[R] Looping through a series of (csv) files
Dear Colleagues. This should be a very common operation and I believe there should be a nice way in R to handle it. I couldn't find it in the manual or by searching online. I am wondering if I could ask for some help in this community. I have 48 csv files; each stores the data for a specific month. The 48 corresponding months are consecutively from January, 2001 to December, 2004. I name the files A200101, A200102, .., A200112, A200201, ,etc. I want to process file A2000101 and store the result to a new file named B200101, process file A200102 and store the result to a new file named B200102 etc. I do not want to manually change a little bit of the code to read a different file and write to a different file every time. I want the program to be able to loop through all the files. My question is, how to loop through the 48 files? Your help will be highly appreciated! Best Wishes Yuchen Luo [[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] Send SMS out of R?
Now I want to send an SMS out of R! Any idea how it could work? Could I send an eMail to a mobile phone number? There are Email to SMS services you can use. Google gives plenty of them (also free ones (which I wouldn't use myself...)). -- Regards, Hans-Peter __ 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] Send SMS out of R?
On 7/14/07, Thomas Schwander [EMAIL PROTECTED] wrote: Hi everyone, Now I read the posting guidelines again; COMPLETELY! ;-) I use Windows XP Professional, R 2.5.1 and I have Blat to send eMails out of R. Works perfect! Thank you for your help! Now I want to send an SMS out of R! Any idea how it could work? Could I send an eMail to a mobile phone number? This might be a good place to start: http://en.wikipedia.org/wiki/SMS_gateways 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.
[R] Mailing List
Please add me to your mailing list. Thank you Alfred Inselberg -- Alfred Inselberg, Professor School of Mathematical Sciences Tel Aviv University Tel Aviv 61390, Israel Tel +972-3-640 5372 http://www.math.tau.ac.il/~aiisreal/ __ 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] Looping through a series of (csv) files
Yuchen Luo wrote:
Dear Colleagues.
This should be a very common operation and I believe there should be a nice
way in R to handle it. I couldn't find it in the manual or by searching
online. I am wondering if I could ask for some help in this community.
I have 48 csv files; each stores the data for a specific month. The 48
corresponding months are consecutively from January, 2001 to December, 2004.
I name the files A200101, A200102,….., A200112, A200201, ……,etc.
I want to process file A2000101 and store the result to a new file named
B200101, process file A200102 and store the result to a new file named
B200102… …etc.
I do not want to manually change a little bit of the code to read a
different file and write to a different file every time. I want the program
to be able to loop through all the files.
My question is, how to loop through the 48 files?
Your help will be highly appreciated!
Best Wishes
Yuchen Luo
This will read each of the 48 csv files, add 50 to the values in each
file, and then write each resulting data frame to its own new file:
for(i in 1:48){
tempdf - read.csv(paste(c:/myfolder/raw, i, .csv, sep=))
tempdf.new - tempdf + 50
write.table(tempdf.new,
file=paste(c:/myfolder/plus50-, i, .csv, sep=),
sep=,, row.names=FALSE)
}
rm(list=c(tempdf, tempdf.new, i))
Of course, you can do something more useful than adding 50.
[[alternative HTML version deleted]]
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Re: [R] unixtime conversion
Prof Brian Ripley [EMAIL PROTECTED] writes: On Sat, 14 Jul 2007, Patrick Drechsler wrote: is there an R function to convert unixtime to one of the R time formats (using chron or POSIXct)? Example data: unixtime year month day hour 1183377301 2007 7 2 13 I would like to only use the first column. See ?as.POSIXct, especially its examples. (Isn't that the very obvious place to look?) It seems you want ISOdatetime(1970,1,1,0,0,0) + unixtime although depending on the 'unix' in 'unixtime' you might have to wrorry about leap seconds (which POSIX-compliant systems ignore). Thank you very much for your reply! Indeed, ISOdatetime was the command I was looking for. Regards Patrick __ 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] Please Please Help me!!!
I posted the message below a few days ago but I have not had any responses. I keep thinking that there must be some easy way to answer the problem I am just not familiar enough with regression to answer the problem myself. If anyone can help me I would be very grateful. I need to fit a gamma curve to a set of data. ie a scatterplot of the data indicates that the curve looks like a truncated gamma density function and I would like to estimate the paramaters so that I can fit a curve to the data points. Its not MLE paramater estimation just a curve fitting exercise. The problem again is I have a set transition intensities and when plotted the curve looks like a gamma density. I want to fit a gamma density curve to these intensities. It is just a curve fitting problem but whats causing the trouble is that I need to use least squares minimization to calculate the parameters for the gamma curve. How do I do this??? The curve will be a truncated gamma function so it will have 3 paramaters a, b, c. I tried to do the following nls(lograte ~ log(c) + a*log(b) + (a-1)*log(age)+b*(age)-lgamma(a), start=list(a=1,b=1,c=1)), where lograte, and age are my data. a,b the gamma parameters and c the parameter we need because we are fitting a truncated distribution. I also tried defining fn = function(p) sum((log(y)-log(dgamma(x,p[1],p[2])*p[3]))^2) a residual sum of squares and using nlm to minimise this and find paramaters but this doesnt work either. Can anyone help me ?? Please :) Original message follows :( Fitting a Gamma Curve by TIMMMAY Jul 12, 2007; 03:38pm :: Rate this Message:(use ratings to moderate[?]) Reply | Reply to Author | Show Only this Message Hi there, I hope someone can help me before I tear all my hair out. I have a set transition intensities and when plotted the curve looks like a gamma density. I want to fit a gamma density curve to these intensities. It is just a curve fitting problem but whats causing the trouble is that I need to use least squares minimization to calculate the parameters for the gamma curve. How do I do this??? The curve will be a truncated gamma function so it will have 3 paramaters a, b, c. I tried to do the following nls(lograte ~ log(c) + a*log(b) + (a-1)*log(age)+b*(age)-lgamma(a), start=list(a=1,b=1,c=1)), where lograte, and age are my data. a,b the gamma parameters and c the parameter we need because we are fitting a truncated distribution. I also tried defining fn = function(p) sum((log(y)-log(dgamma(x,p[1],p[2])*p[3]))^2) a residual sum of squares and using nlm to minimise this and find paramaters but this doesnt work either. Can anyone help me ?? Please :) « Return to forum Start a free forum or mailing list archive on Nabble Help - Powered by - Terms of Use - Jobs at Nabble - Nabble Support -- View this message in context: http://www.nabble.com/Please-Please-Help-me%21%21%21-tf4081887.html#a11601686 Sent from the R help mailing list archive at Nabble.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.
