[R] read txt file - date - no space
Diego Avesani
d|ego@@ve@@n| @end|ng |rom gm@||@com
Thu Aug 2 10:53:22 CEST 2018
Dear Petr,
I have read the file:
MyData <- read.csv(file="obs_prec.csv",header=TRUE, sep=",")
I have used POSIXct to convert properly the date
MyData$date2<-as.POSIXct(MyData$date, format="%m/%d/%Y %H:%M")
creating a second field inside MyDate.
I have converted the -999 to NA:
MyData[MyData== -999] <- NA
dim(MyData):
160008 5
And this is clear because I have 160008 days and 5 field:
date2,date,str1,str2,str3
I have chech the structure of my data:
str(MyData)
'data.frame': 160008 obs. of 5 variables:
$ date : Factor w/ 160008 levels "10/10/1998 0:00",..: 913 914 925 930 931
932 933 934 935 936 ...
$ str1 : num 0.6 0.2 0.6 0 0 0 0 0.2 0.6 0.2 ...
$ str2 : num 0 0.2 0.2 0 0 0 0 0 0.2 0.4 ...
$ str3 : num 0 0.2 0.4 0.6 0 0 0 0 0 0.4 ...
$ date2: POSIXct, format: "1998-10-01 00:00:00" "1998-10-01 01:00:00"
"1998-10-01 02:00:00" "1998-10-01 03:00:00" ...
Almost everything is clear:
str1,str2,str3 are mumbers,
date2 are date in the format according to POSIXct: Y-m-d h:m:s
date has 160008 Factor, i.e. 160008 factors which are the number of
category.
I do not understand "913 914 925 930" are the possibilitiues in levels?
I have no NA in date2:
which(MyData$date2 == NA)
integer(0)
as well in date.
At this point I have applied:
daily_mean1<-aggregate(MyData$str1, list(format(MyData$date, "%Y-%m-%d")),
mean)
which seems to be correct:
I have
dim(daily_mean1):
6667 2
str(daily_mean1)
'data.frame': 6667 obs. of 2 variables:
$ Group.1: chr "1998-10-01" "1998-10-02" "1998-10-03" "1998-10-04" ...
$ x : num 0.1667 0.0583 0.0417 0.3417 0.3333 ...
Really Really thanks:
You not only taught me R but also how to dealwith learning.
Can I ask you anover question about aggregate?
Again thanks
Diego
On 2 August 2018 at 10:10, PIKAL Petr <petr.pikal using precheza.cz> wrote:
> Hi
>
>
>
> *From:* Diego Avesani <diego.avesani using gmail.com>
> *Sent:* Thursday, August 2, 2018 10:03 AM
> *To:* PIKAL Petr <petr.pikal using precheza.cz>
> *Subject:* Re: [R] read txt file - date - no space
>
>
>
> Thanks,
>
> I have just send you a e-mail, before reading this one.
>
> Let's me read your last mail and go carefully through it.
>
>
>
> Thanks again, really really,
>
> I mean it
>
>
>
> P.S.
>
> Do you wand my *.csv file?
>
>
>
> Not necessarily, you should better learn things yourself if you really
> want to use R. Only if after you tested all suggested ways and did not get
> desired result.
>
>
>
> Cheers
>
> Petr
>
>
>
>
> Diego
>
>
>
> On 2 August 2018 at 09:56, PIKAL Petr <petr.pikal using precheza.cz> wrote:
>
> Well,
>
>
>
> you followed my advice only partly. Did you get rid of your silly -999
> values before averaging? Probably not.
>
> Did you tried aggregating by slightly longer construction
>
> aggregate(test[,-1], list(format(test$date, "%Y-%m-%d")), mean)
>
> which keeps difference in month and year? Probably not.
>
>
>
> We do not have your data, we do not know what exactly you want to do so it
> is really difficult to give you a help.
