date:20100227

Re: [R] somebody help me about this error message...

2010-02-27 Thread Allan Engelhardt


You forgot the assign the second time:

assign(paste(a,2,sep=), 4)

does what you want.

Hope this helps a little

Allan.

On 27/02/10 05:13, Joseph Lee wrote:

I created variables automatically like this way

for(i in 1:5){
nam- paste(a,i,sep=)
assign(nam,1:i)
}

and then, i want to insert a new data into a2 variable. so, i did next
sentence

paste(a,2,sep=)- 4

so, i got this error message

Error in get(paste(a, 2, sep = ))[1]- 4 :
   target of assignment expands to non-language object

anyone knows abou this error message and tell me how to solve thie problem,
please..



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Re: [R] Help Computing Probit Marginal Effects

2010-02-27 Thread Ted Harding

On 27-Feb-10 03:52:19, Cardinals_Fan wrote:
 
 Hi,  I am a stata user trying to transition to R.  Typically I
 compute marginal effects plots for (example) probit models by
 drawing simulated betas by using the coefficient/standard error
 estimates after I run a probit model. I then use these simulated
 betas to compute first difference marginal effects.  My question
 is, can I do this in R?  Specifically, I was wondering if anyone
 knows how R stores the coefficient/standard error estimates after
 you estimate the model?  I assume it's  vector, but what is it
 called?
 
 Cheers
 --

Here is an example which sets up (X,Y) data using a probit mechaism,
then fits a probit model, and then extracts the information which
you seek.

  set.seed(54321)
  X - 0.2*(-10:10)
  U - rnorm(21)
  Y - 1*(U = X)  ## binary outcome 0/1, = 1 if N(0,1) = X
  GLM  - glm(Y ~ X, family=binomial(link=probit)) ## fit a probit
  Coef - summary(GLM)$coef  ## apply summary() to the fit

GLM is a list with a large number of components: enter the command

  str(GLM)

and have a look at what you get! Only a few of these are displayed
when you apply print() to it:

  print(GLM)
  # Call:  glm(formula = Y ~ X, family = binomial(link = probit)) 
  # Coefficients:
  # (Intercept)X  
  # 0.08237  0.56982  
  # 
  # Degrees of Freedom: 20 Total (i.e. Null);  19 Residual
  # Null Deviance:  29.06 
  # Residual Deviance: 23.93AIC: 27.93 

Note that you do *not* get Standard Errors from this.

However, all the information in GLM is available for processing
by other functions. In particular, summary(GLM) produces another
list with several components -- have a look at the output from

  str(summary(GLM))

One of these components (listed near the end of this output)
is coef, and it can be accessed as summary(GLM)$coef as in the
above command

  Coef - summary(GLM)$coef

This is a matrix (in this case 2 named rows, 4 named columns):

  Coef
  #  Estimate Std. Error   z value   Pr(|z|)
  # (Intercept) 0.0823684  0.2974595 0.2769063 0.78185207
  # X   0.5698200  0.2638657 2.1595076 0.03081081

So there is one row for each coefficient in the model (here 2,
one for Intercept, one for variable X), and four columns
(for the Estimate itself of the coefficient, for its Standard
Error, for the z-value (Est/SE), and for the P-value).

Hence you can access the estimates as

  Coef[,1]   # (the first column of the matrix)
  # (Intercept)   X 
  #   0.0823684   0.5698200 

and their respective SEs as

  Coef[,2]   # (the second column of the matrix)
  # (Intercept)   X 
  #   0.2974595   0.2638657 

I have spelled this out in detail to demonstrate that the key
to accessing information in objects constructed by R lies in
its structures (especially lists, vectors and matrices). You
can find out what is involved for any function by looking for
the section Value in its help page. For instance, the function
summary() when applied to a GLM uses the method summary.glm(),
so you can enter the command

  ?summary.glm

and then read what is in the section Value. This shows that
it is a list with components whose names are

  call, family, deviance, ... , coefficients, ... , symbolic.cor

and a component with name Name can be accessed using $Name as
in GLM$coef (you can use coef instead of coefficients since
the first four letters are [more than] enough to identify the name
uniquely).

Once you get used to this, things become straightforward!
Ted.