[R] Complex surveys, properly computed SEs and non-parametric analyses
Can someone direct me to an R function that properly computes standard errors
of data obtained from a complex survery design, i.e. perform alnalyses similiar
to those that can be performed with SUDAAN, particularly for a non-parametric
one-way ANOVA, e.g. signed rank test?
Thanks,
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
Baltimore VA Medical Center GRECC,
University of Maryland School of Medicine Claude D. Pepper OAIC,
University of Maryland Clinical Nutrition Research Unit, and
Baltimore VA Center Stroke of Excellence
University of Maryland School of Medicine
Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
[EMAIL PROTECTED]
Confidentiality Statement:
This email message, including any attachments, is for the so...{{dropped}}
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[R] Area selection agorithms
Hi, Various algorithms have been developed for selecting efficient sets of areas for biodiversity conservation, given a list of the areas and the different species they contain (e.g. the algorithms used within the WorldMap software that is distributed by the Natural History Museum in London). Does anyone know if these algorithms are available through an R package please? Thanks very much in advance for any help. Best wishes, Des [[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] Complex surveys, properly computed SEs and non-parametric analyses
Dear John, Can someone direct me to an R function that properly computes standard errors of data obtained from a complex survery design i.e. perform alnalyses similiar to those that can be performed with SUDAAN, articularly for a non-parametric one-way ANOVA, e.g. signed rank test? The survey package of Thomas Lumley has very broad functionality for the analysis of data from complex sampling designs. Please find below the homepage of the package (which is available on CRAN): http://faculty.washington.edu/tlumley/survey/ I don't think non-parametric one-way ANOVA is implemented, but quoting http://faculty.washington.edu/tlumley/survey/survey-wss.pdf slide 101: Many features of the survey package result from requests from unsatisfied users. For new methods the most important information is a reference that gives sufficient detail for implementation. A data set is nice but not critical. HTH, Tobias -- Tobias Verbeke - Consultant Business Decision Benelux Rue de la révolution 8 1000 Brussels - BELGIUM +32 499 36 33 15 [EMAIL PROTECTED] __ 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] Please Please Help me!!!
The reason no one answered may be that you did not follow the last line to every r-help message or read and follow the posting guide. Its time consuming to develop a test environment and data needed to clarify and test the answer to a question. By providing data and reproducible code you reduce the amount of time it takes responders to answer thereby making it more likely someone will. On 7/15/07, TIMMMAY [EMAIL PROTECTED] wrote: I posted the message below a few days ago but I have not had any responses. I keep thinking that there must be some easy way to answer the problem I am just not familiar enough with regression to answer the problem myself. If anyone can help me I would be very grateful. I need to fit a gamma curve to a set of data. ie a scatterplot of the data indicates that the curve looks like a truncated gamma density function and I would like to estimate the paramaters so that I can fit a curve to the data points. Its not MLE paramater estimation just a curve fitting exercise. The problem again is I have a set transition intensities and when plotted the curve looks like a gamma density. I want to fit a gamma density curve to these intensities. It is just a curve fitting problem but whats causing the trouble is that I need to use least squares minimization to calculate the parameters for the gamma curve. How do I do this??? The curve will be a truncated gamma function so it will have 3 paramaters a, b, c. I tried to do the following nls(lograte ~ log(c) + a*log(b) + (a-1)*log(age)+b*(age)-lgamma(a), start=list(a=1,b=1,c=1)), where lograte, and age are my data. a,b the gamma parameters and c the parameter we need because we are fitting a truncated distribution. I also tried defining fn = function(p) sum((log(y)-log(dgamma(x,p[1],p[2])*p[3]))^2) a residual sum of squares and using nlm to minimise this and find paramaters but this doesnt work either. Can anyone help me ?? Please :) Original message follows :( Fitting a Gamma Curve by TIMMMAY Jul 12, 2007; 03:38pm :: Rate this Message:(use ratings to moderate[?]) Reply | Reply to Author | Show Only this Message Hi there, I hope someone can help me before I tear all my hair out. I have a set transition intensities and when plotted the curve looks like a gamma density. I want to fit a gamma density curve to these intensities. It is just a curve fitting problem but whats causing the trouble is that I need to use least squares minimization to calculate the parameters for the gamma curve. How do I do this??? The curve will be a truncated gamma function so it will have 3 paramaters a, b, c. I tried to do the following nls(lograte ~ log(c) + a*log(b) + (a-1)*log(age)+b*(age)-lgamma(a), start=list(a=1,b=1,c=1)), where lograte, and age are my data. a,b the gamma parameters and c the parameter we need because we are fitting a truncated distribution. I also tried defining fn = function(p) sum((log(y)-log(dgamma(x,p[1],p[2])*p[3]))^2) a residual sum of squares and using nlm to minimise this and find paramaters but this doesnt work either. Can anyone help me ?? Please :) « Return to forum Start a free forum or mailing list archive on Nabble Help - Powered by - Terms of Use - Jobs at Nabble - Nabble Support -- View this message in context: http://www.nabble.com/Please-Please-Help-me%21%21%21-tf4081887.html#a11601686 Sent from the R help mailing list archive at Nabble.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. __ 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] Pairlist of pairlsit assembly howto
Hy guys,
I'm trying something like this
minbins - list()
for (minute in sequence(3)) {
minbins[minute] - list(data=a,variable=b)
}
And it doesn't work ...
Warning messages:
1: number of items to replace is not a multiple of replacement length in:
minbins[minute] - list(data = a, variable = b)
2: number of items to replace is not a multiple of replacement length in:
minbins[minute] - list(data = a, variable = b)
3: number of items to replace is not a multiple of replacement length in:
minbins[minute] - list(data = a, variable = b)
What am I doing wrong and how to do this properly?
Thanks, Joh
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[R] Break during the recursion?
Hi,
Is it possible to break using if-condition during the recursive function?
Here is a function which almost works. It is for inorder-tree-walk.
iotw-function(v,i,Stack,Indexes) # input: a vector and the first index (1),
Stack=c(), Indexes=c().
{
print(Indexes)
# if (sum(i)==0) break # Doesn't work...
if (is.na(v[i])==FALSE is.null(unlist(v[i]))==FALSE)
{Stack=c(i,Stack); i=2*i; iotw(v,i,Stack,Indexes)}
Indexes=c(Indexes,Stack[1])
Stack=pop.stack(Stack)$vector
Indexes=c(Indexes,Stack[1])
i=2*Stack[1]+1
Stack=pop.stack(Stack)$vector
iotw(v,i,Stack,Indexes)
}
v=c(`-`,`+`,1,`^`,`^`,NA,NA,X,3,X,2)
Stack=c()
Indexes=c()
iotw(v,1,Stack,Indexes)
NULL
NULL
NULL
NULL
NULL
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
Error in if (is.na(v[i]) == FALSE is.null(unlist(v[i])) == FALSE) { :
argument is of length zero
Regards,
Atte Tenkanen
University of Turku, Finland
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Re: [R] Pairlist of pairlsit assembly howto
(This isn't important to your question, but those aren't pairlists.
