Multivariate normal distributions with PSD variance matrices occur all the time.
The most common example in practice would probably be the distribution of the
vector of residuals from a normal regression. It has a degenerate distribution
wrt R^n because it is subject to p linear restrictions.
you might try
dummy - with(cleary,
cbind(B4 = as.numeric(D1 == 4),
B6 = as.numeric(D1 == 6),
B7 = as.numeric(D1 == 7)))
and do it all in one go.
___
to fix up your apporach you need to use
if(cleary$D1[i] == 4)
If b1+b2+b3 = 0 can't you put b3 = -b1-b2 and re-write the model in terms of b1
and b2 alone? The model is stil (g)linear.
If the parameters to which you refer corresponds to the levels of a factor,
then you can use the contr.sum contrast matrices. It essentailly does the
above.
It would certainly not be straightforward, unless Tibco has done a lot of work
to make it possible in recent years, (which is not unlikely). They have done a
lot of work in the other direction, i.e. enabling R packages to be used within
S-PLUS.
S-PLUS has the concept of a chapter, which
Bill D. is quite right, I did have in mind conversion the other way round. I
am happy to concede.
The interesting thing for me is the (relatively) automatic conversion from
.sgml to .Rd format. This is likely to speed things up considerably.
Bill.
-Original Message-
From: William
Here is a dodge I often use. This is a mock-up example.
___
bar - data.frame(matrix(rnorm(1001), nrow = 1))
names(bar)[1] - y ## say
head(bar[,1:5])
nbar - names(bar)
form - as.formula(paste(nbar[1], ~, paste(nbar[-1], collapse = +)))
fitModel - substitute(tm - rpart(FORM, data =
sales - read.csv(file=C:/Program Files/R/Test Data/sales.csv,
header=TRUE, row.names = Month)
^^^
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of
You are nearly there.
example(data.frame)
zz - c(aa_bb,bb_cc,cc_dd,dd_ee,ee_ff,
ff_gg,gg_hh,ii_jj,jj_kk,kk_ll)
ddd - cbind(dd, group = zz)
ddd - within(ddd, {
group - as.character(group)
tmp - do.call(rbind, strsplit(group, _))
group_b - tmp[,2]
To take it one step further:
x - as.complex(-4)
cx - x^(1/3)
r - complex(modulus = Mod(cx), argument = Arg(cx)*c(1,3,5))
r
[1] 0.793701+1.37473i -1.587401+0.0i 0.793701-1.37473i
r^3
[1] -4+0i -4+0i -4+0i
So when you ask for the cube root of -4, R has a choice of three possible
It's because in this instance R (and S before it) behaves a bit like Microsoft
and tries to guess what you really wanted rather than listen to what you asked
for.
If you ask for a single column from a data frame, e.g.
df[,1]
then by default this behaves like df[[1]] and you get a single
Your outlier has row.names 1. If this is selected in the bootstrap sample
once, it will also have row.names 1. If it is selected more than once the
row.names of the successive entries will begin with 1.
Here is a possibility you may wish to consider.
txt - textConnection(
+ y
Another approach would be
Y - list(sqrt, sin, function(u) u/2)
Ybar - function(u) rowMeans(sapply(Y, function(fun) fun(u)))
integrate(Ybar, 0, 1)
0.4587882 with absolute error 5.6e-05
i.e. make the function vectorized directly.
Note, however, that if you had
Y[[4]] - function(u) 1
That was my thought, and if you are going to be integrating it, you do need to
be concerned with efficiency to some extent, I would imagine.
My experience is that Vecotrize() is theoretically interesting and it is great
for getting you out of a tight spot like this, but if you can avoid it
The function is not exported from the package. You have to get tough with it,
e.g.
library(nlme)
summary.lme
Error: object 'summary.lme' not found
But if you insist:
nlme:::summary.lme
function (object, adjustSigma = TRUE, verbose = FALSE, ...)
