be a problem, but to avoid it
you could use exactRankTests::wilcox.exact() which, I believe, was
written by the same author. It uses the same syntax as wilcox.test().
Note, though, that the package is no longer
being developed.
Peter Ehlers
On Thu, May 30, 2013 at 6:21 PM, Greg Snow 538
is there some way to look for what function is used to calculate
the statistics?
Just look at the code of perm.test.default (it's not complicated).
Type
perm.test.default
to see the code or get the package source and study the code.
Peter Ehlers
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$Depth, ylim = rev(range(0:100)), xlab=CHAO, ylab=Depth,
pch=15, las=2, main=Sep12-RNA, cex.main=1)
abline(lmR)
lines(cd$CHAOsep12RNA, a[,2], lty=2)
But I see both cases kind of zigzags. What can it be the reason? thank you!
Sort your dataframe by x-values (CHAOsep12RNA).
Peter Ehlers
in the calculation, see any
intro stats text or look at the code of cor.test.default().
Peter Ehlers
Prof. José Iparraguirre
Chief Economist
Age UK
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Elaine Kuo
Sent: 17 May 2013 10:40
To: r
#[1] tol14 tol15
To construct the formula:
rhs - paste(use, collapse = + )
form - paste(exposure ~, rhs)
And then use it:
fit_multi - lm(formula = form, data = tolerance)
Peter Ehlers
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Carmo
According to the plyr NEWS file, mutate was introduced in
Version 1.3 (2010-12-28). I would hope that your version is
newer than that. You should tell us what the error message is.
Anyway, you can always use R's within() function instead;
or use transform() as Jean suggested.
Peter Ehlers
in the plot:
boxplot(residual ~ firm, data = newdata)
Peter Ehlers
On Wed, Apr 3, 2013 at 3:38 AM, Cecilia Carmo cecilia.ca...@ua.pt wrote:
Hi R-helpers,
My real data is a panel (unbalanced and with gaps in years) of thousands
of firms, by year and industry, and with financial information
'.
But try it for an nls model. Mynlsmodel$coefficients won't work (well,
it won't give an error but it will yield NULL).
That's why there are special extractor functions such as coef.nls,
coef.Arima, etc. For lm models, coef.default is used.
Peter Ehlers
For suggestion 2, thanks! I'm new
and then do findFn(mode) to see what's available.
E.g. packages, pracma, asbio, dprep, rattle and many others.
Do note that they handle the multimodal situation differently.
Or, write your own, perhaps using table() and which.max().
Peter Ehlers
all1 - ddply(all,ACT_NAME, summarise, mean=mean
calculations easier.
This, however,
is an implementation issue and is not guaranteed to hold in all
implementations of R.
Hint:
f - factor(sample(5, 10, TRUE))
as.numeric(levels(f))[f]
g - factor(sample(letters[1:5], 10, TRUE))
as.numeric(levels(g))[g]
Peter Ehlers
[[alternative HTML
-change may perhaps
have to be determined empirically).
Isn't this set by the 'bufbytes' and 'buflines' specifications in the
Rconsole file?
Anyway, it's probably best to use 'View' to inspect data.
Peter Ehlers
With thanks,
Ted.
-
E-Mail: (Ted
))
and I usually find it much more flexible to add the title with a
separate title(titt) call.
Peter Ehlers
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()
function.
Peter Ehlers
On Mon, Apr 1, 2013 at 2:58 PM, Bert Gunter gunter.ber...@gene.com
mailto:gunter.ber...@gene.com wrote:
Yup. Note also:
as.character.factor
function (x, ...)
levels(x)[x]
But of course this is OK, since this can change if the implementation
does
On 2013-04-01 15:06, Ted Harding wrote:
On 01-Apr-2013 21:26:07 Robert Baer wrote:
On 4/1/2013 4:08 PM, Peter Ehlers wrote:
On 2013-04-01 13:37, Ted Harding wrote:
Greetings All.
This is a somewhat generic query (I'm really asking on behalf
of a friend who uses R on Windows, whereas I'm
On 2013-04-01 19:23, arun wrote:
gsub(\\,.*,,x)
#[1] foo bar qux
A.K.
No big deal, but does , have to be escaped?
sub(,.*, , x)
Peter Ehlers
- Original Message -
From: Gundala Viswanath gunda...@gmail.com
To: r-h...@stat.math.ethz.ch r-h...@stat.math.ethz.ch
Cc:
Sent: Monday
to check is
range(x[, 17])
It's also _always_ a good idea to check str(x) before you do anything
further with x.
Peter Ehlers
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To the OP:
Sooner or later most R beginners are bitten by this all too
convenient shortcut. As an R newbie, think of R as your
bank account: overuse of $-extraction can lead to undesirable
consequences. It's best to acquire the '[[' and '[' habit early.
Peter Ehlers
On 2013-03-25 12:43, Bert
how to achieve that?
