Re: [R] get is.na !is.na count of various combinations of columns
Hi,
I just noticed that you also have all is.na() or !is.na() subsets. To do that:
mydatatest - structure(list(CaseID = structure(1:8, .Label = c(1605928,
1605943, 1605945, 1605947, 1605949, 1605951, 1605952,
1605953), class = factor), Structure = structure(c(1L, 1L,
3L, 2L, 2L, 3L, 3L, 4L), .Label = c(corp, LLC, prop, LAC
), class = factor), Poster = c(1, 1, 1, NA, NA, 1, 1, NA),
Records = c(1, 1, 1, 1, 1, NA, 1, NA), MWBW = c(NA, NA, 495.1,
NA, NA, NA, 425, NA), OT = c(NA, NA, NA, NA, 411.25, NA,
432.5, NA), CL = c(NA, NA, NA, 13.52, NA, NA, 14.72, NA)), .Names =
c(CaseID,
Structure, Poster, Records, MWBW, OT, CL), row.names = c(NA,
8L), class = data.frame)
vec1 - names(mydatatest)[-(1:2)]
res - lapply(c(0,seq(length(vec1))),function(i) {x1 -
as.data.frame(combn(vec1,i),stringsAsFactors=FALSE); lapply(x1, function(x)
{indx - vec1 %in% x; nisna - if(length(vec1[!indx]) 0 )
paste(paste0(!is.na,(, vec1[!indx],)),collapse= );isna -
if(length(vec1[indx]) 0 ) paste(paste0(is.na,(, vec1[indx],)),collapse=
); nisna_isna - gsub(^ | $,, paste(nisna, isna, sep= ));
subset(mydatatest, eval(parse(text=nisna_isna)))})})
names1 - unlist(lapply(c(0,seq(length(vec1))),function(i) {x1 -
as.data.frame(combn(vec1,i),stringsAsFactors=FALSE); unlist(lapply(x1,
function(x) {indx - vec1 %in% x; nisna - if(length(vec1[!indx]) 0 )
paste(paste0(!is.na,(, vec1[!indx],)),collapse= );isna -
if(length(vec1[indx]) 0 ) paste(paste0(is.na,(, vec1[indx],)),collapse=
); nisna_isna - gsub(^ | $,, paste(nisna, isna, sep=
))}))}),use.names=FALSE)
res1 - unlist(res,recursive=FALSE)
names(res1) - names1
res2 - res1[sapply(res1,nrow)!=0]
A.K.
On Wednesday, March 19, 2014 12:59 AM, arun smartpink...@yahoo.com wrote:
If you want to identify the combination of columns:
names1 - unlist(lapply(seq(length(vec1)-1),function(i) {x1 -
as.data.frame(combn(vec1,i),stringsAsFactors=FALSE); unlist(lapply(x1,
function(x) {indx - vec1 %in% x; paste(paste(paste0(!is.na,(,
vec1[!indx],)),collapse= ), paste(paste0(is.na,(,
vec1[indx],)),collapse= ), sep= ) }))}),use.names=FALSE)
res1 - unlist(res,recursive=FALSE)
names(res1) - names1
res1
length(res1)
#[1] 30
res1[30]
#$`!is.na(Poster) is.na(Records) is.na(MWBW) is.na(OT) is.na(CL)`
# CaseID Structure Poster Records MWBW OT CL
#6 1605951 prop 1 NA NA NA NA
res2 - res1[sapply(res1,nrow)!=0]
res2[4]
#$`!is.na(Poster) !is.na(Records) is.na(MWBW) is.na(OT) is.na(CL)`
# CaseID Structure Poster Records MWBW OT CL
#1 1605928 corp 1 1 NA NA NA
#2 1605943 corp 1 1 NA NA NA
Hope this helps.
A.K.
On , arun smartpink...@yahoo.com wrote:
Hi,
May be this helps:
res - lapply(seq(length(vec1)-1),function(i) {x1 -
as.data.frame(combn(vec1,i),stringsAsFactors=FALSE); lapply(x1, function(x)
{indx - vec1 %in% x; nisna - paste(paste0(!is.na,(,
vec1[!indx],)),collapse= );isna - paste(paste0(is.na,(,
vec1[indx],)),collapse= );subset(mydatatest, eval(parse(text=nisna))
eval(parse(text=isna)))})})
A.K.
On Tuesday, March 18, 2014 11:45 PM, bcrombie bcrom...@utk.edu wrote:
I'm trying to count the number of combinations of columns containing data
not containing data as described below. Im not sure how to do this in R
and need some help.
#get TRUE/FALSE count of various combinations of columns per CaseID or per
Structure
mydatatest - data.frame (CaseID = c(1605928, 1605943, 1605945,
1605947, 1605949, 1605951),
Structure = c(corp, corp, prop, LLC, LLC,
prop),
Poster = c(1, 1, 1, NA, NA, 1),
Records = c(1, 1, 1, 1, 1, NA),
MWBW = c(NA, NA, 495.10, NA, NA, NA),
OT = c(NA, NA, NA, NA, 411.25, NA),
CL = c(NA, NA, NA, 13.52, NA, NA))
combo1 - subset(mydatatest, !is.na(Poster Records MWBW OT CL))
combo2 - subset(mydatatest, !is.na(Poster Records MWBW OT)
is.na(CL))
combo3 - subset(mydatatest, !is.na(Poster Records MWBW) is.na(OT
CL))
combo4 - subset(mydatatest, !is.na(Poster Records) is.na(MWBW OT
CL))
combo5 - subset(mydatatest, !is.na(Poster) is.na(Records MWBW OT
CL))
combo6 - subset(mydatatest, !is.na(Records) is.na(Poster MWBW OT
CL))
combo7 - subset(mydatatest, !is.na(MWBW) is.na(Poster Records OT
CL))
combo8 - subset(mydatatest, !is.na(OT) is.na(Poster Records MWBW
CL))
combo9 - subset(mydatatest, !is.na(CL) is.na(Poster Records MWBW
OT))
etc.
--
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Re: [R] Data file verification protocol
On Wed, Mar 19, 2014 at 4:03 AM, Wolf, Steven wolfs...@msu.edu wrote: Hi R users, This isn't a R-specific issue, per-se, but I thought that this list would have some helpful input on this topic. First, a bit of background. I am working on a project which is interested in following approx 1000 students each semester, and collects about 15 different measurements about each student. These are both numeric and text, for example grades in a course, race, gender, etc. I am looking for a verification protocol which can look at a data file and see if it has been modified. Ideally, this should be something that I can check the file with to see if the file has been changed or corrupted and incorporate into my analysis workflow. (i.e., every time I look at my data, I can run this protocol to ensure the file hasn't changed.) Operating systems will keep the last modification time of a file, and you can use the file.info function in R to check that. However, if someone just opens and re-saves the file without changing it that will usually trigger an update of the modification time. The big question you haven't answered is has the file been changed since when?. Since you last ran your analysis? This then looks like a job for the 'make' utility. You specify rules in a 'Makefile' that specify how to create targets based on dependencies. For example: results.txt: data.dat process.R Rscript process.R - says that results.txt depends on data.dat (your input data) and process.R (your R code that creates results.txt from data.dat), and would run Rscript process.R if data.dat or process.R have a newer modification time than results.txt. Run twice in rapid succession, this makefile wouldn't run R the second time because results.txt would be newer then its dependencies since it was just created. Objects within R don't have timestamps, so its not possible to conditionally run an R function if its parameter objects are newer than the result object. But if you save R objects as .RData files, you can use make based on the timestamps of the .RData files. Alternatively you can just keep recent versions of the file hanging around (1000x15 is pretty small, even multiplied by another 1000 is still not exactly Big Data) and compare them. In a unix environment the cmp command will quickly test two files for equality, or if you don't want to store copies of your file you simply compute a checksum or digest and compare digests. In a unix environment you'd typically use the md5sum command which spits out a 128-bit (32 character) checksum for its arguments. If the checksum is different, then the file is different. Your use case is still a bit vague - for example you haven't said what the file format is, or how its being updated. Barry [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] New package metap on CRAN
This package provides meta-analysis of significance values. Although the preferred way of performing meta-analysis uses effect sizes sometimes these are unavailable and only p-values can be obtained from the primary studies. The metap package provides several ways of synthesising p-values and a graphical display. There exist a number of packages on CRAN which use one or other of these techniques in the context of genetic studies but metap is designed for general purpose use. Comments welcome. Michael Dewey i...@aghmed.fsnet.co.uk http://www.aghmed.fsnet.co.uk/home.html ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data file verification protocol
Wolf, Steven wolfs...@msu.edu [2014-03-18 21:05]: I am looking for a verification protocol which can look at a data file and see if it has been modified. Ideally, this should be something that I can check the file with to see if the file has been changed or corrupted and incorporate into my analysis workflow. (i.e., every time I look at my data, I can run this protocol to ensure the file hasn?t changed.) http://dirk.eddelbuettel.com/code/digest.html Overview digest provides `hash' function summaries for GNU R objects. The md5, sha-1, sha-256 and crc32 hash functions are available. The md5 algorithm by Ron Rivest is specified in RFC 1321, the SHA-1 and SHA-256 algorithm is specified in FIPS-180-1 and FIPS-180-2, respectively, and the crc32 algorithm is described in here. For md5, sha-1 and sha-256, this packages uses small standalone C implementations that were provided by by Christophe Devine. For crc32, code from the zlib library is used. For sha-512, a routine by Aaron Gifford is used. Please note that this package is not meant to be deployed for cryptographic purposes for which more comprehensive (and widely tested) libraries such as OpenSSL should be used. Example The following verbatim R session loads digest and runs the example() from the corresponding help page: library(digest) example(digest) digest md5Input - c(, a, abc, message digest, abcdefghijklmnopqrstuvwxyz, ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789, paste(12345678901234567890123456789012345678901234567890123456789012, 34567890123456 ... ... [TRUNCATED] [...] HTH, -rex -- ...I paid a visit to Schrodinger in his Vienna apartment before his death... There were no cats. I was told he did not like cats. -quantam leaps, bernstein. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] write.csv: set the name of the column containing the row.names
Hi everybody, I am sorry if this is a silly question (I have tried to search it at http://tolstoy.newcastle.edu.au/R/ but I get a 404 page). I have a script writing some data.frames to csv files. I write the csv files using the write.csv function and the option row.names=TRUE. Everything works fine, except that I am not able to set the name of the column holding the row.names. For example I tried to set the column name to ID like this: df = data.frame(a=c(1,2,3), b=c(3,4,5)) rownames(df) = c(A,B,C) write.csv(df, row.names=TRUE, colnames=c(ID, colnames(df))) This fails with the following error: Error in write.table(df, row.names = TRUE, colnames = c(ID, colnames(df)), : unused argument (colnames = c(ID, colnames(df))) How can I do this? I have tried to use write.table() as well, but it seems I never get to set the options right. Thanks a lot in advance for the help, Cheers, Luca __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] write.csv: set the name of the column containing the row.names
