?unlist
On Mon, Jul 14, 2008 at 11:49 AM, <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am writing a very long program that deals with multiple large databases
> which
> often have missing data cells and other quirks. I've been testing it on
> small
> samples of the data so it doesn't take too long, but
try at=seq(from=26, to=32, by=2) in the axis statment
hope this helps
Stephen
On Mon, Jul 14, 2008 at 11:21 AM, Zroutik Zroutik <[EMAIL PROTECTED]> wrote:
> Dear R-users,
>
> I'm tackling with a problem which causing me a head-ache for a long time. I
> would like to create a nice x-axis to my pl
if it is a time series the interpolation methods in zoo are an option.
On Thu, Jul 10, 2008 at 6:41 AM, Daniel Malter <[EMAIL PROTECTED]> wrote:
>
> Please do read the posting guide. Please provide self-contained code (e.g.
> to
> randomly generate data) and illustrate (e.g. in a small table) wha
I am plotting a twelve panel plot of a zoo object. I have tried to raise
the cex from 0.6 to 1 to 2 and it does not seem to do anything. I am trying
to output this to a tiff file
tiff()
how do I get the labels of a larger font size
Stephen
--
Let's not spend our time and resources thinking abo
I am going to assume your data.frame is called x
#this transposes the matrix
x.t <- t(x)
rollmean(x.t)
On Wed, Jul 9, 2008 at 12:50 PM, Rheannon <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> I am trying to calculate a 31 day running mean in some temperature data
> along ROWS. Rollmean() works great a
# I would like to outline the squares in the legend with a black line. Does
anyone know how to do this?
x.t <- structure(c(5987.387, 4354.516, 3685.789, 6478.592, 5924.315,
NA, 8386, 5559.468, NA, 4651.273, 3967.5, NA, 4339.167, 5053.56,
NA, 4631.978, 4808.694, NA, 5217.306, 4017.632, NA, 5846.90
a.BHC",
"Cadmium", "Calcium", "Chromium", "Cobalt", "Copper", "Dalapon",
"delta.BHC", "Dicamba", "Dichloroprop", "Dieldrin", "Dinoseb",
"Dissolved.Aluminum", "Dissol
e in R and so it gives
> similar results:
>
> # uses axis.Date
> plot(aggregate(x.zoo[, 25], as.Date, force))
>
> You may wish to try a custom axis:
>
> plot(x.zoo[, 25], xaxt = "n")
> rng <- range(time(x.zoo))
> axis(1, at = seq(rng[1], rng[2], 1/12),
Grothendieck <[EMAIL PROTECTED]>
wrote:
> On Tue, Jul 8, 2008 at 2:59 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> > x.zoo <- zoo(x,as.yearmon(as.character(x$Yearmonth), "%Y-%m"))
> > plot(x.zoo[,25])
>
>
> 1. You are trying to pass data f
ou are left with a character date which you
> > may not want:
> >
> > zoo(, x$Yearmonth)
>
> This last one should have been:
>
> zoo(, as.character(x$Yearmon))
>
> since your data frame holds a factor rather than character column.
>
>
> >
> >
>
legend will accept locator() which would not automate it but would get
closer
On Tue, Jul 8, 2008 at 2:31 PM, <[EMAIL PROTECTED]> wrote:
> Dear R-Users,
>
> I am looking for a way to get legends placed automagically in an empty
> spot on a graph. Additional complication comes through my useage of
rsenic", "beta.BHC", "Cadmium", "Calcium",
"Chromium", "Cobalt", "Copper", "Dalapon", "delta.BHC", "Dicamba",
"Dichloroprop", "Dieldrin", "Dinoseb", "Dissolved.Aluminum&quo
are any of the subsets all NA?
On Tue, Jul 8, 2008 at 1:39 PM, Paul Adams <[EMAIL PROTECTED]> wrote:
> Hello everyone,
> I am trying to plot an MvA plot with the following code:
> dat<-read.table(file="C:\\Documents and Settings\\.txt",header=T)
> file.show(file="C:\\Documents and Settings\\O
#this is a subset of a larger data frame and I am okay with subsetting it as
there are redundant time stamps, but I would like to create a zoo object out
of this and I am having a hard #time figuring out how to do this the date
structure is year and then month
x <- structure(list(Yearmonth = str
I don't know if this will help, but look at the zoo, chron, and Posix Date
Time packages/classes.