{
fixed - fixef(object)
...
You
The conventional view used to be that S is the language and that R and S-PLUS
are implementations of it. R is usually described as 'a programming
environment for data analysis and graphics' (as was S-PLUS before it). However
as the language that R implements diverges inexorably from the
What you show below is only a representation of the matrix to 7dp. If you look
at that, though, the condition number is suspiciously large (i.e. the matrix is
very ill-conditioned):
txt - textConnection(
+ 0.99252358 0.93715047 0.7540535 0.4579895
+ 0.01607797 0.09616267 0.2452471
It's actually not too difficult to write the density function itself as
returning a function rather than a list of x and y values. Here is a no frills
(well, few frills) version:
### cut here ###
densityfun - local({
normd - function(value, bw) {
force(value); force(bw)
function(z)
perhaps you need something like this.
par(yaxs = i)
plot(runif(10), type = h, ylim = c(0, 1.1))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Sebastian Rudnick
Sent: Tuesday, 23 November 2010 10:37 AM
To: r-help@r-project.org
I think all you need is
?split
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Eduardo de Oliveira Horta
Sent: Sunday, 28 November 2010 8:02 AM
To: r-help@r-project.org
Subject: [R] Two time measures
Hello!
I have a csv file
Here is one way
a_clip - sub(^([0-9]+\\.[0-9]+\\.[0-9]+).*$, \\1, a)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Lorenzo Cattarino
Sent: Tuesday, 13 July 2010 5:20 PM
To: r-help@r-project.org
Subject: [R] modifying vector
As it says, (at least) one of your rows is all zeros. CCA cannot handle such
sites.
You might want to try something like this:
m.cca -
cca(WinterM_ratio ~ topo_mean + coast + prec_max +
temp_min + evi_min, data = datam,
subset = rowSums(WinterM_ratio) 0) ### Note.
(Note that you do
As far as I know the answer to your question is No, but there are things you
can do to improve the readability of your code. One thing I find useful is to
avoid using $ as much as possible and to favour things like with() and
within().
The first thing you might do is think about choosing
Is this the kind of thing you mean?
wData - within(wData, {
+ z - (weigth - mean(weigth))/sd(weigth)
+ p-value - 2*pnorm(-abs(z)) ## 2-sided
+ rm(z) })
wData
employee_id weigth p-value
1 100150 0.3763641
2 101200 0.9081403
3 102300 0.1547139
4
Isn't there a danger that what you teach will then be driven by what templates
you receive?
I would have thought this was an easy thing to do yourself, once you have
decided what you want to teach your students. Just write a scropt (e.g. using
the inbuilt script edit in R for Windows if you
It won't happen automatically unless you make it do so yourself.
Fortunately this is not hard. See,
?arrows
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Trying To learn again
Sent: Wednesday, 28 July 2010 7:07 AM
To:
Is this the kind of thing you are looking for?
dat - data.frame(x = 1:3, freq = 2:4)
dat
x freq
1 12
2 23
3 34
newDat - dat[rep(rownames(dat), dat$freq), ]
newDat
x freq
1 12
1.1 12
2 23
2.1 23
2.2 23
3 34
3.1 34
3.2 34
3.3 34
?vcov ### now in the stats package
You would use
V - vcov(my.glm)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Bojuan Zhao
Sent: Thursday, 29 July 2010 9:52 AM
To: r-help@r-project.org
Subject: [R] Variance-covariance
There is a much simpler way.
