[R] CLI Issue

2006-04-07 Thread A Mani
Hello,
 I have a script that must be executed by R at the CLI completely
... without halting execution on encountering an error (object not found).
Rcmdr works fine, but source behaves strangely in R-2.1...

Thanks,

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A. Mani
Member, Cal. Math. Soc

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[R] glm automation

2006-03-07 Thread A Mani
Hello,
I have two problems in automating multiple glm(s) operations.
The data file is tab delimited file with headers and two columns. like

ABC  EFG
1  2
2  3
3  4
dat - read.table(FILENAME, header=TRUE, sep=\t, na.strings=NA,
dec=., strip.white=TRUE)
dataf - read.table(FILENAME, header=FALSE, sep=\t, na.strings=NA,
dec=., strip.white=TRUE)
norm1 - glm(dataf[1,1] ~ dataf[1,2], family= normal(log), data=dat)
norm2 - glm(dataf[1,1] ~ dataf[1,2], family= normal(identity), data=dat)
and so on.

But glm does not work on the data unless I write ABC and EFG there... I want
to automate the script for multiple files.

The other problem is to write the plot(GLM) to a file without  displaying it
at stdout.

Thanks,


A. Mani
Member, Cal. Math. Soc

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[R] Strange Expression

2006-03-03 Thread A Mani
Hello,
 I have a dataframe with a column 'col' of expressions of the form
2*3+4*5+6*7 and substrings thereof.
Now
eval(2*3+4*5+6*7) is ok.
But I want it done on that column... nothing seems to work
Even
 cil - dataframe$col
eval(cil[1])
2*3+4*5+6*7

I want the elements of the column evaluated.

Thanks,

A. Mani
Member, Cal. Math. Soc

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Re: [R] Strange Expression

2006-03-03 Thread A Mani
On 3/3/06, ¨Tariq Khan [EMAIL PROTECTED] wrote:

 Did you try eval(parse(text = cil[1])) ?

 yes, function (x) eval(parse(x)) works within the *apply  commands.

thanks,
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Member, Cal. Math. Soc

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[R] Help with ldBands(Hmisc)

2006-02-02 Thread Lakshminarayanan, Mani
Hello My Fellow Users,

Under the details it is given that the ld98 executable should be in a 
subdirectory that is in the system.  Should ld98.exe be included under Hmisc 
(where it is stored) or under the lib subdirectory within Hmisc?  I am getting 
the following error message: Error in (head + 1):length(w) : NA/NaN argument.  
Does this mean that I didn't store the ld98.exe in the right place or something 
else?  Please help.

Mani

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[R] Clustering Large Data Sets

2005-10-03 Thread A Mani
Hello,
Suppose the problem is to cluster each document in a set of
'n' documents (.csv) separately by some clustering method. If 'n' is
very large, then is there a method of increasing speed and efficiency
by treating the set as a document vector or as a set of document
vectors ?

Thanks

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[R] Unpaste Problem

2005-08-26 Thread A Mani
Hello,
Easy ways to unpaste?
 xp - paste(x2, x3) # x2, x3 are two non-numeric columns.
.
.
xfg - data.frame(xp,sc1, sc2, sc3) # sc1,sc2, sc3 are numeric cols.

I want xp to be split up to form a new dataframe of the form (x3, sc1,
sc2, sc3).
IMPORTANT info : elements of xp have the form abcspaceefg, with abc
in x2 and efg in x3.

Thanks in advance, 
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Member, Cal. Math. Soc

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[R] A. Mani : Tapply

2005-08-25 Thread A Mani
Hello,
Is it safe to use tapply when the result will be of dim 2
x 1 or more ? In my PC R crashes. The code used was on a 3-col
data frame with two factor cols and a numeric column. The fn was diff
.
data form being A, B, Num
tapply(data$A, list(data$A, data$B), diff)

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Member, Cal. Math. Soc

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[R] Re : A. Mani : Avoiding Loops

2005-08-19 Thread A Mani
On Friday 19 August 2005 11:54, Sean O'Riordain wrote:
 Hi,
 I'm not sure what you actually want from your email (following the
 posting guide is a good way of helping you explain things to the rest
 of us in a way we understand - it might even answer your question!

 I'm only a beginner at R so no doubt one of our expert colleagues will
 help me...

  fred - data.frame()
  fred - edit(fred)
  fred

   A B C D E
 1 1 2 X Y 1
 2 2 3 G L 1
 3 3 1 G L 5

  fred[,3]

 [1] X G G
 Levels: G X

  fred[fred[,3]==G,]

   A B C D E
 2 2 3 G L 1
 3 3 1 G L 5

 so at this point I can create a new dataframe with column 3 (C) ==
 G; either explicitly or implicitly...

 and if I want to calculate the sum() of column E, then I just say
 something like...

  sum(fred[fred[,3]==G,][,5])

 [1] 6


 now naturally being a bit clueless at manipulating stuff in R, I
 didn't know how to do this before I started... and you guys only get
 to see the lines that I typed in and got a successful result...

 according to section 6 of the Introduction to R manual which comes
 with R, I could also have said

  sum(fred[fred$C==G,]$E)

 [1] 6

 Hmmm I wonder would it be reasonable to put an example of this
 type into section 2.7 of the Introduction to R?


 cheers!
 Sean

 On 18/08/05, A. Mani [EMAIL PROTECTED] wrote:
  Hello,
  I want to avoid loops in the following situation. There is a
  5-col dataframe with col headers alone. two of the columns are
  non-numeric. The problem is to calculate statistics(scores) for each
  element of one column. The functions depend on matching in the other
  non-numeric column.
 
