Dear Jeff, Thank you so much for your time. I tried your code. It successfully assigned NA to the zeros. But the main code seems not to work with the NAs. The mean, for example, resulted in NA. I am attaching the data for a period of one year and the code which I use in plotting the data. Maybe it might be easier for you to spot where I run into error (my plot was just empty). Thanks again. Best regards Ogbos On Wed, Jun 3, 2020 at 8:47 PM Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:> df[[ 5 ]][ 0 == df[[ 5 ]] ] <- NA > > On June 3, 2020 1:59:06 AM PDT, Ogbos Okike <giftedlife2014 at gmail.com> > wrote: > >Dear R-Experts, > >I have a cosmic ray data that span several years. The data frame is of > >the > >form: > >03 01 01 00 3809 > >03 01 01 01 3771 > >03 01 01 02 3743 > >03 01 01 03 3747 > >03 01 01 04 3737 > >03 01 01 05 3751 > >03 01 01 06 3733 > >03 01 01 07 3732. > >where the columns 1 to 5 stand for year, month, day, hour and counts. > >Some hours when the station does not have data are assigned zero, > >implying > >there could be several zeros in column 5. Since my aim is to plot the > >hourly mean for all the years, I started learning with one year - year > >2003. > > > >I carefully went through the data, removing any day that contains zero > >for > >any of the hours. Instead of the 365 days in the year 2003, I ended up > >with 362 days. > > > >I saved that as CLMX1C (now stored in Ogbos2 with dput function, see > >attached please). > > > >If I run the data with my script, it gives me what I am expecting. My > >script is: > >d<-read.table("CLMX1C",col.names=c("h","count")) > >y<-d$count > >data<-(y-mean(y))/mean(y)*100 > > > >A<-matrix(rep(1:24,362)) > >B<-matrix(data) > > > > oodf<-data.frame(A,B) > > oodf<-data.frame(A,B) > >library(plotrix) > >std.error<-function(x) return(sd(x)/(sum(!is.na(x)))) > >oomean<-as.vector(by(oodf$B,oodf$A,mean)) > >oose<-as.vector(by(oodf$B,oodf$A,std.error)) > >plot(1:24,oomean,type="b",ylim=c(-0.4,0.5), > > xlab="Hours",ylab="CR count",main="CR daily variation for 2004") > >dispersion(1:24,oomean,oose,arrow.cap=.01) > > > >Now, instead of foraging through the big data removing the day for > >which > >there is a missing data for any hour, I wish to try to replace the > >missing > >data with NA and hoping that it will do the job for me. > > > >I added just three lines in the script above: > >d<-read.table("2003",col.names=c("y","m","d","h","count")) > >y<-d$count > >df<-data.frame(y)#line 1 > >library('dplyr') # line 2 > >y<-na_if(df, 0) #line 3 > >data<-(y-mean(y))/mean(y)*100. > >Then I started getting error messages: > >Error in is.data.frame(x) : > > (list) object cannot be coerced to type 'double' > >In addition: There were 26 warnings (use warnings() to see them). > > > >I hope you will assist me to deal with the issues of replacing zeros > >with > >NA in column 5 in such a way that my code will run. > > > >Iam ever indebted!! > >Best regards > >Ogbos > >______________________________________________ > >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. > > -- > Sent from my phone. Please excuse my brevity. >
Perhaps read ?mean... On June 3, 2020 6:15:11 PM PDT, Ogbos Okike <giftedlife2014 at gmail.com> wrote:>Dear Jeff, >Thank you so much for your time. >I tried your code. It successfully assigned NA to the zeros. > >But the main code seems not to work with the NAs. The mean, for >example, >resulted in NA. I am attaching the data for a period of one year and >the >code which I use in plotting the data. Maybe it might be easier for >you to >spot where I run into error (my plot was just empty). >Thanks again. >Best regards >Ogbos > > >On Wed, Jun 3, 2020 at 8:47 PM Jeff Newmiller ><jdnewmil at dcn.davis.ca.us> >wrote: > >> df[[ 5 ]][ 0 == df[[ 5 ]] ] <- NA >> >> On June 3, 2020 1:59:06 AM PDT, Ogbos Okike ><giftedlife2014 at gmail.com> >> wrote: >> >Dear R-Experts, >> >I have a cosmic ray data that span several years. The data frame is >of >> >the >> >form: >> >03 01 01 00 3809 >> >03 01 01 01 3771 >> >03 01 01 02 3743 >> >03 01 01 03 3747 >> >03 01 01 04 3737 >> >03 01 01 05 3751 >> >03 01 01 06 3733 >> >03 01 01 07 3732. >> >where the columns 1 to 5 stand for year, month, day, hour and >counts. >> >Some hours when the station does not have data are assigned zero, >> >implying >> >there could be several zeros in column 5. Since my aim is to plot >the >> >hourly mean for all the years, I started learning with one year - >year >> >2003. >> > >> >I carefully went through the data, removing any day that contains >zero >> >for >> >any of the hours. Instead of the 365 days in the year 2003, I ended >up >> >with 362 days. >> > >> >I saved that as CLMX1C (now stored in Ogbos2 with dput function, see >> >attached please). >> > >> >If I run the data with my script, it gives me what I am expecting. >My >> >script is: >> >d<-read.table("CLMX1C",col.names=c("h","count")) >> >y<-d$count >> >data<-(y-mean(y))/mean(y)*100 >> > >> >A<-matrix(rep(1:24,362)) >> >B<-matrix(data) >> > >> > oodf<-data.frame(A,B) >> > oodf<-data.frame(A,B) >> >library(plotrix) >> >std.error<-function(x) return(sd(x)/(sum(!is.na(x)))) >> >oomean<-as.vector(by(oodf$B,oodf$A,mean)) >> >oose<-as.vector(by(oodf$B,oodf$A,std.error)) >> >plot(1:24,oomean,type="b",ylim=c(-0.4,0.5), >> > xlab="Hours",ylab="CR count",main="CR daily variation for 2004") >> >dispersion(1:24,oomean,oose,arrow.cap=.01) >> > >> >Now, instead of foraging through the big data removing the day for >> >which >> >there is a missing data for any hour, I wish to try to replace the >> >missing >> >data with NA and hoping that it will do the job for me. >> > >> >I added just three lines in the script above: >> >d<-read.table("2003",col.names=c("y","m","d","h","count")) >> >y<-d$count >> >df<-data.frame(y)#line 1 >> >library('dplyr') # line 2 >> >y<-na_if(df, 0) #line 3 >> >data<-(y-mean(y))/mean(y)*100. >> >Then I started getting error messages: >> >Error in is.data.frame(x) : >> > (list) object cannot be coerced to type 'double' >> >In addition: There were 26 warnings (use warnings() to see them). >> > >> >I hope you will assist me to deal with the issues of replacing zeros >> >with >> >NA in column 5 in such a way that my code will run. >> > >> >Iam ever indebted!! >> >Best regards >> >Ogbos >> >______________________________________________ >> >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. >> >> -- >> Sent from my phone. Please excuse my brevity. >>-- Sent from my phone. Please excuse my brevity.
Dear Jeff, It worked!!! I took a second look at "?mean" as you suggested. I then adjusted my code, inserting rm.na here and there until it was fine. Thank you very much. Warm regards. Ogbos On Thu, Jun 4, 2020 at 3:14 AM Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:> Perhaps read ?mean... > > On June 3, 2020 6:15:11 PM PDT, Ogbos Okike <giftedlife2014 at gmail.com> > wrote: > >Dear Jeff, > >Thank you so much for your time. > >I tried your code. It successfully assigned NA to the zeros. > > > >But the main code seems not to work with the NAs. The mean, for > >example, > >resulted in NA. I am attaching the data for a period of one year and > >the > >code which I use in plotting the data. Maybe it might be easier for > >you to > >spot where I run into error (my plot was just empty). > >Thanks again. > >Best regards > >Ogbos > > > > > >On Wed, Jun 3, 2020 at 8:47 PM Jeff Newmiller > ><jdnewmil at dcn.davis.ca.us> > >wrote: > > > >> df[[ 5 ]][ 0 == df[[ 5 ]] ] <- NA > >> > >> On June 3, 2020 1:59:06 AM PDT, Ogbos Okike > ><giftedlife2014 at gmail.com> > >> wrote: > >> >Dear R-Experts, > >> >I have a cosmic ray data that span several years. The data frame is > >of > >> >the > >> >form: > >> >03 01 01 00 3809 > >> >03 01 