con - textConnection(
A 1 101 M 55 5 A 1 104 F 27 0 A 1 106 M 31 35
A 1 107 F 44 21 A 1 109 M 47 15 A 1 111 F 69 70
A 1 112 F 31 10 A 1 114 F 50 0 A 1 116 M 32 20
A 1 118 F 39 25 A 1 119 F 54 0 A 1 121 M 70 38
A 1 123 F 57 55 A 1 124 M 37 18 A
Just to add to the confusion, for (nearly) all practical purposes they are the
same:
all.equal(df1, df2)
[1] TRUE
do.call(paste, df1)
[1] 1 4 2 5 3 6
do.call(paste, df2)
[1] 1 4 2 5 3 6
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
with(a, Samples[apply(a[,-1], 1, any)])
[1] S2 S4
Levels: S1 S2 S3 S4 S5
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Alexander Eggel
Sent: Wednesday, 11 August 2010 12:05 PM
To: r-help@r-project.org
Subject: [R] TRUE FALSE
For huge cases this might be a whiff faster
with(a, Samples[rowSums(a[, -1], na.rm = TRUE) 0])
[1] S2 S4
Levels: S1 S2 S3 S4 S5
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of bill.venab...@csiro.au
Sent: Wednesday, 11 August
The types of bitmap available depend on the ghostscript you have installed.
Have you checked that yours handles jpeg?
What happens if you use the jpeg driver directly?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of kayj
Sent:
I take it you have the data in character form. If the DMS data is already in a
data frame it is probably in factor form, so you will need to coerce it to
character first. Here is how you could do it with the DMS data in a data frame:
m ## an example with 3 entries
DMS
1 122:45:45
2
Do you expect this to be easy? It may be, but I can't see a particularly
graceful way to do it. Here is one possible solution.
dat
StandID PlotNum HerbNum Woody
1 001 1 1low
2 001 2 2 medium
3 001 3 1low
4 001 4 3
Here is a suggestion you may care to develop
func - function(a, b) {
plot(1:10)
title(main = bquote(a == .(a)*','~ b == .(b)))
invisible()
}
try with
func(1,2)
func(36, 2^10)
c
-Original Message-
From: r-help-boun...@r-project.org
Eh?
## First calculate the means:
mns - with(alldata, tapply(param1, list(Pair, group), mean))
## now put the results into a data frame
res - data.frame(mns, dif = mns[, D] - mns[, R])
## Then write it out if that's your thing.
-Original Message-
From: r-help-boun...@r-project.org
Is this all you want?
j - matrix(nrow=10,ncol=10)
k - matrix(seq(1:50), ncol=10)
row.names(k) - seq(2,10,by=2)
row.names(j) - 1:10
j[row.names(k), ] - k
j
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
1NA NA NA NA NA NA NA NA NANA
2 16 11 16
colorRampPalette returns a function, not a list of colours. You might want to
try something like:
ascols -
colorRampPalette(c(gray, yellow, darkgoldenrod1, orange, red),
interpolate=spline)
x - 10*(1:nrow(volcano))
y - 10*(1:ncol(volcano))
image(x, y, volcano, col = ascols(300), axes =
sseq - c(1, seq(5, 40, by = 5))
for(i in 1:length(sseq))
assign(paste(arima, i, sep=), arima(data.ts[sseq[i]:(sseq[i]+200)],
order=c(1,1,1)))
...but why would you want to do so?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Is this the kind of thing you are talking about?
### 8 cut here 8 ###
A - rep(NA, 100)
B - sort(runif(25))
C - sort(sample(1:100, 25))
A[C] - B
B
C
A
### 8 cut here 8 ###
(The sorting is not necessary. It's only there to make checking what happened
easier.)
-Original Message-
In R, the glm families poisson and quasipoisson will give you the same
estimates. Their standard errors will (usually) be different, though, and
family = quasipoisson does not give you an AIC (since it does not maximise a
true likelihood; it uses quasi-likelihood estimation).
I hope you are
Most likely.
If you care to be more specific, we might be able to give you a more specific
answer.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of lord12
Sent: Wednesday, 15 September 2010 6:30 AM
To: r-help@r-project.org
An envorniment is a hash table.