  A  B  C  E  F
  1  2  X  Y  1
  2  3  G  L  1
  3  1  G  L  5
  and so on ...3+ entries.
 
  I need scores for col E entries which depend on conditional implications.
 
 
  Thanks,
 
Hello,
  Sorry about the incomplete problem. Here is a better version for the
problem: (the measure is not simple)
The data frame is like
  col1   col2col3   col4col5
  num  nonum   nonum  num   num
   A   B C  E   F   
There are repeated strings in col3, col2.
Problem : Calculate  Measure(Ci) = [No. of repeats of Ci *100] + [If (Bi, Ci) 
is same as (Bj, Cj) and 6= Ej - Ei =3 then add 100 else  10] .


Actually it is to stretched further by adding similar blocks.

 How do we use *apply or
something else in the situation  ?


In prolog it is extremely easy, but here it is not quite...


A. Mani
Member, Cal. Math. Soc

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A. Mani
Member, Cal. Math. Soc

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[R] A. Mani : Avoiding loops

2005-08-18 Thread A. Mani
Hello,
I want to avoid loops in the following situation. There is a 5-col 
dataframe with col headers alone. two of the columns are non-numeric. The 
problem is to calculate statistics(scores) for each element of one column. 
The functions depend on matching in the other non-numeric column.

A  B  C  E  F
1  2  X  Y  1
2  3  G  L  1
3  1  G  L  5
and so on ...3+ entries.

I need scores for col E entries which depend on conditional implications.


Thanks,

A. Mani
Member, Cal. Math. Soc

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Re: [R] R-help Digest, Vol 28, Issue 28

2005-06-28 Thread A. Mani
On Tuesday 28 June 2005 15:30, [EMAIL PROTECTED] wrote:
Re :   37. Re: A. Mani : colours in Silhouette (Mulholland, Tom)
   
 Message: 37
 Date: Tue, 28 Jun 2005 09:08:24 +0800
 From: Mulholland, Tom [EMAIL PROTECTED]
 Subject: Re: [R] A. Mani : colours in Silhouette
 To: [EMAIL PROTECTED], r-help@stat.math.ethz.ch
 Message-ID:
  [EMAIL PROTECTED]
 Content-Type: text/plain; charset=iso-8859-1

 It's not so much a problem, as not working the way you expected.
 cluster:::plot.partition is used to do the plotting. If you look at the
 code for this you can see the difficulty in putting every possible
 permutation into the code. If for example you want the silhouette plot to
 be red using col = red is not intuitive as the cluster plot (which comes
 up first) has more than one colour. If you have a look at methods(plot)
 (assuming that you have loaded the cluster package) you will see that there
 is a specific piece of code in the form of plot.silhouette. It has an
 asterisk next to it so you need to use cluster:::plot.silhouette to see the
 code. It has what you need.

 args(cluster:::plot.silhouette)

  function (x, nmax.lab = 40, max.strlen = 5, main = NULL, sub = NULL,

 xlab = expression(Silhouette width  * s[i]), col = gray,
 do.col.sort = length(col)  1, border = 0, cex.names = par(cex.axis),
 do.n.k = TRUE, do.clus.stat = TRUE, ...)


  data(ruspini)
   pr4 - pam(ruspini, 4)
   si - silhouette(pr4)
   plot(si,col = red)

I tried that before with many more options and got a blank image. It must have 
been due to the options.
 The issue is that whenever code is written there is always a choice as to
 what functionality is put in place. Just because something can be done,
 does not mean it will or in some cases should be done. In this case the
 help for plot.partition notes that For more flexibility, use
 'plot(silhouette(x), ...)', see 'plot.silhouette'.

 Tom

 Thanks for that I found out something I will find useful in the future.

  -Original Message-
  From: [EMAIL PROTECTED]
  [mailto:[EMAIL PROTECTED] Behalf Of A. Mani
  Sent: Tuesday, 28 June 2005 4:30 AM
  To: r-help@stat.math.ethz.ch
  Subject: [R] A. Mani : colours in Silhouette
 
 
  Hello,
 In cluster analysis with cluster, how does one colour
  the silhouette
  plots ? For example in using pam. There seems to be some
  problem there.
  Everything else can be coloured.
 
  Thanks,
 

 A. Mani
 Member, Cal. Math. Soc

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[R] A. Mani : colours in Silhouette

2005-06-27 Thread A. Mani
Hello,
   In cluster analysis with cluster, how does one colour the silhouette 
plots ? For example in using pam. There seems to be some problem there. 
Everything else can be coloured.

Thanks,

A. Mani
Member, Cal. Math. Soc

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[R] A. Mani : Read

2005-06-23 Thread A. Mani
Hello,
   What is the best way to read a .rdat file with data in the following 
form :

$xyz1
column names
columns(numeric)

$xyz2
column names
columns(numeric)

$xyz3
column names
columns(numeric)

and so on.

A. Mani
Member, Cal. Math. Soc

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