01 01 3771 > >> >03 01 01 02 3743 > >> >03 01 01 03 3747 > >> >03 01 01 04 3737 > >> >03 01 01 05 3751 > >> >03 01 01 06 3733 > >> >03 01 01 07 3732. > >> >where the columns 1 to 5 stand for year, month, day, hour and > >counts. > >> >Some hours when the station does not have data are assigned zero, > >> >implying > >> >there could be several zeros in column 5. Since my aim is to plot > >the > >> >hourly mean for all the years, I started learning with one year - > >year > >> >2003. > >> > > >> >I carefully went through the data, removing any day that contains > >zero > >> >for > >> >any of the hours. Instead of the 365 days in the year 2003, I ended > >up > >> >with 362 days. > >> > > >> >I saved that as CLMX1C (now stored in Ogbos2 with dput function, see > >> >attached please). > >> > > >> >If I run the data with my script, it gives me what I am expecting. > >My > >> >script is: > >> >d<-read.table("CLMX1C",col.names=c("h","count")) > >> >y<-d$count > >> >data<-(y-mean(y))/mean(y)*100 > >> > > >> >A<-matrix(rep(1:24,362)) > >> >B<-matrix(data) > >> > > >> > oodf<-data.frame(A,B) > >> > oodf<-data.frame(A,B) > >> >library(plotrix) > >> >std.error<-function(x) return(sd(x)/(sum(!is.na(x)))) > >> >oomean<-as.vector(by(oodf$B,oodf$A,mean)) > >> >oose<-as.vector(by(oodf$B,oodf$A,std.error)) > >> >plot(1:24,oomean,type="b",ylim=c(-0.4,0.5), > >> > xlab="Hours",ylab="CR count",main="CR daily variation for 2004") > >> >dispersion(1:24,oomean,oose,arrow.cap=.01) > >> > > >> >Now, instead of foraging through the big data removing the day for > >> >which > >> >there is a missing data for any hour, I wish to try to replace the > >> >missing > >> >data with NA and hoping that it will do the job for me. > >> > > >> >I added just three lines in the script above: > >> >d<-read.table("2003",col.names=c("y","m","d","h","count")) > >> >y<-d$count > >> >df<-data.frame(y)#line 1 > >> >library('dplyr') # line 2 > >> >y<-na_if(df, 0) #line 3 > >> >data<-(y-mean(y))/mean(y)*100. > >> >Then I started getting error messages: > >> >Error in is.data.frame(x) : > >> > (list) object cannot be coerced to type 'double' > >> >In addition: There were 26 warnings (use warnings() to see them). > >> > > >> >I hope you will assist me to deal with the issues of replacing zeros > >> >with > >> >NA in column 5 in such a way that my code will run. > >> > > >> >Iam ever indebted!! > >> >Best regards > >> >Ogbos > >> >______________________________________________ > >> >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. > >> > >> -- > >> Sent from my phone. Please excuse my brevity. > >> > > -- > Sent from my phone. Please excuse my brevity. >[[alternative HTML version deleted]]
Dear Jeff,
Yes. It worked. But I am still fine turning the script.
Please permit me to ask again.
Since I can handle one year. I wish to take on more years. I added extra
two years and now have 2000 to 2002.
I wish to plot the same graph for those years such that the x-axis will be
showing 2000, 2001 and 2002. I tried to manipulate replicate function with
times and length.
I have a result but it does seem like what I am looking for. For each year,
I should get a similar graph. So the three years will just look like
similar plot repeated with some small differences. When I did per year and
added, I got it. But I could not add the error bars by doing them
separately for each year.
Here is part of my adjusted code:
d<-read.table("3years",col.names=c("y","m","d","h","count"))
y<-d$count
data<-(y-mean(y,na.rm=TRUE))/mean(y,na.rm=TRUE)*100
A<-matrix(rep(1:24,366))
A1<-matrix(rep(25:49,365))
A2<-matrix(rep(50:72,365))
A<-matrix(c(A,A1,A2),ncol=1)
I have also attached the complete code and data for the three years.
I am thanking you for any pointer.
Best regards
Ogbos
On Thu, Jun 4, 2020 at 3:14 AM Jeff Newmiller <jdnewmil at
dcn.davis.ca.us>
wrote:
> Perhaps read ?mean...