There are several packages that you might care to look at, e.g. filehasn
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of lord12
Sent: Wednesday, 15 September 2010 7:06 AM
To: r-help@r-project.org
Here is one possibility:
set.seed(1)
dat - matrix(rnorm(4*16), 4, 16)
dim(dat) - c(4,4,4)
dat
, , 1
[,1] [,2] [,3][,4]
[1,] -0.6264538 0.3295078 0.5757814 -0.62124058
[2,] 0.1836433 -0.8204684 -0.3053884 -2.21469989
[3,] -0.8356286 0.4874291 1.5117812
It means that two of the packages on which TeachingDemos depends were not
installed correctly, namely rgl and tkplot.
The problem could be that you do not have the system dependencies in place.
For rgl these are listed as:
SystemRequirements:OpenGL, GLU Library, zlib (optional),
no.
Your best way is to use
write.csv(table_list, file = paste(file, x, .csv, sep=))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of lord12
Sent: Friday, 17 September 2010 12:55 PM
To: r-help@r-project.org
Subject: Re: [R]
This is an ESS question, not an R one.
The problem is that the argument names to the function help() have changed in
recent releases of R, and htmlhelp is no longer an accepted argument. The
replacement argumennt is now help_type (with default value
getOption(help_type)).
This may have been
have - list(a=7,b=3,c=1)
have2 - lapply(have, rep, 2)
have2
$a
[1] 7 7
$b
[1] 3 3
$c
[1] 1 1
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Derek Ogle
Sent: Sunday, 19 September 2010 11:20 AM
To: R (r-help@R-project.org)
You could do most of this with the function lmList in the nlme package, but
since you want both plots and summaries, you might as well do it in a more
flexible loop.
How about something like this:
Code:
## This makes a single factor to define your groups
BCI - within(BCI,
That's if the variables are visible. If they are only in the data frame it's
not much more difficult
d - data.frame(group = rep(1:5, each=5),
variable = rnorm(25))
with(d, tapply(variable, group, max))
(Tip: avoid using attach().)
Bill Venables.
-Original Message-
library(help = survival)
help(survreg, package = survival)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Halabi, Anan
Sent: Tuesday, 21 September 2010 3:12 PM
To: r-help@r-project.org
Subject: [R] Estimating Weibull parameters
You left out the subscript. Why not just do
d - within(data.frame(group = rep(1:5, each = 5), variable = rnorm(25)),
scaled - variable/tapply(variable, group, max)[group])
and be done with it?
(Warning: if you replace the second '-' above by '=', it will not work.
It is NOT true
?par
See col.lab, col.main, col.sub, c.
To see what colours are available you can use
colours()
(or colors() if you don't speak English).
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of tooblue
Sent: Thursday, 23 September
I was looking for an example of complex variables in R. This one is trivial,
but rather cute (though World War II aficionados may 'come over all funny').
See if you can guess the image before you try the function. It's not difficult.
jif - function(res = 100) {
z - sample(do.call(complex,
This is worse than before and getting pretty silly. Now you are calling outer
with a function that only takes one argument.
R might be hard for you, but mind reading is even harder for most of us. To
get help you need to explain clearly and sensibly what it is you want to do.
Look at your
It would help if you were to tell up which package you are talking about. I
take it this is the survival package and this is the proportioanl hazards
fitting function for case-cohort data.
You would have seen earlier in the code that
method - match.arg(method)
so at that stage 'method'
dump(x, file = x.R)
file.show(x.R)
will get you most of the way.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Steven Kang
Sent: Wednesday, 29 September 2010 3:11 PM
To: r-help@r-project.org
Subject: [R] String split and
R packages do not solve modelling problems. They can be used to solve
computational problems.
Wouldn't it be a better plan to sort out what kind of models you want to
investigate to address your research questions first? Then, and only then, can
you sensibly look around for packages that
?par
lood at the 'las' parameter.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of ANJAN PURKAYASTHA
Sent: Wednesday, 6 October 2010 8:13 AM
To: r-help@r-project.org
Subject: [R] Plotting x-axis labels perpendicular to the axis
R-bugs is the appropriate place to report this.