The argument is col.names, not colnames. Best, Ista On Mar 19, 2014 6:14 AM, Luca Cerone luca.cer...@gmail.com wrote: Hi everybody, I am sorry if this is a silly question (I have tried to search it at http://tolstoy.newcastle.edu.au/R/ but I get a 404 page). I have a script writing some data.frames to csv files. I write the csv files using the write.csv function and the option row.names=TRUE. Everything works fine, except that I am not able to set the name of the column holding the row.names. For example I tried to set the column name to ID like this: df = data.frame(a=c(1,2,3), b=c(3,4,5)) rownames(df) = c(A,B,C) write.csv(df, row.names=TRUE, colnames=c(ID, colnames(df))) This fails with the following error: Error in write.table(df, row.names = TRUE, colnames = c(ID, colnames(df)), : unused argument (colnames = c(ID, colnames(df))) How can I do this? I have tried to use write.table() as well, but it seems I never get to set the options right. Thanks a lot in advance for the help, Cheers, Luca __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] write.csv: set the name of the column containing the row.names
Thanks Ista, sorry it was a typo in my email, but in the code I have the right option. On Wed, Mar 19, 2014 at 11:58 AM, Ista Zahn istaz...@gmail.com wrote: The argument is col.names, not colnames. Best, Ista On Mar 19, 2014 6:14 AM, Luca Cerone luca.cer...@gmail.com wrote: Hi everybody, I am sorry if this is a silly question (I have tried to search it at http://tolstoy.newcastle.edu.au/R/ but I get a 404 page). I have a script writing some data.frames to csv files. I write the csv files using the write.csv function and the option row.names=TRUE. Everything works fine, except that I am not able to set the name of the column holding the row.names. For example I tried to set the column name to ID like this: df = data.frame(a=c(1,2,3), b=c(3,4,5)) rownames(df) = c(A,B,C) write.csv(df, row.names=TRUE, colnames=c(ID, colnames(df))) This fails with the following error: Error in write.table(df, row.names = TRUE, colnames = c(ID, colnames(df)), : unused argument (colnames = c(ID, colnames(df))) How can I do this? I have tried to use write.table() as well, but it seems I never get to set the options right. Thanks a lot in advance for the help, Cheers, Luca __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] write.csv: set the name of the column containing the row.names
Thanks Duncan, I knew that, I wanted to know if there was some way not involving having to recreate the df. It is not a big issue in this case (they are not very big), but it might be problematic for big data frames. Thanks for advice on using col.names=NA. Cheers, Luca Luca Cerone Tel: +34 692 06 71 28 Skype: luca.cerone On Wed, Mar 19, 2014 at 11:44 AM, Duncan Murdoch murdoch.dun...@gmail.com wrote: On 14-03-19 6:12 AM, Luca Cerone wrote: Hi everybody, I am sorry if this is a silly question (I have tried to search it at http://tolstoy.newcastle.edu.au/R/ but I get a 404 page). I have a script writing some data.frames to csv files. I write the csv files using the write.csv function and the option row.names=TRUE. Everything works fine, except that I am not able to set the name of the column holding the row.names. For example I tried to set the column name to ID like this: df = data.frame(a=c(1,2,3), b=c(3,4,5)) rownames(df) = c(A,B,C) write.csv(df, row.names=TRUE, colnames=c(ID, colnames(df))) This fails with the following error: Error in write.table(df, row.names = TRUE, colnames = c(ID, colnames(df)), : unused argument (colnames = c(ID, colnames(df))) How can I do this? I have tried to use write.table() as well, but it seems I never get to set the options right. You can't get a column heading for the row names. If you want a blank entry there, use col.names=NA and row.names=TRUE. If you want the heading to be ID, then you need to write the row names as a regular column of the dataframe, e.g. df2 - data.frame(ID=rownames(df), df) and then write df2 *without* rownames (so they don't appear twice). Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] write.csv: set the name of the column containing the row.names
Hello, write.csv(df, row.names=TRUE, col.names=c(ID, colnames(df))) ,a,b A,1,3 B,2,4 C,3,5 Warning message: In write.csv(df, row.names = TRUE, col.names = c(ID, colnames(df))) : attempt to set 'col.names' ignored This gives a warning, not an error message. I've also tried setting col.names = NA, and the same warning message is given. Why I don't know, according to the help page ?write.csv the default behavior should be overriden. Hope this helps, Rui Barradas Em 19-03-2014 11:01, Luca Cerone escreveu: Thanks Ista, sorry it was a typo in my email, but in the code I have the right option. On Wed, Mar 19, 2014 at 11:58 AM, Ista Zahn istaz...@gmail.com wrote: The argument is col.names, not colnames. Best, Ista On Mar 19, 2014 6:14 AM, Luca Cerone luca.cer...@gmail.com wrote: Hi everybody, I am sorry if this is a silly question (I have tried to search it at http://tolstoy.newcastle.edu.au/R/ but I get a 404 page). I have a script writing some data.frames to csv files. I write the csv files using the write.csv function and the option row.names=TRUE. Everything works fine, except that I am not able to set the name of the column holding the row.names. For example I tried to set the column name to ID like this: df = data.frame(a=c(1,2,3), b=c(3,4,5)) rownames(df) = c(A,B,C) write.csv(df, row.names=TRUE, colnames=c(ID, colnames(df))) This fails with the following error: Error in write.table(df, row.names = TRUE, colnames = c(ID, colnames(df)), : unused argument (colnames = c(ID, colnames(df))) How can I do this? I have tried to use write.table() as well, but it seems I never get to set the options right. Thanks a lot in advance for the help, Cheers, Luca __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] computationally singular
Hi Weiwei, I found your post on Mahalanobis distances five years ago. / Error in solve.default(cov, ...) : system is computationally singular: // reciprocal condition number = 1.09501e-25 /I have the same problem now - and I don't know how to deal with it. I would be thankful if you could help me with that. All the best, Matthias // [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Fwd: help
Topic: importing data from a web page Dear Sir/ Madam Hi. First I apologize for this question. But this also shows I'm inexperienced in using R. I need to import some tables from a web page (https) into R and analyze them through the readHTMLTable function. After loading XML package, the following commands are run: url - https://.../TopUp/CdrReport.aspx?lang=DK; mydata - readHTMLTable(url) and there is a warning: XML content does not seem to be XML: ' https:// .../TopUp/CdrReport.aspx?lang=DK' Would you do me a favor on this issue, please? Kind regards Mehrshad Koleini [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Row names at the end of the record
HI all How to add row names at the end of the record after transposing a matrix Sample code Data = read.table(set_a_2.txt,header = TRUE, sep = \t) Data.T - t(Data[,0:ncol(fooData)]) write.table(Data.T, file =set_a_2_format.tab,row.names = TRUE, col.names = TRUE, quote = FALSE) example of Transposed matrix class rs1 rs2 rs3 rs4 rs5 LG_1 BB AB BB BB BB LG_2 BB AA BB BB BB LG_3 BB BB AB AB BB LG_4 AA BB BB BB BB LG_5 BB AB BB BB BB how can i add row headers (LG_1) at the end of rows also Regards, Ankush Sharma [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] RSelenium package
I would like to announce the release of version 1.2.2 of the RSelenium package on CRAN. RSelenium is an R package that provides a set of R bindings for the Selenium 2.0 webdriver. Selenium automates browsers. Using RSelenium you can automate browsers locally or remotely. The set of supported browsers includes Chrome, Firefox, Internet Explorer, Opera, Safari, PhantomJS and HtmlUnit running on operating systems including Linux, Mac, Windows, Android, iOS, Firefox OS and Windows Phone. RSelenium has a project page at http://johndharrison.github.io/RSelenium/ http://cran.r-project.org/web/packages/RSelenium/index.html The package comes with several vignettes which contain overviews on basic operation and a few examples linking to projects such as Shiny and SauceLabs. Selenium uses include automated application testing, load testing and web scraping. Comments and suggestions are greatly appreciated. John Harrison [[alternative HTML version deleted]] ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Row names at the end of the record
Hi Not sure what you exactly want but some comments in line -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Ankush Sharma Sent: Wednesday, March 19, 2014 11:36 AM To: r-help@r-project.org Subject: [R] Row names at the end of the record HI all How to add row names at the end of the record after transposing a matrix Sample code Data = read.table(set_a_2.txt,header = TRUE, sep = \t) Data.T - t(Data[,0:ncol(fooData)]) Indexing by 0 although in this case it probably does not matter. write.table(Data.T, file =set_a_2_format.tab,row.names = TRUE, col.names = TRUE, quote = FALSE) example of Transposed matrix class rs1 rs2 rs3 rs4 rs5 LG_1 BB AB BB BB BB LG_2 BB AA BB BB BB LG_3 BB BB AB AB BB LG_4 AA BB BB BB BB LG_5 BB AB BB BB BB how can i add row headers (LG_1) at the end of rows also trans.matrix$whatever - transmatrix$class Or you want something different? Petr Regards, Ankush Sharma [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou určeny pouze jeho adresátům. Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze svého systému. Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat. Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či zpožděním přenosu e-mailu. V případě, že je tento e-mail součástí obchodního jednání: - vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření smlouvy, a to z jakéhokoliv důvodu i bez uvedení důvodu. - a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout; Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce s dodatkem či odchylkou. - trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným dosažením shody na všech jejích náležitostech. - odesílatel tohoto emailu informuje, že není oprávněn uzavírat za společnost žádné smlouvy s výjimkou případů, kdy k tomu byl písemně zmocněn nebo písemně pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu případně osobě, kterou adresát zastupuje, předloženy nebo jejich existence je adresátovi či osobě jím zastoupené známá. This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fwd: help