E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 27-Feb-10   Time: 08:38:41
-- XFMail --

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Re: [R] Preserving lists in a function

2010-02-27 Thread baptiste auguie

Hi,

I think I would follow this approach too, using updatelist() from the
reshape package,


updatelist - function (x, y)
{
common - intersect(names(x), names(y))
x[common] - y[common]
x
}

myfunction=function(list1=NULL, list2=NULL, list3=NULL){
   list1=updatelist(list(variable1=1,
 variable2=2,
 variable3=3), list1)

   list2=updatelist(list(variable1=variable1,
 variable2=variable2,
 variable3=variable3), list2)

   list3=updatelist(list(variable1=character,
 variable2=24,
 variable3=c(0.1,0.1,0.1,0.1),
 variable4=TRUE), list3)

   return(list(list1=list1,list2=list2,list3=list3))

 }


Best regards,

baptiste

On 27 February 2010 01:51, Don MacQueen m...@llnl.gov wrote:
 Barry explained your first puzzle, but let  me add some explanation and
 examples.


  tmpfun - function( a =3 ) {a}
  tmpfun()

 [1] 3

  tmpfun(a='x')

 [1] x

 Inside the function, the value of the argument is whatever the user
 supplied. The default is replaced by what the user supplies. There is no
 mechanism for retaining the default structure and filling in any missing
 parts. R never preserves the defaults when the user supplies something other
 than the default.

 For example, and using your function,

  myfunction(list1='x')

 $list1
 [1] x

 $list2
 $list2$variable1
 [1] variable1

 $list2$variable2
 [1] variable2

 $list2$variable3
 [1] variable3


 $list3
 $list3$variable1
 [1] character

 $list3$variable2
 [1] 24

 $list3$variable3
 [1] 0.1 0.1 0.1 0.1

 $list3$variable4
 [1] TRUE


  myfunction(list1=data.frame(a=1:2, b=c('x','y')))

 $list1
  a b
 1 1 x
 2 2 y

 $list2
 $list2$variable1
 [1] variable1

 $list2$variable2
 [1] variable2

 $list2$variable3
 [1] variable3


 $list3
 $list3$variable1
 [1] character

 $list3$variable2
 [1] 24

 $list3$variable3
 [1] 0.1 0.1 0.1 0.1

 $list3$variable4
 [1] TRUE

 What you put in is what you get out.

 I don't know that I would deal with this the way Barry did. I would probably
 write code to examine the structure of what the user supplies, compare it to
 the required structure, and then fill in.

 myf - function(l1, l2, l3) {
  if (missing(l1)) {
   ## user did not supply l1, so set it = to the default
    l1 - list(v1=1, v2=2, v3=3)
  }  else if (!is.list(l1)) {
   ## user must supply a list, if not, it's an error
   stop('l1 must be a list')
 } else {
   ## user has at least supplied a list
   ## now write code to check the names of the list that the user supplied
   ## make sure the names that the user supplied are valid, if not, stop()
   ## if the user supplied too few elements, fill in the missing ones
   ## if the user supplied too many elements stop()
   ## if the user supplied all the correct elements, with all the correct
 names, use what the user supplied
 }

 Looks complicated; maybe Barry's way is better...

 -Don

 At 5:56 PM -0500 2/26/10, Shang Gao wrote:

 Dear R users,

 A co-worker and I are writing a function to facilitate graph plotting in
 R. The function makes use of a lot of lists in its defaults.

 However, we discovered that R does not necessarily preserve the defaults
 if we were to input them in the form of list() when initializing the
 function. For example, if you feed the function codes below into R:

 myfunction=function(
    list1=list  (variable1=1,
                variable2=2,
                variable3=3),

    list2=list  (variable1=variable1,
                variable2=variable2,
                variable3=variable3),

    list3=list  (variable1=character,
                variable2=24,
                variable3=c(0.1,0.1,0.1,0.1),
                variable4=TRUE))

 {return(list(list1=list1,list2=list2,list3=list3))}

 By definition, the values associated with each variable in the lists would
 be the default unless the user impute a different value while executing the
 function. But a problem arises when a variable in the list is left out
 completely (not imputed at all). An example is shown below:

 myfunction( list1=list  (variable1=1,
                        variable2=2), #variable 3 deliberately left out

            list2=list  (variable1=variable1,
                        variable3=position changed,
                        variable2=variable2),

            list3=list  (variable1=character,
                        variable2=24,
                        variable4=FALSE)) #variable 3 deliberately left out