Pairlists are rarely used in R code, except implicitly in the way R
stores parsed code.)
On 15/07/2007 10:00 AM, Johannes Graumann wrote:
Hy guys,
I'm trying something like this
minbins - list()
for (minute in sequence(3)) {
minbins[minute] - list(data=a,variable=b)
}
You want to use
minbins[[minute]] - list(data=a,variable=b)
The difference between [[ ]] and [ ] is that the former works on the
element, the latter works on a subset. So your version tried to change
a subset of length 1 into a subset of length 2, which generates the
warnings. You want to assign a list of length 2 as an element of minbins.
Duncan Murdoch
And it doesn't work ...
Warning messages:
1: number of items to replace is not a multiple of replacement length in:
minbins[minute] - list(data = a, variable = b)
2: number of items to replace is not a multiple of replacement length in:
minbins[minute] - list(data = a, variable = b)
3: number of items to replace is not a multiple of replacement length in:
minbins[minute] - list(data = a, variable = b)
What am I doing wrong and how to do this properly?
Thanks, Joh
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Re: [R] Break during the recursion?
Oh. I forgot one extra-function:
pop.stack-function(v){
if(length(v)==0){x=NA}
if(length(v)==1){x=v[1]; v=c()}
if(length(v)1){x=v[1]; v=v[2:length(v)]}
return(list(vector=v,x=x))
}
Atte
Hi,
Is it possible to break using if-condition during the recursive
function?
Here is a function which almost works. It is for inorder-tree-walk.
iotw-function(v,i,Stack,Indexes) # input: a vector and the first
index (1), Stack=c(), Indexes=c().
{
print(Indexes)
# if (sum(i)==0) break # Doesn't work...
if (is.na(v[i])==FALSE is.null(unlist(v[i]))==FALSE)
{Stack=c(i,Stack); i=2*i; iotw(v,i,Stack,Indexes)}
Indexes=c(Indexes,Stack[1])
Stack=pop.stack(Stack)$vector
Indexes=c(Indexes,Stack[1])
i=2*Stack[1]+1
Stack=pop.stack(Stack)$vector
iotw(v,i,Stack,Indexes)
}
v=c(`-`,`+`,1,`^`,`^`,NA,NA,X,3,X,2)
Stack=c()
Indexes=c()
iotw(v,1,Stack,Indexes)
NULL
NULL
NULL
NULL
NULL
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
Error in if (is.na(v[i]) == FALSE is.null(unlist(v[i])) == FALSE)
{ :
argument is of length zero
Regards,
Atte Tenkanen
University of Turku, Finland
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Break during the recursion?
On 15/07/2007 10:06 AM, Atte Tenkanen wrote:
Hi,
Is it possible to break using if-condition during the recursive function?
You can do
if (condition) return(value)
Here is a function which almost works. It is for inorder-tree-walk.
iotw-function(v,i,Stack,Indexes) # input: a vector and the first index (1),
Stack=c(), Indexes=c().
{
print(Indexes)
# if (sum(i)==0) break # Doesn't work...
if (sum(i)==0) return(NULL)
should work.
Duncan Murdoch
if (is.na(v[i])==FALSE is.null(unlist(v[i]))==FALSE)
{Stack=c(i,Stack); i=2*i; iotw(v,i,Stack,Indexes)}
Indexes=c(Indexes,Stack[1])
Stack=pop.stack(Stack)$vector
Indexes=c(Indexes,Stack[1])
i=2*Stack[1]+1
Stack=pop.stack(Stack)$vector
iotw(v,i,Stack,Indexes)
}
v=c(`-`,`+`,1,`^`,`^`,NA,NA,X,3,X,2)
Stack=c()
Indexes=c()
iotw(v,1,Stack,Indexes)
NULL
NULL
NULL
NULL
NULL
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
Error in if (is.na(v[i]) == FALSE is.null(unlist(v[i])) == FALSE) { :
argument is of length zero
Regards,
Atte Tenkanen
University of Turku, Finland
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Re: [R] Break during the recursion?
On 15/07/2007 10:06 AM, Atte Tenkanen wrote:
Hi,
Is it possible to break using if-condition during the recursive
function?
You can do
if (condition) return(value)
Here is a function which almost works. It is for inorder-tree-
walk.
iotw-function(v,i,Stack,Indexes) # input: a vector and the first
index (1), Stack=c(), Indexes=c().
{
print(Indexes)
# if (sum(i)==0) break # Doesn't work...
if (sum(i)==0) return(NULL)
should work.
Duncan Murdoch
Hmm - - - I'd like to save the Indexes-vector (in the example
c(8,4,9,2,10,5,11,1,3)) and stop, when it is ready.
The results are enclosed to the end.
Atte
if (is.na(v[i])==FALSE is.null(unlist(v[i]))==FALSE)
{Stack=c(i,Stack); i=2*i; iotw(v,i,Stack,Indexes)}
Indexes=c(Indexes,Stack[1])
Stack=pop.stack(Stack)$vector
Indexes=c(Indexes,Stack[1])
i=2*Stack[1]+1
Stack=pop.stack(Stack)$vector
iotw(v,i,Stack,Indexes)
}
v=c(`-`,`+`,1,`^`,`^`,NA,NA,X,3,X,2)
Stack=c()
Indexes=c()
iotw(v,1,Stack,Indexes)
NULL
NULL
NULL
NULL
NULL
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
Error in if (is.na(v[i]) == FALSE is.null(unlist(v[i])) ==
FALSE) { :
argument is of length zero
Regards,
Atte Tenkanen
University of Turku, Finland
iotw(v,1,Stack,Indexes)
NULL
NULL
NULL
NULL
NULL
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2 5 1 3
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2 5 1 3
[1] 4 2
[1] 4 2
[1] 4 2
[1] 4 2 10 5
[1] 4 2 10 5
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1 3
[1] 4 2 10 5 11 1 3
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1 3
[1] 4 2 10 5 11 1 3
[1] 4 2 10 5
[1] 4 2 10 5
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1 3
[1] 4 2 10 5 11 1 3
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1 3
[1] 4 2 10 5 11 1 3
[1] 4 2 5 1
[1] 4 2 5 1
[1] 4 2 5 1 3
[1] 4 2 5 1 3
[1] 2 1
[1] 2 1
[1] 2 1 3
[1] 2 1 3
[1] 1
[1] 1
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PLEASE do read the posting guide
Re: [R] Break during the recursion?
