Re: [R] ts instead of xts object
Hello Pascal, Yes that is what I was worried about. The date-stamps are there and I would like to use that information but I think using as.ts will not do this. Does anyone know how this is done? Thank you. On Wed, Mar 12, 2014 at 1:02 AM, Pascal Oettli kri...@ymail.com wrote: Hello, On Tue, Mar 11, 2014 at 8:45 PM, Joshua Ulrich josh.m.ulr...@gmail.com wrote: On Tue, Mar 11, 2014 at 12:14 AM, Bill william...@gmail.com wrote: Hello. I have a dataframe that has a date column. The intervals between dates vary. I want to convert this to a ts object. I was able to convert it to an xts object but the package I want to analyse this data with (called 'changepoint') does not seem to want to deal with xts. In the example they give they use the following: data(discoveries) dis.pelt=cpt.meanvar(discoveries,test.stat='Poisson',method='PELT') plot(dis.pelt,cpt.width=3) cpts.ts(dis.pelt) and if I check: str(discoveries) Time-Series [1:100] from 1860 to 1959: 5 3 0 2 0 3 2 3 6 1 ... If I try with my data str(testTSRad) An 'xts' object on 2011-07-16 07:08:02/2013-09-20 01:25:48 containing: Data: num [1:501, 1] 76 77 79 86 79 79 85 86 89 88 ... Indexed by objects of class: [POSIXct,POSIXt] TZ: xts Attributes: NULL where I used this: testTSRad=xts(radSampPerRegion[[2]][ ,2],order.by=as.POSIXct(radSampPerRegion[[2]][ ,1])) I get this: testt=cpt.mean(testTSRad) Error in single.mean.norm(data, penalty, pen.value, class, param.estimates) : Data must have atleast 2 observations to fit a changepoint model. This is because of what ?cpt.mean says about the data argument: data: A vector, ts object or matrix containing the data within which you wish to find a changepoint. If data is a matrix, each row is considered a separate dataset. An xts object is a matrix (with an index attribute), so each row is considered a separate data set. Your object only has one column, hence only one observation per data set. Things will work if you drop the dimensions of your single-column xts object: testt - cpt.mean(drop(testTSRad)) My data is below. Is there a way to convert it to ts? Yes, as is generally the case, use the as method: as.ts(testTSRad) But in this case, the time serie will have a frequency of 1, which is inconsistent with irregular sampling. This probably will lead to inaccurate results Best, -- Joshua Ulrich | about.me/joshuaulrich FOSS Trading | www.fosstrading.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. Regards, Pascal -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan [[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] ts instead of xts object
On Sun, Mar 16, 2014 at 6:27 AM, Bill william...@gmail.com wrote: Hello Pascal, Yes that is what I was worried about. The date-stamps are there and I would like to use that information but I think using as.ts will not do this. Does anyone know how this is done? It cannot be done. The ts class is used to represent data which has been sampled at equispaced points in time. You need to look at the changepoint package code (or ask the author(s)) to determine whether or not the frequency attribute is used. Thank you. On Wed, Mar 12, 2014 at 1:02 AM, Pascal Oettli kri...@ymail.com wrote: Hello, On Tue, Mar 11, 2014 at 8:45 PM, Joshua Ulrich josh.m.ulr...@gmail.com wrote: On Tue, Mar 11, 2014 at 12:14 AM, Bill william...@gmail.com wrote: Hello. I have a dataframe that has a date column. The intervals between dates vary. I want to convert this to a ts object. I was able to convert it to an xts object but the package I want to analyse this data with (called 'changepoint') does not seem to want to deal with xts. In the example they give they use the following: data(discoveries) dis.pelt=cpt.meanvar(discoveries,test.stat='Poisson',method='PELT') plot(dis.pelt,cpt.width=3) cpts.ts(dis.pelt) and if I check: str(discoveries) Time-Series [1:100] from 1860 to 1959: 5 3 0 2 0 3 2 3 6 1 ... If I try with my data str(testTSRad) An 'xts' object on 2011-07-16 07:08:02/2013-09-20 01:25:48 containing: Data: num [1:501, 1] 76 77 79 86 79 79 85 86 89 88 ... Indexed by objects of class: [POSIXct,POSIXt] TZ: xts Attributes: NULL where I used this: testTSRad=xts(radSampPerRegion[[2]][ ,2],order.by=as.POSIXct(radSampPerRegion[[2]][ ,1])) I get this: testt=cpt.mean(testTSRad) Error in single.mean.norm(data, penalty, pen.value, class, param.estimates) : Data must have atleast 2 observations to fit a changepoint model. This is because of what ?cpt.mean says about the data argument: data: A vector, ts object or matrix containing the data within which you wish to find a changepoint. If data is a matrix, each row is considered a separate dataset. An xts object is a matrix (with an index attribute), so each row is considered a separate data set. Your object only has one column, hence only one observation per data set. Things will work if you drop the dimensions of your single-column xts object: testt - cpt.mean(drop(testTSRad)) My data is below. Is there a way to convert it to ts? Yes, as is generally the case, use the as method: as.ts(testTSRad) But in this case, the time serie will have a frequency of 1, which is inconsistent with irregular sampling. This probably will lead to inaccurate results Best, -- Joshua Ulrich | about.me/joshuaulrich FOSS Trading | www.fosstrading.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. Regards, Pascal -- 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] ts instead of xts object
Joshua, Thanks. I am going to check. On Mon, Mar 17, 2014 at 11:10 AM, Joshua Ulrich josh.m.ulr...@gmail.comwrote: On Sun, Mar 16, 2014 at 6:27 AM, Bill william...@gmail.com wrote: Hello Pascal, Yes that is what I was worried about. The date-stamps are there and I would like to use that information but I think using as.ts will not do this. Does anyone know how this is done? It cannot be done. The ts class is used to represent data which has been sampled at equispaced points in time. You need to look at the changepoint package code (or ask the author(s)) to determine whether or not the frequency attribute is used. Thank you. On Wed, Mar 12, 2014 at 1:02 AM, Pascal Oettli kri...@ymail.com wrote: Hello, On Tue, Mar 11, 2014 at 8:45 PM, Joshua Ulrich josh.m.ulr...@gmail.com wrote: On Tue, Mar 11, 2014 at 12:14 AM, Bill william...@gmail.com wrote: Hello. I have a dataframe that has a date column. The intervals between dates vary. I want to convert this to a ts object. I was able to convert it to an xts object but the package I want to analyse this data with (called 'changepoint') does not seem to want to deal with xts. In the example they give they use the following: data(discoveries) dis.pelt=cpt.meanvar(discoveries,test.stat='Poisson',method='PELT') plot(dis.pelt,cpt.width=3) cpts.ts(dis.pelt) and if I check: str(discoveries) Time-Series [1:100] from 1860 to 1959: 5 3 0 2 0 3 2 3 6 1 ... If I try with my data str(testTSRad) An 'xts' object on 2011-07-16 07:08:02/2013-09-20 01:25:48 containing: Data: num [1:501, 1] 76 77 79 86 79 79 85 86 89 88 ... Indexed by objects of class: [POSIXct,POSIXt] TZ: xts Attributes: NULL where I used this: testTSRad=xts(radSampPerRegion[[2]][ ,2],order.by=as.POSIXct(radSampPerRegion[[2]][ ,1])) I get this: testt=cpt.mean(testTSRad) Error in single.mean.norm(data, penalty, pen.value, class, param.estimates) : Data must have atleast 2 observations to fit a changepoint model. This is because of what ?cpt.mean says about the data argument: data: A vector, ts object or matrix containing the data within which you wish to find a changepoint. If data is a matrix, each row is considered a separate dataset. An xts object is a matrix (with an index attribute), so each row is considered a separate data set. Your object only has one column, hence only one observation per data set. Things will work if you drop the dimensions of your single-column xts object: testt - cpt.mean(drop(testTSRad)) My data is below. Is there a way to convert it to ts? Yes, as is generally the case, use the as method: as.ts(testTSRad) But in this case, the time serie will have a frequency of 1, which is inconsistent with irregular sampling. This probably will lead to inaccurate results Best, -- Joshua Ulrich | about.me/joshuaulrich FOSS Trading | www.fosstrading.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. Regards, Pascal -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan [[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] ts instead of xts object
