# [R] [R studio] Plotting of line chart for each columns at 1 page

Subhamitra Patra @ubh@mitr@@p@tr@ @ending from gm@il@com
Sun Dec 16 07:54:41 CET 2018

```Thank you very much sir. Actually, I excluded all the non-trading days.
Therefore, Each year will have 226 observations and total 6154 observations
for each column. The data which I plotted is not rough data. I obtained the
rolling observations of window 500 from my original data. So, the no. of
observations for each resulted column is (6154-500)+1=5655. So, It is not
accurate as per the days of calculations of each year.

Ok, Sir, I will go through your suggestion, obtain the results for each
column of my data and would like to discuss the results with you. After
solving of this problem, I would like to discuss another 2 queries.

Thank you very much Sir for educating a new R learner.

[image: Mailtrack]
<https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&>
Sender
notified by
Mailtrack
<https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&>
12/16/18,
12:20:17 PM

On Sun, Dec 16, 2018 at 8:10 AM Jim Lemon <drjimlemon using gmail.com> wrote:

> Hi Subhamitra,
> Thanks. Now I can provide some assistance instead of just complaining.
> Your first problem is the temporal extent of the data. There are 8613 days
> and 6512 weekdays between the two dates you list, but only 5655
> observations in your data. Therefore it is unlikely that you have a
> complete data series, or perhaps you have the wrong dates. For the moment
> I'll assume that there are missing observations. What I am going to do is
> to match the 24 years (1994-2017) to their approximate positions in the
> time series. This will give you the x-axis labels that you want, close
> enough for this illustration. I doubt that you will need anything more
> accurate. You have a span of 24.58 years, which means that if your missing
> observations are uniformly distributed, you will have almost exactly 226
> observations per year. When i tried this, I got too many intervals, so I
> increased the increment to 229 and that worked. To get the positions for
> the middle of each year in the indices of the data:
>
> year_mids<-seq(182,5655,by=229)
>
> Now I suppress the x-axis by adding xaxt="n" to each call to plot. Then I
> add a command to display the years at the positions I have calculated:
>
> axis(1,at=year_mids,labels=1994:2017)
>
> Also note that I have added braces to the "for" loop. Putting it all
> together:
>
> year_mids<-seq(182,5655,by=229)
> pdf("EMs.pdf",width=20,height=20)
> par(mfrow=c(5,4))
> # import your first sheet here (16 columns)
> ncolumns<-ncol(EMs1.1)
> for(i in 1:ncolumns) {
>   plot(EMs1.1[,i],type="l",col = "Red", xlab="Time",
>        ylab="APEn", main=names(EMs1.1)[i],xaxt="n")
>  axis(1,at=year_mids,labels=1994:2017)
> }
> #import your second sheet here, (1 column)
> ncolumns<-ncol(EMs2.1)
> for(i in 1:ncolumns) {
>   plot(EMs2.1[,i],type="l",col = "Red", xlab="Time",
>        ylab="APEn", main=names(EMs2.1)[i],xaxt="n")
>  axis(1,at=year_mids,labels=1994:2017)
> }
> # import your Third sheet here, (1 column)
> ncolumns<-ncol(EMs3.1)
> for(i in 1:ncolumns) {
>   plot(EMs3.1[,i],type="l",col = "Red", xlab="Time",
>        ylab="APEn", main=names(EMs3.1)[i],xaxt="n")
>  axis(1,at=year_mids,labels=1994:2017)
> }
> # import your fourth sheet here, (1 column)
> ncolumns<-ncol(EMs4.1)
> for(i in 1:ncolumns) {
>   plot(EMs4.1[,i],type="l",col = "Red", xlab="Time",
>        ylab="APEn", main=names(EMs4.1)[i],xaxt="n")
>  axis(1,at=year_mids,labels=1994:2017)
> }
> # finish plotting
> dev.off()
>
> With any luck, you are now okay. Remember, this is a hack to deal with
> data that are not what you think they are.
>
> Jim
>
>

--
*Best Regards,*
*Subhamitra Patra*
*Phd. Research Scholar*
*Department of Humanities and Social Sciences*
*Indian Institute of Technology, Kharagpur*
*INDIA*

[[alternative HTML version deleted]]

```