Dear all, I have found the error, my fault. Sorry. There was an extra come in the headers line. Thanks again. If I can I would like to ask you another questions about the imported data. I would like to compute the daily average of the different date. Basically I have hourly data, I would like to ave the daily mean of them. Is there some special commands? Thanks a lot. Diego On 31 July 2018 at 10:40, Diego Avesani <diego.avesani at gmail.com> wrote:> Dear all, > I move to csv file because originally the date where in csv file. > In addition, due to the fact that, as you told me, read.csv is a special > case of read.table, I prefer start to learn from the simplest one. > After that, I will try also the *.txt format. > > with read.csv, something strange happened: > > This us now the file: > > date,st1,st2,st3, > 10/1/1998 0:00,0.6,0,0 > 10/1/1998 1:00,0.2,0.2,0.2 > 10/1/1998 2:00,0.6,0.2,0.4 > 10/1/1998 3:00,0,0,0.6 > 10/1/1998 4:00,0,0,0 > 10/1/1998 5:00,0,0,0 > 10/1/1998 6:00,0,0,0 > 10/1/1998 7:00,0.2,0,0 > 10/1/1998 8:00,0.6,0.2,0 > 10/1/1998 9:00,0.2,0.4,0.4 > 10/1/1998 10:00,0,0.4,0.2 > > When I apply: > MyData <- read.csv(file="obs_prec.csv",header=TRUE, sep=",") > > this is the results: > > 10/1/1998 0:00 0.6 0.00 0.0 NA > 2 10/1/1998 1:00 0.2 0.20 0.2 NA > 3 10/1/1998 2:00 0.6 0.20 0.4 NA > 4 10/1/1998 3:00 0.0 0.00 0.6 NA > 5 10/1/1998 4:00 0.0 0.00 0.0 NA > 6 10/1/1998 5:00 0.0 0.00 0.0 NA > 7 10/1/1998 6:00 0.0 0.00 0.0 NA > 8 10/1/1998 7:00 0.2 0.00 0.0 NA > > I do not understand why. > Something wrong with date? > > really really thanks, > I appreciate a lot all your helps. > > Diedro > > > Diego > > > On 31 July 2018 at 01:25, MacQueen, Don <macqueen1 at llnl.gov> wrote: > >> Or, without removing the first line >> dadf <- read.table("xxx.txt", stringsAsFactors=FALSE, skip=1) >> >> Another alternative, >> dadf$datetime <- as.POSIXct(paste(dadf$V1,dadf$V2)) >> since the dates appear to be in the default format. >> (I generally prefer to work with datetimes in POSIXct class rather than >> POSIXlt class) >> >> -Don >> >> -- >> Don MacQueen >> Lawrence Livermore National Laboratory >> 7000 East Ave., L-627 >> Livermore, CA 94550 >> 925-423-1062 >> Lab cell 925-724-7509 >> >> >> >> ?On 7/30/18, 4:03 PM, "R-help on behalf of Jim Lemon" < >> r-help-bounces at r-project.org on behalf of drjimlemon at gmail.com> wrote: >> >> Hi Diego, >> You may have to do some conversion as you have three fields in the >> first line using the default space separator and five fields in >> subsequent lines. If the first line doesn't contain any important data >> you can just delete it or replace it with a meaningful header line >> with five fields and save the file under another name. >> >> It looks as thought you have date-time as two fields. If so, you can >> just read the first field if you only want the date: >> >> # assume you have removed the first line >> dadf<-read.table("xxx.txt",stringsAsFactors=FALSE >> dadf$date<-as.Date(dadf$V1,format="%Y-%m-%d") >> >> If you want the date/time: >> >> dadf$datetime<-strptime(paste(dadf$V1,dadf$V2),format="%Y-%m-%d >> %H:%M:%S") >> >> Jim >> >> On Tue, Jul 31, 2018 at 12:29 AM, Diego Avesani < >> diego.avesani at gmail.com> wrote: >> > Dear all, >> > >> > I am dealing with the reading of a *.txt file. >> > The txt file the following shape: >> > >> > 103001930 103001580 103001530 >> > 1998-10-01 00:00:00 0.6 0 0 >> > 1998-10-01 01:00:00 0.2 0.2 0.2 >> > 1998-10-01 02:00:00 0.6 0.2 0.4 >> > 1998-10-01 03:00:00 0 0 0.6 >> > 1998-10-01 04:00:00 0 0 0 >> > 1998-10-01 05:00:00 0 0 0 >> > 1998-10-01 06:00:00 0 0 0 >> > 1998-10-01 07:00:00 0.2 0 0 >> > >> > If it is possible I have a coupe of questions, which will sound >> stupid but >> > they are