On 15/07/2007 10:33 AM, Atte Tenkanen wrote:
On 15/07/2007 10:06 AM, Atte Tenkanen wrote:
Hi,
Is it possible to break using if-condition during the recursive
function?
You can do
if (condition) return(value)
Here is a function which almost works. It is for inorder-tree-
walk.
iotw-function(v,i,Stack,Indexes) # input: a vector and the first
index (1), Stack=c(), Indexes=c().
{
print(Indexes)
# if (sum(i)==0) break # Doesn't work...
if (sum(i)==0) return(NULL)
should work.
Duncan Murdoch
Hmm - - - I'd like to save the Indexes-vector (in the example
c(8,4,9,2,10,5,11,1,3)) and stop, when it is ready.
This seems more like a problem with the design of your function than a
question about R. I can't really help you with that, because your
description of the problem doesn't make sense to me. What does it mean
to do an inorder tree walk on something that isn't a tree?
Duncan Murdoch
The results are enclosed to the end.
Atte
if (is.na(v[i])==FALSE is.null(unlist(v[i]))==FALSE)
{Stack=c(i,Stack); i=2*i; iotw(v,i,Stack,Indexes)}
Indexes=c(Indexes,Stack[1])
Stack=pop.stack(Stack)$vector
Indexes=c(Indexes,Stack[1])
i=2*Stack[1]+1
Stack=pop.stack(Stack)$vector
iotw(v,i,Stack,Indexes)
}
v=c(`-`,`+`,1,`^`,`^`,NA,NA,X,3,X,2)
Stack=c()
Indexes=c()
iotw(v,1,Stack,Indexes)
NULL
NULL
NULL
NULL
NULL
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
Error in if (is.na(v[i]) == FALSE is.null(unlist(v[i])) ==
FALSE) { :
argument is of length zero
Regards,
Atte Tenkanen
University of Turku, Finland
iotw(v,1,Stack,Indexes)
NULL
NULL
NULL
NULL
NULL
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2 5 1 3
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2 5 1 3
[1] 4 2
[1] 4 2
[1] 4 2
[1] 4 2 10 5
[1] 4 2 10 5
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1 3
[1] 4 2 10 5 11 1 3
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1 3
[1] 4 2 10 5 11 1 3
[1] 4 2 10 5
[1] 4 2 10 5
[1] 4 2 10 5 11 1
[1] 4 2 10 5 11 1
[1] 4 2 10 5
Re: [R] Break during the recursion?
On 15/07/2007 10:33 AM, Atte Tenkanen wrote:
On 15/07/2007 10:06 AM, Atte Tenkanen wrote:
Hi,
Is it possible to break using if-condition during the recursive
function?
You can do
if (condition) return(value)
Here is a function which almost works. It is for inorder-tree-
walk.
iotw-function(v,i,Stack,Indexes) # input: a vector and the
first
index (1), Stack=c(), Indexes=c().
{
print(Indexes)
# if (sum(i)==0) break # Doesn't work...
if (sum(i)==0) return(NULL)
should work.
Duncan Murdoch
Hmm - - - I'd like to save the Indexes-vector (in the example
c(8,4,9,2,10,5,11,1,3)) and stop, when it is ready.
This seems more like a problem with the design of your function
than a
question about R. I can't really help you with that, because your
description of the problem doesn't make sense to me. What does it
mean
to do an inorder tree walk on something that isn't a tree?
Duncan Murdoch
The symbols in vector v have been originally derived from tree. See
http://users.utu.fi/attenka/Tree.jpg
But perhaps there's another way to do this, for instance by using loops and
if-conditions?
Atte
The results are enclosed to the end.
Atte
if (is.na(v[i])==FALSE is.null(unlist(v[i]))==FALSE)
{Stack=c(i,Stack); i=2*i; iotw(v,i,Stack,Indexes)}
Indexes=c(Indexes,Stack[1])
Stack=pop.stack(Stack)$vector
Indexes=c(Indexes,Stack[1])
i=2*Stack[1]+1
Stack=pop.stack(Stack)$vector
iotw(v,i,Stack,Indexes)
}
v=c(`-`,`+`,1,`^`,`^`,NA,NA,X,3,X,2)
Stack=c()
Indexes=c()
iotw(v,1,Stack,Indexes)
NULL
NULL
NULL
NULL
NULL
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
Error in if (is.na(v[i]) == FALSE is.null(unlist(v[i])) ==
FALSE) { :
argument is of length zero
Regards,
Atte Tenkanen
University of Turku, Finland
iotw(v,1,Stack,Indexes)
NULL
NULL
NULL
NULL
NULL
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2 5 1 3
[1] 8 4
[1] 8 4
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2 5 1 3
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 10 5 11 1 3
[1] 8 4 9 2 5 1
[1] 8 4 9 2 5 1
[1] 8
Re: [R] Break during the recursion?
On 15/07/2007 11:36 AM, Atte Tenkanen wrote:
On 15/07/2007 10:33 AM, Atte Tenkanen wrote:
On 15/07/2007 10:06 AM, Atte Tenkanen wrote:
Hi,
Is it possible to break using if-condition during the recursive
function?
You can do
if (condition) return(value)
Here is a function which almost works. It is for inorder-tree-
walk.
iotw-function(v,i,Stack,Indexes) # input: a vector and the
first
index (1), Stack=c(), Indexes=c().
{
print(Indexes)
# if (sum(i)==0) break # Doesn't work...
if (sum(i)==0) return(NULL)
should work.
Duncan Murdoch
Hmm - - - I'd like to save the Indexes-vector (in the example
c(8,4,9,2,10,5,11,1,3)) and stop, when it is ready.
This seems more like a problem with the design of your function
than a
question about R. I can't really help you with that, because your
description of the problem doesn't make sense to me. What does it
mean
to do an inorder tree walk on something that isn't a tree?
Duncan Murdoch
The symbols in vector v have been originally derived from tree. See
http://users.utu.fi/attenka/Tree.jpg
But perhaps there's another way to do this, for instance by using loops and
if-conditions?
Or perhaps by doing the tree walk on the tree, before you collapse it
into a vector.
Duncan Murdoch
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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] Text with differents colors in a plot / Detecting if text exists
Hi everybody, I want to write some text in a plot. That's simple I know. But I want to make use of different colors. Eg. text(x,y,paste(Sunderland,high)). Then Sunderland should be black and high red. Has anyone an idea? By the way. I'm looking for a function or something similar, that can check whether there is text in some regions on the plot. Because I don't want to overwrite an existing text with a new one. Thank you so much for your help. I appologize in advance if my question is very stupid but I have really no idea. Yours, Maja -- __ 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] Break during the recursion?