 #The outcome of the above execution is shown below:

 $list1
 $list1$variable1
 [1] 1

 $list1$variable2
 [1] 2
 #list1$variable3 is missing. Defaults in function not assigned in this
 execution

 $list2
 $list2$variable1
 [1] variable1

 $list2$variable3
 [1] position changed

 $list2$variable2
 [1] variable2


 $list3
 $list3$variable1
 [1] character

 $list3$variable2
 [1] 24

 $list3$variable4
 [1] FALSE
 #list3$variable3 is missing. Defaults in function not assigned in this
 execution

 We later realized that the problem lies in list() commands. Hence, we
 tried to enforce the defaults on

Re: [R] Error in mvpart example

2010-02-27 Thread Gavin Simpson

These functions (rpart, the mvpart wrapper, and summary.rpart) are
fairly complex doing many things.

For contributed packages you'd be best served by contacting the
author/maintainer. I've CC'd Glenn (the maintainer) here.

HTH

G

On Fri, 2010-02-26 at 13:55 +, Wearn, Oliver wrote:
 Dear all,
 
 I'm getting an error in one of the stock examples in the 'mvpart'
 package. I tried:
 
 require(mvpart)
 data(spider)
 fit3 - rpart(gdist(spider[,1:12],meth=bray,full=TRUE,sq=TRUE)~water
 +twigs+reft+herbs+moss+sand,spider,method=dist) #directly
 from ?rpart
 summary(fit3)
 
 ...which returned the following:
 
 Error in apply(formatg(yval, digits - 3), 1, paste, collapse = ,,
 sep = ) : 
   dim(X) must have a positive length
 
 This seems to be a problem with the cross-validation, since the
 xerror and xstd columns are missing from the summary table as
 well.
 
 Using the mpart() wrapper results in the same error:
 
 fit4-mvpart(gdist(spider[,1:12],meth=bray,full=TRUE,sq=TRUE)~water
 +twigs+reft+herbs+moss+sand,spider,method=dist)
 summary(fit4)
 
 Note, changing the 'method' argument to =mrt seems, superficially,
 to solve the problem. However, when the dependent variable is a
 dissimilarity matrix, shouldn't method=dist be used (as per the
 examples)?
 
 Thanks, in advance, for any help on this error.
 
 Oliver
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 Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/
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Re: [R] counting the number of ones in a vector

2010-02-27 Thread Gavin Simpson

On Fri, 2010-02-26 at 10:43 -0800, David Reinke wrote:
 The length will remain the same no matter what expression appears in
 the subscript.

No it won't! x == 1 evaluates to logical and when used to *subset* x, it
*will* return the required answer. As observed with this example:

 set.seed(1)
 x - sample(rep(1:3, times = 20))
 x
 [1] 1 1 1 1 3 2 3 3 3 1 2 3 3 1 2 2 2 1 3 2 2 1 1 2 1 2 1 1 1 1
[31] 3 3 2 3 2 2 2 3 3 3 1 3 3 1 3 2 1 1 2 2 1 1 3 2 2 3 2 2 3 3
 
 ## compare
 sum(x == 1, na.rm = TRUE)
[1] 20
 length(x)
[1] 60
 length(x[x == 1])
[1] 20

G

  I suggest this:
 
 sum(x == 1)
 
 David Reinke
 
 Senior Transportation Engineer/Economist
 Dowling Associates, Inc.
 180 Grand Avenue, Suite 250
 Oakland, California 94612-3774
 510.839.1742 x104 (voice)
 510.839.0871 (fax)
 www.dowlinginc.com
 
  Please consider the environment before printing this e-mail.
 
 Confidentiality Notice:  This e-mail message, including any attachments, is 
 for the sole use of the intended recipient(s), and may contain confidential  
 and privileged information. Any unauthorized review, use, disclosure or 
 distribution is prohibited. If you are not the intended recipient, please 
 contact the sender by reply e-mail and destroy all copies of the original 
 message.
 
 
 
 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf Of Randall Wrong
 Sent: Friday, February 26, 2010 6:44 AM
 To: r-help@r-project.org
 Subject: [R] counting the number of ones in a vector
 
  Dear R users,
 
 I want to count the number of ones in a vector x.
 