It may be particular to your system. Here's what happens on mine (Windows XP,
R 2.11.1)
f - factor(1:5)
g - reorder(f, rnorm(5))
is.ordered(g)
[1] FALSE
g
[1] 1 2 3 4 5
attr(,scores)
1 2 3 4 5
I was unaware of this discussion till now, so I'm not up with what's been said
already.
Here is how I would go about the problem I think you are trying to solve.
m - matrix(1:16, 4, 4)
require(PolynomF)
(lop - as.polylist(lapply(1:nrow(m), function(i) polynom(m[i,]
List of polynomials:
PS ... actually, now I think about it...
as.polylist(apply(m, 1, polynom))
List of polynomials:
[[1]]
1 + 5*x + 9*x^2 + 13*x^3
[[2]]
2 + 6*x + 10*x^2 + 14*x^3
[[3]]
3 + 7*x + 11*x^2 + 15*x^3
[[4]]
4 + 8*x + 12*x^2 + 16*x^3
This works with the PolynomF package because in that case the
Try !(group %in% C(A, ...))
A slightly cuter way is to define your own operator.
`%ni%` - Negate(`%in%`)
letters[1:3] %ni% letters[3:5]
[1] TRUE TRUE FALSE
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Emily
One possible way to treat parameters as nuisance parameters is to model them
as random. This gives allows them to have a reduced parametric load.
There are many packages with funcitons to fit glmms. One you may wish to look
at is lme4, which has the lmer fitting function
library(lme4)
fm -
AIC is only defined up to an additive constant (as is log-likelihood).
It should not surprise you that the values for AIC differ between packages.
The real question is whether the change in AIC when going form one model to
anoth is the same. If not, one is wrong (at least).
-Original
What are you trying to find out from your analysis and what techniques do you
have in mind for doing it?
It might be as well to sort out this first (and tell us if you have) rather
than look round for a package first and then decide on what you want to do in
the analysis.
Having said all
This is pretty standard.
varList - 1:4
subData - subset(dataset, var %in% varList)
Should do it.
W.
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
JustinNabble [justinmmcgr...@hotmail.com]
Sent: 11 October 2009 16:13
To:
Dear R_Help Help,
The critical questions are
a) how many parameters do you have
b) how pathological is the log-likelihood function
c) how good are your initial values and
d) how efficiently have you coded your objective function?
Of these, the last is most likely the critical one, and the one
glm is not, and never was. part of the MASS package. It's in the stats package.
Have you sorted out why there is a big difference between the results you get
using glm and lrm?
Are you confident it is due to the algorithms used and not your ability to use
the software?
To be helpful, if
Try using
source(P_Value)
before anything else.
Also, P_Value can be written as a one-liner:
P_Value - function(Table) fisher.test(Table)$p.value
so you don't really need a separate function at all.
Bill Venables
CSIRO/CMIS Cleveland Laboratories
-Original Message-
From:
This is one way to do it. Suppose your data is in the file rainfall.txt, as
set out below. Then
dat - read.table(rainfall.txt, header = TRUE)
dat - within(dat, {
+ date - as.Date(paste(year, month, day, sep=-))
+ week - factor(as.numeric(date - date[1]) %/% 7)
+ })
wRain - with(dat,
Here is one way. Suppose your data frame is called 'dat'.
o - with(dat, order(as.Date(paste(year, month, day, sep=-
newDat - dat[o, c(year, month, day, rain)]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Roslina Zakaria
xyzs - matrix(rnorm(3*10,0,1),ncol=3)
V - c(2,3,4)
system.time(vx - apply(t(xyzs) * V, 2 ,sum))
user system elapsed
1.060.021.08
system.time(wx - as.vector(xyzs %*% V))
user system elapsed
0 0 0
all.equal(vx, wx)
[1] TRUE
?