On 7/15/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 15/07/2007 11:36 AM, Atte Tenkanen wrote:
On 15/07/2007 10:33 AM, Atte Tenkanen wrote:
On 15/07/2007 10:06 AM, Atte Tenkanen wrote:
Hi,
Is it possible to break using if-condition during the recursive
function?
You can do
if (condition) return(value)
Here is a function which almost works. It is for inorder-tree-
walk.
iotw-function(v,i,Stack,Indexes) # input: a vector and the
first
index (1), Stack=c(), Indexes=c().
{
print(Indexes)
# if (sum(i)==0) break # Doesn't work...
if (sum(i)==0) return(NULL)
should work.
Duncan Murdoch
Hmm - - - I'd like to save the Indexes-vector (in the example
c(8,4,9,2,10,5,11,1,3)) and stop, when it is ready.
This seems more like a problem with the design of your function
than a
question about R. I can't really help you with that, because your
description of the problem doesn't make sense to me. What does it
mean
to do an inorder tree walk on something that isn't a tree?
Duncan Murdoch
The symbols in vector v have been originally derived from tree. See
http://users.utu.fi/attenka/Tree.jpg
But perhaps there's another way to do this, for instance by using loops and
if-conditions?
Or perhaps by doing the tree walk on the tree, before you collapse it
into a vector.
If it's a binary tree with n levels, I think you should be able to
generate the positions more directly, depending on how the tree has
been flattened. Binary heaps work this way, so that might be a good
place to start. See http://en.wikipedia.org/wiki/Binary_heap,
particularly heap implementation.
Hadley
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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] Complex surveys, properly computed SEs and non-parametric analyses
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Re: [R] Choosing the number of colour breaks in ggplot2
Fredag 13. juli 2007 skreiv hadley wickham: There's no official way to do it, but you can hack the colour gradient scale to do what you want: # Create a modified scale gr - scale_fill_gradient2()$clone() gr$breaks - function(.) seq(-100, 100, by=10) Thank you so much. It works perfectly. -- Karl Ove Hufthammer __ 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.
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I noticed a decent guide on how to sort data.frames is somewhat lacking. To fill the gap I have written a quick post on the subject, which is here: http://www.andrewprendergast.com/2007/07/sorting_a_dataframe_in_r.html Regards, ap. __ 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 load a dataset
Hi: Since I didn't get any answers, I'll refresh my question. I have a dataset called Chinook Run saved in Excel and I want it to be loaded everytime R starts, so I can call it with a statement like the one below: qplot(color, Year/Forklength, data = Chinook Run) I wonder how can I do that. I went to the rprofile.site and copied the path to my dataset there and it seems to work but I am wondering if that's how's done. Felipe Felipe D. Carrillo Fishery Biologist US Fish Wildlife Service Red Bluff, California 96080 - [[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] Break during the recursion?
Here is now more elegant function for inorder tree walk, but I still can't save
the indexes!? This version now prints them ok, but if I use return, I get only
the first v[i].
leftchild-function(i){return(2*i)}
rightchild-function(i){return(2*i+1)}
iotw-function(v,i)
{
if (is.na(v[i])==FALSE is.null(unlist(v[i]))==FALSE)
{
iotw(v,leftchild(i))
print(v[i]) # return doesn't work here
iotw(v,rightchild(i))
}
}
iotw(1:11,1)
[1] 8
[1] 4
[1] 9
[1] 2
[1] 10
[1] 5
[1] 11
[1] 1
[1] 6
[1] 3
[1] 7
Yours faithfully ;-)
Atte
- Original Message -
From: hadley wickham [EMAIL PROTECTED]
Date: Sunday, July 15, 2007 6:58 pm
Subject: Re: [R] Break during the recursion?
On 7/15/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 15/07/2007 11:36 AM, Atte Tenkanen wrote:
On 15/07/2007 10:33 AM, Atte Tenkanen wrote:
On 15/07/2007 10:06 AM, Atte Tenkanen wrote:
Hi,
Is it possible to break using if-condition during the
recursive function?
You can do
if (condition) return(value)
Here is a function which almost works. It is for inorder-
tree-
walk.
iotw-function(v,i,Stack,Indexes) # input: a vector and the
first
index (1), Stack=c(), Indexes=c().
{
print(Indexes)
# if (sum(i)==0) break # Doesn't work...
if (sum(i)==0) return(NULL)
should work.
Duncan Murdoch
Hmm - - - I'd like to save the Indexes-vector (in the example
c(8,4,9,2,10,5,11,1,3)) and stop, when it is ready.
This seems more like a problem with the design of your function
than a
question about R. I can't really help you with that, because
your description of the problem doesn't make sense to me. What
does it
mean
to do an inorder tree walk on something that isn't a tree?
Duncan Murdoch
The symbols in vector v have been originally derived from
tree. See
http://users.utu.fi/attenka/Tree.jpg
But perhaps there's another way to do this, for instance by
using loops and if-conditions?
Or perhaps by doing the tree walk on the tree, before you
collapse it
into a vector.
If it's a binary tree with n levels, I think you should be able to
generate the positions more directly, depending on how the tree has
been flattened. Binary heaps work this way, so that might be a good
place to start. See http://en.wikipedia.org/wiki/Binary_heap,
particularly heap implementation.
Hadley
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Re: [R] Break during the recursion?
On 7/15/07, Atte Tenkanen [EMAIL PROTECTED] wrote:
Here is now more elegant function for inorder tree walk, but I still can't
save the indexes!? This version now prints them ok, but if I use return, I
get only the first v[i].
leftchild-function(i){return(2*i)}
rightchild-function(i){return(2*i+1)}
iotw-function(v,i)
{
if (is.na(v[i])==FALSE is.null(unlist(v[i]))==FALSE)
{
iotw(v,leftchild(i))
print(v[i]) # return doesn't work here
iotw(v,rightchild(i))
}
}
Shouldn't you return:
c(iotw(v, leftchild(i)), v[i], iotw(v, rightchild(i)))
(and rewrite the conditition to return null if the node doesn't exist,
I think it reads clearer that way)
Hadley
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Re: [R] Break during the recursion?
TNX Hadley!
Atte
- Original Message -
From: hadley wickham [EMAIL PROTECTED]
Date: Sunday, July 15, 2007 11:04 pm
Subject: Re: [R] Break during the recursion?