 That's what I did : length( x[x==1] )
 
 Is that a good solution ?
 
 Thank you very much,
 Randall
 
   [[alternative HTML version deleted]]
 
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 ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
 Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/
 UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
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Re: [R] Preserving lists in a function

2010-02-27 Thread Gabor Grothendieck

Or use modifyList which is in the core of R.

On Sat, Feb 27, 2010 at 5:22 AM, baptiste auguie
baptiste.aug...@googlemail.com wrote:
 Hi,

 I think I would follow this approach too, using updatelist() from the
 reshape package,


 updatelist - function (x, y)
 {
    common - intersect(names(x), names(y))
    x[common] - y[common]
    x
 }

 myfunction=function(list1=NULL, list2=NULL, list3=NULL){
   list1=updatelist(list(variable1=1,
     variable2=2,
     variable3=3), list1)

   list2=updatelist(list(variable1=variable1,
     variable2=variable2,
     variable3=variable3), list2)

   list3=updatelist(list(variable1=character,
     variable2=24,
     variable3=c(0.1,0.1,0.1,0.1),
     variable4=TRUE), list3)

   return(list(list1=list1,list2=list2,list3=list3))

  }


 Best regards,

 baptiste

 On 27 February 2010 01:51, Don MacQueen m...@llnl.gov wrote:
 Barry explained your first puzzle, but let  me add some explanation and
 examples.


  tmpfun - function( a =3 ) {a}
  tmpfun()

 [1] 3

  tmpfun(a='x')

 [1] x

 Inside the function, the value of the argument is whatever the user
 supplied. The default is replaced by what the user supplies. There is no
 mechanism for retaining the default structure and filling in any missing
 parts. R never preserves the defaults when the user supplies something other
 than the default.

 For example, and using your function,

  myfunction(list1='x')

 $list1
 [1] x

 $list2
 $list2$variable1
 [1] variable1

 $list2$variable2
 [1] variable2

 $list2$variable3
 [1] variable3


 $list3
 $list3$variable1
 [1] character

 $list3$variable2
 [1] 24

 $list3$variable3
 [1] 0.1 0.1 0.1 0.1

 $list3$variable4
 [1] TRUE


  myfunction(list1=data.frame(a=1:2, b=c('x','y')))

 $list1
  a b
 1 1 x
 2 2 y

 $list2
 $list2$variable1
 [1] variable1

 $list2$variable2
 [1] variable2

 $list2$variable3
 [1] variable3


 $list3
 $list3$variable1
 [1] character

 $list3$variable2
 [1] 24

 $list3$variable3
 [1] 0.1 0.1 0.1 0.1

 $list3$variable4
 [1] TRUE

 What you put in is what you get out.

 I don't know that I would deal with this the way Barry did. I would probably
 write code to examine the structure of what the user supplies, compare it to
 the required structure, and then fill in.

 myf - function(l1, l2, l3) {
  if (missing(l1)) {
   ## user did not supply l1, so set it = to the default
    l1 - list(v1=1, v2=2, v3=3)
  }  else if (!is.list(l1)) {
   ## user must supply a list, if not, it's an error
   stop('l1 must be a list')
 } else {
   ## user has at least supplied a list
   ## now write code to check the names of the list that the user supplied
   ## make sure the names that the user supplied are valid, if not, stop()
   ## if the user supplied too few elements, fill in the missing ones
   ## if the user supplied too many elements stop()
   ## if the user supplied all the correct elements, with all the correct
 names, use what the user supplied
 }

 Looks complicated; maybe Barry's way is better...

 -Don

 At 5:56 PM -0500 2/26/10, Shang Gao wrote:

 Dear R users,

 A co-worker and I are writing a function to facilitate graph plotting in
 R. The function makes use of a lot of lists in its defaults.

