is a problem.
for ( i in levels(df[,1])) {
+ name = paste('df',i,sep='')
+ name - df[df[,1]==i,]
+ }
name
aa bb
3 c 21.33470
6 c 20.38979
ls()
[1] df iname
i
[1] c
there is not data frames df.a, df.b,etc.
Could you please give me some suggestion?
I have found that write a function in R for a beginner is difficult. Is there
any tutorial on writing the functions in R?
Furthermore, someone also said that loop is not used as frequently as in other
script language (e.g. bash, perl). So, If you have any other smart means do
this more efficiently, please let me know, I would appreciate your kindness.
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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 transform string to variable name in a fuction?
try ?assign
On 12/26/06, jingjiangyan [EMAIL PROTECTED] wrote:
there is a data frame, like this:
df
aa bb
1 a 20.27802
2 b 22.10664
3 c 21.33470
4 a 22.32898
5 b 19.73760
6 c 20.38979
.(suppressed)
what I want to do is to copy the data frame's rows into different data frames
according to the levels of 'aa' column,
df.a - df[df[,1]=='a',] ; df.b - df[df[,1]=='b',] ;
df.a
aa bb
1 a 20.27802
4 a 22.32898
...
So, when completed, there should be df.a, df.b,df.c, etc.
If we could do this by hand, it is pretty fine. But could I write a loop to
do this ?
when I tried this using a funciton, there is a problem.
for ( i in levels(df[,1])) {
+ name = paste('df',i,sep='')
+ name - df[df[,1]==i,]
+ }
name
aa bb
3 c 21.33470
6 c 20.38979
ls()
[1] df iname
i
[1] c
there is not data frames df.a, df.b,etc.
Could you please give me some suggestion?
I have found that write a function in R for a beginner is difficult. Is there
any tutorial on writing the functions in R?
Furthermore, someone also said that loop is not used as frequently as in
other script language (e.g. bash, perl). So, If you have any other smart
means do this more efficiently, please let me know, I would appreciate your
kindness.
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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.
--
WenSui Liu
A lousy statistician who happens to know a little programming
(http://spaces.msn.com/statcompute/blog)
__
R-help@stat.math.ethz.ch 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] plotting time series with zoo pckg
Please read the last line of every message to r-help and follow that. On 12/26/06, ahmad ajakh [EMAIL PROTECTED] wrote: Dear Gabor, Thank you for your quick reply. This solution works for my univariate zoo class time series. I first tried it for a timeseries with 4 columns of data, it did not plot the labels nor the ticks, I tried it on a one dim timeseries (one column zoo class data as the example in the question) and it worked! is there something that I am missing? Thanks again. AA. - Original Message From: Gabor Grothendieck [EMAIL PROTECTED] To: ahmad ajakh [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Tuesday, December 26, 2006 8:31:07 PM Subject: Re: [R] plotting time series with zoo pckg Try this: # test data library(zoo) z - structure(c(21, 34, 33, 41, 39, 38, 37, 28, 33, 40), index = structure(c(8044, 8051, 8058, 8065, 8072, 8079, 8086, 8093, 8100, 8107), class = Date), class = zoo) z # plot without X axis plot(z, xaxt = n) # unlabelled tick at each point axis(1, time(z), lab = FALSE) # labelled tick every third point dd - time(z)[seq(1, length(z), 3)] axis(1, dd, as.character(dd), cex.axis = 0.7, tcl = -0.7) On 12/26/06, ahmad ajakh [EMAIL PROTECTED] wrote: Hi all, I am using the zoo package to plot time series. I have a problem with formatting the axes. my zoo object (z) looks like the following. c1 1992-01-10 21 1992-01-17 34 1992-01-24 33 1992-01-31 41 1992-02-07 39 1992-02-14 38 1992-02-21 37 1992-02-28 28 1992-03-06 33 1992-03-13 40 plot.zoo(z) produces a plot with the labels on the x-axis that I cannot control. I want a an xtick every 10 data points with corresponding date labels. I have tried different combination of axis command without success any idea? Thanks __ R-help@stat.math.ethz.ch 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. __ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com __ R-help@stat.math.ethz.ch 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 transform string to variable name in a fuction?
In the following the components of ss are the data frames in question:
ss - split(df, df$aa)
On 12/26/06, jingjiangyan [EMAIL PROTECTED] wrote:
there is a data frame, like this:
df
aa bb
1 a 20.27802
2 b 22.10664
3 c 21.33470
4 a 22.32898
5 b 19.73760
6 c 20.38979
.(suppressed)
what I want to do is to copy the data frame's rows into different data frames
according to the levels of 'aa' column,
df.a - df[df[,1]=='a',] ; df.b - df[df[,1]=='b',] ;
df.a
aa bb
1 a 20.27802
4 a 22.32898
...
So, when completed, there should be df.a, df.b,df.c, etc.
If we could do this by hand, it is pretty fine. But could I write a loop to
do this ?
when I tried this using a funciton, there is a problem.
for ( i in levels(df[,1])) {
+ name = paste('df',i,sep='')
+ name - df[df[,1]==i,]
+ }
name
aa bb
3 c 21.33470
6 c 20.38979
ls()
[1] df iname
i
[1] c
there is not data frames df.a, df.b,etc.
Could you please give me some suggestion?
I have found that write a function in R for a beginner is difficult. Is there
any tutorial on writing the functions in R?
Furthermore, someone also said that loop is not used as frequently as in
other script language (e.g. bash, perl). So, If you have any other smart
means do this more efficiently, please let me know, I would appreciate your
kindness.
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] McNemar test in R SPSS
Peter, I now see the original E H table was based on matched pairs not the raw counts. I now understand this much better and have the syntax generates results that correspond with your results (and SPSS), Thanks again, Bob __ R-help@stat.math.ethz.ch 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] Rating competitors
Spencer Graves wrote: Have you considered Bradley-Terry models? RSiteSearch(bradley, functions) just returned 31 hits for me. Hope this helps. Spencer Graves Thanks to everyone who responded... this was very helpful. I have a bit of reading and investigation to do. I think the Bradley-Terry model is going to be sufficient for my purposes, if I can figure out how to model handicaps. The eba library mentioned by Kubovy seems more complex than what I need now, but it does look interesting and if I can obtain a copy of the Tversky paper I will read it. The SIS link mentioned by Berry didn't seem to have much, but the article on Bridging Different Eras in Sports is quite interesting. Jeff Newmiller wrote: I am looking for hints on how to estimate ratings for competitors in an ongoing pairwise competition using R... my particular area of interest being the game of Go, but the idea of identifying ratings (on a continuous scale) rather than relative rankings seems easily generalized to other competitions so I thought someone might be studying something related already. I presume the rating of a competitor would be best modeled as a random variate on the rating scale, and an encounter between two competitors would be represented by a binary result. Logistic regression seems promising, but I am at a loss how to represent the model since the pairings are arbitrary and not necessarily repeated often. I have read about some approaches to estimating ratings for Go, but they seem to involve optimization using assumed distributions rather than model fitting which characterizes analysis in R. Does any of this sound familiar? Suggestions for reading, anyone? -- --- Jeff NewmillerThe . . Go Live... DCN:[EMAIL PROTECTED]Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k __ R-help@stat.math.ethz.ch 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.
