Dear All,
I am using the code below to calculate error bars. I note that the length of
the error bars can be varied by varying the constant (0.975). It does
appear that any number can be substituted for 0.975, making it confusing for
me to know how to quantify the error bars.
I wish to quantify
Hello,
The R version on my system is R version 2.8.1 (2008-12-22). I have
previously installed all the desired packages for my work - and they have
been working fine. However, somebody played with my system and removed
almost all the stuff. I am installing the packages again.
But I am finding
, ogbos okike wrote:
Hello,
The R version on my system is R version 2.8.1 (2008-12-22). I have
previously installed all the desired packages for my work - and they have
been working fine. However, somebody played with my system and removed
almost all the stuff. I am installing the packages
Dear All,
I have just installed a new version of R (Version R-2.11.0) and did install
other packages such as raster with ease. However, I could not start the
plotting device x11(). I remembered that somewhere at the stage of
installation, an error occurred : 'configure: error: --with-x=yes
Hey,
Just decided to install the latest version of R and the device is supported.
Thanks
Ogbos
On 7 October 2010 13:49, ogbos okike ogbos.ok...@gmail.com wrote:
Dear All,
I have just installed a new version of R (Version R-2.11.0) and did install
other packages such as raster with ease
did.
On Thu, Oct 7, 2010 at 7:49 AM, ogbos okike ogbos.ok...@gmail.com wrote:
Dear All,
I have just installed a new version of R (Version R-2.11.0) and did
install
other packages such as raster with ease. However, I could not start the
plotting device x11(). I remembered that somewhere
Hi All,
Using maps library, one can generate world map in geographic coordinates.
However, my data is organized in magnetic coordinates and requires world map
in magnetic coordinates.
I am not sure how to generate world map in magnetic coordinates in R.
I would appreciate any help.
Thanks
Ogbos
Hi everybody,
I have a huge longitude and latitude data. I have used raster package to
get the since of its global distribution. But I wish to display the
information using a few points on the world map.
The data is of the form:
95.2156 0.8312
-65.3236 -3.3851
-65.2364 -3.2696
-65.2349 -3.2679
Hi all,
I can't understand why grid package failed to installed on my machine. Can
anybody have a look and advice on where I am missing it.
My R version is 2.9.2. After untarring: tar xzvf grid_0.7-4.tar.gz, I
tried to configure or make but none worked, 'no such files'. I have also
tried using
Good morning. I wish to plot two data on the same axis. I tried plot(x,y,
type = l) for the first and tried to use lines or points(x,y, lty = 2, col
= 4) to add or plot the second data on alongside the first. However, what I
got was not encouraging.
I have attached the two data and would be
Good morning to you. I have about 4 different lines in one plot. I have used
legend to indicate the colour of each plot. But the box contain the legend
covers part of the lines thereby blurring the legend. There are some spaces
in the plot that are empty and large enough to accommodate the legend
I write to thank all those who have helped me to fix this problem.
Best regards
Ogbos
2009/10/7 ogbos okike ogbos.ok...@gmail.com
Good morning. I wish to plot two data on the same axis. I tried plot(x,y,
type = l) for the first and tried to use lines or points(x,y, lty = 2, col
= 4) to add
I've shared a document with you:
plotw5.png
http://docs.google.com/Doc?docid=0AfJ5_yv8GrERZGY3NmhuMnNfMWZtcGhjc2d3hl=eninvite=CPzN4ZsF
It's not an attachment -- it's stored online at Google Docs. To open this
document, just click the link above.
Greetings to you all.
I have two datasets - Time and magnitude. For a particular location, the
magnitude of the parameter varies with time. I wish to obtain a polar
coordinate distribution of time (0-24h) and magnitudes so as to visualize
how magnitude varies with different times of the day
Hi,
I am learning how to do principal component analysis in R. However, since I
am family with only a few built-in functions like prcomp, sd, cor, I started
manually with examples in text books while trying to use the few functions I
know to manipulate what they have in the text. From the example
Hello.
