Perhaps read ?mean...
On June 3, 2020 6:15:11 PM PDT, Ogbos Okike <giftedlife2...@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
><jdnew...@dcn.davis.ca.us>
>wrote:
>
>> df[[ 5 ]][ 0 == df[[ 5 ]] ] <- NA
>>
>> On June 3, 2020 1:59:06 AM PDT, Ogbos Okike
><giftedlife2...@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@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.
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