Hi Although I am not experienced with XML package the warning seems pretty straightforward to me. The object you want to load is not recognised as XML. Most probably readHTMLTable requires some format as input and your url is not accepted. Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Mehrshad Koleini Sent: Wednesday, March 19, 2014 11:10 AM To: r-help@r-project.org Subject: [R] Fwd: help Topic: importing data from a web page Dear Sir/ Madam Hi. First I apologize for this question. But this also shows I'm inexperienced in using R. I need to import some tables from a web page (https) into R and analyze them through the readHTMLTable function. After loading XML package, the following commands are run: url - https://.../TopUp/CdrReport.aspx?lang=DK; mydata - readHTMLTable(url) and there is a warning: XML content does not seem to be XML: ' https:// .../TopUp/CdrReport.aspx?lang=DK' Would you do me a favor on this issue, please? Kind regards Mehrshad Koleini [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou určeny pouze jeho adresátům. Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze svého systému. Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat. Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či zpožděním přenosu e-mailu. V případě, že je tento e-mail součástí obchodního jednání: - vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření smlouvy, a to z jakéhokoliv důvodu i bez uvedení důvodu. - a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout; Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce s dodatkem či odchylkou. - trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným dosažením shody na všech jejích náležitostech. - odesílatel tohoto emailu informuje, že není oprávněn uzavírat za společnost žádné smlouvy s výjimkou případů, kdy k tomu byl písemně zmocněn nebo písemně pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu případně osobě, kterou adresát zastupuje, předloženy nebo jejich existence je adresátovi či osobě jím zastoupené známá. This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help
On 19 Mar 2014, at 11:10 , Mehrshad Koleini mehrkole...@gmail.com wrote: I need to import some tables from a web page (https) into R and analyze them through the readHTMLTable function. After loading XML package, the following commands are run: url - https://.../TopUp/CdrReport.aspx?lang=DK; mydata - readHTMLTable(url) and there is a warning: XML content does not seem to be XML: ' https:// .../TopUp/CdrReport.aspx?lang=DK' Would you do me a favor on this issue, please? Maybe, but you're not giving us much to go on... If the page really isn't HTML or XML, then readHTMLTable() is not going to help you. So perhaps first see what is in there; either by opening the URL in a browser and using View Source, or by using con - url(url) ## you might want to reconsider the naming... readLines(con) close(con) -- Peter Dalgaard, Professor Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Row names at the end of the record
Is this what you're trying to do? mtrx - structure(c(BB, AB, BB, BB, BB, BB, AA, BB, BB, BB, BB, BB, AB, AB, BB, AA, BB, BB, BB, BB, BB, AB, BB, BB, BB), .Dim = c(5L, 5L), .Dimnames = list( c(rs1, rs2, rs3, rs4, rs5), c(LG_1, LG_2, LG_3, LG_4, LG_5))) mtrx - cbind(mtrx, dimnames(mtrx)[[1]]) write.table(mtrx, file =set_a_2_format.tab,row.names = TRUE, col.names = TRUE, quote = FALSE) Before saving simply cbind the row names onto the end of your matrix. NM On 2014-03-19 06:36, Ankush Sharma wrote: HI all How to add row names at the end of the record after transposing a matrix Sample code Data = read.table(set_a_2.txt,header = TRUE, sep = \t) Data.T - t(Data[,0:ncol(fooData)]) write.table(Data.T, file =set_a_2_format.tab,row.names = TRUE, col.names = TRUE, quote = FALSE) example of Transposed matrix class rs1 rs2 rs3 rs4 rs5 LG_1 BB AB BB BB BB LG_2 BB AA BB BB BB LG_3 BB BB AB AB BB LG_4 AA BB BB BB BB LG_5 BB AB BB BB BB how can i add row headers (LG_1) at the end of rows also Regards, Ankush Sharma [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fwd: R basic data manipulation Queries
HI, It seems like you used ?attach(). Better would be to use ?with. set.seed(45) dat1 - as.data.frame(matrix(sample(LETTERS,80,replace=TRUE),20,4)) lapply(dat1,levels) ##get the levels of all the dataset variables in a list with(dat1,levels(V1)) A.K. I have another question, at the moment I am looking at the levels for some of my variables in my dataset. Instead of typing levels statement everytime, is there a shortcut for doing that? So at the moment I am writing levels(time) levels(hour) level(date) levels(qrtr) and so on... can I do all this in one step? -Original Message- From: pavnee...@yahoo.co.uk Sent: Tue, 18 Mar 2014 10:10:38 + To: r-help@r-project.org Subject: [R] Fwd: R basic data manipulation Queries Sent from Samsung Mobile Original message Subject: R basic data manipulation Queries From: Pavneet Arora pavneet.ar...@uk.rsagroup.com To: pavnee...@yahoo.co.uk CC: Hello Guys I am new in R, so please excuse the really basic questions. I have tried reading numeral tutorials, but I am still stuck. Question 1: If I perform correlation on my data [cor(nums2)] or try to produce variance matrix [var(nums2)]. The output comes in R console. Is there any way I can make it go directly to excel somehow? Not exactly but fairly easily. Have a look at the R-site and go to the Manuals section. There is a R Import and Export manual there. Question 2: Also is there any way I can permanently change the variable name in a data frame. I basically imported my dataset from SAS using read.ssd function in library(foreign). However, this package cuts off the variable names and only allows 8 characters! So that means I will have to rename some of my variable names, so they make more sense as to what it is. part (a): At the moment, I did the following to change the variable names, using library(plyr). First of all, is this the most succint way of doing this? new_acc - rename(acc_mod,c(NAME_OF_=weekName, ACCIDENT=AccSev, YEARMONH=YrMonHr, X_1ST_ROA=1RdCls.N, ROAD_TYP=RdType.N, LOCATION=LocEast)) part (b): Secondly, I wanted to do the above, to change the name permanently - but I don't think it worked, because when I do the following, I get: class(LocEast) # New name of LOCATION class(LOCATION) R Output: class(LocEast) Error: object 'LocEast' not found class(LOCATION) [1] numeric Thanks so much!B [[alternative HTML version deleted]] I have never used plyr for thhis but just names(aac.mod) - c(Newname1, Newname2) and so on for a new name for each column in the data.frame should do it. Protect your computer files with professional cloud backup. Get PCRx Backup and upload unlimited files automatically. Learn more at http://backup.pcrx.com/mail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Does a survival probability(the probability not experiencing an event) have to be non-increasing?
My question is related to a cox model with time-dependent variable. When I think about it more, I get a little confused about non-increasing assumption for survival probability for an individual. For example, for a time-dependent ,say x, assuming increasing x increases the risk of event. Assume,time t1 t2. If at x at t1 x at t2, obviously, hazard at t1 will less than hazard at t2, assuming no other covariaates. But is it possible that s(t2|x at t2) s(t1|x at t1), since at t2, an individual is at greater risk. This is kind of confusing to me. Thanks for any helpful insights! Zhiyuan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] fitting
given the data below x1-c(-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1) y-c(-1,-1,-1,0,0,0,1,1,1,-1,-1,-1,0,0,0,1,1,1,-1,-1,-1,0,0,0,1,1,1) z-c(-1,-1,-1,-1,-1,-1,-1,-1,-1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1) using maple i can fit as below and get (1) #PolynomialFit(3, x1, y1, z) ; Fit(a+b*(x^2+y^2+z^2), Matrix([x1,y1,z1]),m1,[x,y,z]); (1) my question, is there a way we can do the same in R (i.e to obtain (1) using R) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ZOO objects: Creating, Plotting, Analyzing
I need to plot and analyze irregular time series. Having read the JSS article on zoo (2005), ?zoo, ?plot.zoo, and the quickref vignette I'm still not doing this correctly. My work flow is to read the *.csv file to create a dataframe, convert the df to a matrix using as.matrix() (see s95.ec.txt attached), then cr⎈eate a zoo object specifying 'sampdate' as the order.by index. The result is the second attachment (s95.ec.z.txt) which does not appear correct to me. Of the several zoo methods involving NA values, none of the imputations seem appropriate for my data as there is neither pattern or sufficiently short gaps between observed values, so I chose 'na.omit.' Putting that in the command, e.g., s95.ec.z - zoo(s95.ec, order.by = 'sampdate', na.omit) produced an error: s95.ec.z - zoo(s95.ec, order.by = 'sampdate', na.omit) Error in frequency 1 : comparison (6) is possible only for atomic and list types I suspect that the first hurdle I need to overcome is correctly creating a zoo object from the matrix. All the data I analyze have irregular collection dates and I need to become fluent with zoo objects so I can correctly analyze and plot the time series of these data. Rich structure(c(1992-03-12, 1992-04-22, 1992-05-22, 1992-06-03, 1992-11-06, 1993-01-28, 1993-02-18, 1993-03-03, 1993-03-15, 1993-04-08, 1993-04-30, 1993-05-11, 1993-05-27, 1993-06-10, 1993-06-25, 1993-07-09, 1993-07-31, 1994-03-28, 1994-04-07, 1994-04-26, 1994-05-27, 1994-06-23, 1995-04-11, 1995-04-29, 1995-05-30, 1995-06-23, 1995-07-26, 1995-12-04, 1996-01-16, 1996-05-07, 1996-06-05, 1996-07-02, 1997-04-30, 1997-05-14, 1997-06-16, 1997-07-29, 1998-05-25, 1998-05-28, 1998-07-27, 1998-09-15, 1998-10-21, 1998-11-24, 1999-05-10, 1999-05-25, 1999-06-22, 1999-07-22, 1999-11-09, 1999-12-01, 2000-03-04, 2000-03-10, 2000-04-21, 2000-05-09, 2000-06-28, 2000-11-27, 2000-12-07, 2001-02-16, 2001-03-07, 2001-04-30, 2001-05-11, 2001-06-25, 2002-05-04, 2002-06-21, 2003-05-03, 2003-06-15, 2004-06-28, 2004-07-01, 85, 80, 85, NA, 90, 110, 90, 100, 875, 85, 83, 75, 49, 65, 57, 55, 64, 60, 80, 80, 45, 65, 83, 60, 40, 62, 80, 119, 55, 65, 44, 48, 68, 43, 50, 70, 52, 45, 66, 92, 92, 134, 77, 52, 382, 72, 83, 78, 78, 77, 62, 51, 59, 79, 65, 77, 81, 59, 55, 67, 67, 16, 70, NA, 42, 42), .Dim = c(66L, 2L), .Dimnames = list( c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66), c(sampdate, EC))) structure(c(1992-03-12, 85), .Dim = 1:2, .Dimnames = list( 1, c(sampdate, EC)), index = sampdate, class = zoo) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with finding mean at 1 second interval
Hi Satish, Pascal's suggestion is better. Please use ?dput() to show the example data. op - options(digits.secs=6) library(zoo) ##added a couple more rows dat - read.zoo(text=V1 V2 V3 V4 V5 V6 '2014-03-14 22:41:46.988804' 10 2 8 3 14 '2014-03-14 22:41:46.991126' 13 4 9 5 15 '2014-03-14 22:41:46.993506' 12 4 8 3 14 '2014-03-14 22:41:46.993755' 19 4 15 5 22 '2014-03-14 22:41:46.997780' 21 5 16 7 24 '2014-03-14 22:41:47.000154' 18 5 13 3 21 '2014-03-14 22:41:47.002376' 21 5 16 6 23 '2014-03-14 22:41:47.011106' 12 4 8 3 14 '2014-03-14 22:41:47.012691' 12 4 8 3 16 '2014-03-14 22:41:47.017579' 11 2 9 3 12 '2014-03-14 22:41:47.019463' 12 5 7 3 15 '2014-03-14 22:41:47.020247' 14 6 8 3 17 '2014-03-15 22:41:47.017579' 11 2 9 3 12 '2014-03-15 22:41:47.019463' 12 5 7 3 15 '2014-03-15 22:41:47.020247' 14 6 8 3 17, index.column=1,sep=,header=TRUE, format= %Y-%m-%d %H:%M:%OS,FUN=as.POSIXct) options(op) library(xts) period.apply(dat,endpoints(dat,seconds),mean) # V2 V3 V4 V5 V6 #2014-03-14 22:41:46 15.0 3.80 11.20 4.60 17.8 #2014-03-14 22:41:47 14.28571 4.428571 9.857143 3.428571 16.85714 #2014-03-15 22:41:47 12.3 4.33 8.00 3.00 14.7 A.K. On Tuesday, March 18, 2014 7:27 PM, Pascal Oettli kri...@ymail.com wrote: Hello, Please have a look at the xts package. Please don't post in HTML. Regards, Pascal On Wed, Mar 19, 2014 at 12:26 AM, Satish Anupindi Rao satish.anupindi@ericsson.com wrote: Hi, I have a zoo object with the first column as index. The columns have not been named yet... but that I can change. It looks like this : V2 V3 V4 V5 V6 2014-03-14 22:41:46.988804 10 2 8 3 14 2014-03-14 22:41:46.991126 13 4 9 5 15 2014-03-14 22:41:46.993506 12 4 8 3 14 2014-03-14 22:41:46.993755 19 4 15 5 22 2014-03-14 22:41:46.997780 21 5 16 7 24 2014-03-14 22:41:47.000154 18 5 13 3 21 2014-03-14 22:41:47.002376 21 5 16 6 23 2014-03-14 22:41:47.011106 12 4 8 3 14 2014-03-14 22:41:47.012691 12 4 8 3 16 2014-03-14 22:41:47.017579 11 2 9 3 12 2014-03-14 22:41:47.019463 12 5 7 3 15 2014-03-14 22:41:47.020247 14 6 8 3 17 I would like to find the mean of the V2 to V6 columns on a per second interval. Would anyone please be able to help me with a function and implementation for that please? Thanks so much! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] package environment versus namespace environment