>
>
>
> If I calculate correctly there are 24 hour in one day and you have data
> for 18 years which gives me approximately 158000 distinct values.
>
>
>
> I can get either 18 values (averaging years) or aproximately 6600 values
> (averaging days).
>
>
>
> So my advice is:
>
>
>
> Read your data to R
>
> Change date column to POSIX but store it in different column
>
> Change NA values from -999 to real NA values
>
> Check dimension of your data ?dim
>
> Check structure of your data ?str
>
> Check if all dates are changed to POSIX correctly, are some of them NA?
>
> Aggregate your values (not by lubridate function day) and store them in
> another object
>
>
>
> Cheers
>
> Petr
>
>
>
>
>
> *From:* Diego Avesani <diego.avesani using gmail.com>
> *Sent:* Thursday, August 2, 2018 9:31 AM
> *To:* jim holtman <jholtman using gmail.com>; PIKAL Petr <petr.pikal using precheza.cz
> >
> *Cc:* R mailing list <r-help using r-project.org>
> *Subject:* Re: [R] read txt file - date - no space
>
>
>
> Dear all,
>
>
>
> I have found and error in the date conversion. Now it looks like:
>
>
>
> MyData <- read.csv(file="obs_prec.csv",header=TRUE, sep=",")
>
> # change date to real
>
> MyData$date<-as.POSIXct(MyData$date, format="%*m*/%*d*/%Y %H:%M")
>
>
>
> After that I apply the PIKAL's suggestions:
>
>
>
> aggregate(MyData[,-1], list(day(MyData$date)), mean)
>
>
>
> And this is the final results:
>
>
>
> 1 -82.43636 -46.12437 -319.2710
>
> 2 2 -82.06105 -45.74184 -319.2696
>
> 3 3 -82.05527 -45.52650 -319.2416
>
> 4 4 -82.03535 -47.59191 -319.2275
>
> 5 5 -77.44928 -50.05953 -320.5798
>
> ...
>
> 31 -86.10234 -47.06247 -340.0968
>
>
>
> However, it is not correct.
>
> This because I have not made myself clear about my purpose. As I told you
> some days ago, I have a *.csv file with hourly data from 10/21/1998
> to 12/31/2016. I would like to compute the daily means. Basically, I would
> like to have the mean of the hourly date for each day from 10/21/1998
> to 12/31/2016 and not 31 values.
>
>
>
> Really really thanks again,
>
> Diego
>
>
>
>
> Diego
>
>
>
> On 2 August 2018 at 08:55, Diego Avesani <diego.avesani using gmail.com> wrote:
>
> Dear
>
>
>
> I have check the one of the line that gives me problem. I mean, which give
> NA after R processing. I think that is similar to the others:
>
>
>
> 10/12/1998 10:00,0,0,0
>
> 10/12/1998 11:00,0,0,0
>
> 10/12/1998 12:00,0,0,0
>
> 10/12/1998 13:00,0,0,0
>
> 10/12/1998 14:00,0,0,0
>
> 10/12/1998 15:00,0,0,0
>
> 10/12/1998 16:00,0,0,0
>
> 10/12/1998 17:00,0,0,0
>
>
>
> @jim: It seems that you suggestion is focus on reading data from the
> terminal. It is possible to apply it to a *.csv file?
>
>
>
> @Pikal: Could it be that there are some date conversion error?