From help(ifelse):
ifelse returns a value with the same shape as test
i.e. in your case, the same 'shape' as 'x 0', a single value.
You _could_ make ifelse() work with, e.g., ifelse(rep(x, 6) 0, )
but you probably want if() instead.
Peter Ehlers
= fun, filename = filename, ...) :
cannot use this formula, probably because it is not vectorized
[...snip...]
This suggests that it might be useful to read ?Vectorize.
Peter Ehlers
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https
!
?axis will show you how to use the 'at' and 'labels' arguments of
the axis() function.
Peter Ehlers
On Tue, Mar 19, 2013 at 2:00 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Try setting the argument xaxt (x axis type) to n (no x axis) and then
use ?axis.
plot(breaks, cumfreq0
a shift to accommodate coef k for given k is simple.
Peter Ehlers
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Forgot to mention: You might find the nlmrt package helpful but
I have no experience with that (yet).
Peter Ehlers
On 2013-03-15 07:57, Shane McMahon wrote:
I have a question regarding robust nonlinear regression with nlrob. I
would like to place lower bounds on the parameters, but when I
):
[1] grid_2.15.3 lattice_0.20-13 tools_2.15.3
Peter Ehlers
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and provide commented, minimal
, if you need to do this a lot, you could
create an expression vector:
x - -8:-3
z - vector(expression, 6)
for(i in 1:6) z[[i]] - bquote(10^.(x[i]))
axis(1, x, z)
Peter Ehlers
On Wed, 13 Mar 2013, Berry Boessenkool wrote:
Hi all,
I want to label an axis with exponents, but can't get it done
(slope, na.rm = TRUE)
idx - which.max(slope)
Obviously, this can be extended to cover more than a 24-hour period.
Now, let's wait for Gabor to show us the trivial way with zoo::rollapply.
Peter Ehlers
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https
.
[... rest of code sample snipped ...]
Peter Ehlers
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained
for ?update.formula where you can read all about it.
Peter Ehlers
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and provide commented, minimal, self-contained
, it should be possible to
use locator() to place each plot.
Peter Ehlers
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and provide commented
ate kids.
Peter Ehlers
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and provide commented, minimal, self-contained, reproducible code.
could set the
argument 'aspect' to 'fill' instead of 'iso'.
Peter Ehlers
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On 2013-03-11 06:07, Jorgen Harmse wrote:
identical(df[1,],df[2,]) is FALSE because of the row names. all( == ) is just a
work-around that I attempted.
Jorgen.
I would just wrap the elements in c():
identical( c(df[1,]), c(df[2,]) )
Peter Ehlers
On Mar 11, 2013, at 02:53 , PIKAL Petr
' and pch-vector 'ch':
with(dat, plot(x, y, pch = ch[z]))
with(dat, text(x, y, lab=z, pos=3, cex=.7))
Try also pos=1,2,4 to see what suits you best.
Peter Ehlers
Date: Mon, 11 Mar 2013 21:33:33 +
From: ruipbarra...@sapo.pt
To: eliza_bo...@hotmail.com
CC: r-help@r-project.org
Subject: Re
factor vary most rapidly, then the next, etc.
I think that's documented somewhere, but I don't know where.
Peter Ehlers
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, victory points) each
obeying a given set of rules.
The purpose of the statistical analysis is to come up with a measure of
the strength of each card.
Any idea if out there there is any R package/script which can help me?
Any suggestion is welcome
Lorenzo
Why not simulate?
Peter Ehlers
.
Peter Ehlers
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with negative subscripts
Looks to me like 'which(newNb 0) - Ntip' evaluates to a
vector that has both positive and negative elements.
Like this:
x - 1:5
x[c(-2,-4)] ## ok
x[c(-2, 0)] ## ok
x[c(-2, 4)] ## generates your error
Peter Ehlers
--
Nicole A Thompson
E3B Columbia
, at=seq(1,40,4), labels=seq(1,10,1))
Thanks in advance for any help!
Have a look at what
par(usr)
gives to see that your at setting makes no sense.
Peter Ehlers
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PLEASE
or another designed for functional data
analysis.
Thanks,
Zoe Richards
What does your question mean? Possibly, you could 'invert' a mean
function, but I have no idea what that would accomplish. Can you
provide an example of just what you want to do?
Peter Ehlers
.
Peter Ehlers
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for approval)
Peter Ehlers
[...snip...]
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On 2013-03-04 12:30, David Winsemius wrote:
On Mar 4, 2013, at 12:18 PM, Peter Ehlers wrote:
On 2013-03-04 12:04, Ista Zahn wrote:
On Mon, Mar 4, 2013 at 2:57 PM, Rolf Turner rolf.tur...@xtra.co.nz wrote:
I never saw the original note nor its resubmission. Nor could I find it
in the R
that this isn't totally out to lunch.