As the author of that post, I must respond... # - first comment --- Ducan: The article that Henrik cited gives a reasonable description up until near the end Suraj: Given the complexity of the topics at hand, the many accolades I've received for the post, and how thoroughly I've approached the topic and carefully navigated the user through rough terrain, I feel that characterizing my description as reasonable substantially shortchanges it. I wrote the post because the available descriptions (including those from well-established R community members) were confusing and filled with holes. # - next comment --- Duncan: ...up until near the end, where (in my opinion) it makes things unnecessarily complicated Suraj: I think Duncan is confusing iterative teaching, repetition, and thoroughness with complexity. These topics ARE complex. The need for a resource such as mine is a testament to that. My approach to complex topics is to break them down fully; to be diligent in the explanation so there are no leaps in logic. It's far more difficult to do that than to assume a reader can connect the dots and default to brevity. # - next comment --- Duncan: I'd recommend that you stop reading around where he tries to explain the dotted lines. Suraj: It's a disservice to yourself to stop here. This section provides a light bulb moment on the path to understanding the search/find mechanism. As stated in the blog post Function execution is the most complex piece of the puzzle. I specifically encourage readers to re-read that section. Other resources will provide the same understanding, but it's likely a cumbersome process for the average reader. I synthesized these insights after laboring over words in books and articles, tinkering in R, and asking lots of questions. # - next comment --- Duncan: In particular, ignore the second version of the Map of the World; the first one is accurate, the second is just misleading Suraj: A reader that doesn't understand the second Map cannot fully grasp the search/find mechanism. Everything in the post builds up to the second map. The second map encompasses function execution. I challenge anyone (in a friendly way) to create a better visualization of enclosing environments and function execution that allows a reader to easily navigate environment-space and figure out how R searches for symbols and where it finds them. I'll replace my map with yours if you can do that - we will all be better off for it! # - last comment --- Duncan: Gupta's article misses the possibility of packages that are loaded but not in the search path... Suraj: I'm pretty sure I discuss all of this in Imports v Depends section. I have a diagram that shows package PLYR as loaded but not in the search path and I discuss it. On Wed, Mar 19, 2014 at 1:12 PM, Mark Leeds marklee...@gmail.com wrote: Hi Everyone: Suraj will respond to Duncan's comments below promptly. Suraj doesn't have the original thread so I am just helping out by commenting here so that he can respond and the thread can be kept continuous. Mark On Sun, Mar 9, 2014 at 9:09 AM, Duncan Murdoch murdoch.dun...@gmail.com wrote: On 14-03-08 6:42 PM, Benjamin Tyner wrote: Duncan, Thank you for the informative link. So, do the loaded namespaces have an ordering akin to the package search path that determines that functions in the base namespace can see objects in the utils namespace? (I noticed that loadedNamespaces() just comes back in alphabetical order.) No. The article that Henrik cited gives a reasonable description up until near the end, where (in my opinion) it makes things unnecessarily complicated. I'd recommend that you stop reading around where he tries to explain the dotted lines. In particular, ignore the second version of the Map of the World; the first one is accurate, the second is just misleading. In answer to your question: Gupta's article misses the possibility of packages that are loaded but not in the search path. In the notation of the first part of that article, loading a namespace just puts it in the middle two columns (i.e. creates the namespace and imports environments) without putting it in the search list. That happens when you import or load a package without attaching it. The search path imposes an ordering, things that aren't in it aren't ordered. Duncan Murdoch Regards Ben On 03/07/2014 11:46 AM, Duncan Murdoch wrote: On 07/03/2014 10:16 AM, Benjamin Tyner wrote: Hello, I realize that a function in environment: base (for example, function head1 below) is unable to see (without resorting to ::, anyway) objects in utils (for example, head below), since package:base is after package:utils on the search path. However, I'm wondering what is the machinery that allows a function in environment: namespace:base (for example, function head2
Re: [R] package environment versus namespace environment
Hi Everyone: Suraj will respond to Duncan's comments below promptly. Suraj doesn't have the original thread so I am just helping out by commenting here so that he can respond and the thread can be kept continuous. Mark On Sun, Mar 9, 2014 at 9:09 AM, Duncan Murdoch murdoch.dun...@gmail.comwrote: On 14-03-08 6:42 PM, Benjamin Tyner wrote: Duncan, Thank you for the informative link. So, do the loaded namespaces have an ordering akin to the package search path that determines that functions in the base namespace can see objects in the utils namespace? (I noticed that loadedNamespaces() just comes back in alphabetical order.) No. The article that Henrik cited gives a reasonable description up until near the end, where (in my opinion) it makes things unnecessarily complicated. I'd recommend that you stop reading around where he tries to explain the dotted lines. In particular, ignore the second version of the Map of the World; the first one is accurate, the second is just misleading. In answer to your question: Gupta's article misses the possibility of packages that are loaded but not in the search path. In the notation of the first part of that article, loading a namespace just puts it in the middle two columns (i.e. creates the namespace and imports environments) without putting it in the search list. That happens when you import or load a package without attaching it. The search path imposes an ordering, things that aren't in it aren't ordered. Duncan Murdoch Regards Ben On 03/07/2014 11:46 AM, Duncan Murdoch wrote: On 07/03/2014 10:16 AM, Benjamin Tyner wrote: Hello, I realize that a function in environment: base (for example, function head1 below) is unable to see (without resorting to ::, anyway) objects in utils (for example, head below), since package:base is after package:utils on the search path. However, I'm wondering what is the machinery that allows a function in environment: namespace:base (for example, function head2 below) to be able to see head just fine, without needing to resort to ::. See Luke Tierney's article in R News, Name space management for R. Luke Tierney, R News, 3(1):2-6, June 2003 http://cran.r-project.org/doc/Rnews/Rnews_2003-1.pdf There's a link to it from the R help system. Run help.start(), then look at Technical papers in the Miscellaneous Material section. I believe most of what it says is still current; the only thing I can see at a glance that is no longer correct is that in those days namespaces were optional in packages. Now all packages have namespaces. Duncan Murdoch I'm also wondering more generally, why there is a need (practically speaking) for a distinction between the environment associated with a package and the environment associated with the namespace. $ export R_PROFILE=/home/btyner/Rprofile.site $ cat /home/btyner/Rprofile.site sys.source(/home/btyner/head1.R, envir = baseenv()) sys.source(/home/btyner/head2.R, envir = .BaseNamespaceEnv) $ cat /home/btyner/head1.R head1 - function(x) head(x) $ cat /home/btyner/head2.R head2 - function(x) head(x) $ Rscript -e head1(letters) Error in head1(letters) : could not find function head Execution halted $ Rscript -e head2(letters) [1] a b c d e f $ Rscript -e sessionInfo() R version 3.0.1 (2013-05-16) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=en_US.UTF-8LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=C LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets base Regards Ben __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- // __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/ posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reshaping Data in R - Transforming Two Columns Into One
Hi
Isn't melt just what you want?
melt(mydt)
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Abraham Mathew
Sent: Tuesday, March 18, 2014 1:59 AM
To: r-help@r-project.org (r-help@r-project.org)
Subject: [R] Reshaping Data in R - Transforming Two Columns Into One
I have the following data frame. Using the stringr package, I've
attempted
to map the url's to some specific elements that are in each url. I then
used the reshape package to join two different data frames. The next
step
is to transform the two columns in the mydt data frame (forester and
customer_support) into one column which specified whether or not it
contained that element or not (0 or 1).
url =
data.frame(u=c(http://www.subaru.com/vehicles/impreza/index.html;,
http://www.subaru.com/index.html?s_kwcid=subaruk_clickid=214495e6-
dbe0-6668-9222-3d7cd876prid=87k_affcode=76602
,
http://www.subaru.com/customer-support.html;,
http://www.subaru.com/;,
http://www.subaru.com/vehicles/forester/index.html;))
url
cs = c(customer-support)
f = c(forester)
one_match - str_c(cs, collapse = |)
two_match - str_c(f, collapse = |)
main - function(df) {
df$customer_support - as.numeric(str_detect(url$u, one_match))
df
}
d1 = main(url)
d1
main - function(df) {
df$forester - as.numeric(str_detect(url$u, two_match))
df
}
d2 = main(url)
d2
mydt = join(d1, d2)
str(mydt)
reshape(mydt, direction=long, idvar=u, varying=2:3, sep=)
reshape(mydt, varying=2:3, direction =long, idvar = u)
I've messed around with the reshape package but I'm not able to figure
it
out. Is there an alternative package or function I can use to get the
desired result.