-Original
Or..
unlist(data)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Nick Matzke
Sent: Saturday, 5 June 2010 11:17 AM
To: r-help@r-project.org
Subject: Re: [R] 380x380 dataframe to list
That's the answer, thanks!!
Nick
Peter
I think what you are groping for is something like this
my_matrix - matrix(1:60, nrow = 6)
id_a - seq(10,60,by=10)
id_b - seq(100,1000,by=100)
my_database - cbind(
expand.grid(id_a = id_a, id_b = id_b),
mat = as.vector(my_matrix)
)
-Original Message-
From:
Here is one way
...
DF4 - cast(formula=Date~V2,data=DF3,value=X1,fill=0)
d - with(DF4, seq(min(Date), max(Date), by = 1)) ### full set
m - as.Date(setdiff(d, DF4$Date)) ### missing dates
if(length(m) 0) {
extras - cbind(data.frame(Date = m), cat = 0, dog = 0, tree = 0)
Here is an alternative
with(iris, rowsum(iris[, -5], Species)/table(Species))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Peter Langfelder
Sent: Thursday, 10 June 2010 12:27 PM
To: SH.Chou
Cc: r-help@r-project.org
Subject:
Here is one way.
f - function(d) {
y - names(d)[1]
xs - names(d)[-1]
nx - length(xs)
xs - sort(sample(xs, sample(1:nx, 1)))
form - as.formula(paste(y, ~, paste(xs, collapse=+)))
Call - substitute(lm(FORM, data = d), list(FORM = form))
eval(Call)
}
d - within(data.frame(y =
This is a lazy way, and a slightly extravagant way if your memory is limited
and you are dealing with large numbers of rows.
NCols - 5
NRows - 7
myMat - matrix(runif(NCols*NRows), ncol=NCols)
d - dist(myMat)
dm - as.matrix(d)
diag(dm) - Inf
ij - which(dm == min(dm), arr.ind = TRUE)[1,]
ij
Is this the kind of thing you want?
test - matrix(rep(letters[1:3],6),nrow=6,byrow=T)
test[2,] - letters[5:7]
test[4,] - c('e','f','i')
test[1,3] - 'i'
colnames(test) - c('leave','arrive','line')
test - data.frame(test)
tab - with(test, table(paste(leave, arrive, sep=), line))
tab -
The usual method is either
lis - vector(list)
or, nearly equivalently,
lis - list()
If you know in advance how many components the list will have, there can be a
slight advantage in using
lis25 - vector(list, 25)
if e.g. you know the list will be of length 25.
Bill Venables.
Here is an alternative:
lst - list(m = c('a','b','c'), n = c('c','a'), l = c('a','bc'))
set - c(a, c)
(w - sapply(lst, function(x) all(set %in% x)))
m n l
TRUE TRUE FALSE
names(w)[w]
[1] m n
-Original Message-
From: r-help-boun...@r-project.org
Why does the naming have to be done inside the cbind()?
How about
dataTest - data.frame(col1 = c(1,2,3))
new.data - c(1,2)
name - test
length(new.data) - nrow(dataTest)
newDataTest - cbind(dataTest, new.data)
names(newDataTest)[[ncol(newDataTest)]] - name
newDataTest
col1 test
11
You ask:
# How can I use predict here, 'newdata' crashes
predict(m1,newdata=wolf$predicted);wolf # it doesn't work
To use predict() you need to give a fitted model object (here m1) and a *data
frame* to specify the values of the predictors for which you want predictions.
Here wolf$predicted
Your intuition is wrong and R is right.
Why should the product of two probability density functions be a normalized pdf
also?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Carrie Li
Sent: Saturday, 26 June 2010 1:28 PM
To:
Here is another one that works:
do.call(subset, list(dat, subsetexp))
x y
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Vadim Ogranovich
Sent: Saturday, 26 June 2010 11:13
Why do you use *double* square brackets on the left side of the replacement?
From the help info for [[:
The most important distinction between [, [[ and $ is that the [ can select
more than one element whereas the other two select a single element.
