1:3]
A<-matrix(c(
0.21, 0.21,0.03,
0.55, 0.58, 0.09,
1.30, 1.35, 0.22), nrow=3, byrow=TRUE, dimnames=list(stages,stages) )
## I can get a surface plot, but that's about it.
persp3d(A, col="red", alpha=0.7,
xlab="fate", ylab="stage", zlab="Sen
.
Parent.ProjectID ProjectID
11973313694 Global Ocean Sampling
Expedition Metagenome
220823 13696 5-Way (CG) Acid Mine Drainage
Biofilm Metagenome
3-13699Waseca County
atrix(numeric(1000*3), ncol=3)
for(i in 1:1000)
{
z <- sample(nrow(y), replace = TRUE)
y1<-y[z,]
n2[i,] <-prop.table( table(y1[,2], y1[,1]), 2) %*% n
}
# the CIs are much wider here
apply(n2, 2, quantile, c(0.025,0.975))
[,1] [,2] [,3]
2.5% 55.89959 132.6868 361.
0. 147.8957 0.0 0.0 136.36804
220.7370
[5,]0 0. 0. 0. 96.92715 0.00.0
108.6551
[6,]0 0. 0. 0. 0.0 69.875270.0
0.
[7,]0 0. 0. 0. 0.0 0.0 65.86229
0.
Chris St
Shawn Morrison-2 wrote:
>
> # The paper reports a 95% CI of 0.79 - 1.10
> # "My" reproduced result for the CIs is much larger, especially on the
> upper end. Why would this be?
> # The authors report using the 'delta' method (Caswell, 2001) to
> calculate the CI in which the
>
>
Shawn,
I p
"
## This prints what I'd want, but I'm not sure why its different.
print.diag(d)
[1] "A diagonal matrix d"
Thanks for any help,
Chris Stubben
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Sent fro
q[i]+1) with sample( 1:(n-q[i]+1),
replace=TRUE) ?
Also, if you have a list of matrices, then maybe combine the code above into
a function and then apply it to your list using lapply.
Chris Stubben
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longch)<-"100 ABCs"
str(longch) # not truncated
Thanks,
Chris Stubben
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Sent from the R help mai
Mark W. Miller wrote:
>
> I have a list of scientific names in a data set. I would like to split
> the names into genus, species and subspecies. Not all names include a
> subspecies. Could someone show me how to do this?
>
strsplit should work for your example...
data.frame(
genus=sappl
- function(x) chartr("ACGT", "TGCA", x)
n<-cbind(key, rev=reverse(comp(key)))
n<-t(apply(n,1, sort))
do.call(rbind, by(a, n[,1], colSums))
V1 V2 V3 V4
AA 5 13 21 29
AT 2 6 10 14
TA 3 7 11 15
Chris Stubben
jholtman wrote:
>
> Try this:
>
>&
f (!all(n == dims[1, ]) || !all(p == dims[2, ]))
stop("the matrices must have the same dimensions")
mat <- matrix(unlist(x), n * p, length(x))
mm <- matrix(rowMeans(mat, ...), n, p)
dimnames(mm) <- dimnames(x[[1]])
mm
}
Chris Stubben
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View this messa
Europa']", xmlValue)
[1] "30.4 4 95.1 113 3.77 1.513 16.90 1 152"
#OR?
xpathSApply(doc, "//record[@id='Lotus Europa']")
getNodeSet(doc, "//record[@id='Lotus Europa']")
# And to list all Ids...
xpathSApply(doc, &qu
Is there an easy way to convert character strings with comma-separated
numbers and ranges to a numeric vector?
x<- "2,5-7,10,12-15"
[1] 2 5 6 7 10 12 13 14 15
Thanks,
Chris
--
Chris Stubben
Los Alamos National Lab
Bioscience Division
MS M888
Los Ala
w.ncbi.nlm.nih.gov/pmc/articles/PMC3544749/table/T1";, which=1)
Thanks,
Chris
t1<-readHTMLTable(
"http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3544749/table/T1";, which=1)
--
Chris Stubben
Los Alamos National Lab
Bioscience Division
MS M888
Los Alamos, NM 87545
> I want to merge dat1 and dat2 based on "ID" in order
Have you tried merge(dat1, dat2) ?