>
>
>
> Thanks again,
>
> Diego
>
>
>
>
> Diego
>
>
>
> On 1 August 2018 at 17:01, jim holtman <jholtman using gmail.com> wrote:
>
>
> Try this:
>
>
>
> > library(lubridate)
>
> > library(tidyverse)
>
> > input <- read.csv(text = "date,str1,str2,str3
>
> + 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", as.is = TRUE)
>
> > # convert the date and add the "day" so summarize
>
> > input <- input %>%
>
> + mutate(date = mdy_hm(date),
>
> + day = floor_date(date, unit = 'day')
>
> + )
>
> >
>
> > by_day <- input %>%
>
> + group_by(day) %>%
>
> + summarise(m_s1 = mean(str1),
>
> + m_s2 = mean(str2),
>
> + m_s3 = mean(str3)
>
> + )
>
> >
>
> > by_day
>
> # A tibble: 1 x 4
>
> day m_s1 m_s2 m_s3
>
> <dttm> <dbl> <dbl> <dbl>
>
> 1 1998-10-01 00:00:00 0.200 0.0500 0.150
>
>
> Jim Holtman
> *Data Munger Guru*
>
>
> *What is the problem that you are trying to solve? Tell me what you want
> to do, not how you want to do it.*
>
>
>
>
>
> On Tue, Jul 31, 2018 at 11:54 PM Diego Avesani <diego.avesani using gmail.com>
> wrote:
>
> Dear all,
> I am sorry, I did a lot of confusion. I am sorry, I have to relax and stat
> all again in order to understand.
> If I could I would like to start again, without mixing strategy and waiting
> for your advice.
>
> I am really appreciate you help, really really.
> Here my new file, a *.csv file (buy the way, it is possible to attach it in
> the mailing list?)
>
> date,str1,str2,str3
> 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
>
>
> I read it as:
> MyData <- read.csv(file="obs_prec.csv",header=TRUE, sep=",")
>
> at this point I would like to have the daily mean.
> What would you suggest?
>
> Really Really thanks,
> You are my lifesaver
>
> Thanks
>
>
>
> Diego
>
>
> On 1 August 2018 at 01:01, Jeff Newmiller <jdnewmil using dcn.davis.ca.us>
> wrote:
>
> > ... and the most common source of NA values in time data is wrong
> > timezones. You really need to make sure the timezone that is assumed when
> > the character data are converted to POSIXt agrees with the data. In most
> > cases the easiest way to insure this is to use
> >
> > Sys.setenv(TZ="US/Pacific")
> >
> > or whatever timezone from
> >
> > OlsonNames()
> >
> > corresponds with your data. Execute this setenv function before the
> > strptime or as.POSIXct() function call.
> >
> > You can use
> >
> > MyData[ is.na(MyData$datetime), ]
> >
> > to see which records are failing to convert time.
> >
> > [1] https://github.com/jdnewmil/eci298sp2016/blob/master/QuickHowto1
> >
> > On July 31, 2018 3:04:05 PM PDT, Jim Lemon <drjimlemon using gmail.com> wrote:
> > >Hi Diego,
> > >I think the error is due to NA values in your data file. If I extend
> > >your example and run it, I get no errors:
> > >
> > >MyData<-read.table(text="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
> > >1998-10-01 08:00:00 0.6 0 0
> > >1998-10-01 09:00:00 0.2 0.2 0.2
> > >1998-10-01 10:00:00 0.6 0.2 0.4
> > >1998-10-01 11:00:00 0 0 0.6
> > >1998-10-01 12:00:00 0 0 0
> > >1998-10-01 13:00:00 0 0 0
> > >1998-10-01 14:00:00 0 0 0
> > >1998-10-01 15:00:00 0.2 0 0
> > >1998-10-01 16:00:00 0.6 0 0
> > >1998-10-01 17:00:00 0.2 0.2 0.2
> > >1998-10-01 18:00:00 0.6 0.2 0.4
> > >1998-10-01 19:00:00 0 0 0.6
> > >1998-10-01 20:00:00 0 0 0
> > >1998-10-01 21:00:00 0 0 0
> > >1998-10-01 22:00:00 0 0 0
> > >1998-10-01 23:00:00 0.2 0 0
> > >1998-10-02 00:00:00 0.6 0 0