On Tue, Jul 8, 2008 at 10:25 AM, collonil <[EMAIL PROTECTED]> wrote:
>
> hello,
>
> i cant find a solution on this (might be) easy problem:
>
> i have a time serie by carlandar weeks, so for every ca
This is what I would like to do and it works just fine. Is there a way to
shorten this code so I don't have to subset a subset of a subset?
d<-subset(subset(subset(subset(x, River.Mile<=202), River.Mile>3),
Lagrangian=="Yes"), EventType=="Regular")
Stephen
--
Let's not spend our time and resour
have you tried package(zoo) it works very well and should do what you want
On Wed, Jul 2, 2008 at 10:54 AM, Kerpel, John <[EMAIL PROTECTED]>
wrote:
> Hi all:
>
>
>
> I'm trying to plot two time series created in Rmetrics and label the
> x-axis with dates. I tried the following:
>
>
>
> dates <-
oo")$Version
> [1] "1.5-2"
>
> On Wed, Jul 2, 2008 at 10:08 AM, stephen sefick <[EMAIL PROTECTED]> wrote:
> > R 2.7.1 windows xp and version of zoo upgraded with new installation of R
> > 2.7.1 yesterday from CA1 mirror
> > window.zoo is
the case of chron, the axes are done by chron:::axis.times in
> the chron package.
>
> On Wed, Jul 2, 2008 at 9:32 AM, stephen sefick <[EMAIL PROTECTED]> wrote:
> > Like for instance that the xlim is small enough where the plot is showing
> > the day instead of the year (I
one other thing- should I read the data in with the argument FUN=as.chron
? if this is the case how do I tell chron that the data is not in
-mm-dd HH:MM:SS but in the format m/d/y HH:MM:SS ?
thanks
stephen
On Wed, Jul 2, 2008 at 9:32 AM, stephen sefick <[EMAIL PROTECTED]> wrote:
9:25 AM, stephen sefick <[EMAIL PROTECTED]> wrote:
> I have a matrix with data that runs from 1/1/06 00:01:00-1/31/08 23:46:00.
> I have read in the data with this
>
> fmt.chron <- function(x) {
>chron(sub(" .*", "", x), gsub(".* (.*)",
I have a matrix with data that runs from 1/1/06 00:01:00-1/31/08 23:46:00.
I have read in the data with this
fmt.chron <- function(x) {
chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x))
}
x <- read.zoo(file.choose(), sep=",", header=T, FUN=fmt.chron)
plotted with this
plot(x[,(seq(3, by=
xaxt="n" and yaxt="n" should do the trick if you just want to suppress the
labels. check
?par for further refinement
On Tue, Jul 1, 2008 at 5:25 PM, <[EMAIL PROTECTED]> wrote:
> I would like to know if there is a way to use plot(zoo) and suppress the
> label text to increase the amount of the sc
I would like to integrate the area under a curve without any smoothing or
the like- just on the raw numbers. I looked at integrate() but it requires
a function which I assume means something like x+x^2+x^3
is there a built in function in R for this?
#let's say
x <- seq(1:50)
y <- seq(1:50)
plot(
I would like to know the answer to this question now that I know what we are
getting at. integrate() looks like it is the right thing, but it has to use
a function- I would like to know how to just integrate the area under a
curve with just an input of x and y coordinates.
Stephen
On Tue, Jul 1
;t know how would I be able to get the integral.
>
>
> On 7/1/08, stephen sefick <[EMAIL PROTECTED]> wrote:
>>
>> ?lm
>> lm(x[,1]~x[,2])
>>
>> On Tue, Jul 1, 2008 at 2:28 PM, Shirin Safa <[EMAIL PROTECTED]> wrote:
>>
>>> Hi
>>&
?lm
lm(x[,1]~x[,2])
On Tue, Jul 1, 2008 at 2:28 PM, Shirin Safa <[EMAIL PROTECTED]> wrote:
> Hi
>
> I have a set of data like this:
>
>*Time of Day* *Pct of Daily Volume* 9:45 7.50% 10 6.25% 10:15 4.45%
> 10:30 4.80% 10:45 4.45% 11:00 4.20% 11:15 2.50% 11:30 2.30% 11:45 2.25%
> 12:00 2.45% 12
does it really getting kicked down to another row or is this just an
artifact of printing in the command window?