I am working with a set of variables which are in columns. Three of them
are of the same length while one has a different length. Typing
'data-matrix(c(ca$value, mo$value,b2$value, y1), ncol = 3)'
appears to read any three columns out of the four, though I can say exactly
which of these
read your text book to the point where it mentions the eigenvectors of
the ccorrelation (or covariance) matrix
ogbos okike ogbos.ok...@gmail.com 01/30/10 7:09 PM
Hi,
I am learning how to do principal component analysis in R. However,
since I
am family with only a few built-in functions like
Hi,
I have a data of size 981.1MB(1028707715) and I intend to calculate the
length of the data using tapply function in R. I was able to read the data
into R but when I tried to use the factor function, I had an error message
Error: cannot allocate vector of size 2.0 Gb.
Can anybody tell me what
Hi all,
I have three columns of lightning data. The first column is dates, the
second and third are latitude and longitude respectively. The data is
extremely large as many observations are made every second. Using the factor
function reduces the activity to a few number per a day.
But factor
Hi Uwe,
I am pleased to inform you that this problem has, following your hint, been
resolved. I installed 64-bit version of R and I was alright.
Thanks again for your time.
Ogbos
2009/6/24 Uwe Ligges lig...@statistik.tu-dortmund.de
ogbos okike wrote:
Hi,
I have a data of size 981.1MB
Good day to you all,
I have lightning data containing date, time, latitude and longitude. I hope
that distribution of latitude and longitude will give number of lightning
occurrence in a region. I have used factor function to sum up the number of
events on latitude and longitude axis and saved as
Hi,
I have a question to post. But I have a problem finding the appropriate
forum. Some of the FAQ on pastecs are answered by
r-h...@stat.math.ethz.chmailing list. I searched to see if this
mailing list requires registration
or not. There was no information about that. I am a registered member of
Good evening to everybody. I have a data of four columns - three of the
columns represent date while the fourth is counts. Setting 4000 as my
minimum point/value, I wish to find out the number of counts that are less
or equal to 4000 and the dates they occur. I have installed pastecs and have
Good morning once more. My problem of yesterday has been addressed. Having
learned a few tricks from that, I wish to ask another question in connection
with that. My data is a cosmic ray data consisting of dates and counts. When
I plot a graph of counts versus dates, the resultant signal shows a
Hello,
To me as a beginner, every problem looks big. Below is what I was asked to
do as part of a code that will solve my problem. I have used read.table to
read my data into R and assigned the column names with colnames(dat)- ... .
But to go from txt- to the last at the bottom of the table below
Hi,
I have used sm package ( sm.sphere(k$x, k$y, kappa = 20, hidden = FALSE,
sphim = FALSE, addpoints = FALSE) to obtain global lightning distribution.
But I am still in trouble with the color image - everything is thick black
and therefore, difficult to note the regions of higher density. Could
Hi,
I have a tar.gz file. Using $tar -xzvf 'filename', I untarred the file.
However, I found that the content is .PGC files (e.g.
ISCCP.D2.0.GLOBAL.1983.07.99..GPC ).
I have tried to search for analysis of gpc data file in R. I have not been
successful to get a good tip.
I would be pleased
Hi Everybody,
I have a matrix of about 45 columns. Some of the rows contain zeros. Using
data1-data[complete.cases(data),], I can remove the NA rows. But I am
unable to tackle that of zeros.
Can anybody give me an idea of how to remove rows containing zeros in a
matrix.
Thanks so much
Best
Ogbos
and hope it helps,
Paul
I hope it helps.
Best,
Dimitris
On 3/9/2010 11:05 AM, ogbos okike wrote:
Hi Everybody,
I have a matrix of about 45 columns. Some of the rows contain zeros.
Using
data1-data[complete.cases(data),], I can remove the NA rows. But
I am
unable to tackle
Hi All,
I have been running pca smoothly for some time now. However, the error
message below makes progress difficult for me.
It will be much appreciated if anybody can hint me on the possible source of
this error.
Thanks
Ogbos
The error:
Error in pca$rotation %*% sqrt(data.cor.eigen.matrix) :
I've shared a document with you:
plot.png
http://docs.google.com/leaf?id=0B_J5_yv8GrERYWRhZjU4NDItMmMyOS00ZGIxLWE2ZDUtMDFiYTY3MzliMzhkhl=eninvite=CI2a3aYC
It's not an attachment -- it's stored online at Google Docs. To open this
document, just click the link above.