On Tue, Mar 11, 2014 at 12:14 AM, Bill william...@gmail.com wrote: Hello. I have a dataframe that has a date column. The intervals between dates vary. I want to convert this to a ts object. I was able to convert it to an xts object but the package I want to analyse this data with (called 'changepoint') does not seem to want to deal with xts. In the example they give they use the following: data(discoveries) dis.pelt=cpt.meanvar(discoveries,test.stat='Poisson',method='PELT') plot(dis.pelt,cpt.width=3) cpts.ts(dis.pelt) and if I check: str(discoveries) Time-Series [1:100] from 1860 to 1959: 5 3 0 2 0 3 2 3 6 1 ... If I try with my data str(testTSRad) An 'xts' object on 2011-07-16 07:08:02/2013-09-20 01:25:48 containing: Data: num [1:501, 1] 76 77 79 86 79 79 85 86 89 88 ... Indexed by objects of class: [POSIXct,POSIXt] TZ: xts Attributes: NULL where I used this: testTSRad=xts(radSampPerRegion[[2]][ ,2],order.by=as.POSIXct(radSampPerRegion[[2]][ ,1])) I get this: testt=cpt.mean(testTSRad) Error in single.mean.norm(data, penalty, pen.value, class, param.estimates) : Data must have atleast 2 observations to fit a changepoint model. This is because of what ?cpt.mean says about the data argument: data: A vector, ts object or matrix containing the data within which you wish to find a changepoint. If data is a matrix, each row is considered a separate dataset. An xts object is a matrix (with an index attribute), so each row is considered a separate data set. Your object only has one column, hence only one observation per data set. Things will work if you drop the dimensions of your single-column xts object: testt - cpt.mean(drop(testTSRad)) My data is below. Is there a way to convert it to ts? Yes, as is generally the case, use the as method: as.ts(testTSRad) Best, -- Joshua Ulrich | about.me/joshuaulrich FOSS Trading | www.fosstrading.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] ts instead of xts object
Hello, On Tue, Mar 11, 2014 at 8:45 PM, Joshua Ulrich josh.m.ulr...@gmail.com wrote: On Tue, Mar 11, 2014 at 12:14 AM, Bill william...@gmail.com wrote: Hello. I have a dataframe that has a date column. The intervals between dates vary. I want to convert this to a ts object. I was able to convert it to an xts object but the package I want to analyse this data with (called 'changepoint') does not seem to want to deal with xts. In the example they give they use the following: data(discoveries) dis.pelt=cpt.meanvar(discoveries,test.stat='Poisson',method='PELT') plot(dis.pelt,cpt.width=3) cpts.ts(dis.pelt) and if I check: str(discoveries) Time-Series [1:100] from 1860 to 1959: 5 3 0 2 0 3 2 3 6 1 ... If I try with my data str(testTSRad) An 'xts' object on 2011-07-16 07:08:02/2013-09-20 01:25:48 containing: Data: num [1:501, 1] 76 77 79 86 79 79 85 86 89 88 ... Indexed by objects of class: [POSIXct,POSIXt] TZ: xts Attributes: NULL where I used this: testTSRad=xts(radSampPerRegion[[2]][ ,2],order.by=as.POSIXct(radSampPerRegion[[2]][ ,1])) I get this: testt=cpt.mean(testTSRad) Error in single.mean.norm(data, penalty, pen.value, class, param.estimates) : Data must have atleast 2 observations to fit a changepoint model. This is because of what ?cpt.mean says about the data argument: data: A vector, ts object or matrix containing the data within which you wish to find a changepoint. If data is a matrix, each row is considered a separate dataset. An xts object is a matrix (with an index attribute), so each row is considered a separate data set. Your object only has one column, hence only one observation per data set. Things will work if you drop the dimensions of your single-column xts object: testt - cpt.mean(drop(testTSRad)) My data is below. Is there a way to convert it to ts? Yes, as is generally the case, use the as method: as.ts(testTSRad) But in this case, the time serie will have a frequency of 1, which is inconsistent with irregular sampling. This probably will lead to inaccurate results Best, -- Joshua Ulrich | about.me/joshuaulrich FOSS Trading | www.fosstrading.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. Regards, Pascal -- 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.