On Mon, Jun 30, 2008 at 2:16 PM, <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I have a CSV file where each row has at least 20 columns and some rows have
> up
> to 30 columns of data. When I u
x <- as.matrix(data.raw)
x[,2]
is this what you want?
BOBMARY JOHN CLIFF PAM
"C/C" NA "C/A""C/C""C/A"
On Sun, Jun 29, 2008 at 3:24 PM, Stephane Bourgeois <[EMAIL PROTECTED]>
wrote:
> Hi,
>
> I'd like to get an argument (I think it's the right term) dynamically from
> a li
kage was installed that fixed it. Whatever the case. Thanks!
>
> Prof Brian Ripley wrote:
> >
> > On Fri, 27 Jun 2008, stephen sefick wrote:
> >
> >> pdf( "yourfile.pdf", height=22, width=17)
> >> #yourcode
> >> dev.off()
> >>
>
pdf( "yourfile.pdf", height=22, width=17)
#yourcode
dev.off()
then use the gimp (free) to transform it to .png or whatever else (pdf makes
good graph)
On Thu, Jun 26, 2008 at 8:59 PM, Daniel Folkinshteyn <[EMAIL PROTECTED]>
wrote:
> not sure why it doesn't work, but try the following:
> first, p
t if you have some data and your intention is simply to
> create the index for it so you can create a zoo or zooreg object
> then the zooreg constructor can do it for you. For example,
> a series of 10 values starting at t1 with successive points
> spaced 15 minutes apart is:
>
I would like a sequence of dates with a time step of 15 minutes
starting:
1/1/2006 00:00:00 - 12/31/2006 23:45:00
function(x) {
chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x))
}
this is the piece of code I use to read in zoo objects
for any help I would be grateful I have tried sequenc
looks like you need to have a look at the package description. If you want
help from this list you probably need to look at the posting guidelines, and
then do a little poking around to figure out what your specific problems
are.
Stephen
On Thu, Jun 26, 2008 at 10:19 AM, mohammed alawa <[EMAIL P
would take
> care of missing data when doing the aggregation.
>
> On Wed, Jun 25, 2008 at 9:57 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> > #this is a daily series of precipitation data. I would like to condense
> it
> > into weekly means. How can I do this
> >
#this is a daily series of precipitation data. I would like to condense it
into weekly means. How can I do this
#as a side note I would like to do this same thing to two years worth of
fifteen minute interval data and make it into
#a series of daily averages (there are 96 readings per day)
#is ag
Example- I don't know SAS well, or at all at this point. But, if I knew
what you wanted maybe I could help you.
thanks
Stephen
On Wed, Jun 25, 2008 at 3:32 PM, Philip Twumasi-Ankrah <
[EMAIL PROTECTED]> wrote:
> R has wonderful graphics but I am wondering whether there is anything in R
> to pro
st my question in
two places until warranted.
thanks
Stephen
On Wed, Jun 25, 2008 at 9:28 AM, Prof Brian Ripley <[EMAIL PROTECTED]>
wrote:
> That is not the filename at the site you gave. Try using the actual file
> name.
>
> On Wed, 25 Jun 2008, stephen sefick wrote:
>
CLI6, CLI7, gap=list(top=c(8000,28),
> bottom=c(0,250)), range=50, outline=TRUE)
> Error in bxgap$out[bxgap$out > gap$top[2]] <- bxgap$out[bxgap$out >
> gap$top[2]] - :
> NAs are not allowed in subscripted assignments
>
>
> >>> "stephen sefick" <[EM
ckages specified
On Wed, Jun 25, 2008 at 8:58 AM, Richard Pearson <
[EMAIL PROTECTED]> wrote:
> From the (terminal window) command line try:
>
> R CMD INSTALL path.to.file/filename.tar.gz
>
> Does that do it?
>
> Richard
>
>
> stephen sefick wrote:
>
>>
the par(ask=FALSE) doesn't belong in the code (I think).
gap.boxplot(CLI3, CLI4, CLI5, CLI6, CLI7, gap=list(top=c(8000,28),
bottom=c(0,250)), range=50, outline=TRUE)
but I haven't tried this.