You seem to be selecting 20 elements.
Since X is a vector, then
A - sum(X, solve(V, X))
is probably slightly better here.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Dmitrij Kudriavcev
Sent: Tuesday, 29 June 2010 12:29 PM
To: r-help@r-project.org
Subject: [R]
Oops. Try
A - sum(X * solve(V, X))
(too fast!)
-Original Message-
From: Venables, Bill (CMIS, Cleveland)
Sent: Tuesday, 29 June 2010 1:05 PM
To: 'Dmitrij Kudriavcev'; 'r-help@r-project.org'
Subject: RE: [R] Matrix operations
Since X is a vector, then
A - sum(X, solve(V, X))
is
newlist - newlist[sapply(newlist, length) 0]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of song song
Sent: Tuesday, 29 June 2010 2:12 PM
To: r-help@r-project.org
Subject: [R] how to remove numeric(0) component from a list
like
require(lattice)
Loading required package: lattice
xyplot(CASES ~ YEAR | AREA, data = t, type = b)
xyplot(CASES ~ YEAR, data = t, type = b, groups = AREA)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Pablo Cerdeira
Sent:
You need to parse it before you evaluate it.
eval(parse(text = avar))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jeremiah H. Savage
Sent: Wednesday, 30 June 2010 3:45 PM
To: r-help@r-project.org
Subject: [R] evaluate a
It's a bit like gravy. You have to make it, it doesn't just come when you
cook the joint.
Here is a mock-up of your situation:
dat - data.frame(ByMonth = gl(6, 10, labels = paste(2010-0, 1:6, sep=)),
+ r1 = rnorm(60), v = rnorm(60), r = rnorm(60))
head(dat)
ByMonth r1 v
ginv() is slower than solve(). This is the price you pay for more generality.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of song song
Sent: Monday, 5 July 2010 10:21 AM
To: r-help@r-project.org
Subject: [R] if using ginv
Here is one way
checkList - data.frame(Day = c(f.n1, f.n2),
+ FN = rep(c(FN1,FN2),
+ c(length(f.n1), length(f.n2
m - match(DF$Date, checkList$Day)
DF - cbind(DF, Fortnight = checkList$FN[m])
DF
XY Date Fortnight
1
apply(matrix1, 1, quantile, probs = c(0.05, 0.9), na.rm = TRUE)
may be what you are looking for.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jim Bouldin
Sent: Thursday, 8 July 2010 8:52 AM
To: R help
Subject: [R] quantiles on
Here is a way:
###
x - sort(c(seq(-4, 4, length=100), 1.96))
hx - dnorm(x)
par(pty=s)
plot(x, hx, type=l, xlab=z value,
ylab=Density, main=density of N(0,1))
abline(v=1.96, col=red)
abline(h = 0)
top - x = 1.96
tail - rbind(cbind(x[top], hx[top]), c(4, 0), c(1.96, 0))
polygon(tail, col = red)
(id - with(df, id[match(v,Name)]))
[1] 1 2 2 4 2 4 2 2 4 2 1 2 2 1 2 2 1 2 2
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Nathan S. Watson-Haigh [nathan.watson-ha...@csiro.au]
Sent: 27 January 2010 10:31
To:
Here is one way.
dat
V1
1 43-156
2 43-43
3 1267-18
dat - within(dat, {
+ m - do.call(rbind, strsplit(as.character(V1), -))
+ XX - as.numeric(m[,1])
+ YY - as.numeric(m[,2])
+ rm(m)
+ })
dat
V1 YY XX
1 43-156 156 43
2 43-43 43 43
3 1267-18 18 1267
Bill
something like this?
c(1, which(c(0, diff(y)) != 0))
[1] 1 11 21 51 61 71 81 91
Bill Venables
CSIRO/CMIS Cleveland Laboratories
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Tim Clark
Sent: Thursday, 11 February 2010 11:59
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