>
> On June 3, 2020 6:15:11 PM PDT, Ogbos Okike <giftedlife2014 at
gmail.com>
> wrote:
> >Dear Jeff,
> >Thank you so much for your time.
> >I tried your code. It successfully assigned NA to the zeros.
> >
> >But the main code seems not to work with the NAs. The mean, for
> >example,
> >resulted in NA. I am attaching the data for a period of one year and
> >the
> >code which I use in plotting the data. Maybe it might be easier for
> >you to
> >spot where I run into error (my plot was just empty).
> >Thanks again.
> >Best regards
> >Ogbos
> >
> >
> >On Wed, Jun 3, 2020 at 8:47 PM Jeff Newmiller
> ><jdnewmil at dcn.davis.ca.us>
> >wrote:
> >
> >> df[[ 5 ]][ 0 == df[[ 5 ]] ] <- NA
> >>
> >> On June 3, 2020 1:59:06 AM PDT, Ogbos Okike
> ><giftedlife2014 at gmail.com>
> >> wrote:
> >> >Dear R-Experts,
> >> >I have a cosmic ray data that span several years. The data
frame is
> >of
> >> >the
> >> >form:
> >> >03 01 01 00 3809
> >> >03 01 01 01 3771
> >> >03 01 01 02 3743
> >> >03 01 01 03 3747
> >> >03 01 01 04 3737
> >> >03 01 01 05 3751
> >> >03 01 01 06 3733
> >> >03 01 01 07 3732.
> >> >where the columns 1 to 5 stand for year, month, day, hour and
> >counts.
> >> >Some hours when the station does not have data are assigned
zero,
> >> >implying
> >> >there could be several zeros in column 5. Since my aim is to
plot
> >the
> >> >hourly mean for all the years, I started learning with one
year -
> >year
> >> >2003.
> >> >
> >> >I carefully went through the data, removing any day that
contains
> >zero
> >> >for
> >> >any of the hours. Instead of the 365 days in the year 2003, I
ended
> >up
> >> >with 362 days.
> >> >
> >> >I saved that as CLMX1C (now stored in Ogbos2 with dput
function, see
> >> >attached please).
> >> >
> >> >If I run the data with my script, it gives me what I am
expecting.
> >My
> >> >script is:
> >>
>d<-read.table("CLMX1C",col.names=c("h","count"))
> >> >y<-d$count
> >> >data<-(y-mean(y))/mean(y)*100
> >> >
> >> >A<-matrix(rep(1:24,362))
> >> >B<-matrix(data)
> >> >
> >> > oodf<-data.frame(A,B)
> >> > oodf<-data.frame(A,B)
> >> >library(plotrix)
> >> >std.error<-function(x) return(sd(x)/(sum(!is.na(x))))
> >> >oomean<-as.vector(by(oodf$B,oodf$A,mean))
> >> >oose<-as.vector(by(oodf$B,oodf$A,std.error))
> >> >plot(1:24,oomean,type="b",ylim=c(-0.4,0.5),
> >> > xlab="Hours",ylab="CR
count",main="CR daily variation for 2004")
> >> >dispersion(1:24,oomean,oose,arrow.cap=.01)
> >> >
> >> >Now, instead of foraging through the big data removing the day
for
> >> >which
> >> >there is a missing data for any hour, I wish to try to replace
the
> >> >missing
> >> >data with NA and hoping that it will do the job for me.
> >> >
> >> >I added just three lines in the script above:
> >>
>d<-read.table("2003",col.names=c("y","m","d","h","count"))
> >> >y<-d$count
> >> >df<-data.frame(y)#line 1
> >> >library('dplyr') # line 2
> >> >y<-na_if(df, 0) #line 3
> >> >data<-(y-mean(y))/mean(y)*100.
> >> >Then I started getting error messages:
> >> >Error in is.data.frame(x) :
> >> > (list) object cannot be coerced to type 'double'
> >> >In addition: There were 26 warnings (use warnings() to see
them).
> >> >
> >> >I hope you will assist me to deal with the issues of replacing
zeros
> >> >with
> >> >NA in column 5 in such a way that my code will run.
> >> >
> >> >Iam ever indebted!!
> >> >Best regards
> >> >Ogbos
> >> >______________________________________________
> >> >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.
> >>
> >> --
> >> Sent from my phone. Please excuse my brevity.
> >>
>
> --
> Sent from my phone. Please excuse my brevity.
>