HI Jim,
This is great!! It is also tricky!!! The problem lies in the choice of
ylim. And looking at the data and choosing ylim based on the maximum and
minimum values of y is a waste of time. And choosing it by other means was
yet much more difficult.
I had to start plotting part of the data with incremental step of 80 data
points and manually varying ylim till I got to the last data point 1136,
where I finally used ylim=c(150000,162000) which has nothing to do with the
raw data.
Many, many thanks.
Best wishes
Ogbos
On Sun, Jun 24, 2018 at 9:51 PM, Jim Lemon <drjimlemon at gmail.com>
wrote:
> Hi Ogbos,
> The problem is almost certainly with the data. I get the plot I expect
> with the sample data that you first posted, so I know that the code
> works. If you try thIs what do you get?
>
> oodf<-read.table(text="S/N A B
> 1 -5 64833
> 2 -4 95864
> 3 -3 82322
> 4 -2 95591
> 5 -1 69378
> 6 0 74281
> 7 1 103261
> 8 2 92473
> 9 3 84344
> 10 4 127415
> 11 5 123826
> 12 6 100029
> 13 7 76205
> 14 8 105162
> 15 9 119533
> 16 10 106490
> 17 -5 82322
> 18 -4 95591
> 19 -3 69378
> 20 -2 74281
> 21 -1 103261
> 22 0 92473
> 23 1 84344
> 24 2 127415
> 25 3 123826
> 26 4 100029
> 27 5 76205
> 28 6 105162
> 29 7 119533
> 30 8 106490
> 31 9 114771
> 32 10 55593
> 33 -5 85694
> 34 -4 65205
> 35 -3 80995
> 36 -2 51723
> 37 -1 62310
> 38 0 53401
> 39 1 65677
> 40 2 76094
> 41 3 64035
> 42 4 68290
> 43 5 73306
> 44 6 82176
> 45 7 75566
> 46 8 89762
> 47 9 88063
> 48 10 94395
> 49 -5 80651
> 50 -4 81291
> 51 -3 63702
> 52 -2 70297
> 53 -1 64117
> 54 0 71219
> 55 1 57354
> 56 2 62111
> 57 3 42252
> 58 4 35454
> 59 5 33469
> 60 6 38899
> 61 7 64981
> 62 8 85694
> 63 9 79452
> 64 10 85216
> 65 -5 71219
> 66 -4 57354
> 67 -3 62111
> 68 -2 42252
> 69 -1 35454
> 70 0 33469
> 71 1 38899
> 72 2 64981
> 73 3 85694
> 74 4 79452
> 75 5 85216
> 76 6 81721
> 77 7 91231
> 78 8 107074
> 79 9 108103
> 80 10 7576",
> header=TRUE)
> 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(-5:10,oomean,type="b",ylim=c(50000,110000),
> xlab="days (epoch is the day of
Fd)",ylab="strikes/km2/day")
> dispersion(-5:10,oomean,oose)
>
> I get the attached plot;
>
> Jim
>
> On Mon, Jun 25, 2018 at 1:58 AM, Ogbos Okike <giftedlife2014 at
gmail.com>
> wrote:
> > Hi Jim
> >
> > Thanks again for returning to this.
> > please not that the line
"oomean<-as.vector(by(oodf$B,oodf$A,mean))" was
> > omitted (not sure whether deliberate) after you introduced the
standard
> > error function.
> > When I used it, empty plot window with the correct axes were generated
> but
> > no data was displayed. No error too.
> >
> > library(plotrix)
> > std.error<-function(x) return(sd(x)/(sum(!is.na(x))))
> > oose<-as.vector(by(oodf$B,oodf$A,std.error))
> > plot(-5:10,oomean,type="b",ylim=c(50000,110000),
> > xlab="days (epoch is the day of
Fd)",ylab="strikes/km2/day")
> > dispersion(-5:10,oomean,oose)
> >
> > When I included the line, the same empty graph window was generated
but
> with
> > the former error "Error in FUN(X[[1L]], ...) : could not find
function
> > "FUN""
> > 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(-5:10,oomean,type="b",ylim=c(50000,110000),
> > xlab="days (epoch is the day of
Fd)",ylab="strikes/km2/day")
> > dispersion(-5:10,oomean,oose)
> >
> > I am sure am missing something but can't place it. Please have a
look
> again
> > to track my mistake.
> >
> > Warmest regards
> > Ogbos
> >
>
[[alternative HTML version deleted]]