On 7/15/07, Atte Tenkanen [EMAIL PROTECTED] wrote:
Here is now more elegant function for inorder tree walk, but I
still can't save the indexes!? This version now prints them ok, but
if I use return, I get only the first v[i].
leftchild-function(i){return(2*i)}
rightchild-function(i){return(2*i+1)}
iotw-function(v,i)
{
if (is.na(v[i])==FALSE is.null(unlist(v[i]))==FALSE)
{
iotw(v,leftchild(i))
print(v[i]) # return doesn't work here
iotw(v,rightchild(i))
}
}
Shouldn't you return:
c(iotw(v, leftchild(i)), v[i], iotw(v, rightchild(i)))
(and rewrite the conditition to return null if the node doesn't exist,
I think it reads clearer that way)
Hadley
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Re: [R] HELP FOR BUGS
You might find the 'arm' package useful. For a good introduction to heirarchical modeling, using 'arm' and also WinBUGS and R2WinBUGS, read Gelman, A; J Hill 2007. Data analysis using regression and multilevel/hierarchical models. Cambridge University Press. Cheers, Mike. Ali raza-4 wrote: Hi Sir I am very new user of R for the research project on multilevel logistic regression. There is confusion about bugs() function in R and BUGS software. Is there any relation between these two? Is there any comprehensive package for Multilevel Logistic modelling in R? Please guide in this regard. Thank You RAZA - Boardwalk for $500? In 2007? Ha! [[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. -- View this message in context: http://www.nabble.com/HELP-FOR-BUGS-tf4078749.html#a11605645 Sent from the R help mailing list archive at Nabble.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.
[R] NNET re-building the model
Hello,
I've been working with nnet and now I'd like to use the weigths, from
the fitted model, to iterpret some of variables impornatce.
I used the following command:
mts - nnet(y=Y,x=X,size =4, rang = 0.1,
decay = 5e-4, maxit = 5000,linout=TRUE)
X is (m x n) Y is (m x 1)
And then I get the coeficients by:
Wts-coef(mts)
b-h1i1-h1i2-h1i3-h1i4-h1
i5-h1i6-h1i7-h1 ...
b-o h1-o h2-o h3-o h4-o ...
I understood that I should get the predicted Y (hat(Y)) by:
hat(Y)=
(b-o)+((b-h1)sum(X[,1:m]*Wts[i1:m-h1]+X[,1:m]*Wts[i1:m-h2]+...+X[,1:m]
*Wts[i1:m-h4]))*(h1-o)+
((b-h2)sum(X[,1:m]*Wts[i1:m-h2]+X[,1:m]*Wts[i1:m-h2]+...+X[,1:m]*Wts[i1
:m-h4]))*(h2-o)+...)
Am I missed some point on this definition of NNET.
Thanks in advance,
Best Regards,
Marlon !!!
===
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Re: [R] spatstat - Fitting a Strauss model with trend determined by kernel density smoother
On 14/07/2007, at 2:51 AM, Alejandro Veen wrote:
Dear r-help,
I would like to use the 'ppm' function of the 'spatstat' package to
fit a Strauss inhibition model. I understand that I can specify a
parametric model for the background trend, but how would I specify a
trend which is estimated using a Kernel density smoother?
In particular, I would like to use the 'kde' function of the 'ks'
package to estimate the background intensity and then use this as
the trend for a Strauss inhibition process.
Questions about a specific contributed package should usually be
directed to the maintainers of
that package rather than to r-help.
To attempt to answer your question:
You need to convert your estimate of the background trend to an
***image***; see the
function im() in the spatstat package.
Or instead of using kde, you could use the ppp method for density()
which is provided in
spatstat; this methop returns an image. See the help for density.ppp
().
Now suppose that your point pattern is ``X'' and your estimate of
the trend is ``bgim''.
You can then fit the model you want via
fit - ppm(X,~bgim,inter=Strauss(42),covariates=list(bgim=bgim))
Note that you have to specify the interaction radius for the Strauss
model (I have specified this
radius to be 42 in the forgoing example); this radius is an
``irregular'' parameter --- i.e. it does
not appear in exponential family form --- and hence is not estimated
by ppm(), at least not
directly.
cheers,
Rolf Turner
##
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Re: [R] Please Please Help me!!!
Followings someones advice on a previous post I am reposting my question again. Here is is and I would appreciate any help with my problem. This question is basically a mathematical question, but I am sure there must be an easy way to achieve the answer to my problem using R as, my problem seems to me to be quite straight forward and something that people must use quite a lot. So if there is an easy way to do this I would appreciate the help. I have a transition intensities at different ages in a markov model. I want to estimate this intensity by fitting a curve to the known intensities. The data is as follows Age Intensity 22 0.0002 27 0.0011 32 0.0074 37 0.0159 42 0.0292 47 0.0428 52 0.0265 57 0.0301 62 0.0270 67 0.0296 When plotted as intensity vs age, the data looks like a gamma curve. I want to fit a gamma density curve to this data, so I need to estimate the paramaters for the gamma curve,It is just a curve fitting problem but whats causing the trouble is that I need to use least squares minimization to calculate the parameters for the gamma curve. How do I do this??? The curve will be a truncated gamma function so it will have 3 paramaters a, b, c. I tried to do the following nls(lograte ~ log(c) + a*log(b) + (a-1)*log(age)+b*(age)-lgamma(a), start=list(a=1,b=1,c=1)), where lograte, and age are my data. a,b the gamma parameters and c the parameter we need because we are fitting a truncated distribution. I also tried defining fn = function(p) sum((log(y)-log(dgamma(x,p[1],p[2])*p[3]))^2) a residual sum of squares and using nlm to minimise this and find paramaters but I cant get this to work either. Can anyone help me ?? Please :) Thanks TIMMMAY wrote: I posted the message below a few days ago but I have not had any responses. I keep thinking that there must be some easy way to answer the problem I am just not familiar enough with regression to answer the problem myself. If anyone can help me I would be very grateful. I need to fit a gamma curve to a set of data. ie a scatterplot of the data indicates that the curve looks like a truncated gamma density function and I would like to estimate the paramaters so that I can fit a curve to the data points. Its not MLE paramater estimation just a curve fitting exercise. The problem again is I have a set transition intensities and when plotted the curve looks like a gamma density. I want to fit a gamma density curve to these intensities. It is just a curve fitting problem but whats causing the trouble is that I need to use least squares minimization to calculate the parameters for the gamma curve. How do I do this??? The curve will be a truncated gamma function so it will have 3 paramaters a, b, c. I tried to do the following nls(lograte ~ log(c) + a*log(b) + (a-1)*log(age)+b*(age)-lgamma(a), start=list(a=1,b=1,c=1)), where lograte, and age are my data. a,b the gamma parameters and c the parameter we need because we are fitting a truncated distribution. I also tried defining fn = function(p) sum((log(y)-log(dgamma(x,p[1],p[2])*p[3]))^2) a residual sum of squares and using nlm to minimise this and find paramaters but this doesnt work either. Can anyone help me ?? Please :) Original message follows :( Fitting a Gamma Curve by TIMMMAY Jul 12, 2007; 03:38pm :: Rate this Message:(use ratings to moderate[?]) Reply | Reply to Author | Show Only this Message Hi there, I hope someone can help me before I tear all my hair out. I have a set transition intensities and when plotted the curve looks like a gamma density. I want to fit a gamma density curve to these intensities. It is just a curve fitting problem but whats causing the trouble is that I need to use least squares minimization to calculate the parameters for the gamma curve. How do I do this??? The curve will be a truncated gamma function so it will have 3 paramaters a, b, c. I tried to do the following nls(lograte ~ log(c) + a*log(b) + (a-1)*log(age)+b*(age)-lgamma(a), start=list(a=1,b=1,c=1)), where lograte, and age are my data. a,b the gamma parameters and c the parameter we need because we are fitting a truncated distribution. I also tried defining fn = function(p) sum((log(y)-log(dgamma(x,p[1],p[2])*p[3]))^2) a residual sum of squares and using nlm to minimise this and find paramaters but this doesnt work either. Can anyone help me ?? Please :) « Return to forum Start a free forum or mailing list archive on Nabble Help - Powered by - Terms of Use - Jobs at Nabble - Nabble Support -- View this message in context: http://www.nabble.com/Please-Please-Help-me%21%21%21-tf4081887.html#a11606008 Sent from the R help mailing list archive at Nabble.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,
Re: [R] NNET re-building the model
This is almost unreadable (is your space bar broken?) but you seem to be
missing the non-linearity of the hidden units.