If ID is the common column (and no others), then that should be all you
need to join (see ?merge). And then order if needed.
Chris
--
Chris Stubben
Los Alamos National Lab
Bioscienc
mple(L3, 10, replace=TRUE))
Chris
--
Chris Stubben
Los Alamos National Lab
Bioscience Division
MS M888
Los Alamos, NM 87545
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide h
raw
datasets
are pretty important, especially if observations are split into different files and
you are trying to join them later. How do you know for ID 0001 and obs 1 that height
is 3.2 and not 2.6, especially if order in the two files are "not exactly the same".
Chris
--
Chr
Methoxy mycolic acid synthase 4
3 Mycolic acid synthesis pcaA Rv0470cMycolic acid
synthase (cyclopropane synthase)
4 Mycolic acid synthesis pcaA Rv0470cMycolic acid
synthase (cyclopropane synthase)
--
Chris Stubben
Los Alamos National Lab
Bioscience Divis
the 90 degree counter clockwise rotation
image(t(x[nrow(x):1,]), axes=FALSE)
## add 100 column names
y<-paste("column", 1:100)
text( seq(0,1,length=100) , 1.01, y, pos = 2, srt = 270, offset=0, cex=.7)
Chris Stubben
Lara Poplarski wrote:
>
> I have a large (1600*1600) mat
Just increase the margins on the left side and add the rownames
x <- cor(matrix(rnorm(600), 60, 100))
rownames(x)<-paste("row", 1:100)
op<-par(mar=c(1,5,1,1), xpd=TRUE)
image(t(x[nrow(x):1,]), axes=FALSE)
text(-0.01, seq(0,1,length=nrow(x) ), rownames(x), pos = 2, offset = 0,
cex = .7)
Another
Gene names often have single quotes like
5'-methylthioadenosine phosphorylase
ATP synthase B' chain
ppGpp 3'-pyrophosphohydrolase
so maybe try adding quote="" to the read table options.
Chris Stubben
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To keep the table dimensions the same, try changing the columns to factors...
boxes$y97<-factor(boxes$y97, 1:4)
boxes$y98<-factor(boxes$y98, 1:4)
boxes$y99<-factor(boxes$y99, 1:4)
...
> table(boxes$y98, boxes$y97)
1 2 3 4
1 1 0 1 0
2 0 0 0 0
3 0 0 0 0
4 1 0 0 1
> table(boxes$y99,
ct above, but I'd
rather just work with the results from htmlParse if possible (and not use
readLines to load raw HTML first).
Thanks,
Chris Stubben
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Sorry if I was not clear. I wanted to remove the superscripts using xpath
queries if possible. For example this will get p nodes with superscripts,
but how do I remove the superscripts if there are many matching nodes and
different superscripts?
xpathSApply(doc, "//p[sup]", xmlValue)
[1] "Cata"
1
12700 8874.441
7 61 2 9 940 1
22883 8874.441
8 61 2 8 642 1
33016 8874.441
9 71 1 7 9 39 1
y is located, if the user has
permission to save to that directory, and probably some other complications
I'm missing. Any suggestions?
Thanks,
Chris Stubben
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Sent from the R hel
Greg,
If you have a MySQL table with an auto_increment field, you could just
insert a NULL value into that column and the database will increment the key
(it may not work in SQL STRICT mode, I'm not sure). I don't think there's
any way to specify which columns you want to load data into using
db
Sorry, I wasn't sure what you meant. This way will return more than one
answer, right?
N<-c(1,2,1,3)
R<-c(1.75,3.5,1.75,1.3125)
## get all 126 combinations of five 0's and four 1's for matrix
cbn<-as.matrix(expand.grid( rep( list(0:1), 9)))
cbn<- cbn[rowSums(cbn)==4,]
ans<-list()
ctr<-0
## lo
You could use a combination of rle, cumsum and append.
