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

Hi Subhamitra,
As I said, the code I sent is an approximation to get your year labels in
about the correct places. You are welcome to improve the calculations.

182 days is about half a year, so that the first "tick" will fall around
the end of June (i.e. the middle of the year). If you specify the increment
as 226, you get one too many labels. 229 is what is known as a kludge (a
clumsy solution that works)

Yes, I mistakenly thought that the observations were the same throughout
the four files. As you know this (and I didn't) you can do a better job of
placing the year labels by changing the sequence for each of the CSV (not
Excel) files. The best method of all would be to have a date for each
observation. You could then discard all these approximations I have made to
get the plots to work.

No, the arguments of the axis function are:

axis(<side of plot>, <position of ticks>, <labels for the ticks>)

The first argument is; 1=bottom, 2=left, 3=top, 4=right. The next two
arguments must be the same length. If not, you will get an error. As you
can see, only every other tick has a label to avoid crowding. There are
ways to get more tick labels on an axis.

Jim


On Sun, Dec 16, 2018 at 7:03 PM Subhamitra Patra <
subhamitra.patra at gmail.com> wrote:

Hello Sir,

I have three queries regarding your suggested code.

*1. *In my last email, I mentioned why there are missing observations in
my data series. In the line, *year_mids<-seq(182,5655,by=229), *

*A. what 182 indicates and what is the logic behind the consideration of
229 increments, although there are 226 observations per year?*
*B.  Each excel file is having different observations depending on the
variation of starting dates. So, is it required to add  **year_mids in
the loop? I think I need to justify **year_mids object each time after
importing the individual excel files. If I am wrong, kindly correct me.*

2. Further, in the command* axis(1,at=year_mids,labels=1994:2017), 1
indicates the no. of increments of year name, right?*

Kindly clarify my queries Sir for which I shall be always grateful to you.

Thank you very much.

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On Sun, Dec 16, 2018 at 12:24 PM Subhamitra Patra <
subhamitra.patra at gmail.com> wrote:

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.

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12:20:17 PM

On Sun, Dec 16, 2018 at 8:10 AM Jim Lemon <drjimlemon at 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)
EMs1.1<-read.csv("EMs1.1.csv")
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)
EMs2.1<-read.csv("EMs2.1.csv")
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)
EMs3.1<-read.csv("EMs3.1.csv")
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)
EMs4.1<-read.csv("EMs4.1.csv")
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*

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