> > >1998-10-02 01:00:00 0.2 0.2 0.2
> > >1998-10-02 02:00:00 0.6 0.2 0.4
> > >1998-10-02 03:00:00 0 0 0.6
> > >1998-10-02 04:00:00 0 0 0
> > >1998-10-02 05:00:00 0 0 0
> > >1998-10-02 06:00:00 0 0 0
> > >1998-10-02 07:00:00 0.2 0 0
> > >1998-10-02 08:00:00 0.6 0 0
> > >1998-10-02 09:00:00 0.2 0.2 0.2
> > >1998-10-02 10:00:00 0.6 0.2 0.4
> > >1998-10-02 11:00:00 0 0 0.6
> > >1998-10-02 12:00:00 0 0 0
> > >1998-10-02 13:00:00 0 0 0
> > >1998-10-02 14:00:00 0 0 0
> > >1998-10-02 15:00:00 0.2 0 0
> > >1998-10-02 16:00:00 0.6 0 0
> > >1998-10-02 17:00:00 0.2 0.2 0.2
> > >1998-10-02 18:00:00 0.6 0.2 0.4
> > >1998-10-02 19:00:00 0 0 0.6
> > >1998-10-02 20:00:00 0 0 0
> > >1998-10-02 21:00:00 0 0 0
> > >1998-10-02 22:00:00 0 0 0
> > >1998-10-02 23:00:00 0.2 0 0",
> > >skip=1,stringsAsFactors=FALSE)
> > >names(MyData)<-c("date","time","st1","st2","st3")
> > >MyData$datetime<-strptime(paste(MyData$date,MyData$time),
> > > format="%Y-%m-%d %H:%M:%S")
> > >MyData$datetime
> > >st1_daily<-by(MyData$st1,MyData$date,mean)
> > >st2_daily<-by(MyData$st2,MyData$date,mean)
> > >st3_daily<-by(MyData$st3,MyData$date,mean)
> > >st1_daily
> > >st2_daily
> > >st3_daily
> > >
> > >Try adding na.rm=TRUE to the "by" calls:
> > >
> > >st1_daily<-by(MyData$st1,MyData$date,mean,na.rm=TRUE)
> > >st2_daily<-by(MyData$st2,MyData$date,mean,na.rm=TRUE)
> > >st3_daily<-by(MyData$st3,MyData$date,mean,na.rm=TRUE)
> > >
> > >Jim
> > >
> > >On Tue, Jul 31, 2018 at 11:11 PM, Diego Avesani
> > ><diego.avesani using gmail.com> wrote:
> > >> Dear all,
> > >>
> > >> I have still problem with date.
> > >> Could you please tel me how to use POSIXct.
> > >> Indeed I have found this command:
> > >> timeAverage, but I am not able to convert MyDate to properly date.
> > >>
> > >> Thank a lot
> > >> I hope to no bother you, at least too much
> > >>
> > >>
> > >> Diego
> > >>
> > >>
> > >> On 31 July 2018 at 11:12, Diego Avesani <diego.avesani using gmail.com>
> > >wrote:
> > >>>
> > >>> Dear Jim, Dear all,
> > >>>
> > >>> thanks a lot.
> > >>>
> > >>> Unfortunately, I get the following error:
> > >>>
> > >>>
> > >>> st1_daily<-by(MyData$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$str1,na.rm=TRUE)
> > >>>
> > >>>
> > >>> it works
> > >>>
> > >>>
> > >>> Sorry, I have to learn a lot.
> > >>> You are really boosting me
> > >>>
> > >>> Diego
> > >>>
> > >>>
> > >>> On 31 July 2018 at 11:02, Jim Lemon <drjimlemon using gmail.com> wrote:
> > >>>>
> > >>>> Hi Diego,
> > >>>> One way you can get daily means is:
> > >>>>
> > >>>> st1_daily<-by(MyData$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 using 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 using 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 using 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 using r-project.org on behalf of
> > >drjimlemon using 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 using 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 using 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 using 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 using 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.
> >
> > --
> > Sent from my phone. Please excuse my brevity.
> >
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help using 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.
>
>
>
>
>
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