Stephen
On Wed, Jun 25, 2008 at 10:49 AM, Megan J Bellamy <[EMAIL PROTECTED]>
wrote:
> Hello,
>
> Wh
install.packages(file.choose(), repos=NULL)
thought this would work, but it didn't
the package is the sowas package - this doesn't seem to be a CRAN package,
and it can be found at :
http://tocsy.agnld.uni-potsdam.de/wavelets/
On Wed, Jun 25, 2008 at 8:25 AM, stephen sefick <[EM
I can't figure this one out- I am the administrator, The file that I want
to install is a .tar.gz which is located on my desktop. How do I get it
into my packages directory- through the GUI or through brute force?
thanks
stephen
--
Let's not spend our time and resources thinking about things t
I would like to filter using a periodogram produced by spec.pgram() then
pick the frequencies that I would like to filter out and then get the
inverse back into the time domain. Any thoughts.
Stephen
--
Let's not spend our time and resources thinking about things that are so
little or so large
A comment from the peanut gallery- It is always good to write an explicative
title because it facilitates searching in the forums.
thanks
Stephen
On Tue, Jun 24, 2008 at 6:49 AM, Gustaf Rydevik <[EMAIL PROTECTED]>
wrote:
> dear Xu,
>
> does:
> >library(urca)
> >example(ur.ers)
> >ers.gnp
> >str(
?cor
you can't calculate a correlation on a missing data point. I think you need
to use the argument use=pairwise.complete.obs in the cor call, but look at
the function description it will tell you what you need to know (I think).
Also, search the list-R nabble, Rsitsearch, or ...
Good Luck
Steph
http://www.rforge.net/JRI/
how about this?
that is about as far as I can help (assuming I did in the first place).
good luck
Stephen
On Mon, Jun 23, 2008 at 11:10 AM, Manjit Barman <[EMAIL PROTECTED]>
wrote:
> Dear Stephen,
> First of all thank for your response.
> But i am having the issue with
not having a clue It looks like you are missing the java library. I would
imagine that you can get this from sun relatively easily.
Stephen
On Mon, Jun 23, 2008 at 10:36 AM, Manjit Barman <[EMAIL PROTECTED]>
wrote:
> Hi ,
> I am getting these errors in Solaris 10. Can anyone help me on these? I
This is a lot like music- a good musician knows better when not to play than
when to (guitar teacher I had 15+ years ago).
stephen
On Mon, Jun 23, 2008 at 8:33 AM, jim holtman <[EMAIL PROTECTED]> wrote:
> To quote Jon Bentley (Programming Pearls):
>
> "The fastest, cheapest, most reliable piece
Is there anyway you could include the data so we can see what the problem
is- I am getting better at this code reading stuff, but it reduces my time
if I can just copy it into my R session and then fiddle.
thanks
Stephen
On Fri, Jun 20, 2008 at 4:39 AM, Thomas Pedersen <[EMAIL PROTECTED]> wrote:
#is there a way to get NA in the table of descriptive statistics instead of
the function stopping Thank you in advance
#data
x.f <- structure(list(Site = structure(c(9L, 9L, 9L, 9L, 9L, 9L, 9L,
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L,
9L, 9L, 9L, 9L), .Label = c("BC", "HC",
This discussion has already occurred- to my knowledge at least once. I
would suggest searching the archived list, and see what you get. If you
still have questions after you have a look then fire a couple of specifics,
but speaking for myself I don't know anything about matlab only S (and even
t
an introduction to R
On Wed, Jun 18, 2008 at 2:11 PM, Marvin Lists <[EMAIL PROTECTED]>
wrote:
> Dear all,
>
> Can anyone recommend a good book or an online tutorial for using data
> frames
> in R?
>
> I want to do statistical analysis on some survey data and I can specify
> what
> I would like to
#is this what you want?
t <- matrix(rnorm(120), ncol=12)
(colnames(t) <- paste(rep(LETTERS[1:4], each=3), 1:3, sep="."))
f<-as.matrix(cbind(c(t[,1:3]), c(t[,4:6]), c(t[7:9]), c(t[10:12])))
colnames(f)<-paste(rep(LETTERS[1:4]))
library(prettyR)
describe(f, num.desc=c("mean", "sd"))
On Wed, Jun 18,
give use a dummy (or the whole thing if it is not too large) and the code
you are using for the boxplot in copy and paste into R format (dput() the
data), and it may be easier.
Stephen
On Mon, Jun 16, 2008 at 2:51 PM, Thomas Adams <[EMAIL PROTECTED]> wrote:
> I have a problem where I need to lab
methodology to exhibit a periodic distance
> between my data? If, true, what could you recommend me to do this?
> Thanks in advance for your answers.