important to me in order to understand ho R deal with file >> or date. >> > >> > 1) Do I have to convert it to a *csv file? >> > 2) Can a deal with space and not "," >> > 3) How can I read date? >> > >> > thanks a lot to all of you, >> > Thanks >> > >> > >> > Diego >> > >> > [[alternative HTML version deleted]] >> > >> > ______________________________________________ >> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> > https://stat.ethz.ch/mailman/listinfo/r-help >> > PLEASE do read the posting guide http://www.R-project.org/posti >> ng-guide.html >> > and provide commented, minimal, self-contained, reproducible code. >> >> ______________________________________________ >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posti >> ng-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> >> >> >[[alternative HTML version deleted]]
Hi Diego, One way you can get daily means is: st1_daily<-by(MyData$st1,MyData$date,mean) st2_daily<-by(MyData$st2,MyData$date,mean) st3_daily<-by(MyData$st3,MyData$date,mean) Jim On Tue, Jul 31, 2018 at 6:51 PM, Diego Avesani <diego.avesani at gmail.com> wrote:> Dear all, > I have found the error, my fault. Sorry. > There was an extra come in the headers line. > Thanks again. > > If I can I would like to ask you another questions about the imported data. > I would like to compute the daily average of the different date. Basically I > have hourly data, I would like to ave the daily mean of them. > > Is there some special commands? > > Thanks a lot. > > > Diego > > > On 31 July 2018 at 10:40, Diego Avesani <diego.avesani at gmail.com> wrote: >> >> Dear all, >> I move to csv file because originally the date where in csv file. >> In addition, due to the fact that, as you told me, read.csv is a special >> case of read.table, I prefer start to learn from the simplest one. >> After that, I will try also the *.txt format. >> >> with read.csv, something strange happened: >> >> This us now the file: >> >> date,st1,st2,st3, >> 10/1/1998 0:00,0.6,0,0 >> 10/1/1998 1:00,0.2,0.2,0.2 >> 10/1/1998 2:00,0.6,0.2,0.4 >> 10/1/1998 3:00,0,0,0.6 >> 10/1/1998 4:00,0,0,0 >> 10/1/1998 5:00,0,0,0 >> 10/1/1998 6:00,0,0,0 >> 10/1/1998 7:00,0.2,0,0 >> 10/1/1998 8:00,0.6,0.2,0 >> 10/1/1998 9:00,0.2,0.4,0.4 >> 10/1/1998 10:00,0,0.4,0.2 >> >> When I apply: >> MyData <- read.csv(file="obs_prec.csv",header=TRUE, sep=",") >> >> this is the results: >> >> 10/1/1998 0:00 0.6 0.00 0.0 NA >> 2 10/1/1998 1:00 0.2 0.20 0.2 NA >> 3 10/1/1998 2:00 0.6 0.20 0.4 NA >> 4 10/1/1998 3:00 0.0 0.00 0.6 NA >> 5 10/1/1998 4:00 0.0 0.00 0.0 NA >> 6 10/1/1998 5:00 0.0 0.00 0.0 NA >> 7 10/1/1998 6:00 0.0 0.00 0.0 NA >> 8 10/1/1998 7:00 0.2 0.00 0.0 NA >> >> I do not understand why. >> Something wrong with date? >> >> really really thanks, >> I appreciate a lot all your helps. >> >> Diedro >> >> >> Diego >> >> >> On 31 July 2018 at 01:25, MacQueen, Don <macqueen1 at llnl.gov> wrote: >>> >>> Or, without removing the first line >>> dadf <- read.table("xxx.txt", stringsAsFactors=FALSE, skip=1) >>> >>> Another alternative, >>> dadf$datetime <- as.POSIXct(paste(dadf$V1,dadf$V2)) >>> since the dates appear to be in the default format. >>> (I generally prefer to work with datetimes in POSIXct class rather than >>> POSIXlt class) >>> >>> -Don >>> >>> -- >>> Don MacQueen >>> Lawrence Livermore National Laboratory >>> 7000 East Ave., L-627 >>> Livermore, CA 94550 >>> 925-423-1062 >>> Lab cell 925-724-7509 >>> >>> >>> >>> ?On 7/30/18, 4:03 PM, "R-help on behalf of Jim Lemon" >>> <r-help-bounces at r-project.org on behalf of drjimlemon at gmail.com> wrote: >>> >>> Hi Diego, >>> You may have to do some conversion as you have three fields in the >>> first line using the default space separator and five