> x <- c(1,1,1,2,2,3,3,3,3,3,4)
> y<-rle(x)$lengths
> y
[1] 3 2 5 1
> z<-cumsum(y)[y>1]
> z
[1] 3 5 10
>
> for(i in rev(z)) x <- append(x, c(0,0,0), after = i)
> x
[1] 1 1 1 0 0 0 2 2 0 0 0 3 3 3 3 3 0 0 0 4
Chris
Serguei Kaniovski-3
Also try the odiag function in the demogR package
odiag( 1:5, -1)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]000000
[2,]100000
[3,]020000
[4,]003000
[5,]000400
[6,]000
privalan wrote:
>
> However, I cannot figure out the way to compute the asymptotic stochastic
> population growth rate using “popbio”. I would like to perform a
> stochastic model in which my demographic rates are sampled from beta
> distribution with known mean and SD. For example, rather than
-Halcyon- wrote:
>
> My question is : How do I arbitrarily assign the animals being "FALSE"
> ***according to the uniformal chance generation*** to the array
> Dead.Animals, there being "TRUE"?
>
Maybe I'm missing something, but aren't dead animals just the ones that
aren't alive? Why do you
99 40 77 -- Chc -- -- --
3 50 43 44 17 SW99 40 77 -- Chc -- -- --
4 48 43 42 16 SW99 40 83 -- Chc -- -- --
5 45 43 38 15WSW 100 40 93 -- Chc -- -- --
6 42 42 35 14 W
Michelle,
I would probably run a loop as well and save all matrices to a single list
(see hudsonia and calathea on working with lists of matrices).
First, run the example(test.census) to get the stage-fate data frame "trans"
and then run this code to save the matrices into a list "all".
years<-
2[j],
y2[j+1]), border=0, col = rev(heat.colors(100))[j])
}
}
Chris Stubben
Van Dongen Stefan wrote:
>
> Hi All,
>
> I would like to fill the area under a curve with a gradient of colors. Are
> there any packages or trick I could use
>
> Thanks
>
> Stefan
&g
hadley wickham gmail.com> writes:
>
> Why do you want a 3d barchart? They are generally a bad way to
> present information as tall bars can obscure short bars, and it is
> hard to accurately read off the height of a bar. While adding
> rotation can reduce some of these problems, why not create
Duncan Murdoch stats.uwo.ca> writes:
> That demo gives you the basics of the code, so it shouldn't be too hard
> to put your own together: just strip out the counting part.
>
Thanks, I did download the source to check the hist3d demo, but honestly it
didn't look very easy to simplify. I sw
>Transition matrices are Markov transition matrices among different
> life stages of organisms -- in the simplest case (Leslie matrices,
Ben,
Thanks for your clear explanation and plot examples. I like the dotplots alot
and added a few modifications below. Since I often compare rows and col
Adam Wilson-4 wrote:
>
> I am running R 2.5.1, RMySQL 0.6 , and DBI 0.2-3 on Windows XP
>
> Like others, I am having trouble with NA/Null value conversions between R
> and a MySQL database via DBI, but I could not find my exact problem in the
> archives. Most of the time NA values in R get tra
> On 10/12/07, Ben Bolker ufl.edu> wrote:
> >
> > Trying to find a quick/slick/easily interpretable way to
> > collapse a data set.
> >
Another alternative for SQL fans is the sqldf package. I used the MySQL driver
here since SQLite does not support standard deviation.
sqldf("select BROOD,
> I would like to calculate elasticities and sensitivities of each parameters
> involved in the following transition matrix:
>
> A <- matrix(c(
> sigma*s0*f1, sigma*s0*f2,
>s, v
> ), nrow=2, byrow=TRUE,dimnames=list(stage,stage))
>
> The command "eigen.analysis" a
Sorry, I didn't think carefully about your question on my first reply, so
please ignore it.
Chapter 9 in Morris and Doak (2002) cover vital rate elasticities and
sensitivities (see Box 9.1 for Matlab code
http://www.sinauer.com/PVA/vitalsens.m). Simon, thanks for posting your
code, I will add
Maybe something like this?
N<-c(1,2,1,3)
## create empty matrix
x<-diag(0,3)
## fill off diagonal
x[row(x)==col(x)+1]<-N[1:2]
# fill 3rd column
x[1:2,3]<-N[3:4]
Or create a function and return both x and y
mat3<-function(N)
{
x<-diag(0,3)
# fill each element separately
x[2,1]<-N[1]
x[3,2]<-N
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