> Best regards,
>
> Anthony
>
> On Tue, Jun 10, 2008 at 6:13 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
>
>> I fro
RSEIS - I think may have a piper diagram.
On Thu, Jun 12, 2008 at 8:39 PM, Michael Grant <[EMAIL PROTECTED]> wrote:
> Sorry no previous message text or addresses, but I just cleaned my mailbox
> and then found something relevant. Regarding the Piper diagram. I just
> noticed the 'hydrogeo' packa
I had to to get the data as a matrix and fix a couple of other things- If
you would like for me to post my solution then I can. I appoligize my
fingers went faster than my brain.
Stephen
On Wed, Jun 11, 2008 at 1:55 PM, Marc Schwartz <[EMAIL PROTECTED]>
wrote:
> on 06/11/2008 12:44 P
#I am having trouble figuring out this one. I have read the help and I am
at a loss. what am I missing
x <- structure(list(X = structure(c(6L, 5L, 9L, 2L, 10L, 8L, 7L, 3L,
13L, 12L, 11L, 4L, 1L, 1L, 1L), .Label = c("", "April", "August",
"December", "February", "January", "July", "June", "March"
I can not seem to get a ylim argument into plotMeans() is there anyway to do
this.
thanks in advance
Stephen
--
Let's not spend our time and resources thinking about things that are so
little or so large that all they really do for us is puff us up and make us
feel like gods. We are mammals, and
I have had good sucess with the par(mfrow=c(#,#)) for formating graphs and
they look good to me. I have seen a lot of use of the lattice package and
thought I would go fishing on the list for y'all's comments. Is there a
time when lattice would be easier more appropriate for certain graphics over
does lty not work? please read the posting guide.
On Tue, Jun 10, 2008 at 12:13 PM, Tariq Perwez <[EMAIL PROTECTED]>
wrote:
> Hi,
> I am trying to plot multiple lines on one plot such that all lines are of
> the same color (black) and continuous. I need to distinguish these multiple
> plots from
#try this cex.main=1.1 This is a par parameter ?par
plot(x,y,main="p=0.05:A-B=3,C-D=10,D-E=100,A-F=2,AFR-E=3,ACE-D=1,ADEF-M=0,AED-E=10,DE-F=3,AB-J=4,AC-J=10,ED-F=1,ED-B=4,AF-B=10,CD-S=10,AM-C=4",
cex.main=1.1)
On Tue, Jun 10, 2008 at 12:07 PM, Hua Li <[EMAIL PROTECTED]> wrote:
> Hi All,
>
> I
I from a first thought I would say that you are apply this wrong! The
fourier transform convolves a function (cos(x)+isin(x) (this may not be the
exact formula but I don't have my books near)) to the data and then
integrates over -1/2 to 1/2 takes the modulus and plots this- the
periodogram. The
I would like to add another axis on side 4 (see code below)
#order the box plot any damn way I want too
order1 <- factor(as.character(x$Site),
levels=c("Betty's Branch", "Stevens Creek",
"North Augusta", "520", "Horse Creek", "Stan's", "place","Downstream",
"IP", "Vo
I am aware of the inherent risks of having plots with more than two axes,
but I am trying to produce the graphs that I have been tasked with. That
being said I am having a hard time figuring out how to have two axes onto a
boxplot. below is the sample code.I would like BC on the plot produce
Good point. Thanks
On Fri, Jun 6, 2008 at 9:05 AM, Daniel Folkinshteyn <[EMAIL PROTECTED]>
wrote:
> should work - don't even have to put them in quotes, if your field
> separator is not space. why don't you just try it and see what comes out? :)
>
> on 06/06/2008
if I wanted to use a name for a column with two words say Dick Cheney and
George Bush
can I put these in quotes "Dick Cheney" and "George Bush" to get them to
read into R using both read.table and read.zoo to recognize this.
thanks
Stephen
--
Let's not spend our time and resources thinking about
I have looked at math plot and still can not figure out how to get a degree
symbol into the y label. I would like degrees C
thanks
Stephen
--
Let's not spend our time and resources thinking about things that are so
little or so large that all they really do for us is puff us up and make us
feel
If you want power spectral density then spec.pgram() and a wrapper to this
spectrum should do the trick (MASS which is a part of R base I think)
add the argument log="no" if you don't want to see the power on a log scale
also remember that the smoothed periodogram is a consitent estimator of the
sp
Rnabble, search on the CRAN site, and some others
On Thu, May 29, 2008 at 12:03 PM, Axel Etzold <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> I am fairly new to R and this list (this is my first post), so I am
> wondering whether
> there is a possibility to view posts on this list conveniently on a
Is there a way to write and analysis to disk and then reconstruct the whole
thing back into an object.
wavCWT() #wmtsa package
I am running out of memory on my computer and I was wondering if there was a
way to iterate through this process (as it is an iterative process anyway-
it just stores the
is there a way to plot the modulus graph with a different color scheme. The
contrast is very bad (for my eyes).