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[R] Calculations with aggregate data: mean +/- standard deviation
Dear all, I have obtained the averages (means) and standard deviations (SD) of different variable using the aggregate function, which I have found very useful for these kind of computation. However I would like to calculate the lower and upper ends of the data (that is mean - SD and mean + SD) varible-wise, but I don't know how to cope with this aggregate data. I cannot simply subtract the aggregated results (which I have called AVG and SD) and the aggregate function is a bit too complicated for me. Does anybody knows how to add and subtract the means and standard deviations obtained using the aggregate function? Best regards, Luigi my.data-structure(list( column_1 = 1:120, column_2 = structure(c( 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8), .Label = c(Unstimulated, ESAT6, CFP10, Rv3615c, Rv2654, Rv3879, Rv3873, PHA), class = factor), column_3 = c( 192.0519108,183.6403531,53.46798757,83.60638077,69.60749873,159.4706861,256.8765622,499.2899303, 2170.799076,1411.349719,2759.472348,2098.973397,2164.739515,1288.676574,1611.486543,6205.229575, 870.7424981,465.9967135,191.8962375,864.0937485,2962.693675,1289.259137,2418.651212,7345.712517, 0,168.1198893,674.4342961,101.1575401,47.81596237,0,0,1420.793922, 142.6871331,5.466468742,291.9564635,80.73914133,73.02239621,64.47806871,144.3543635,3167.959757, 3164.748333,1092.634557,28733.20269,1207.87783,729.6090973,151.8706088,241.2466141,9600.963594, 1411.718287,12569.96285,1143.254476,6317.378481,16542.27718,79.68025792,1958.495138,7224.503437, 208.4382941,69.48609769,656.691151,0.499017582,7114.910926,187.6296174,41.73980805,8930.784541, 4.276752185,0.432300363,60.89228665,1.103924786,0.490686366,1.812993239,7.264531581,1518.610307, 2172.051528,595.8513744,17141.84336,589.6565971,1340.287628,117.350942,593.7034054,24043.61463, 0,81.83292179,1539.864321,36.41722958,8.385131047,161.7647376,65.21615696,7265.573875, 97.84753179,154.051827,0.613835842,10.06138851,45.04879285,176.8284258,18795.75462,3067686.769, 5780.34957,944.2200834,2398.235596,1083.393165,2541.714557,1251.670895,1547.178549,1792.679176, 3067.988416,8117.210173,23676.02226,8251.937547,17360.80494,18563.61561,16941.865,31453.96708, 2767.493803,4796.33016,12292.93705,3864.657567,9380.673835,14886.44683,8457.88646,26050.47191)), .Names = c(row, stimulation, copy), row.names = c(NA, -120L), class = data.frame) attach(my.data) AVG-aggregate(copy ~ stimulation , my.data, mean, na.rm = T) SD-aggregate(copy ~ stimulation , my.data, sd, na.rm = T) # ??? question: AVG-SD and AVG + SD ??? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ZOO objects: Creating, Plotting, Analyzing
I need to plot and analyze irregular time series. Having read the JSS article on zoo (2005), ?zoo, ?plot.zoo, and the quickref vignette I'm still not doing this correctly. This did not post to the list with the dput() outputs attached; they are at the end of this message. My work flow is to read the *.csv file to create a data.frame, subset the df to yield a date/data file for each constituent, convert the subset df to a matrix using as.matrix() (see s95.ec.txt attached), then create a zoo object specifying 'sampdate' as the order.by index. The result is the second attachment (s95.ec.z.txt) which does not appear correct to me. Of the several zoo methods involving NA values, none of the imputations seem appropriate for my data as there is neither pattern or sufficiently short gaps between observed values, so I chose 'na.omit.' Putting that in the command, e.g., s95.ec.z - zoo(s95.ec, order.by = 'sampdate', na.omit) produced an error: s95.ec.z - zoo(s95.ec, order.by = 'sampdate', na.omit) Error in frequency 1 : comparison (6) is possible only for atomic and list types I suspect that the first hurdle I need to overcome is correctly creating a zoo object from the matrix. All the data I analyze have irregular collection dates and I need to become fluent with zoo objects so I can correctly analyze and plot the time series of these data. Rich s95.ec.txt++ structure(c(1992-03-12, 1992-04-22, 1992-05-22, 1992-06-03, 1992-11-06, 1993-01-28, 1993-02-18, 1993-03-03, 1993-03-15, 1993-04-08, 1993-04-30, 1993-05-11, 1993-05-27, 1993-06-10, 1993-06-25, 1993-07-09, 1993-07-31, 1994-03-28, 1994-04-07, 1994-04-26, 1994-05-27, 1994-06-23, 1995-04-11, 1995-04-29, 1995-05-30, 1995-06-23, 1995-07-26, 1995-12-04, 1996-01-16, 1996-05-07, 1996-06-05, 1996-07-02, 1997-04-30, 1997-05-14, 1997-06-16, 1997-07-29, 1998-05-25, 1998-05-28, 1998-07-27, 1998-09-15, 1998-10-21, 1998-11-24, 1999-05-10, 1999-05-25, 1999-06-22, 1999-07-22, 1999-11-09, 1999-12-01, 2000-03-04, 2000-03-10, 2000-04-21, 2000-05-09, 2000-06-28, 2000-11-27, 2000-12-07, 2001-02-16, 2001-03-07, 2001-04-30, 2001-05-11, 2001-06-25, 2002-05-04, 2002-06-21, 2003-05-03, 2003-06-15, 2004-06-28, 2004-07-01, 85, 80, 85, NA, 90, 110, 90, 100, 875, 85, 83, 75, 49, 65, 57, 55, 64, 60, 80, 80, 45, 65, 83, 60, 40, 62, 80, 119, 55, 65, 44, 48, 68, 43, 50, 70, 52, 45, 66, 92, 92, 134, 77, 52, 382, 72, 83, 78, 78, 77, 62, 51, 59, 79, 65, 77, 81, 59, 55, 67, 67, 16, 70, NA, 42, 42), .Dim = c(66L, 2L), .Dimnames = list( c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66), c(sampdate, EC))) s95.ex.z.txt+++ structure(c(1992-03-12, 85), .Dim = 1:2, .Dimnames = list( 1, c(sampdate, EC)), index = sampdate, class = zoo) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] reading row title
Dear R-Family, I have a data-set of the following format. I am only presenting a part of it. DateA B C D 1-Jan-61 0.00 1.27 8.128 0.25 2-Jan-61 6.10 9.144 94.742 15.49 3-Jan-61 0.00 0.508 1.27 0.00 4-Jan-61 0.00 0 NA 0.00 5-Jan-61 0.00 0 0 0.00 6-Jan-61 0.00 NA 0 0.00 7-Jan-61 0.00 0 0 0.00 8-Jan-61 0.00 NA 0 0.00 9-Jan-61 0.00 NA 0 NA 10-Jan-61 0.00 4.064 4.826 0.76 You can see that each column has some NAs in it, What i want to do is to learn that which date has NA for each column. Here it should be something like the following A BC D 6-Jan-61 4-Jan-61 9-Jan-61 8-Jan-61 9-Jan-61 Thankyou very much in advance Eliza [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] reading row title
Dear Arun and Jim,
Thanks for your replies. I will be extremely careful next time.Anyway, here is
the dput format of dat
dput(dat)
structure(list(Date = c(1-Jan-61, 2-Jan-61, 3-Jan-61, 4-Jan-61,
5-Jan-61, 6-Jan-61, 7-Jan-61, 8-Jan-61, 9-Jan-61, 10-Jan-61), A =
c(0, 6.1, 0, 0, 0, 0, 0, 0, 0, 0), B = c(1.27, 9.144, 0.508, 0, 0, NA, 0, NA,
NA, 4.064), C = c(8.128, 94.742, 1.27, NA, 0, 0, 0, 0, 0, 4.826), D = c(0.25,
15.49, 0, 0, 0, 0, 0, 0, NA, 0.76)), .Names = c(Date, A, B, C, D),
class = data.frame, row.names = c(NA, -10L))
dput(res)
structure(list(A = c(, , ), B = c(6-Jan-61, 8-Jan-61, 9-Jan-61), C
= c(4-Jan-61, , ), D = c(9-Jan-61, , )), .Names = c(A, B, C,
D), row.names = c(NA, -3L), class = data.frame) Eliza
Date: Wed, 19 Mar 2014 12:11:43 -0700
From: smartpink...@yahoo.com
Subject: Re: [R] reading row title
To: eliza_bo...@hotmail.com
BTW, if you had looked at your post, you would understand how HTML mangled
your post.á It is better to dput().á You are posting for sometime now, so I
hope you understand it...
On Wednesday, March 19, 2014 3:04 PM, arun smartpink...@yahoo.com wrote:
Hi,
dat - read.table(text=Dateáá Aááá Báá Cááá D
1-Jan-61áá 0.00á 1.27áá 8.128á 0.25
2-Jan-61 6.10 9.144 94.742 15.49
3-Jan-61á 0.00 0.508á 1.27 0.00
4-Jan-61á 0.00á 0á NA 0.00
5-Jan-61á 0.00á 0á 0 0.00
6-Jan-61áá 0.00á NAá 0 0.00
7-Jan-61á 0.00á 0á 0 0.00
8-Jan-61á 0.00á NAá 0 0.00
9-Jan-61á 0.00á NAá 0 NA
10-Jan-61 0.00 4.064 4.826 0.76,sep=,header=TRUE,stringsAsFactors=FALSE)
res -á as.data.frame(apply(dat[,-1],2,FUN=function(x) {x1
-dat[,1][is.na(x)]; x2 - max(colSums(is.na(dat[,-1]))); if(length(x1) x2)
c(x1, rep(,x2-length(x1))) else x1}),stringsAsFactors=FALSE)
A.K.
On Wednesday, March 19, 2014 2:39 PM, eliza botto eliza_bo...@hotmail.com
wrote:
Dear R-Family,
I have a data-set of the following format. I am only presenting a part of it.
Dateá á á á á á á á Aá á á á á á á áá Bá á á á á á á á á á á á áá Cá á á á á
á á á á á á á á á á á áá D
á 1-Jan-61
á 0.00
á 1.27
á 8.128
á
á 0.25
á
2-Jan-61
á
6.10
á
9.144
á
94.742
á
á
15.49
á 3-Jan-61
á 0.00
á 0.508
á 1.27
á
á
0.00
á 4-Jan-61
á 0.00
á 0
á NA
á
á
0.00
á 5-Jan-61
á 0.00
á 0
á 0
á
á
0.00
á 6-Jan-61
á 0.00
á NA
á 0
á
á
0.00
á 7-Jan-61
á 0.00
á 0
á 0
á
á
0.00
á 8-Jan-61
á 0.00
á NA
á 0
á
á
0.00
á 9-Jan-61
á 0.00
á NA
á 0
á
á
NA
á
10-Jan-61
á
0.00
á
4.064
á
4.826
á
á
0.76
You can see that each column has some NAs in it, What i want to do is to
learn that which date has NA for each column.