The definition is in the book nnet (sic) supports, in a layout I cannot
reproduce in plain text.
On Mon, 16 Jul 2007, dos Reis, Marlon wrote:
Hello,
I've been working with nnet and now I'd like to use the weigths, from
the fitted model, to iterpret some of variables impornatce.
I used the following command:
mts - nnet(y=Y,x=X,size =4, rang = 0.1,
decay = 5e-4, maxit = 5000,linout=TRUE)
X is (m x n) Y is (m x 1)
And then I get the coeficients by:
Wts-coef(mts)
b-h1i1-h1i2-h1i3-h1i4-h1
i5-h1i6-h1i7-h1 ...
b-o h1-o h2-o h3-o h4-o ...
I understood that I should get the predicted Y (hat(Y)) by:
hat(Y)=
(b-o)+((b-h1)sum(X[,1:m]*Wts[i1:m-h1]+X[,1:m]*Wts[i1:m-h2]+...+X[,1:m]
*Wts[i1:m-h4]))*(h1-o)+
((b-h2)sum(X[,1:m]*Wts[i1:m-h2]+X[,1:m]*Wts[i1:m-h2]+...+X[,1:m]*Wts[i1
:m-h4]))*(h2-o)+...)
Am I missed some point on this definition of NNET.
Thanks in advance,
Best Regards,
Marlon !!!
===
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] Restructuring data
Hi folks, I am new to the list and relatively new to R. I am trying to unstack data arraywise and could not find a convenient solution yet. I tried to find a solution for the problem on help archives. I also tried to use the reshape command in R and the reshape package but could not get result. I will illustrate the case below, but the real dataset is quite large so that I would appreciate an easy solution if there is any. The current data structure (variable names): ID, TIME, BUY-A, BUY-B, SELL-A, SELL-B Achieved structure (with the reshape command or the reshape package) ID, TIME, BUY-A ID, TIME, BUY-B ID, TIME, SELL-A ID, TIME, SELL-B This is regular unstacking with two identifier variables. Nothing special though. What I am looking for and did not manage is the following structure: ID, TIME, BUY-A, SELL-A ID, TIME, BUY-B, SELL-B I am quite sure it's pretty easy, but I could not find how to do this. Thanks a bunch, Daniel __ 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] Partial Proportional Odds model using vglm
Hi: I am trying to fit a PPO model using vglm from VGAM, and get an error while executing the code. Here is the data, code, and error: Data = rc13, first row is the column names. a = age, and 1,2,3, 4 and 5 are condition grades. a 1 2 3 4 5 1 1 0 0 0 0 2 84 2 7 10 2 3 16 0 6 6 2 4 13 0 3 4 0 5 0 0 0 1 0 rc13-read.table(icg_rcPivot_group2.txt,header=F) names(rc13)-c(a,1,2,3,4,5) ppo-vglm(cbind(rc13[,2],rc13[,3],rc13[,4],rc13[,5],rc13[,6])~a,family = cumulative(link = logit, parallel = F , reverse = F),na.action=na.pass, data=rc13) summary(ppo) I get the following error: Error in [-(`*tmp*`, , index, value = c(1.13512932539841, 0.533057528200189, : number of items to replace is not a multiple of replacement length In addition: Warning messages: 1: NaNs produced in: log(x) 2: fitted values close to 0 or 1 in: tfun(mu = mu, y = y, w = w, res = FALSE, eta = eta, extra) 3: 19 elements replaced by 1.819e-12 in: checkwz(wz, M = M, trace = trace, wzeps = control$wzepsilon) I will appreciate any help to fix this problem. Thanks Reez You Grad Student University of Waterloo Waterloo, ON Canada __ 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] neural networks function in R
Hi all gurus, I have a few general questions about using neural networks function, i.e. the nnet function. I'm new to this function and still exploring it. So please kindly bear with me. Here are my questions. 1. Is there anyway that I can specify my own objective or loss function for my neural networks? I see that by arguments that we can pass into the networks, we have either square loss or entropy. I'm trying to create an NN that fits ordinal regression data. So my objective function is the likelihood function but is not exactly the entropy function due to the fact that order of the labels matters in my application. Or I'm better of to write my own function? 2. If I need to write my own function, I could see that I can use constOptim to help me maximize my likelihood. Another question here is how can I pass analytical gradient function of each of my network weight into the function? Specifically, my question is that does my gradient function has to be flat representation of each parameter like the following arbitrary example: 2*w_i1_h1+3*w_h1_o1+4 (where w_i1_h1 is weight of input node 1 to hidden layer node 1, and vice versa) Or I can do a nest expression like: sigmoid(node_eval(input_x, j)) here my node_eval is a function that takes in input_x feature value vector and produces output to node j in the next layer. If expression base evaluation like in case 2 is possible, it will significantly simplify the definition of my gradient function output. Any help would be really appreciated on these issues. Thank you. - adschai __ 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] Restructuring data
On 7/15/07, Daniel Malter [EMAIL PROTECTED] wrote: Hi folks, I am new to the list and relatively new to R. I am trying to unstack data arraywise and could not find a convenient solution yet. I tried to find a solution for the problem on help archives. I also tried to use the reshape command in R and the reshape package but could not get result. I will illustrate the case below, but the real dataset is quite large so that I would appreciate an easy solution if there is any. The current data structure (variable names): ID, TIME, BUY-A, BUY-B, SELL-A, SELL-B Achieved structure (with the reshape command or the reshape package) ID, TIME, BUY-A ID, TIME, BUY-B ID, TIME, SELL-A ID, TIME, SELL-B This is regular unstacking with two identifier variables. Nothing special though. What I am looking for and did not manage is the following structure: ID, TIME, BUY-A, SELL-A ID, TIME, BUY-B, SELL-B I am quite sure it's pretty easy, but I