--
Let's not spend our time and resources thinking about things that are so
little or so large that all they really do for us is puff us up and make us
feel like gods. We are mammals,
I can send data along, but it is too large for the list:
I am using wmtsa and the function wavCWT
I would like to use the range.scale to only evaluate a subset of
frequencies.
deltat of the time series that I am trying to analyze is 1
d = wavCWT(RM202.ts, range.scale = c(96, 2688))
and I get the
:48 PM, Julian Burgos <[EMAIL PROTECTED]>
wrote:
> Hi Stephen,
>
> Your link doesn't work. In any case, check out the wavCWT function in the
> wmtsa package.
>
> Julian
>
>
> stephen sefick wrote:
>
>> http://ion.researchsystems.com/cgi-bin/ion-p
>
http://ion.researchsystems.com/cgi-bin/ion-p
I would like a continuous wavelet transform. I have downloaded wavethresh,
Rwave, and waveslim. I would like an output very similar to the above
website. any suggestions?
--
Let's not spend our time and resources thinking about things that are so
li
#try this
plot(survival)
lines(suvival)
On Fri, May 9, 2008 at 2:14 PM, Zhandong Liu <[EMAIL PROTECTED]>
wrote:
> Dear helpers,
> I am trying to plot two survival curves in the same figure.
>
> plot(survival)
> // in matlab, one just need to call "hold on"
> plot(survival2)
>
>
> I am wondering h
t breaking
it up into individual data frames
xx <- prepREIS(GG)
plot(x.ts) is what I want to go into PICK.GEN
Any help would be greatly appreciated- I am sure I am missing something.
thank you very much
Stephen Sefick
--
Let's not spend our time and resources thinking about things t
ephen
On Tue, May 6, 2008 at 12:53 PM, Yasir Kaheil <[EMAIL PROTECTED]> wrote:
>
> cast the two vectors as.matrix-- see here:
> plot(as.matrix(f[2,4:26]), as.matrix(f[1,4:26]))
>
> y
>
>
> stephen sefick wrote:
> >
> > f <- (structure(list(X = struc
f <- (structure(list(X = structure(96:97, .Label = c("119DAmm", "119DN",
"119DNN", "119DO", "119DOC", "119Flow", "119Nit", "119ON", "119OPhos",
"119OrgP", "119Phos", "119TKN", "119TOC", "148DAmm", "148DN",
"148DNN", "148DO", "148DOC", "148Flow", "148Nit", "148ON", "148OPhos",
"148OrgP", "148Phos",
look at the spectrums before you do the cbind - I would not suggest letting
R wrap the data to fill in a data frame. I would suggest using something
that you "know how it acts" in the frequency domain like zero. You are
probably introducing periodicies that are not real, and I would suggest not
t
engths ... ???
>
> Thank you so much.
>
> Maura
>
> On Wed, Apr 30, 2008 at 8:56 AM, stephen sefick <[EMAIL PROTECTED]> wrote:
>
> > $names
> > [1] "freq" "spec" "coh" "phase" "kernel"&q
black and white graph with points at rivermiles- publication quality.
> thanks for all of the help!
>
>
> On Thu, May 1, 2008 at 12:09 PM, Ruben Roa Ureta <[EMAIL PROTECTED]> wrote:
>
> > > Does anyone know of a package to make a map from GIS data, and/or
> > would it
> > > be easier in one of the
Does anyone know of a package to make a map from GIS data, and/or would it
be easier in one of the free GIS programs. I would like to make a map of
the savannah river area with our sampling locations.
thanks
stephen
--
Let's not spend our time and resources thinking about things that are so
lit
$names
[1] "freq" "spec" "coh" "phase" "kernel""df"
[7] "bandwidth" "n.used""orig.n""series""snames""method"
[13] "taper" "pad" "detrend" "demean"
$freq and $spec are used to plot the power spectrum. freq is the x-axis and
spec is the y-axis.