 However, we discovered that R does not necessarily preserve the defaults
 if we were to input them in the form of list() when initializing the
 function. For example, if you feed the function codes below into R:

 myfunction=function(
    list1=list  (variable1=1,
                variable2=2,
                variable3=3),

    list2=list  (variable1=variable1,
                variable2=variable2,
                variable3=variable3),

    list3=list  (variable1=character,
                variable2=24,
                variable3=c(0.1,0.1,0.1,0.1),
                variable4=TRUE))

 {return(list(list1=list1,list2=list2,list3=list3))}

 By definition, the values associated with each variable in the lists would
 be the default unless the user impute a different value while executing the
 function. But a problem arises when a variable in the list is left out
 completely (not imputed at all). An example is shown below:

 myfunction( list1=list  (variable1=1,
                        variable2=2), #variable 3 deliberately left out

            list2=list  (variable1=variable1,
                        variable3=position changed,
                        variable2=variable2),

            list3=list  (variable1=character,
                        variable2=24,
                        variable4=FALSE)) #variable 3 deliberately left out

 #The outcome of the above execution is shown below:

 $list1
 $list1$variable1
 [1] 1

 $list1$variable2
 [1] 2
 #list1$variable3 is missing. Defaults in function not assigned in this
 execution

 $list2
 $list2$variable1
 [1] variable1

 $list2$variable3
 [1] position changed

 $list2$variable2
 [1] variable2


 $list3
 $list3$variable1
 [1] character

 $list3$variable2
 [1] 24

 $list3$variable4
 [1] FALSE
 #list3$variable3 is missing. Defaults in

Re: [R] two questions for R beginners

2010-02-27 Thread Johannes Huesing

Dieter Menne dieter.me...@menne-biomed.de [Fri, Feb 26, 2010 at 08:39:14AM 
CET]:
 
 
 Patrick Burns wrote:
  
  * What were your biggest misconceptions or
  stumbling blocks to getting up and running
  with R?
  
  
 (This derives partly from teaching)
 
[...]
 
 The concept of environment. With S it was worse, though.
 

Agreed, though a beginner shouldn't be exposed to this aspect.
In the beginning you can analyse away before you are drowning
in variable names if you start with simple examples.

Which plotting parameters can be passed with basic plot functions,
and which ones have to be declared with par()? How do I 
set the min and max values for the x and y axis? (This 
aspect is drowned among all the options under ?par.)

Generally, the help pages are built like man pages, where
all options are given more or less equal consideration, even
if one option is used almost always and one only for esoteric
purposes. Given that help() is the most intuitive thing to
look for, it may be nice to include references to other
sources like rwiki if the respective page is good, even if
it may be disruptive wrt display device.

-- 
Johannes Hüsing   There is something fascinating about science. 
  One gets such wholesale returns of conjecture 
mailto:johan...@huesing.name  from such a trifling investment of fact.  
  
http://derwisch.wikidot.com (Mark Twain, Life on the Mississippi)

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help Computing Probit Marginal Effects

2010-02-27 Thread Peter Ehlers



On 2010-02-27 1:38, (Ted Harding) wrote:

On 27-Feb-10 03:52:19, Cardinals_Fan wrote:


Hi,  I am a stata user trying to transition to R.  Typically I
compute marginal effects plots for (example) probit models by
drawing simulated betas by using the coefficient/standard error
estimates after I run a probit model. I then use these simulated
betas to compute first difference marginal effects.  My question
is, can I do this in R?  Specifically, I was wondering if anyone
knows how R stores the coefficient/standard error estimates after
you estimate the model?  I assume it's  vector, but what is it
called?

Cheers
--


Here is an example which sets up (X,Y) data using a probit mechaism,
then fits a probit model, and then extracts the information which
you seek.

   set.seed(54321)
   X- 0.2*(-10:10)
   U- rnorm(21)
   Y- 1*(U= X)  ## binary outcome 0/1, = 1 if N(0,1)= X
   GLM- glm(Y ~ X, family=binomial(link=probit)) ## fit a probit
   Coef- summary(GLM)$coef  ## apply summary() to the fit

GLM is a list with a large number of components: enter the command

   str(GLM)

and have a look at what you get! Only a few of these are displayed
when you apply print() to it:

   print(GLM)
   # Call:  glm(formula = Y ~ X, family = binomial(link = probit))
   # Coefficients:
   # (Intercept)X
   # 0.08237  0.56982
   #
   # Degrees of Freedom: 20 Total (i.e. Null);  19 Residual
   # Null Deviance:  29.06
   # Residual Deviance: 23.93AIC: 27.93

Note that you do *not* get Standard Errors from this.