Here it should be something like the following
Aá á á á á á á á á á á á á áá Bá á á á á á á á á á á á á á á á Cá á á á á á á
á á á á á á á á á á á á á D
á á á á á á á á á á áá 6-Jan-61á á á á á á á á á á á á áá 4-Jan-61á á á á á á
á á á á á áá 9-Jan-61
á á á á á á á á á á áá 8-Jan-61
á á á á á á á á á á áá 9-Jan-61
Thankyou very much in advance
Eliza ááá ááá áá ááá ááá á
ááá [[alternative HTML version deleted]]
__
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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.
[[alternative HTML version deleted]]
__
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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] reading row title
Hi,
dat - read.table(text=Date A B C D
1-Jan-61 0.00 1.27 8.128 0.25
2-Jan-61 6.10 9.144 94.742 15.49
3-Jan-61 0.00 0.508 1.27 0.00
4-Jan-61 0.00 0 NA 0.00
5-Jan-61 0.00 0 0 0.00
6-Jan-61 0.00 NA 0 0.00
7-Jan-61 0.00 0 0 0.00
8-Jan-61 0.00 NA 0 0.00
9-Jan-61 0.00 NA 0 NA
10-Jan-61 0.00 4.064 4.826 0.76,sep=,header=TRUE,stringsAsFactors=FALSE)
res - as.data.frame(apply(dat[,-1],2,FUN=function(x) {x1 -dat[,1][is.na(x)];
x2 - max(colSums(is.na(dat[,-1]))); if(length(x1) x2) c(x1,
rep(,x2-length(x1))) else x1}),stringsAsFactors=FALSE)
A.K.
On Wednesday, March 19, 2014 2:39 PM, eliza botto eliza_bo...@hotmail.com
wrote:
Dear R-Family,
I have a data-set of the following format. I am only presenting a part of it.
Date A B C
D
1-Jan-61
0.00
1.27
8.128
0.25
2-Jan-61
6.10
9.144
94.742
15.49
3-Jan-61
0.00
0.508
1.27
0.00
4-Jan-61
0.00
0
NA
0.00
5-Jan-61
0.00
0
0
0.00
6-Jan-61
0.00
NA
0
0.00
7-Jan-61
0.00
0
0
0.00
8-Jan-61
0.00
NA
0
0.00
9-Jan-61
0.00
NA
0
NA
10-Jan-61
0.00
4.064
4.826
0.76
You can see that each column has some NAs in it, What i want to do is to learn
that which date has NA for each column.
Here it should be something like the following
A B C
D
6-Jan-61 4-Jan-61
9-Jan-61
8-Jan-61
9-Jan-61
Thankyou very much in advance
Eliza
[[alternative HTML version deleted]]
__
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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.
__
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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] Calculations with aggregate data: mean +/- standard deviation
Hello, Maybe something like the following. res - merge(AVG, SD, by = stimulation) names(res)[2:3] - c(AVG, SD) res$lower - res$AVG - res$SD res$upper - res$AVG + res$SD res Hope this helps, Rui Barradas Em 19-03-2014 18:15, Luigi Marongiu escreveu: Dear all, I have obtained the averages (means) and standard deviations (SD) of different variable using the aggregate function, which I have found very useful for these kind of computation. However I would like to calculate the lower and upper ends of the data (that is mean - SD and mean + SD) varible-wise, but I don't know how to cope with this aggregate data. I cannot simply subtract the aggregated results (which I have called AVG and SD) and the aggregate function is a bit too complicated for me. Does anybody knows how to add and subtract the means and standard deviations obtained using the aggregate function? Best regards, Luigi my.data-structure(list( column_1 = 1:120, column_2 = structure(c( 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8), .Label = c(Unstimulated, ESAT6, CFP10, Rv3615c, Rv2654, Rv3879, Rv3873, PHA), class = factor), column_3 = c( 192.0519108,183.6403531,53.46798757,83.60638077,69.60749873,159.4706861,256.8765622,499.2899303, 2170.799076,1411.349719,2759.472348,2098.973397,2164.739515,1288.676574,1611.486543,6205.229575, 870.7424981,465.9967135,191.8962375,864.0937485,2962.693675,1289.259137,2418.651212,7345.712517, 0,168.1198893,674.4342961,101.1575401,47.81596237,0,0,1420.793922, 142.6871331,5.466468742,291.9564635,80.73914133,73.02239621,64.47806871,144.3543635,3167.959757, 3164.748333,1092.634557,28733.20269,1207.87783,729.6090973,151.8706088,241.2466141,9600.963594, 1411.718287,12569.96285,1143.254476,6317.378481,16542.27718,79.68025792,1958.495138,7224.503437, 208.4382941,69.48609769,656.691151,0.499017582,7114.910926,187.6296174,41.73980805,8930.784541, 4.276752185,0.432300363,60.89228665,1.103924786,0.490686366,1.812993239,7.264531581,1518.610307, 2172.051528,595.8513744,17141.84336,589.6565971,1340.287628,117.350942,593.7034054,24043.61463, 0,81.83292179,1539.864321,36.41722958,8.385131047,161.7647376,65.21615696,7265.573875, 97.84753179,154.051827,0.613835842,10.06138851,45.04879285,176.8284258,18795.75462,3067686.769, 5780.34957,944.2200834,2398.235596,1083.393165,2541.714557,1251.670895,1547.178549,1792.679176, 3067.988416,8117.210173,23676.02226,8251.937547,17360.80494,18563.61561,16941.865,31453.96708, 2767.493803,4796.33016,12292.93705,3864.657567,9380.673835,14886.44683,8457.88646,26050.47191)), .Names = c(row, stimulation, copy), row.names = c(NA, -120L), class = data.frame) attach(my.data) AVG-aggregate(copy ~ stimulation , my.data, mean, na.rm = T) SD-aggregate(copy ~ stimulation , my.data, sd, na.rm = T) # ??? question: AVG-SD and AVG + SD ??? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Does a survival probability(the probability not experiencing an event) have to be non-increasing?
Don't really understand your equations but. If X is the (random variable of) time of event, then the survival S(x) = 1 - Pr( X x ) has to be non increasing since: If t1 t2. then S(t1) - S(t2) = Pr(X t2) - Pr( X t1 ) = Pr( t1 X t2) 0 which means that S(t1) S(t2). But you must not confuse this with the conditional probability of survival given the age. On Wed, Mar 19, 2014 at 10:41 AM, Zhiyuan Sun sam.d@gmail.com wrote: My question is related to a cox model with time-dependent variable. When I think about it more, I get a little confused about non-increasing assumption for survival probability for an individual. For example, for a time-dependent ,say x, assuming increasing x increases the risk of event. Assume,time t1 t2. If at x at t1 x at t2, obviously, hazard at t1 will less than hazard at t2, assuming no other covariaates. But is it possible that s(t2|x at t2) s(t1|x at t1), since at t2, an individual is at greater risk. This is kind of confusing to me. Thanks for any helpful insights! Zhiyuan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cforest sampling methods
Hi all, I've been using the randomForest package and I'm trying to make the switch over to party. My problem is that I have an extremely unbalanced outcome (only 1% of the data has a positive outcome) which makes resampling methods necessary. randomForest has a very useful argument that is sampsize which allows me to use a balanced subsample to build each tree in my forest. lets say the number of positive cases is 100, my forest would look something like this: rf-randomForest(y~. ,data=train, ntree=800,replace=TRUE,sampsize = c(100, 100)) so I use 100 cases and 100 controls to build each individual tree. Can I do the same for cforests? I know I can always upsample but I'd rather not. I've tried playing around with the weights argument but I'm either not getting it right or it's just the wrong thing to use. Any advice on how to adapt cforests to datasets with imbalanced outcomes is greatly appreciated... Thanks! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cforest sampling methods
You might look at the 'bag' function in the caret package. It will not do the subsampling of variables at each split but you can bag a tree and down-sample the data at each iteration. The help page has an examples bagging ctree (although you might want to play with the tree depth a little). Max On Wed, Mar 19, 2014 at 3:32 PM, Maggie Makar maggieyma...@gmail.com wrote: Hi all, I've been using the randomForest package and I'm trying to make the switch over to party. My problem is that I have an extremely unbalanced outcome (only 1% of the data has a positive outcome) which makes resampling methods necessary. randomForest has a very useful argument that is sampsize which allows me to use a balanced subsample to build each tree in my forest. lets say the number of positive cases is 100, my forest would look something like this: rf-randomForest(y~. ,data=train, ntree=800,replace=TRUE,sampsize = c(100, 100)) so I use 100 cases and 100 controls to build each individual tree. Can I do the same for cforests? I know I can always upsample but I'd rather not. I've tried playing around with the weights argument but I'm either not getting it right or it's just the wrong thing to use. Any advice on how to adapt cforests to datasets with imbalanced outcomes is greatly appreciated... Thanks! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] package environment versus namespace environment
On 14-03-19 1:18 PM, Suraj Gupta wrote: As the author of that post, I must respond... # - first comment --- Ducan: The article that Henrik cited gives a reasonable description up until near the end Suraj: Given the complexity of the topics at hand, the many accolades I've received for the post, and how thoroughly I've approached the topic and carefully navigated the user through rough terrain, I feel that characterizing my description as reasonable substantially shortchanges it. I wrote the post because the available descriptions (including those from well-established R community members) were confusing and filled with holes. Okay, next time I write about it I'll be sure to make the positive parts more effusive in their praise. I can't remember all the details of my complaints now (I wrote the quoted lines 10 days ago), and I doubt if anyone would be interested in hearing them, so I'll just let them and your responses stand. Duncan Murdoch # - next comment --- Duncan: ...up until near the end, where (in my opinion) it makes things unnecessarily complicated Suraj: I think Duncan is confusing iterative teaching, repetition, and thoroughness with complexity. These topics ARE complex. The need for a resource such as mine is a testament to that. My approach to complex topics is to break them down fully; to be diligent in the explanation so there are no leaps in logic. It's far more difficult to do that than to assume a reader can connect the dots and default to brevity. # - next comment --- Duncan: I'd recommend that you stop reading around where he tries to explain the dotted lines. Suraj: It's a disservice to yourself to stop here. This section provides a light bulb moment on the path to understanding the search/find mechanism. As stated in the blog post Function execution is the most complex piece of the puzzle. I specifically encourage readers to re-read that section. Other resources will provide the same understanding, but it's likely a cumbersome process for the average reader. I synthesized these insights after laboring over words in books and articles, tinkering in R, and asking lots of questions. # - next comment --- Duncan: In particular, ignore the second version of the Map of the World; the first one is accurate, the second is just misleading Suraj: A reader that doesn't understand the second Map cannot fully grasp the search/find mechanism. Everything in the post builds up to the second map. The second map encompasses function execution. I challenge anyone (in a friendly way) to create a better visualization of enclosing environments and function execution that allows a reader to easily navigate environment-space and figure out how R searches for symbols and where it finds them. I'll replace my map with yours if you can do that - we will all be better off for it! # - last comment --- Duncan: Gupta's article misses the possibility of packages that are loaded but not in the search path... Suraj: I'm pretty sure I discuss all of this in Imports v Depends section. I have a diagram that shows package PLYR as loaded but not in the search path and I discuss it. On Wed, Mar 19, 2014 at 1:12 PM, Mark Leeds marklee...@gmail.com mailto:marklee...@gmail.com wrote: Hi Everyone: Suraj will respond to Duncan's comments below promptly. Suraj doesn't have the original thread so I am just helping out by commenting here so that he can respond and the thread can be kept continuous. Mark On Sun, Mar 9, 2014 at 9:09 AM, Duncan Murdoch murdoch.dun...@gmail.com mailto:murdoch.dun...@gmail.comwrote: On 14-03-08 6:42 PM, Benjamin Tyner wrote: Duncan, Thank you for the informative link. So, do the loaded namespaces have an ordering akin to the package search path that determines that functions in the base namespace can see objects in the utils namespace? (I noticed that loadedNamespaces() just comes back in alphabetical order.) No. The article that Henrik cited gives a reasonable description up until near the end, where (in my opinion) it makes things unnecessarily complicated. I'd recommend that you stop reading around where he tries to explain the dotted lines. In particular, ignore the second version of the Map of the World; the first one is accurate, the second is just misleading. In answer to your question: Gupta's article misses the possibility of packages that are loaded but not in the search path. In the notation of the first part of that article, loading a namespace just puts it in the middle two columns (i.e. creates the namespace and imports environments) without putting it in the search list. That happens when you import or load a package without attaching it. The search path imposes an ordering,