could not find how to do this. This seems to work: foo - data.frame(ID = 1:4, TIME=1:4, + BUY-A = rnorm(4), + BUY-B = rnorm(4), + SELL-A = rnorm(4), + SELL-B = rnorm(4), check.names = FALSE) foo ID TIME BUY-A BUY-B SELL-A SELL-B 1 11 0.47022807 1.09573107 0.1977035 -0.08333043 2 22 -0.20672870 0.07397772 1.4959044 -0.98555020 3 33 0.05533779 0.25821758 1.3531913 0.16808307 4 44 -0.11471772 1.27798740 -0.1101390 -0.36937994 reshape(foo, direction=long, + varying = list(c(BUY-A, BUY-B), c(SELL-A, SELL-B)), + v.names=c(BUY, SELL), idvar=ID, + times = c(A, B), timevar=which) ID TIME which BUYSELL 1.A 11 A 0.47022807 0.19770349 2.A 22 A -0.20672870 1.49590443 3.A 33 A 0.05533779 1.35319133 4.A 44 A -0.11471772 -0.11013896 1.B 11 B 1.09573107 -0.08333043 2.B 22 B 0.07397772 -0.98555020 3.B 33 B 0.25821758 0.16808307 4.B 44 B 1.27798740 -0.36937994 -Deepayan __ 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 sort a data.frame
The example could be simplified like this: T - productSalesByStore T - merge(T, rowsum(T$sales, T$store), by.x = 2, by.y = 0) T[order(T$V1, T$sales, decreasing = TRUE), 1:3] If you transfer your data to a data base then this SQL statement would also do it: select store, product, sales from productSalesByStore natural join (select store, sum(sales) storesales from productSalesByStore group by store) order by storesales desc, sales desc On 7/15/07, Andrew Prendergast [EMAIL PROTECTED] wrote: I noticed a decent guide on how to sort data.frames is somewhat lacking. To fill the gap I have written a quick post on the subject, which is here: http://www.andrewprendergast.com/2007/07/sorting_a_dataframe_in_r.html Regards, ap. __ 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] Dates in 'persp' plots
Hello, I would like to have the y axis show dates in a 3D 'persp' plot. The following example works... x - spot y - as.numeric(dates)# class(dates) produces output [1] Date z - t(price) persp(x, y, z, ticktype=detailed) ...however the y axes contains 'meaningless' integers (days since 1970-01-01 is not very intuitive!) Changing the commented line above to y - dates is no good because 'persp' automatically coerces input to numeric types. I would really like the axis to be labeled with dates. A string, eg 13-Jul-2007 would be best, but ISO date integer, eg 20070713, would also be acceptable. I've checked help for 'persp' and 'par' with no luck - any ideas appreciated! Cheers, Josh. __ 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 sort a data.frame
In thinking about this a bit more an even simpler solution based on ave is: storesales - ave(T$sales, T$store, FUN = sum) T[order(storesales, T$sales, decreasing = TRUE), ] On 7/15/07, Gabor Grothendieck [EMAIL PROTECTED] wrote: The example could be simplified like this: T - productSalesByStore T - merge(T, rowsum(T$sales, T$store), by.x = 2, by.y = 0) T[order(T$V1, T$sales, decreasing = TRUE), 1:3] If you transfer your data to a data base then this SQL statement would also do it: select store, product, sales from productSalesByStore natural join (select store, sum(sales) storesales from productSalesByStore group by store) order by storesales desc, sales desc On 7/15/07, Andrew Prendergast [EMAIL PROTECTED] wrote: I noticed a decent guide on how to sort data.frames is somewhat lacking. To fill the gap I have written a quick post on the subject, which is here: http://www.andrewprendergast.com/2007/07/sorting_a_dataframe_in_r.html Regards, ap. __ 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] Restructuring data
On 7/16/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: On 7/15/07, Daniel Malter [EMAIL PROTECTED] wrote: Hi folks, I am new to the list and relatively new to R. I am trying to unstack data arraywise and could not find a convenient solution yet. I tried to find a solution for the problem on help archives. I also tried to use the reshape command in R and the reshape package but could not get result. I will illustrate the case below, but the real dataset is quite large so that I would appreciate an easy solution if there is any. The current data structure (variable names): ID, TIME, BUY-A, BUY-B, SELL-A, SELL-B Achieved structure (with the reshape command or the reshape package) ID, TIME, BUY-A ID, TIME, BUY-B ID, TIME, SELL-A ID, TIME, SELL-B This is regular unstacking with two identifier variables. Nothing special though. What I am looking for and did not manage is the following structure: ID, TIME, BUY-A, SELL-A ID, TIME, BUY-B, SELL-B I am quite sure it's pretty easy, but I could not find how to do this. This seems to work: foo - data.frame(ID = 1:4, TIME=1:4, + BUY-A = rnorm(4), + BUY-B = rnorm(4), + SELL-A = rnorm(4), + SELL-B = rnorm(4), check.names = FALSE) foo ID TIME BUY-A BUY-B SELL-A SELL-B 1 11 0.47022807 1.09573107 0.1977035 -0.08333043 2 22 -0.20672870 0.07397772 1.4959044 -0.98555020 3 33 0.05533779 0.25821758 1.3531913 0.16808307 4 44 -0.11471772 1.27798740 -0.1101390 -0.36937994 reshape(foo, direction=long, + varying = list(c(BUY-A, BUY-B), c(SELL-A, SELL-B)), + v.names=c(BUY, SELL), idvar=ID, + times = c(A, B), timevar=which) ID TIME which BUYSELL 1.A 11 A 0.47022807 0.19770349 2.A 22 A -0.20672870 1.49590443 3.A 33 A 0.05533779 1.35319133 4.A 44 A -0.11471772 -0.11013896 1.B 11 B 1.09573107 -0.08333043 2.B 22 B 0.07397772 -0.98555020 3.B 33 B 0.25821758 0.16808307 4.B 44 B 1.27798740 -0.36937994 It's a little more verbose with the reshape package, but I find it easier to understand what's going on. fm - melt(foo, id=c(ID,TIME)) fm - cbind(fm, colsplit(fm$variable, -, c(direction,type))) fm$variable - NULL cast(fm, ... ~ direction) There's an example like this in the introduction to reshape manual. 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.