[[alternative HTML
I've shared a document with you:
plot.png
http://docs.google.com/leaf?id=0B_J5_yv8GrERYWRhZjU4NDItMmMyOS00ZGIxLWE2ZDUtMDFiYTY3MzliMzhkhl=eninvite=CIa4nqkK
It's not an attachment -- it's stored online at Google Docs. To open this
document, just click the link above.
Hi there,
Using the code
Hi there,
Using the code below I generated the plot attached. The error bars at day =
-3, 2 and 4 appear are larger. I was thinking if there is a way I could make
all the error bars to be of the same size. I don't know if that makes sense.
If not, then, is there a way I can plot only these 3
April 2010 00:28, Jim Lemon j...@bitwrit.com.au wrote:
On 04/07/2010 03:49 AM, ogbos okike wrote:
Hi there,
Using the code below I generated the plot attached. The error bars at day
=
-3, 2 and 4 appear are larger. I was thinking if there is a way I could
make
all the error bars
Dear List,
I am trying to label a plot with the symbol +/- sigma. Using something like
- expression (2*sigma) gives me the symbol 2ó. However, adding +/- to it
beats me.
The code I am using is: plot(x,y,type=l,main= expression(paste(±,
plain(2*ó)),sep=).
Any suggestion will be appreciated.
Best
Hi Yves,
Many thanks. Very straight forward, the way you put.
Best regards
Ogbos
On 27 July 2011 15:13, Yves REECHT yves.ree...@gmail.com wrote:
**
Hi,
You may try something like:
plot(rnorm(10), rnorm(10), main=expression( %+-% 2*sigma))
HTH,
Yves
Le 27/07/2011 14:57, ogbos okike
Dear All,
I am trying to put 10^-8 st km^-2day^-1 on x-axis of my plot. I tried using
: ylab = expression(paste(st / , plain(km)^2, / day)) to see if I can
at least get the unit before thinking about the power of 10 (10^-8).
However, ylab = expression(paste(st / , plain(km)^2, / day)) didn't
Hi Peter,
Many thanks. It worked.
Regards
Ogbos
On 1 August 2011 14:05, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2011-08-01 03:32, ogbos okike wrote:
Dear All,
I am trying to put 10^-8 st km^-2day^-1 on x-axis of my plot. I tried
using
: ylab = expression(paste(st / , plain(km)^2, / day
Dear List,
Following the link below (
http://rgm2.lab.nig.ac.jp/RGM2/func.php?rd_id=plotrix:clock24.plot) I got an
interesting polar plots which displayed my data and the time of observation.
Thank you very much for providing such details.
However, I have two set of data which I wish to display
Dear all,
I am trying to add error bars on a boxplot but have encountered an error as
indicated below. Is there a package I need to install or a library I have to
load before this goes please.
Thanks for any idea.
Ogbos
x-replicate(20,rnorm(50))
boxplot(x,notch=TRUE,main=Notched boxplot with
Dear list,
Indeed, there is no need adding error bars to boxplot. Points with error
bars as suggested will be quite better.
Many thanks to those who have advised me on this.
Regards
Ogbos
On 27 January 2011 12:29, Jim Lemon j...@bitwrit.com.au wrote:
On 01/27/2011 05:04 AM, ogbos okike wrote
Dear List,
I am currently running R-2.10.1. I wish to install R-2.13.0 or update my
R-2.10.1 to the R-2.13.0.
I have downloaded R-2.13.0 but wish to seek your advice before installing
that.
Please is there a better way of updating to the newer version other than
downloading and installing the
/2011 6:02 AM, ogbos okike wrote:
Dear List,
I am currently running R-2.10.1. I wish to install R-2.13.0 or update my
R-2.10.1 to the R-2.13.0.
I have downloaded R-2.13.0 but wish to seek your advice before installing
that.
Please is there a better way of updating to the newer version other
,
On Thu, May 26, 2011 at 10:05 AM, ogbos okike ogbos.ok...@gmail.com
wrote:
Hi Duncan,
Thanks for your time.
Using ./configure as specified in the installation manual, I attempted to
install R-2.13.0 but it reported an error message:
checking for IceConnectionNumber in -lICE
the key following the two ways specified at
http://keyserver.ubuntu.com:11371/ but no success yet, access denied.
I will post to debian list.
Regards
Ogbos
On 26 May 2011 16:51, Steve Lianoglou mailinglist.honey...@gmail.comwrote:
Hi,
On Thu, May 26, 2011 at 10:42 AM, ogbos okike ogbos.ok
.