brevity is refreshing
On Tue, Apr 29, 2008 at 3:40 PM, Mittal Meghna <[EMAIL PROTECTED]> wrote:
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posti
I am not entirely sure what it is that you want to do.
functionname<-function(a,b,c){
+ a<-sum(a)
+ b<-sum(b)
+ c<-sum(c)
+ y<-(a+bx+cx^2)
+ x<-(x)
+ }
for each <- your are defining and object. it doesn't make much sense to
sum(a) unless you have made a something etc. Look at packages- type CRA
merge can only merge two objects at a time- I would like to merge more than
two objects at a time.
s.d <- structure(list(RiverMile = c(202L, 198L, 190L, 185L, 179L, 148L,
119L, 61L)), .Names = "RiverMile", row.names = c(NA, -8L), class =
"data.frame")
#s.d is all of the river miles that can occur
al_, NA_real_, NA_real_)), .Names = c("RiverMile",
"X5.1.06"), row.names = c(29L, 30L, 31L, 32L, 34L), class = "data.frame")
merge(s.d ,c(feb06, may06), all=TRUE, by.x="RiverMile")
#I know they have different row numbers, but I can not figure out how to
fil
This is an example of two months of data from a twenty four month data set
that I would like to apply this too. These data are subsets of the same
stations throught time, but differing ones were included on different
sampling dates. I would like to subset these data and then put them
together as
I have a large data.frame rows=sites, column=species. Being a species
matrix it is very sparse and I have found that there are no site rows that
are = 0 and no species = 0 and I am getting and error that reads
x = read.table("shit.txt", sep="\t",header=T, row.names=1)
x.m <- as.matrix(x)
x.t <- t
d = c(0L, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 0L, 0L, 7375L,
NA, NA, 17092L, 0L, 0L, 32390L, 2326L, 22672L, 13550L, 18285L)
boot.out <-boot(d, mean, R=1000, sim="permutation")
Error in mean.default(data, original, ...) :
'trim' must be numeric of length one
I know that I am missing somethi
Like all learning- you have to learn how to think like the people that
create: language, poem, song, etc. In my own learning curve I have found
that I am too often popping emails off to the list when in an hour or so I
can figure out what was troubling me (using the provided examples). This
may b
boxplot(x[,c(2,15,28,41,54,67,80,93,106)], ylab="mg/s", names=c("RM215",
"RM202", "RM198", "RM190", "RM185", "RM179", "RM148", "RM119", "RM61"))
this is the code I am using to make a standard box plot. Is there a way to
get the number of NA observations plotted onto the graph easily. I can
alwa
do
this? sorry for my lack of understanding. All help is greatly appreciated.
Stephen
On Tue, Apr 15, 2008 at 7:08 PM, Rolf Turner <[EMAIL PROTECTED]>
wrote:
>
> RML: *** Data frames are NOT matrices!!! ***
>
> On 16/04/2008, at 10:35 AM, stephen sefick wrote:
>
I have a species as rows and sites as columns matrix. I would like to
tranpose this matrix so that I can do an NMDS analysis with the vegan
package. I have used
y <- t(x)
and it indeed transposes the matrix, but it changes the whole make up
chr [1:193, 1:288] "oligocha" "0" " 0" " 0" " 0" "0"
"upwd1201" "upwd0502" "upwd0702" ...
..$ : NULL
x is a data frame with a whole bunch of numeric vectors I would like for the
rows and columns to be reversed == transposed but with the same attributes.
I am trying to feed the transposed data frame to metaMDS(vegan).
the below is a three dimensional solution
On Mon, Apr 14, 2008 at 3:09 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> cor(x, dist(mds))^2
> Error in dist(mds) : (list) object cannot be coerced to 'double'
>
>
>
> On Mon, Apr 14, 2008 at 2:40 PM, Sarah Goslee &
I am using the function metaMDS with jaccard distances to ordinate a
set of constituent by site matrix. I can post this data if it would
be helpful, but it is large to include in an email. I can also
provide reproducable code if necessary. I would like to get an R^2
value for the axes of the ord
col="chocolate")
points(site.sc[c(287:294),c(1:2)], col="cornflowerblue")
points(site.sc[295,c(1:2)], col="coral")
points(site.sc[c(296:299),c(1:2)], col="darkgoldenrod")
points(site.sc[c(300:301),c(1:2)], col="darkgreen")
points(site.sc[302,c(1:2)],
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