#Hi all,
#Using the example data that follows, can someone please show me how to get
a scatterplot of points with #error bars in the Y direction. something like
this works for one Y:
xYplot(Cbind(y1, l1, u1) ~x1, data=y)
#but this:
xYplot(Cbind(y1, l1, u1) + Cbind(y2, l2, u2)~x1, data=y)
#
?paste
something like...
paste ("group", i, sep="_")
jiliguala wrote:
>
> Hello R users,
>
> I am new to R and am having difficulty with the output name of my "for"
> loops.
>
> here is the problem:
>
>
> for (i in c(1:100))
> {
> the name of the groups <- which(k1$cluster==i)
> }
>
> ho
?paste
something like...
paste (group, i, sep="_")
jiliguala wrote:
>
> Hello R users,
>
> I am new to R and am having difficulty with the output name of my "for"
> loops.
>
> here is the problem:
>
>
> for (i in c(1:100))
> {
> the name of the groups <- which(k1$cluster==i)
> }
>
> how
Hello,
How can I randomly sample individuals within a sites from a site (row) X
species abundance (column) data frame or matrix? As an example, the matrix
"abund2" made below.
# (sorry, Im a newbie and this is the only way I know to get an example
on here)
abund1 <-c(150, 300, 0, 36
0 022 32 0
If you know, I'd appreciate it.. thanks again for the help.
David Winsemius wrote:
>
>
> On Feb 6, 2011, at 3:25 PM, B77S wrote:
>
>>
>> Hello,
>> How can I randomly sample individuals within a sites from a site
>> (row) X
hehe...
very true sir; I apologize, that was very straightforward. Thank you for
your time.
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Also, I really appreciate you explaining why you used factor. I'm still not
quite sure what set.seed does (i read ?set.seed) or why you chose 123... but
it and the function below work, so that is all that matters. :)
randSub <- function(L1, s.size)
{
set.seed(123)
samptbl <- apply(L1, 1, fun
So, after thinking about this a bit, I realized that the previous solution
wasn't exactly what I needed. I really needed replacement=F and to be able
to choose any sample size (n.sample) less than or equal to the site (row)
with the lowest total abundance.
Anyway, I think this works. Forgiv
It is a lot prettier than mine too.
Thanks Jari.
> rrarefy
function (x, sample)
{
if (length(sample) > 1 && length(sample) != nrow(x))
stop("length of 'sample' and number of rows of 'x' do not match")
sample <- rep(sample, length = nrow(x))
colnames(x) <- colnames(x, do.NULL
## i didn't try this, but I would think it would work
newAB <-data.frame(AB$id, AB$age, AB$sex, AB$area)
colnames(newAB)<-c("id","age", "sex, "area")
uni.newAB <- unique(newAB)
t3<-merge(t2, uni.newAB, by="id", all=FALSE)
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All:
I have been looking through the string of posts regarding this same issue,
but I haven't been able to fix this problem.
I am running Ubuntu 10.4 64bit, R version 2.10.1
I cannot install certain packages (e.g. "vegetarian") and each time it says
basically the same thing (regardless of the pa
also, I have 'r-base-dev' installed as well
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R-he
assuming this is from a list
mydata[[1]][1]
and
mydata[[1]][7]
??
Wonjae Lee wrote:
>
> Hi,
>
> I have a stupid and simple question.
> Please forgive me.
>
> In an example below, please tell me how to call "1947" in mydata.
> Thank you in advance.
>
> Wonjae
>
>> mydata
> [[
ifelse(length(unique(x))==1, "All Equal", "Not All Equal")
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FYI:
The problem I had was that my modified .Rprofile text file was being read
and had lines to load the libraries of packages that were not installed yet
on this Ubuntu machine; I guess I copied the .Rprofile from my laptop.
If you have similar problems, simply do this in command line:
mv .Rpro
That is odd, I noticed some weird sorting with merge() a while back too and
always am careful with it now. Fortunately, sort=FALSE seems to work the
way one would think most of the time.