Christophe
Le 31/05/2011 10:07, ogbos okike a écrit :
Hello,
I recently install ubuntus 10.04 LTS on my desktop. My target is to
install
R from package manager as I learn that that will easier for me. I got the
SECURE APT from Micheal yesterday and successfully installed that.
However, I am
Hello,
I am attempting to randomly select a data of equal length from my dataset.
My dataset is of equal length each ranging from 1 to 16 rows. Since they are
of equal length, I can form a matrix of equal length and rows or concatenate
them into a data of 16n x 2 matrix where n is number of
Hello Everybody,
The way I learn some analyses tools in R is to install the package and play
with some examples.
I am trying to see if declustering algorithm is implemented in R. I have
seen some examples examples of declustering data. However, I could not try
them as I do not know which package
Dear all,
Using the code I got from the link (
http://www.phaget4.org/R/image_matrix.html), I obtained a nice plot that
suits me. However, adding axes labels proved difficult for me . I have
succeeded in adding a few things to the plot function so as to get what a
better plot. Other things work
Dear Jim,
One touch!! Thank you very much
Best regards
Ogbos
On 4 February 2011 13:47, Jim Lemon j...@bitwrit.com.au wrote:
On 02/04/2011 09:14 PM, ogbos okike wrote:
Dear all,
Using the code I got from the link (
http://www.phaget4.org/R/image_matrix.html), I obtained a nice plot
Dear All,
I am trying to generate a circular/radial plot. The script below has a
result I am looking for:
testlen<-rnorm(24)*2+5
testpos<-0:23+rnorm(24)/4
clock24.plot(testlen,testpos,main="Test Clock24 (lines)",show.grid=FALSE,
line.col="green",lwd=3)
if(dev.interactive()) par(ask=TRUE)
#
Dear All,
I have a script that draws longitude and latitude of lightning
occurrence. This script was running fine before. But when I changed my
system and do a fresh install on another laptop, this error persist.
source("script")
Error in eval(expr, envir, enclos) :
could not find function
/16, Ben Tupper <btup...@bigelow.org> wrote:
> Hi,
>
> A terrific resource for this type of issue (and pretty much anything related
> to R) is http://rseek.org/ I'm sure I use it at least daily. Check out
> ...
>
> http://rseek.org/?q=pointsToRaster
>
> The first hit
g:
> http://rstudio-pubs-static.s3.amazonaws.com/3369_998f8b2d788e4a0384ae565c4280aa47.html
>
> On Fri, 22 Apr 2016 at 08:31 Ogbos Okike <giftedlife2...@gmail.com> wrote:
>
>> Dear All,
>> I am trying to generate a circular/radial plot. The script below has a
>
l come from that package.
>
> -
> David L Carlson
> Department of Anthropology
> Texas A University
> College Station, TX 77840-4352
>
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ogbos Okike
Hi Saba,
Your main worry may be that of non- zero status and hence your attempt to
load what the system claims you have. I have encountered such problems
severally. You can try two things: run it several times ( network issues
might play a role) or try different crab mirrors.
Ogbos
On May 19, 2016
Dear All,
This problem is over. Clock24.plot did the job. Thanks to all those who
assisted me.
Ogbos
On Apr 22, 2016 8:34 PM, "Ogbos Okike" <giftedlife2...@gmail.com> wrote:
> Dear All,
> One hand. Many thanks!! The code run as soon as I loaded lubridate.
>
> Pl
most important thing is that Date objects by definition do not include
> time of day. You want to look at ISOdatetime() and as.POSIXct() instead. And
> beware daylight savings time issues.
>
> -pd
>
> On 18 Apr 2016, at 15:09 , Ogbos Okike <giftedlife2...@gmail.com> wr
Dear All,
Many thanks for bailing me out.
Ogbos
On Apr 18, 2016 9:07 PM, "David Winsemius" <dwinsem...@comcast.net> wrote:
>
> > On Apr 18, 2016, at 10:44 AM, Ogbos Okike <giftedlife2...@gmail.com>
> wrote:
> >
> > Dear ALL,
> > Thank you
Dear All,
I have a data set containing year, month, day and counts as shown below:
data <- read.table("data.txt", col.names = c("year", "month", "day", "counts"))
Using the formula below, I converted the data to as date and plotted.
new.century <- data$year < 70
data$year <- ifelse(new.century,
Dear All,
I am using R to do my work and thank you very much for developing,
maintaining and making such excellent software available to anyone
that is interested enough to ask for it.