[R] ZOO objects: Creating, Plotting, Analyzing
I need to plot and analyze irregular time series. Despite reading the 2005 JSS article on zoo, ?zoo, ?plot.zoo, and the quickref vignette I'm still not doing this correctly. dput() output of a matrix and the zoo object are at the end of this message. My work flow is to read the *.csv file to create a data.frame, subset the df to yield a date/data file for each constituent, convert the subset df to a matrix using as.matrix() (see s95.ec.txt attached), then create a zoo object specifying 'sampdate' as the order.by index. The result is the second attachment (s95.ec.z.txt) which does not appear correct to me. Of the several zoo methods involving NA values, none of the imputations seem appropriate for my data as there is neither pattern or sufficiently short gaps between observed values, so I chose 'na.omit.' Putting that in the command, e.g., s95.ec.z - zoo(s95.ec, order.by = 'sampdate', na.omit) produced an error: s95.ec.z - zoo(s95.ec, order.by = 'sampdate', na.omit) Error in frequency 1 : comparison (6) is possible only for atomic and list types I suspect that the first hurdle I need to overcome is correctly creating a zoo object from the matrix. All the data I analyze have irregular collection dates and I need to become fluent with zoo objects so I can correctly analyze and plot the time series of these data. Rich s95.ec.txt++ structure(c(1992-03-12, 1992-04-22, 1992-05-22, 1992-06-03, 1992-11-06, 1993-01-28, 1993-02-18, 1993-03-03, 1993-03-15, 1993-04-08, 1993-04-30, 1993-05-11, 1993-05-27, 1993-06-10, 1993-06-25, 1993-07-09, 1993-07-31, 1994-03-28, 1994-04-07, 1994-04-26, 1994-05-27, 1994-06-23, 1995-04-11, 1995-04-29, 1995-05-30, 1995-06-23, 1995-07-26, 1995-12-04, 1996-01-16, 1996-05-07, 1996-06-05, 1996-07-02, 1997-04-30, 1997-05-14, 1997-06-16, 1997-07-29, 1998-05-25, 1998-05-28, 1998-07-27, 1998-09-15, 1998-10-21, 1998-11-24, 1999-05-10, 1999-05-25, 1999-06-22, 1999-07-22, 1999-11-09, 1999-12-01, 2000-03-04, 2000-03-10, 2000-04-21, 2000-05-09, 2000-06-28, 2000-11-27, 2000-12-07, 2001-02-16, 2001-03-07, 2001-04-30, 2001-05-11, 2001-06-25, 2002-05-04, 2002-06-21, 2003-05-03, 2003-06-15, 2004-06-28, 2004-07-01, 85, 80, 85, NA, 90, 110, 90, 100, 875, 85, 83, 75, 49, 65, 57, 55, 64, 60, 80, 80, 45, 65, 83, 60, 40, 62, 80, 119, 55, 65, 44, 48, 68, 43, 50, 70, 52, 45, 66, 92, 92, 134, 77, 52, 382, 72, 83, 78, 78, 77, 62, 51, 59, 79, 65, 77, 81, 59, 55, 67, 67, 16, 70, NA, 42, 42), .Dim = c(66L, 2L), .Dimnames = list( c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66), c(sampdate, EC))) s95.ex.z.txt+++ structure(c(1992-03-12, 85), .Dim = 1:2, .Dimnames = list( 1, c(sampdate, EC)), index = sampdate, class = zoo) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] 2013-06-28 coverts to 15884?
MaxUpdated_row-NULL MaxUpdated_val- 2013-06-28 rbind(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [,1] [1,] 15884 c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [1] 15884 c(MaxUpdated_row, MaxUpdated_val) [1] 15884 Evidently, I'm again missing something simple, as I would prefer to be able to see the actual date to be shown. I found a work around, but I don't like it, as I have to convert back to date later: c(MaxUpdated_row, as.character(MaxUpdated_val)) Any alternatives suggestions are much appreciated. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 2013-06-28 coverts to 15884?
On 03/20/2014 08:46 AM, Jason Rupert wrote: MaxUpdated_row-NULL MaxUpdated_val- 2013-06-28 rbind(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [,1] [1,] 15884 c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [1] 15884 c(MaxUpdated_row, MaxUpdated_val) [1] 15884 Evidently, I'm again missing something simple, as I would prefer to be able to see the actual date to be shown. I found a work around, but I don't like it, as I have to convert back to date later: c(MaxUpdated_row, as.character(MaxUpdated_val)) Hi Jason, You aren't converting the date to something else, just displaying it as a character vector by using the format function. Whenever you form a vector, R tries to convert everything in it to the same data type. If you _assign_ that vector to something else: mymax-c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) you will have the date type in the MaxUpdated_val and the character type in mymax. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 2013-06-28 coverts to 15884?
Not sure I follow. I would like the date formatting information preserved as is implemented when using as.character: MaxUpdated_row-NULL MaxUpdated_val- 2013-06-28 c(MaxUpdated_row, as.character(MaxUpdated_val)) [1] 2013-06-28 However, when I try to used as.Date(...) I loose the date formatting information: c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [1] 15884 I thought as.Date was an instruction describing how the data element should be represented? Maybe as.character is the only way to preserve the date formatting information in character string format in a vector and then just convert it back later on... Thanks again. On Wednesday, March 19, 2014 5:42 PM, Jim Lemon j...@bitwrit.com.au wrote: On 03/20/2014 08:46 AM, Jason Rupert wrote: MaxUpdated_row-NULL MaxUpdated_val- 2013-06-28 rbind(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [,1] [1,] 15884 c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [1] 15884 c(MaxUpdated_row, MaxUpdated_val) [1] 15884 Evidently, I'm again missing something simple, as I would prefer to be able to see the actual date to be shown. I found a work around, but I don't like it, as I have to convert back to date later: c(MaxUpdated_row, as.character(MaxUpdated_val)) Hi Jason, You aren't converting the date to something else, just displaying it as a character vector by using the format function. Whenever you form a vector, R tries to convert everything in it to the same data type. If you _assign_ that vector to something else: mymax-c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) you will have the date type in the MaxUpdated_val and the character type in mymax. Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 2013-06-28 coverts to 15884?
Instead of making your initial value of MaxUpdated_row NULL, make it a 0-long Date object. Then c() will work as you wish (but not rbind). E.g., MaxUpdated_row - as.Date(character(0)) MaxUpdated_row - c(MaxUpdated_row, as.Date(2013-06-28, %Y-%m-%d)) MaxUpdated_row - c(MaxUpdated_row, as.Date(2013-07-06, %Y-%m-%d)) str(MaxUpdated_row) Date[1:2], format: 2013-06-28 2013-07-06 Bill Dunlap TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jason Rupert Sent: Wednesday, March 19, 2014 2:47 PM To: R-help@r-project.org Subject: [R] 2013-06-28 coverts to 15884? MaxUpdated_row-NULL MaxUpdated_val- 2013-06-28 rbind(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [,1] [1,] 15884 c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [1] 15884 c(MaxUpdated_row, MaxUpdated_val) [1] 15884 Evidently, I'm again missing something simple, as I would prefer to be able to see the actual date to be shown. I found a work around, but I don't like it, as I have to convert back to date later: c(MaxUpdated_row, as.character(MaxUpdated_val)) Any alternatives suggestions are much appreciated. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] package environment versus namespace environment
Hi Duncan: If you think there is anything incorrect in the document, I'd be interested because I refer to that document whenever I'm confused about namespaces, importing etc. Thanks. On Wed, Mar 19, 2014 at 4:31 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote: On 14-03-19 1:18 PM, Suraj Gupta wrote: As the author of that post, I must respond... # - first comment --- Ducan: The article that Henrik cited gives a reasonable description up until near the end Suraj: Given the complexity of the topics at hand, the many accolades I've received for the post, and how thoroughly I've approached the topic and carefully navigated the user through rough terrain, I feel that characterizing my description as reasonable substantially shortchanges it. I wrote the post because the available descriptions (including those from well-established R community members) were confusing and filled with holes. Okay, next time I write about it I'll be sure to make the positive parts more effusive in their praise. I can't remember all the details of my complaints now (I wrote the quoted lines 10 days ago), and I doubt if anyone would be interested in hearing them, so I'll just let them and your responses stand. Duncan Murdoch # - next comment --- Duncan: ...up until near the end, where (in my opinion) it makes things unnecessarily complicated Suraj: I think Duncan is confusing iterative teaching, repetition, and thoroughness with complexity. These topics ARE complex. The need for a resource such as mine is a testament to that. My approach to complex topics is to break them down fully; to be diligent in the explanation so there are no leaps in logic. It's far more difficult to do that than to assume a reader can connect the dots and default to brevity. # - next comment --- Duncan: I'd recommend that you stop reading around where he tries to explain the dotted lines. Suraj: It's a disservice to yourself to stop here. This section provides a light bulb moment on the path to understanding the search/find mechanism. As stated in the blog post Function execution is the most complex piece of the puzzle. I specifically encourage readers to re-read that section. Other resources will provide the same understanding, but it's likely a cumbersome process for the average reader. I synthesized these insights after laboring over words in books and articles, tinkering in R, and asking lots of questions. # - next comment --- Duncan: In particular, ignore the second version of the Map of the World; the first one is accurate, the second is just misleading Suraj: A reader that doesn't understand the second Map cannot fully grasp the search/find mechanism. Everything in the post builds up to the second map. The second map encompasses function execution. I challenge anyone (in a friendly way) to create a better visualization of enclosing environments and function execution that allows a reader to easily navigate environment-space and figure out how R searches for symbols and where it finds them. I'll replace my map with yours if you can do that - we will all be better off for it! # - last comment --- Duncan: Gupta's article misses the possibility of packages that are loaded but not in the search path... Suraj: I'm pretty sure I discuss all of this in Imports v Depends section. I have a diagram that shows package PLYR as loaded but not in the search path and I discuss it. On Wed, Mar 19, 2014 at 1:12 PM, Mark Leeds marklee...@gmail.com mailto:marklee...@gmail.com wrote: Hi Everyone: Suraj will respond to Duncan's comments below promptly. Suraj doesn't have the original thread so I am just helping out by commenting here so that he can respond and the thread can be kept continuous. Mark On Sun, Mar 9, 2014 at 9:09 AM, Duncan Murdoch murdoch.dun...@gmail.com mailto:murdoch.dun...@gmail.comwrote: On 14-03-08 6:42 PM, Benjamin Tyner wrote: Duncan, Thank you for the informative link. So, do the loaded namespaces have an ordering akin to the package search path that determines that functions in the base namespace can see objects in the utils namespace? (I noticed that loadedNamespaces() just comes back in alphabetical order.) No. The article that Henrik cited gives a reasonable description up until near the end, where (in my opinion) it makes things unnecessarily complicated. I'd recommend that you stop reading around where he tries to explain the dotted lines. In particular, ignore the second version of the Map of the World; the first one is accurate, the second is just misleading. In answer to your question: Gupta's article misses the possibility of packages that are loaded but not in the search path.