Although, the following results seem weird too! (adding by="date" makes it
not sort oddly, regardless of so
step # 1
?read.csv
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https
This probably is not ideal, but this works on a list of mine..
## so you can see the structure of my list
> str(srMT)
List of 4
$ mode : chr "discrete"
$ ks.stat : chr "mean"
$ observed :List of 4
..$ filter: num [1:13, 1:4] 0.213 0.207 0.144 0.311 0.24 ...
.. ..- attr(*, "dimnames
I'm thinking this isn't what you want.. but also:
> data.frame((srMT[[3]][1]))[b,][2]
filter.vel2
MK_SP10.2503257
SB1_SP1 0.2075117
SB4_SP1 0.2358855
B77S wrote:
>
> This probably is not ideal, but this works on a list of mine..
>
> ## so you ca
#replace v with whatever
rm(list=(ls()[ls()!="v"]))
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If you don't want your history, why not just do this.
>q()
Save workspace image? [y/n/c]:
>n
??
Am I missing something?
1Rnwb wrote:
>
> Thanks, I thought that removing the list would take care of it. the
> question is I do not see a .Rhistory file in my current working directory,
> so where
If I understand you correctly, see ?paste
and the following to extract the values you require:
summary(res)[[4]][1]
summary(res)[[4]][2]
summary(res)[[8]]
HTH
ashz wrote:
>
> Dear all,
>
> How can I covert lm data to text in the form of "y=ax+b, r2" and how do I
> calculate R-squared(r
I agree with Ken.. if you can, save it as a CSV file. But if you have a
bunch of these, then it isn't very efficient. I use read.xlsx() from the
package "xlsx".
I notice that you are using the full path.. have you tried changing
directories?... I find it is best to compartmentalize my work and
package "caTools"
see ?trapz
. wrote:
>
> Hi all,
>
> is there any alternative to the function integrate?
>
> Any comments are welcome.
>
> Thanks in advance.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
see ?mean
Then avoid other peoples code.
cenae27 wrote:
>
> bob<-read.csv('shi.csv', header=T)
>
> newmean<-matrix(0, test, dim(bob)[2]-6);a<-0; for (i in
> c(4,8:(dim(bob)[2])))
> {a<-a+1;newmean[,a]<-tapply(bob[,i], bob$Exam, mean)}
> colnames(newmean)<-colnames(bob)[c(4,8:(dim(bob)[2]))]
>
I have the following data (see RawData using dput below)
How do I get it in the following 3 column format (CO2 measurements are the
elements of the original data frame). I'm sure the package reshape is where
I should look, but I haven't figured out how.
Thanks ahead of time
Month Year CO2
J
NA
> 48 1958 F NA
> 95 1958 M 315.71
> 142 1958 A 317.45
> 189 1958 M.1 317.50
> 236 1958 J.1 NA
>
>
> On Wed, Sep 7, 2011 at 7:35 AM, B77S <bps0...@auburn.edu> wrote:
>
>> I have the following data (see RawData using dput below)
My guess is that anyone willing to help will want more information.
What version of R do you have(?), for example.
BiodiversityR Depends on R version ≥ 2.13.1, and vegan ≥ 1.17-12
SG wrote:
>
> I have installed R on windows 7 machine.
> In the install packages I can't find BiodiversityR pac
You'll never figure it out if you don't "play around" with your data.
Assuming you have been able to import the data, a good place to start is to
look at what tools you have available.Check out this:
http://cran.r-project.org/doc/contrib/Short-refcard.pdf
check out things like
?which
?max
?