I have registered at Nabble. I was wondering the right forum for me
to send my help request. I have tried
Dear workers,
I have a data of length 1136. Below is the code I use to get the means B.
It worked fine and I had the mean calculated and plotted.
I wish to plot the error bars as well. I already plotted such means with
error bars before. Please see attached for example.
I tried to redo the same
Dear Contributors,
I am surprised that I cannot add legend to a certain plot. Although the
x-axis indicates years, the actual data was in year, month and day format.
I then used as.Date to play around and get what I am looking for. I am,
however, worried that I cannot add legend to the plot no
Dear List,
I am happy to report that the problem is fixed. as.Date("1998-02-10") as
suggested by David handled the problem with easy. Many thanks to everybody.
as.Date(1998-02-10) really resulted in error. It is my oversight. I really
tried many things the day I was working on that and have
t; plot(-5:10,oomean,type="b",ylim=c(5,11),
> xlab="days (epoch is the day of Fd)",ylab="strikes/km2/day")
> dispersion(-5:10,oomean,oose)
>
> I get the expected plot.
>
> Jim
>
>
> On Sat, Jun 23, 2018 at 9:36 PM, Ogbos Okike
8 107074
> 799 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&quo
e(2018,01,22)
> [1] "2018-01-22 12:00:00 GMT"
> > ISOdate(2018,01,22,18,17)
> [1] "2018-01-22 18:17:00 GMT"
>
> Add something like:
>
> if(is.null(data$hour),data$hour<-12
>
> then pass data$hour as it will default to the same value as if you
>
Dear Members,
Compliments of the Season!!
Below is a part of a code I use for Fourier analysis of signals. The code
handles data with the format 05 01 018628 (year, month, day and count)
05 01 028589 (year, month, day and count)
The sample data is attached
Kind R-users,
I run a simple regression. I am interested in using the Monte Carlo to test
the slope parameter.
Here is what I have done:
d1<-read.table("Lightcor",col.names=c("a"))
d2<-read.table("CRcor",col.names=c("a"))
Li<-d1$a
CR<-d2$a
fit<-lm(Li~CR)
a<-summary(fit)
a gives the slope as
t;
> N <- length(Li)
>
> HTH,
> Eric
>
>
> On Wed, Aug 22, 2018 at 6:02 PM, Ogbos Okike
> wrote:
>
>> Kind R-users,
>> I run a simple regression. I am interested in using the Monte Carlo to
>> test
>> the slope parameter.
>> Here is wha
... it should be
>
> Li[sample(1:N, size = S, replace = TRUE)]
>
> i.e. no comma after the closing parenthesis
>
>
>
> On Wed, Aug 22, 2018 at 7:20 PM, Ogbos Okike
> wrote:
>
>> Hello Eric,
>> Thanks for this.
>>
>> I tried it. It went but a
<- sumb2 + coef(mod)[[2]]
> }
> print(sumb2/C, digits = 3)
>
> Best,
> Eric
>
>
>
> On Wed, Aug 22, 2018 at 7:28 PM, Ogbos Okike
> wrote:
>
>> Hello Erick,
>>
>> Thanks again.
>> Another line indicated error:
>>
>> source(&quo
Dear List,
I have a dataset of high variability. I conducted epoch analysis and
attempted to plot the standard error bar alongside.
I am, however, surprised that the error bars are of equal length. I do not
think that the variability in the data is captured, except there is a kind
of averaging
s:
>
> range(oose)
> [1] 1728.234 6890.916
>
> What was the range of oose values for the data in the plot you included
> with your message?
>
>
> David L Carlson
> Department of Anthropology
> Texas A University
> College S
Dear Experts,
I generated the plot attached. Every other thing is OK except the black
horizontal lines which should appear like points or dots as the coloured
ones. I can't understand why.
I tried to change it to look like dots by calling empty plots so that I
will add them as points.
Since I
Hi David,
That's it!!! The outcome is attached.
Many thanks please.