fields in >>> subsequent lines. If the first line doesn't contain any important >>> data >>> you can just delete it or replace it with a meaningful header line >>> with five fields and save the file under another name. >>> >>> It looks as thought you have date-time as two fields. If so, you can >>> just read the first field if you only want the date: >>> >>> # assume you have removed the first line >>> dadf<-read.table("xxx.txt",stringsAsFactors=FALSE >>> dadf$date<-as.Date(dadf$V1,format="%Y-%m-%d") >>> >>> If you want the date/time: >>> >>> dadf$datetime<-strptime(paste(dadf$V1,dadf$V2),format="%Y-%m-%d >>> %H:%M:%S") >>> >>> Jim >>> >>> On Tue, Jul 31, 2018 at 12:29 AM, Diego Avesani >>> <diego.avesani at gmail.com> wrote: >>> > Dear all, >>> > >>> > I am dealing with the reading of a *.txt file. >>> > The txt file the following shape: >>> > >>> > 103001930 103001580 103001530 >>> > 1998-10-01 00:00:00 0.6 0 0 >>> > 1998-10-01 01:00:00 0.2 0.2 0.2 >>> > 1998-10-01 02:00:00 0.6 0.2 0.4 >>> > 1998-10-01 03:00:00 0 0 0.6 >>> > 1998-10-01 04:00:00 0 0 0 >>> > 1998-10-01 05:00:00 0 0 0 >>> > 1998-10-01 06:00:00 0 0 0 >>> > 1998-10-01 07:00:00 0.2 0 0 >>> > >>> > If it is possible I have a coupe of questions, which will sound >>> stupid but >>> > they are important to me in order to understand ho R deal with file >>> or date. >>> > >>> > 1) Do I have to convert it to a *csv file? >>> > 2) Can a deal with space and not "," >>> > 3) How can I read date? >>> > >>> > thanks a lot to all of you, >>> > Thanks >>> > >>> > >>> > Diego >>> > >>> > [[alternative HTML version deleted]] >>> > >>> > ______________________________________________ >>> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >>> > 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 at r-project.org mailing list -- To UNSUBSCRIBE and more, see >>> 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. >>> >>> >> >
Dear Jim, Dear all, thanks a lot. Unfortunately, I get the following error: st1_daily<-by(MyData$st1,MyData$date,mean)Error in tapply(seq_len(0L), list(`MyData$date` = c(913L, 914L, 925L, : arguments must have same length This is particularly strange. indeed, if I apply mean(MyData$str1,na.rm=TRUE) it works Sorry, I have to learn a lot. You are really boosting me Diego On 31 July 2018 at 11:02, Jim Lemon <drjimlemon at gmail.com> wrote:> Hi Diego, > One way you can get daily means is: > > st1_daily<-by(MyData$st1,MyData$date,mean) > st2_daily<-by(MyData$st2,MyData$date,mean) > st3_daily<-by(MyData$st3,MyData$date,mean) > > Jim > > On Tue, Jul 31, 2018 at 6:51 PM, Diego Avesani <diego.avesani at gmail.com> > wrote: > > Dear all, > > I have found the error, my fault. Sorry. > > There was an extra come in the headers line. > > Thanks again. > > > > If I can I would like to ask you another questions about the imported > data. > > I would like to compute the daily average of the different date. > Basically I > > have hourly data, I would like to ave the daily mean of them. > > > > Is there some special commands? > > > > Thanks a lot. > > > > > > Diego > > > > > > On 31 July 2018 at 10:40, Diego Avesani <diego.avesani at gmail.com> wrote: > >> > >> Dear all, > >> I move to csv file because originally the date where in csv file. > >> In addition, due to the fact that, as you told me, read.csv is a special > >> case of read.table, I prefer start to learn from the simplest one. > >> After that, I will try also the *.txt format. > >> > >> with read.csv, something strange happened: > >> > >> This us now the file: > >> > >> date,st1,st2,st3, > >> 10/1/1998 0:00,0.6,0,0 > >> 10/1/1998 1:00,0.2,0.2,0.2 > >> 10/1/1998 2:00,0.6,0.2,0.4 > >> 10/1/1998 3:00,0,0,0.6 > >> 10/1/1998 4:00,0,0,0 > >> 10/1/1998 5:00,0,0,0 > >> 10/1/1998 6:00,0,0,0 > >> 10/1/1998 7:00,0.2,0,0 > >> 10/1/1998 8:00,0.6,0.2,0 > >> 10/1/1998 9:00,0.2,0.4,0.4 > >> 10/1/1998 10:00,0,0.4,0.2 > >> > >> When I apply: > >> MyData <- read.csv(file="obs_prec.csv",header=TRUE, sep=",") > >> > >> this is the results: > >> > >> 10/1/1998 0:00 0.6 0.00 0.0 NA > >> 2 10/1/1998 1:00 0.2 0.20 0.2 NA > >> 3 10/1/1998 2:00 0.6 0.20 0.4 NA > >> 4 10/1/1998 3:00 0.0 0.00 0.6 NA > >> 5 10/1/1998 4:00 0.0 0.00 0.0 NA > >> 6 10/1/1998 5:00 0.0 0.00 0.0 NA > >> 7 10/1/1998 6:00 0.0 0.00 0.0 NA > >> 8 10/1/1998 7:00 0.2 0.00 0.0 NA > >> > >> I do not understand why. > >> Something wrong with date? > >> > >> really really thanks, > >> I appreciate a lot all your helps. > >> > >> Diedro > >> > >> > >> Diego > >> > >> > >> On 31 July 2018 at 01:25, MacQueen, Don <macqueen1 at llnl.gov> wrote: > >>> > >>> Or, without removing the first line > >>> dadf <- read.table("xxx.txt", stringsAsFactors=FALSE, skip=1) > >>> > >>> Another alternative, > >>> dadf$datetime <- as.POSIXct(paste(dadf$V1,dadf$V2)) > >>> since the dates appear to be in the default format. > >>> (I generally prefer to work with datetimes in POSIXct class rather than > >>> POSIXlt class) > >>> > >>> -Don > >>> > >>> -- > >>> Don MacQueen > >>> Lawrence Livermore National Laboratory > >>> 7000 East Ave., L-627 > >>> Livermore, CA 94550 > >>> 925-423-1062 > >>> Lab cell 925-724-7509 > >>> > >>> > >>> > >>> ?On 7/30/18, 4:03 PM, "R-help on behalf of Jim Lemon" > >>> <r-help-bounces at r-project.org on behalf of drjimlemon at gmail.com> > wrote: > >>> > >>> Hi Diego, > >>> You may have to do some conversion as you have three fields in the > >>> first line using the default space separator and five fields in > >>> subsequent lines. If the first line doesn't contain any important > >>> data > >>> you can just delete it or replace it with a meaningful header line > >>> with five fields and save the file under another name. > >>> > >>> It looks as thought you have date-time as two fields. If so, you > can > >>> just read the first field if you only want the date: > >>> > >>> # assume you have removed the first line > >>> dadf<-read.table("xxx.txt",stringsAsFactors=FALSE > >>> dadf$date<-as.Date(dadf$V1,format="%Y-%m-%d") > >>> > >>> If you want the date/time: > >>> > >>> dadf$datetime<-strptime(paste(dadf$V1,dadf$V2),format="%Y-%m-%d > >>> %H:%M:%S") > >>> > >>> Jim > >>> > >>> On Tue, Jul 31, 2018 at 12:29 AM, Diego Avesani > >>> <diego.avesani at gmail.com> wrote: > >>> > Dear all, > >>> > > >>> > I am dealing with the reading of a *.txt file. > >>> > The txt file the following shape: > >>> > > >>> > 103001930 103001580 103001530 > >>> > 1998-10-01 00:00:00 0.6 0 0 > >>> > 1998-10-01 01:00:00 0.2 0.2 0.2 > >>> > 1998-10-01 02:00:00 0.6 0.2 0.4 > >>> > 1998-10-01 03:00:00 0 0 0.6 > >>> > 1998-10-01 04:00:00 0 0 0 > >>> > 1998-10-01 05:00:00 0 0 0 > >>> > 1998-10-01 06:00:00 0 0 0 > >>> > 1998-10-01 07:00:00 0.2 0 0 > >>> > > >>> > If it is possible I have a coupe of questions, which will sound > >>> stupid but > >>> > they are important to me in order to understand ho R deal with > file > >>> or date. > >>> > > >>> > 1) Do I have to convert it to a *csv file? > >>> > 2) Can a deal with space and not "," > >>> > 3) How can I read date? > >>> > > >>> > thanks a lot to all of you, > >>> > Thanks > >>> > > >>> > > >>> > Diego > >>> > > >>> > [[alternative HTML version deleted]] > >>> > > >>> > ______________________________________________ > >>> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, > see > >>> > 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 at r-project.org mailing list -- To UNSUBSCRIBE and more, see > >>> 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]]