If your data is named 'test_file'
then use dput(test_file)
You can copy and paste the results here so people can more easily try and
help you.
see
?dput
sujitha wrote:
>
> Hi group,
> I am trying to right a code to do the following
> This is how the test file looks like:
> Chr start end sa
], rle(test[,5])[[1]]))
names(test2) <- c("Values", "Probes")
> test2
Values Probes
1 0.00 4
2 -0.43 4
3 0.00 2
4 0.00 2
5 -0.35 6
6 0.00 2
# Obviously this is not exactly what you wanted, only the last 2 columns.
Obviously, th
I'm sure Sarah's solution works (and she knows more about R than myself), but
I ran into a similar problem and used:
as.character(start.time)==as.character(expected_start.time)
good luck regardless.
-BS
Sarah Goslee wrote:
>
> Sounds like a case for FAQ 7.31, or, yet another machine precision
** I am not a statistician, but...
The default arguments are:
onet.permutation(x, nsim=2000, plotit=TRUE)
you are making nsim=4 (this parameter is not the size of each sample, it
is the number of samples taken)
You are building the distribution for a statistical test through random
sampling, s
Have you been shown how to save a graph as a JPEG or PNG?
try this:
png("myGraph.png")
plot(your_data)
dev.off()
A png will appear in your working directory, which can be imported into the
Word document.
You can do the same with a JPEG
see ?jpeg or ?png and ?dev.off
HTH
bonnieyuan wrote:
>
This isn't a question about R; more appropriate for stackexchange.
Here is one string that might interest you:
http://stats.stackexchange.com/questions/4997/can-aic-compare-across-different-types-of-model
Tania Sav wrote:
>
> Hello,
>
> I'm using AIC() to choose a better model. I have 3 o
I don't have access to your "alon.txt" file (see ?dput for future posts),
but...
I'm pretty sure info you want isn't in row.names(colon[1:2])
it should just be
text(x,y, label = colon[1:20])
??
HTH
baumeist wrote:
>
> Hi,
> I am new to R.
>
> I have a matrix that I have assigned to the o
Actually, you appear to have re-assigned your object “colon” (from
"c:\\alon.txt") with a character vector of intended row.names.
so use
row.names(colon) <-paste("g",c(1:nrow(colon)),sep="")
B77S wrote:
>
> I don't have access to your "a
#This should work (again, without your data ??)
colon<-read.table("c:\\alon.txt",header=T,row.names=1)
row.names(colon) <-paste("g",c(1:nrow(colon)),sep="")
with(colon[1:20,], plot(norm1, norm2, type='n',xlab='x norm1 sample',ylab='y
norm2 sample',main='Norm1 vs Norm2 - 20 genes'))
with(co
1st of all Dr Utz, thanks for your recent pubs on regional differences
between the piedmont and coastal plain streams.
and to add to Daniels post (giving your binary yes/no):
df<-merge(x,y,all.x=T,all.y=F)
df["exceed"] <- ifelse(df$Q>df$Threshold_Q , 1, 0)
## now look at df
df
Daniel Mal
Hi,
How can I discern which elements in x (see below) are in 'order', but more
specifically.. only the 1st 'ordered run'?
I would like for it to return elements 1:8... there may be ordered values
after 1:8, but those are not of interest.
x <- c(1, 2, 3, 4, 5, 6, 7, 8, 20, 21, 22, 45)
Thanks fo
well.. the following works, but if you have another idea I am still
interested.
1:(which(diff(x)!=1)[1])
B77S wrote:
>
> Hi,
>
> How can I discern which elements in x (see below) are in 'order', but more
> specifically.. only the 1st 'ordered run
:32 PM, B77S <bps0...@auburn.edu> wrote:
>> Hi,
>>
>> How can I discern which elements in x (see below) are in 'order', but
>> more
>> specifically.. only the 1st 'ordered run'?
>> I would like for it to return elements 1:8... there may be
Have you looked at the manual for any of these?
?get
?sprintf
?assign
#and so on...
?