Best
Ogbos
On Wed, Sep 19, 2018 at 11:34 PM David Winsemius
wrote:
>
> > On Sep 19, 2018, at 7:55 AM, Ogbos Okike
> wrote:
> >
> > Dear Experts,
> > I generated the plot attached. Every othe
Dear Jim,
Good news to me!! Welcome.
I am fine. The code elegantly displayed the color.
I also tried to adjust the line:
draw.circle(lonmids[lon],latmids[lat],radius=sqrt(counts[lat,lon])/100,
border=countcol[lat,lon],col=countcol[lat,lon]) in order to reduce
the radius of the circle in
ec 11, 2018 at 11:14 AM Jim Lemon wrote:
>
> Hi Ogbos,
> I have been off the air for a couple of days. Look at the color.legend
> function in the plotrix package.
>
> Jim
> On Tue, Dec 11, 2018 at 12:39 PM Ogbos Okike wrote:
> >
> > Dear Jim,
> > I a
Dear Contributors,
I have a data of the form:
Lat Lon
30.1426 104.7854
30.5622 105.0837
30.0966 104.6213
29.9795 104.8430
39.2802 147.7295
30.2469 104.6543
26.4428 157.7293
29.4782 104.5590
32.3839 105.3293
26.4746 157.8411
25.1014 159.6959
25.1242 159.6558
30.1607 104.9100
31.4900
ounts,extremes=c("blue","red"))
> map("world",xlim=c(-90,160),ylim=c(20,45))
> for(lon in 1:length(lonmids)) {
> for(lat in 1:length(latmids)) {
> if(counts[lat,lon] > 0)
>draw.circle(lonmids[lon],latmids[lat],radius=sqrt(counts[lat,lon]),
> border=countcol[
Dear Contributors,
I have a data of the form:
4 8 10 8590 12516
4 8 11 8641 98143
4 8 12 8705 98916
4 8 13 8750 89911
4 8 14 8685 104835
4 8 15 8629 121963
4 8 16 8676 77655
4 8 17 8577 81081
4 8 18 8593 83385
4 8 19 8642 112164
4 8 20 8708 103684
4 8 21 8622 83982
4 8 22 8593 75944
4 8 23 8600
h you could modify or you
> are free to modify the code itself.
>
> Cheers
> Petr
>
> > -Original Message-
> > From: Ogbos Okike
> > Sent: Thursday, November 29, 2018 5:17 PM
> > To: PIKAL Petr
> > Cc: r-help
> > Subject: Re: [R] Correct
if (linky)
> lines(x, yleft, col = col[2], lty = 2, ...)
> if (smooth != 0)
> lines(supsmu(x, yleft, span = smooth), col = col[2],lty = 2, lwd
> = lwds, ...)
> }
>
> something like
>
> plot.yy(Year, Li, CR)
>
> Cheers
> Petr
>
> > -Origin
Dear List,
I have three data-column data. The data is of the form:
1 8590 12516
2 8641 98143
3 8705 98916
4 8750 89911
5 8685 104835
6 8629 121963
7 8676 77655
1 8577 81081
2 8593 83385
3 8642 112164
4 8708 103684
5 8622 83982
6 8593 75944
7 8600 97036
1 8650 104911
2 8730 114098
3 8731 99421
4
1)/28)
> col2means<-by(oodf[,2],oodf[,4],mean)
> col3means<-by(oodf[,3],oodf[,4],mean)
>
> Jim
>
> On Wed, Nov 28, 2018 at 2:06 PM Ogbos Okike
> wrote:
> >
> > Dear List,
> > I have three data-column data. The data is of the form:
> > 1 8590 12516
>
(d4$CR-mean(d4$CR))/mean(CR))*100
a5<-((d5$CR-mean(d5$CR))/mean(CR))*100
a6<-((d6$CR-mean(d6$CR))/mean(CR))*100
a7<-((d7$CR-mean(d7$CR))/mean(CR))*100
a1-a7 actually gives percentage change in the data.