However, all the information in GLM is available for processing
by other functions. In particular, summary(GLM) produces another
list with several components -- have a look at the output from

   str(summary(GLM))

One of these components (listed near the end of this output)
is coef, and it can be accessed as summary(GLM)$coef as in the
above command

   Coef- summary(GLM)$coef

This is a matrix (in this case 2 named rows, 4 named columns):

   Coef
   #  Estimate Std. Error   z value   Pr(|z|)
   # (Intercept) 0.0823684  0.2974595 0.2769063 0.78185207
   # X   0.5698200  0.2638657 2.1595076 0.03081081

So there is one row for each coefficient in the model (here 2,
one for Intercept, one for variable X), and four columns
(for the Estimate itself of the coefficient, for its Standard
Error, for the z-value (Est/SE), and for the P-value).

Hence you can access the estimates as

   Coef[,1]   # (the first column of the matrix)
   # (Intercept)   X
   #   0.0823684   0.5698200

and their respective SEs as

   Coef[,2]   # (the second column of the matrix)
   # (Intercept)   X
   #   0.2974595   0.2638657

I have spelled this out in detail to demonstrate that the key
to accessing information in objects constructed by R lies in
its structures (especially lists, vectors and matrices). You
can find out what is involved for any function by looking for
the section Value in its help page. For instance, the function
summary() when applied to a GLM uses the method summary.glm(),
so you can enter the command

   ?summary.glm

and then read what is in the section Value. This shows that
it is a list with components whose names are

   call, family, deviance, ... , coefficients, ... , symbolic.cor

and a component with name Name can be accessed using $Name as
in GLM$coef (you can use coef instead of coefficients since
the first four letters are [more than] enough to identify the name
uniquely).


I would just add one suggestion to Ted's excellent tutorial:
R has the extractor function(s) coef() for getting the coefficients
(and SEs) for various types of models.

coef(GLM)
coef(summary(GLM))

While these will produce precisely the same output in the above
example, they may be the better way to go with, say, nonlinear
models. Using the first example in ?nls:

DNase1 - subset(DNase, Run == 1)
fm1DNase1 - nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1)

fm1DNase1$coef
# NULL  # - probably not what was expected

coef(fm1DNase1)
# Asym xmid scal
# 2.345180 1.483090 1.041455

Of course, looking at str(fm1DNase1) would show that there is no
component called coefficients, but it might take a bit of head
scratching to realize that component m has as a subcomponent
the getAllPars() function which produces the ouput given by
coef(fm1DNase1).

I would recommend using extractor funtions like coef(), resid(),
etc. where available.

  -Peter Ehlers


Once you get used to this, things become straightforward!
Ted.


E-Mail: (Ted Harding)ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 27-Feb-10   Time: 08:38:41
-- XFMail --

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[R] reading data from web data sources

2010-02-27 Thread Tim Coote

Hullo
I'm trying to read some time series data of meteorological records
that are available on the web (eg http://climate.arm.ac.uk/calibrated/soil/dsoil100_cal_1910-1919.dat)
. I'd like to be able to read in the digital data directly into R.
However, I cannot work out the right function and set of parameters to
use. It could be that the only practical route is to write a parser,
possibly in some other language, reformat the files and then read
these into R. As far as I can tell, the informal grammar of the file is:

comments terminated by a blank line
[year number on a line on its own
daily readings lines ]+

and the daily readings are of the form:
whitespace day number [whitespace reading on day of month] 12

Readings for days in months where a day does not exist have special
values. Missing values have a different special value.

And then I've got the problem of iterating over all relevant files to
get a whole timeseries.

Is there a way to read in this type of file into R? I've read all of
the examples that I can find, but cannot work out how to do it. I
don't think that read.table can handle the separate sections of data
representing each year. read.ftable maybe can be coerced to parse the
data, but I cannot see how after reading the documentation and
experimenting with the parameters.

I'm using R 2.10.1 on osx 10.5.8 and 2.10.0 on Fedora 10.

Any help/suggestions would be greatly appreciated. I can see that this
type of issue is likely to grow in importance, and I'd also like to
give the data owners suggestions on how to reformat their data so that
it is easier to consume by machines, while being easy to read for
humans.

The early records are a serious machine parsing challenge as they are
tiff images of old notebooks ;-)

tia

Tim
Tim Coote
t...@coote.org
vincit veritas

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