Bazman76 wrote:
>
> m0<-epxression((4*theta1*theta2-theta3^2)/(2*x*theta3^2)-0.5*theta1*x)
>
> params<-all.vars(m0) this reads all the
> params from m0 so theta1,2 and 3 corre
## Hello.. I have asked a similar question, but this is not fixed as before.
## I am running the following using Ubuntu OS:
R version 2.13.1 (2011-07-08)
Copyright (C) 2011 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-pc-linux-gnu (64-bit)
## when I do this:
day doesn't exist?
That would be the 1st problem.
Johannes Egner wrote:
>
> Dear all,
>
> how come the first loop in the below fails, but the second performs as
> expected?
>
> days <- as.Date( c("2000-01-01", "2000-01-02") )
>
> for(day in days)
> {
> as.POSIXct(day)
> }
>
> for( n in
If this is the same as geometric mean regression (aka: line of organic
correlation-- Kruskal 1953), I ended up writing my own function (although
one may exist and I didn't see it).
Bill Hyman wrote:
>
> Dear all,
>
> Does any one know if any R package or function can do Ordinary Least
> Pro
Steffen,
Did you ever have luck getting rid of the tick marks?...
I like your idea and have modified it, but yes, the tick marks need to go.
Steffen Fleischer wrote:
>
> Dear all,
>
> I want to draw a graph that contains the scatterplot matrix in the lower
> panel and coefficients in the u
In addition to getting help from others, I always find that seeking answers
through reading to be helpful. I would suggest a basic stats book and this
paper regarding issues indices like species richness:
http://www.amazon.com/Primer-Ecological-Statistics-Nicholas-Gotelli/dp/0878932690
Here is
Hello,
I would like to find out if a function already exists that returns only
pairwise correlations above/below a certain threshold (e.g, -.90, .90)
Thank you.
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hi,
If i have a list of things, like this
var.names <- c("a", "b", "c", "d", "e", "f")
how can i get this:
"a, b", "c, d", "e, f"
thanks ahead of time.
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Sent from
ke this:
>
> BigCorrelation <- function(X){
>
> return(which(abs(cor(X)) > 0.9, arr.ind = T))
> }
>
> but it hardly seems worth its own function call.
>
> On Thu, Nov 17, 2011 at 12:42 AM, B77S <bps0002@> wrote:
>> Hello,
>>
>> I woul
p 0.8324475
4 cyl drat -0.6999381
7 disp hp 0.7909486
8 disp drat -0.7102139
I'm not sure how to make melt() quit giving the "Using as id variables"
warning, but I don't really care either.
B77S wrote:
>
> Thanks Michael,
>
> I just started on the fo
i <- which(abs(x) >= r, arr.ind = TRUE)
> data.frame(matrix(colnames(x)[as.vector(i)], ncol = 2), value = x[i])
> }
>
> spec.cor(mtcars[, 2:5], .6)
>
> Cheers,
>
> Josh
>
> On Wed, Nov 16, 2011 at 9:58 PM, B77S <bps0002@> wrote:
>> Thanks M
This is ugly, but it gets what you want.
dat[which(dat[,1] %in% unique((dat[duplicated(dat[,1], fromLast = T),
1]))),]
AC Del Re wrote
>
> Hi,
>
> Is there an easy way to remove dataframe rows without duplicated values of
> a specified column ('id')? e.g.,
>
> dat <- data.frame(id = c(1
out <- vector("list")
Ylab <- for(i in 1:length(BndY))
{
out[i] <- paste(BndY[i]," to ",BndY[i],"mN")
}
Ylab <- do.call(c, out)
markm0705 wrote
>
> Dear R helpers
>
> I'm trying to make up some labels for plot from this vector
>
> BndY<-seq(from = 18900,to= 19700, by = 50)
>
> using
>
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R-he
Try the daisy() function from the package "cluster", it seems to be able to
handle NAs and non-dummy coded character variables
metaMDS(daisy(df, metric="gower"))
Edwin Lebrija Trejos wrote
>
> Hi, First I should note I am relatively new to R so I would appreciate
> answers that take this into
Someone is bound to know a better way, but...