Instead of doing this one after the other, can you please give an
indication on h
1)/28)
> col2means<-by(oodf[,2],oodf[,4],mean)
> col3means<-by(oodf[,3],oodf[,4],mean)
>
> Jim
>
> On Wed, Nov 28, 2018 at 2:06 PM Ogbos Okike
> wrote:
> >
> > Dear List,
> > I have three data-column data. The data is of the form:
> > 1 8590 12516
>
9.4889267
> #> 39 4 8561 122195 6 -0.028359802 16.6179943
> #> 40 5 8532 100945 6 -0.367009209 -3.6621512
> #> 41 6 8560 108552 6 -0.040037368 3.5976637
> #> 42 7 8634 108707 6 0.824102496 3.7455895
> #> 43 1 8646 117420 7 -0.816125860 14.4890796
> #> 44 2 8633 113
Dear Volunteers,
I have a table involving many decimal places:
2005-01-04 -2.13339817688037
2005-01-19 -6.86312349695117
2005-01-22 -4.33662370554386
2005-02-10 -1.40789214441639
2005-02-13 -1.1334121785854
2005-02-19 -1.28411233010119
2005-05-09 -1.6895978161324
2005-05-16 -3.07664523496947
been looking for.
xtable(data, digits=12) and I am fine.
Thank you so much.
Ogbos
On Tue, Sep 18, 2018 at 6:27 AM Jeff Newmiller
wrote:
> Have you read
>
> ?xtable
>
>
> On September 17, 2018 7:24:37 PM PDT, Ogbos Okike <
> giftedlife2...@gmail.com> wrote:
> &
your
> confusion. Beware that if you don't read them first and mention why they
> didn't answer your question then you may get less helpful responses when
> when you do ask questions.
>
> On September 17, 2018 10:49:43 PM PDT, Ogbos Okike <
> giftedlife2...@gmail.com> wrot
Dear Contributors,
I have two data frame of different column lengths. I am trying to have them
in one data frame.
Using
A<-d1$date
B<-d2$date
a<-data.table(A )[ , I := .I][data.table(B )[ , I := .I], on = "I"]
I got
1: 2005-01-04 1 2005-01-04
2: 2005-01-19 2 2005-01-19
3: 2005-01-22 3
t( n )
> a <- data.frame( d1_date=d1$date[ix], d2_date=d2$date[ix],
> d3_date=d3$date[ix] )
>
> On September 17, 2018 12:17:03 AM PDT, Ogbos Okike <
> giftedlife2...@gmail.com> wrote:
> >Dear Contributors,
> >
> >I have two data frame of different col
Dear Contributors,
I conducting epoch analysis. I tried to test the significance of my
result using randomization test.
Since I have 71 events, I randomly selected another 71 events, making
sure that none of the dates in the random events corresponds with the
ones in the real event.
Following
rstanding
> of how the data was generated, it is impossible for you (and anyone else) to
> say how such a test might be designed.
>
> Michael
>
>
> > -Original Message-
> > From: Ogbos Okike
> > Sent: Mittwoch, 27. Februar 2019 22:53
> > To: Meyners, Mic
ated numbers to test the
statistical significance level of the signal generated by
plot(-5:10,oomean,type="l",ylim=c(8890,9100), )?
I wish to test for 90% and 99% percentile.
I am sorry that this is too long.
Many thanks for your kind contributions
Best
Ogbos
On Sun, Feb 10, 201
ot going to
> work.
>
> Jim
>
> On Sun, Feb 17, 2019 at 12:55 PM Ogbos Okike wrote:
> >
> > Dear Jim,
> > Thank you and welcome back. It seems you have been away as you have not
> > been responding to people's questions as before.
> >
> > I have
ctual minimal working example of what you had working
> before you changed to POSIXct.
>
> On February 16, 2019 1:08:38 PM PST, Ogbos Okike
> wrote:
> >Dear Jeff,
> >One more problem please.
> >
> >When I used as.Date(ISOdate(dta$year, dta$month, dta$day,dta$hour)) to
Dear List,
I have a simple code with which I convert year, month, and day to a date format.
My data looks like:
67 01 2618464
67 01 2618472
67 01 2618408
67 01 2618360
67 01 2618328
67 01 2618320
67 01 2618296
while my code is:
data <- read.table("CALG.txt",
Dear Jeff,
I am alright now
Please accept my indebtedness!!!
Warmest regards
Ogbos
On Fri, Feb 15, 2019 at 8:25 AM Ogbos Okike wrote:
>
> Dear Jeff,
>
> Please hold.
> It is begging to work. There was an error somewhere. One ")" is
> missing and as I went back t
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