[R] How to fix the warning message the condition has length 1 and only the first element will be used?
I'm trying to create a function to return the date x months in the past.
With the code below I'm getting the warning message:
Warning message:
In if (MonthsBack = CurrentMonth) { :
the condition has length 1 and only the first element will be used
##
DateBack - function(CurrentYear, CurrentMonth, MonthsBack){
if(MonthsBack 13){
if(MonthsBack=CurrentMonth){
month-12+CurrentMonth-MonthsBack
datet-paste(CurrentYear-1,month, sep=-)
}
else{
month-CurrentMonth-MonthsBack
datet-paste(CurrentYear-1, month, sep=-)
}
}
else{
Years - trunc(MonthsBack/12,0)
if((MonthsBack-12*Years)=CurrentMonth){
month-12+CurrentMonth-MonthsBack
}
else{
month-CurrentMonth-MonthsBack
}
datet-paste(CurrentYear-Years, month, sep=-)
}
return(datet)
}
CurrentYear- c(2000, 2000, 2003, 2004)
CurrentMonth-c(1, 2, 6, 12)
df- data.frame(CurrentYear, CurrentMonth)
df$MonthsBack - DateBack(df$CurrentYear,df$CurrentMonth,1)
--
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Re: [R] 2013-06-28 coverts to 15884?
Hello, It works for me: R MaxUpdated_row - NULL R MaxUpdated_val - 2013-06-28 R R R rbind(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [,1] [1,] 15884 R c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [1] 15884 R c(MaxUpdated_row, MaxUpdated_val) [1] 2013-06-28 R Regards, Pascal On Thu, Mar 20, 2014 at 6:46 AM, Jason Rupert jasonkrup...@yahoo.com wrote: MaxUpdated_row-NULL MaxUpdated_val- 2013-06-28 rbind(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [,1] [1,] 15884 c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [1] 15884 c(MaxUpdated_row, MaxUpdated_val) [1] 15884 Evidently, I'm again missing something simple, as I would prefer to be able to see the actual date to be shown. I found a work around, but I don't like it, as I have to convert back to date later: c(MaxUpdated_row, as.character(MaxUpdated_val)) Any alternatives suggestions are much appreciated. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to fix the warning message the condition has length 1 and only the first element will be used?
Hi jcrosbie ,
Re:
I'm trying to create a function to return the date x months in the past.
With the code below I'm getting the warning message:
Warning message:
In if (MonthsBack = CurrentMonth) { :
the condition has length 1 and only the first element will be used
Use ifelse(), that's for vectors. If is intended for single elements only.
Best wishes,
Franklin
-
Franklin Bretschneider
Dept of Biology
Utrecht University
brets...@xs4all.nl
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Re: [R] ZOO objects: Creating, Plotting, Analyzing
On Wed, 19 Mar 2014, Rich Shepard wrote: I need to plot and analyze irregular time series. Despite reading the 2005 JSS article on zoo, ?zoo, ?plot.zoo, and the quickref vignette I'm still not doing this correctly. dput() output of a matrix and the zoo object are at the end of this message. It's sufficient to send such messages once even if others don't reply within minutes. My work flow is to read the *.csv file to create a data.frame, subset the df to yield a date/data file for each constituent, convert the subset df to a matrix using as.matrix() (see s95.ec.txt attached), then create a zoo object specifying 'sampdate' as the order.by index. The result is the second attachment (s95.ec.z.txt) which does not appear correct to me. Please have a look at read.zoo() and the accompanying read.zoo vignette which does exactly what you are looking for (reading .csv files and turning it into a zoo series). Of the several zoo methods involving NA values, none of the imputations seem appropriate for my data as there is neither pattern or sufficiently short gaps between observed values, so I chose 'na.omit.' Putting that in the command, e.g., s95.ec.z - zoo(s95.ec, order.by = 'sampdate', na.omit) produced an error: s95.ec.z - zoo(s95.ec, order.by = 'sampdate', na.omit) Error in frequency 1 : comparison (6) is possible only for atomic and list types I suspect that the first hurdle I need to overcome is correctly creating a zoo object from the matrix. Yes. The first argument should be a numeric vector or matrix and the second argument should be something suitable as a time index. In your case s95.ec is a data.frame with two character columns which is not suitable directly. This works: s95.ec.z - zoo(as.numeric(s95.ec[,2]), order.by = as.Date(s95.ec[,1])) plot(s95.ec.z) But the route via read.zoo should be more convenient, I guess. All the data I analyze have irregular collection dates and I need to become fluent with zoo objects so I can correctly analyze and plot the time series of these data. Rich s95.ec.txt++ structure(c(1992-03-12, 1992-04-22, 1992-05-22, 1992-06-03, 1992-11-06, 1993-01-28, 1993-02-18, 1993-03-03, 1993-03-15, 1993-04-08, 1993-04-30, 1993-05-11, 1993-05-27, 1993-06-10, 1993-06-25, 1993-07-09, 1993-07-31, 1994-03-28, 1994-04-07, 1994-04-26, 1994-05-27, 1994-06-23, 1995-04-11, 1995-04-29, 1995-05-30, 1995-06-23, 1995-07-26, 1995-12-04, 1996-01-16, 1996-05-07, 1996-06-05, 1996-07-02, 1997-04-30, 1997-05-14, 1997-06-16, 1997-07-29, 1998-05-25, 1998-05-28, 1998-07-27, 1998-09-15, 1998-10-21, 1998-11-24, 1999-05-10, 1999-05-25, 1999-06-22, 1999-07-22, 1999-11-09, 1999-12-01, 2000-03-04, 2000-03-10, 2000-04-21, 2000-05-09, 2000-06-28, 2000-11-27, 2000-12-07, 2001-02-16, 2001-03-07, 2001-04-30, 2001-05-11, 2001-06-25, 2002-05-04, 2002-06-21, 2003-05-03, 2003-06-15, 2004-06-28, 2004-07-01, 85, 80, 85, NA, 90, 110, 90, 100, 875, 85, 83, 75, 49, 65, 57, 55, 64, 60, 80, 80, 45, 65, 83, 60, 40, 62, 80, 119, 55, 65, 44, 48, 68, 43, 50, 70, 52, 45, 66, 92, 92, 134, 77, 52, 382, 72, 83, 78, 78, 77, 62, 51, 59, 79, 65, 77, 81, 59, 55, 67, 67, 16, 70, NA, 42, 42), .Dim = c(66L, 2L), .Dimnames = list( c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66), c(sampdate, EC))) s95.ex.z.txt+++ structure(c(1992-03-12, 85), .Dim = 1:2, .Dimnames = list( 1, c(sampdate, EC)), index = sampdate, class = zoo) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ZOO objects: Creating, Plotting, Analyzing
On Thu, 20 Mar 2014, Achim Zeileis wrote: It's sufficient to send such messages once even if others don't reply within minutes. Achim, I sent it but my postfix spam filter would not let it out. There should have been only a single copy. Please have a look at read.zoo() and the accompanying read.zoo vignette which does exactly what you are looking for (reading .csv files and turning it into a zoo series). Thank you, I'll do that. I missed that help file and vignette. Yes. The first argument should be a numeric vector or matrix and the second argument should be something suitable as a time index. In your case s95.ec is a data.frame with two character columns which is not suitable directly. This works: s95.ec.z - zoo(as.numeric(s95.ec[,2]), order.by = as.Date(s95.ec[,1])) plot(s95.ec.z) But the route via read.zoo should be more convenient, I guess. That clarifies my mis-understanding. I'll read about read.zoo, too. Thank you, Rich __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 2013-06-28 coverts to 15884?
On 03/20/2014 09:57 AM, Jason Rupert wrote: Not sure I follow. I would like the date formatting information preserved as is implemented when using as.character: MaxUpdated_row-NULL MaxUpdated_val- 2013-06-28 c(MaxUpdated_row, as.character(MaxUpdated_val)) [1] 2013-06-28 However, when I try to used as.Date(...) I loose the date formatting information: c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d)) [1] 15884 I thought as.Date was an instruction describing how the data element should be represented? Maybe as.character is the only way to preserve the date formatting information in character string format in a vector and then just convert it back later on... As I wrote before, all of the data types in a vector object in R must be the same. So when you catenate (c) two things that are different data types, R coerces them to the same type. Because you are just displaying the values in the vector you have created, it doesn't change the data type of MaxUpdated_val: MaxUpdated_val-as.Date(2013-06-28,%Y-%m-%d) # prints as a formatted date MaxUpdated_val [1] 2013-06-28 # make a vector with a numeric and MaxUpdated_val c(0,MaxUpdated_val) # R coerces the value to numeric and prints both [1] 0 15884 # now assign it to something mynewmax-c(0,MaxUpdated_val) # in mynewmax the value is numeric mynewmax [1] 0 15884 # but MaxUpdated_val is still a date type on its own MaxUpdated_val [1] 2013-06-28 Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