subset(unlist(Version1_), subset=names(unlist(Version1_))=="First")
LCOG1 wrote
>
> Hi everyone,
>I looked around the list for a while but couldn't find a solution to my
> problem. I am storing some results to a simulation in a list and fo
# need zoo to use rollapply()
# your data (I called df)
df <- structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L),
e = c(1L, 5L), f = c(4, 7)), .Names = c("a", "b", "c", "d",
"e", "f"), class = "data.frame", row.names = c(NA, -2L))
# transpose and make a zoo object
df2 <- zoo(t(df
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https://stat.ethz.ch/ma
Sorry for that, and thanks Gabor,
I could have sworn that it wouldn't.
Gabor Grothendieck wrote
>
> On Thu, Dec 1, 2011 at 7:13 PM, B77S <bps0002@> wrote:
>> # need zoo to use rollapply()
>>
>> # your data (I called df)
>> df <- structure(list(a
A simple way to determine if it is NOT is to see if the mean (the single
parameter of a poisson: lambda) and variance are the same.
This really has nothing to do with R (other than the data source), and since
it is homework, you will likely get no further help here.
Good luck.
RToss wrote
>
>
Hi Michael,
How would you do this with lapply to return a list?
I can't seem to get that to work (I haven't used these much and am trying to
learn).
Thanks
Brad
Michael Weylandt wrote
>
> ? replicate
>
> or a for loop
>
> or do all one hundred simulations at once
>
> x <- matrix(rnorm(100^2,
Interesting and thank you; I'm confused as to why this doesn't work with:
lapply(rep(1,6), FUN=rnorm, n=10, mean=1.0, sd=1)
andrija djurovic wrote
>
> Hi Brad. Maybe something like this:
>
> lapply(rep(1,6), function(x) rnorm(10,0,1))
>
> Andrija
>
> On
y on a plain matrix faster than on a
> zoo object? I am asking since I wanted to apply the provided averaging
> code to a larger matrix (2500 rows x 200 cols), which is quite time
> consuming...
>
> Thanks again!
>
> Chega
>
>
> From: B77S
use a < ? > to get help on a function; example:
?read.table
If you do this you will see an option called "header"...
use header=T if your top row contains column names.
Learn how to read these help pages. Also, read thru a few beginner R
manuals and see this website:
http://www.statmethods.
As Stephen pointed out, this is easy to do. The word file the OP posted has
all the necessary formulae. Now you just need to learn how to convert those
formulae into R functions
Stephen gave you an example of how to create a function for CV. Now run
with it.
perhaps something like this is w
Please use dput() to post your example data sets.
dput(Adata)
dput(Bdata)
**then copy and paste the results of each so that we can "play" around with
it easily.
Miriam -2 wrote
>
> I have been trying to merge datasets, one of which has a long format
> (Adata) and one has a (different) long f
just use indexing.
without doing it all for you...
df <- structure(list(AA = c(0.3, 0.1, 0.6), BB = c(0.9, 0.4, 0.2),
CC = c(1, 0.8, 0.6), DD = c(0.7, 0.5, 0.5)), .Names = c("AA",
"BB", "CC", "DD"), class = "data.frame", row.names = c(NA, -3L
))
write.csv(df[,1], paste(colnames(df[1]), "csv
I apologize to the list and you if I am mis-understanding something, but...
As an example:
ind2 occurs with ind1 only in cluster#3, so why does it get a value of 4 in
your "similarity" matrix?
Also, if this isn't a recognized similarity algorithm, perhaps you should at
the very least put quotes a
That really all depends on what you need; and I can't tell you what you need.
set wrote
>
> I'm sorry, I made a mistake in my example. you're right. I don't really
> know how a similarity alogrithm worksbut I'm willing to try that...are
> there any good examples available?
> Thank you
>
Why can't you figure this out?
I think you already know the answer: "I don't speak computer".
Time to learn, or get another job I suppose. I hope you at least speak
math-statistics if you are going to attempt to understand this.
The only way you are going to be able to figure that code ou
Set,
This is the same post as your "Similarity Matrix" post.
I'm not trying to be a smart ass here, but ... ?Can you fit a square peg in
a round hole?... yes, but it doesn't mean it belongs there.
I suggest you get a piece of paper and a pencil and figure out 1) what you
are trying to do and w
If all else fails, open it in Excel... save as .csv
read.csv()
Nikhil Joshi wrote
>
> I am trying to read an xlsx spreadsheet (1506 rows, 501columns) all
> populated but getting the following error:
> Please advise as to how to get around this issue.
>
>> res <- read.xlsx("c:\\BSE_v2.xlsx",1
Try posting this question at least 1 more time.
plocq wrote
>
> Hi,
>
> I try to use the function profile() of the SpatialExtremes' package to
> obtain the profile likelihood of parameters for an extreme values fit
> based on Poisson process :
>
> fit<-fpot(data, threshold, model="pp", npp=
343GS
You should consider dropping out of college; I don't think you belong there.
McDonalds is hiring.
343GS wrote
>
> Thanks for nothing, jerk! DENMARK IS POO
>
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Sent fro
I don't know why you had "c in a(1:2000))"
c is a function see ?c ...
and you want "a" (the row number in SQL_Code) to change with each iteration
in the loop.
Perhaps this might work (I'm not saying this is the best option, just a
potential fix for what you have):
for (a in 1:2000) {
Dataset
Where is dropvalue(s) mentioned?
> ?subset
subset: logical expression indicating elements or rows to keep:
missing values are taken as false.
select: expression, indicating columns to select from a data frame.
drop: passed on to ‘[’ indexing operator.
...: further arg
It is not simply the answer, it is a list
str(integrate(dnorm, -1.96, 1.96))
List of 5
$ value : num 0.95
$ abs.error : num 1.05e-11
$ subdivisions: int 1
$ message : chr "OK"
$ call: language integrate(f = dnorm, lower = -1.96, upper = 1.96)
- attr(*, "class")= chr "i
I have never used that function, but I know that with read.csv() you can do
the following to select only the columns you want:
chosen_vars <- read.csv("Workbook1.csv", header=T)[c("variable1",
"variable3")]
HTH
sassorauk wrote:
>
> Is it possible to import only certain variables from a SPSS fi
I see what you mean.
Sorry and thanks for pointing that out to me Ben.
bbolker wrote:
>
> B77S auburn.edu> writes:
>
>>
>> I have never used that function, but I know that with read.csv() you can
>> do
>> the following to select only the columns you
The first suggestion would be to use dput() to allow people on here to access
your data.
see ?dput
I think you will want to post the output from:
dput(QInflAvgbyPlot) ## or whatever object contains your data
kelseyann wrote:
>
> Hello, I am using the following script to run an anova fo
Your question as answered by Timothy in your previous thread
http://r.789695.n4.nabble.com/Re-Creating-the-mean-using-algebra-matrix-td3895689.html
flokke wrote:
>
> Dear all,
> Sorry to bother you with such a stupid question, but I just cannot find
> the solution to my problem.
>
> I'd lik
This is just scientific notation, so
8.15e-01 is the same as:
> 8.15*10^-1
[1] 0.815
niki wrote:
>
> Dear all,
>
> i have done some regression analyses but i do not understand the p value.
> These are the results
>
>t-value p value
> geno.1
I know in my experience "Cond" (conductivity??) doesn't vary much within a
stream except for during high flow events, and I would imagine the same is
true for TDS. If these are all low flow values, you could possibly
determine a mean/median value to use for the missing data points. Obviously
this
here is one way
df1 <- data.frame(c(1:20), c(21:40), c(31:50))
list1 <- c(3, 6, 20)
df2 <- df1[-list1,]
hanansela wrote:
>
> Hello
> I have a list of row names that needs to be deleted from a data frame. How
> do i do that?
> one of the columns in the data frame contains the row names as nu
Why not format the data like this:
site sampledate SO4 TDS NA Mg Cond Cl Ca
i.e. with a column for each parameter?
It seems to me that you summary doesn't make any sense. Those quantiles are
meaningless as they encompass all the parameters. Am I missing something?
Rich Shepard wrote:
>
>
sampdate, param as id variables
site sampdate Cond SO4 TDS
1 BC-0.5 1996-06-02 NA 194 530
2BC-2 1994-07-25 826 NA NA
3BC-2 1994-10-19 848 NA NA
4BC-2 1995-03-16 1795 NA NA
5BC-2 1995-06-21 640 NA NA
6BC-3 1994-01-20 406 NA NA
7BC-3 1994-02-17 401 NA N
the following is a more appropriate forum for your question, seeing as this
has nothing to do with R (per se).
http://stats.stackexchange.com/questions
good luck.
peak99 wrote:
>
> Hello,
>
> I have some patient data for my masters thesis with three groups (n=16, 19
> & 20)
>
> I have com
If no one has a better solution, split it, take a sample of size X from both
and put it back together.
hgwelec wrote:
>
> Dear members,
>
> Consider the following data frame (first 4 rows shown)
>
>
> age sex class
> 15 m low
> 20 f high
> 15 f low
> 10 m low
>
> in
# Perhaps I misunderstand your original need, but
## I added a few lines to your data and used dput() to get the below data (I
named "df")
df<- structure(list(age = c(15L, 20L, 15L, 10L, 10L, 12L, 17L, 17L,
11L, 12L, 16L, 20L, 23L, 14L, 22L, 16L, 10L, 11L, 21L, 10L, 13L,
17L), sex = struct
I would start by reading one or more of the introduction manuals available
here:
http://mirrors.ibiblio.org/pub/mirrors/CRAN/
wizi wrote:
>
> Hi everyone,
>
> I am new to R-project. I did search through the list for my problem but i
> can't find it. I am sorry if this question has been
Please see ?dput
use dput(your data) and paste the output into a reply, thanks.
This way we know what you are working with.
Rich Shepard wrote:
>
> I would appreciate pointers on what I should read to understand this
> output:
>
> summary(lm(TDS ~ Cond + Ca + Cl + Mg + Na + SO4))
>
> Ca
There is only one row with a complete set of observations; I think lm() is
throwing out the rest.
Rich Shepard wrote:
>
> On Wed, 9 Nov 2011, John C Frain wrote:
>
>> As far as I know if there is an NA in any variable in an observation the
>> default is to drop the entire observation. Thus the
,
156L, 140L, 128L, 160L, 215L, 230L, 316L, 163L)), .Names = c("Ca",
"Cl", "Cond", "Mg", "Na", "SO4", "TDS"), class = "data.frame", row.names =
c(NA,
-64L))
B77S wrote:
>
> Please see ?dput
>
> use d
Not sure if this helps, but did you try Google?
http://www.jstatsoft.org/v35/i09/paper
http://cran.r-project.org/web/packages/lcmm/lcmm.pdf
http://www.warwick.ac.uk/statsdept/useR-2011/abstracts/010411-liquetbenoit.pdf
yurirouge wrote:
>
> HI,
>
> I am relatively new to R and would appreci
I know this isn't what you are asking, but have you considered examining the
relationship between dA and the community density excluding dA?
JulieV wrote
>
> Hi Josh,
>
> Thanks for your response !
>
> Actually, I already tried to plot it with a "classical" regression and I
> know the rel
> 3x4
Error: unexpected symbol in "3x4"
R has no idea that you equate "x" as multiplication.. use an astrix
> 3*4
[1] 12
dominic wrote
>
> This is basically my code:
>
> library(MASS)
> lmsreg(formula = b0 ~ b1 + b3 + b1xb2, data=mydata)
>
> b1xb2 is an interaction but it was the centere
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