Peter Ehlers
On 2013-03-02 09:20, Frank Harrell wrote:
Whoops - these 2 lines should have been omitted from the program:
n - sprintf('%s (n%s=%g, n%s=%g)', v, nam[1],n[1], nam[2],n[2])
vn[var == v] - n
Frank Harrell wrote
I would like to have a lattice
Duncan's comment may not qualify as a fortune, but it did make
me chuckle.
Peter Ehlers
On 2013-03-02 03:01, Duncan Murdoch wrote:
On 13-03-01 8:35 PM, C W wrote:
[...snip...]
pie is a function, but all it does is draw pie charts, so who cares if
you mask it? :-).
Duncan Murdoch
[...snip
- realroots(po.lm, 3)
predict(po.lm, newdata = data.frame(b = r)) # confirm
1
1.69
So I think there's a calculation error somehwere.
You need to replace the following line
if(names(model)[1] == (Intercept))
with
if(names(coef(model))[1] == (Intercept))
Peter Ehlers
On 3/1/13, arun
remaining when resizing
panels.
Thank you.
Elaine
[[alternative HTML version deleted]]
I think that you should provide (minimal) code to illustrate the problem.
Peter Ehlers
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(xleft, ybottom, xright, ytop, fill = bisque)
panel.text(x = (xleft + xright) / 2,
y = (ybottom + ytop) / 2,
labels = lab)
})
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it on your search path.
This should do it:
library(parallel)
--
Peter Ehlers
computing on windows?
Thanks for your help.
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in the top strips, using the factor.levels argument.
But perhaps your real use is more involved than the example.
Peter Ehlers
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~ height, data = singer, fill = mycol)
Peter Ehlers
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to specify the variable types correctly?
Thank you.
Joanna
Peter Ehlers
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on the levelplot help page.
Peter Ehlers
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in the TeachingDemos package.
Peter Ehlers
(B) Suppose I have 100 pairs of (x, y ). then is it possible to display in the
graph (irrespective of the curosr position) the values of (x, y) corresponding
to say 10th, 20th, 30th, 40th etc. observations in the graph.
?text
' dimensions. I don't see why, but since the
'residuals' argument to plot.gam() can be an array
_of the correct length_ (see ?plot.gam), it just might be
that you have an object called 'T' hanging around, in which
case using 'T' in place of 'TRUE' is a bad idea.
Actually, it's _always_ a bad idea.
Peter
: see MASS (the book, 4ed) page 191; also found
in the ch07.R file in the /library/MASS/scripts folder. I seem to
recall that this is mentioned somewhere in the docs, but put my finger
on it now.
One additional comment about the analysis: overdispersion might be
a problem.
Peter Ehlers
you're in the habit of processing
your data by column _number_ rather than variable _name_.
If so, I would strongly discourage that habit.
And a cursory look at str.default() suggests that it may not be
all that trivial a code change.
Peter Ehlers
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.
And
!drop_var
doesn't work because you need something that evaluates to a logical
value if you want to ! it.
This will do it:
df[df$X1=8,] [, !names(df) %in% drop_var]
Or use the subset() function, as Jorge suggests.
Peter Ehlers
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),
distance =c(2, 5, 8)))
Adjust as you prefer.
Peter Ehlers
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of that attribute.
Peter Ehlers
- Original Message -
From: Andre Zege az...@yahoo.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Thursday, November 15, 2012 9:36 PM
Subject: [R] lubridate concatenation issue
I took a look at Hadley's lubridate which seems a very neat
use of the semicolon to place two R expressions on one line.
He neither needs nor wants the ';dx2dx' in his own
expressions.
Peter Ehlers
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analyses.
Just use the formula version of t.test().
Peter Ehlers
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are not Normal, I don't
see why a t-distribution would be expected.
I seem to recall that Welch included some simulation results in his
Biometrika paper (1947? 1953?; I'm getting senile). Shouldn't be
difficult to generate in R. Maybe Greg Snow's TeachingDemos package
has something.
Peter Ehlers
' attribute.
This is mentioned on the help page for POSIXct.
The only way I know to 'fix' this is to reassign the
attribute:
e - c(d)
attr(e, tzone) - UTC
But there might a better solution to whatever the real problem is.
Peter Ehlers
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guess that 'Groups' is a factor variable and that
you have not set its levels explicitly to be what you want.
See ?levels
Peter Ehlers
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updating your R version.
Peter Ehlers
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)
(type g indicates a grid, but the 'grid' argument is preferred.)
Peter Ehlers
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that will permit different colours for the
circles/radii is left as an exercise.
Peter Ehlers
On Fri, Oct 26, 2012 at 10:59 AM, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2012-10-25 14:42, Bernie Wone wrote:
Hi Peter,
Thanks for the help. Being a beginner R user with little programming
experience, once
it easiest to
start with a matrix of zeros, fill in the upper.tri part as you
have done, then just add newmat and t(newmat), then fix the diagonal.
Peter Ehlers
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PLEASE do
.
thanks in advanceeliza
I hope that you're doing _exploratory_ data analysis.
Have a look at the 'leaps' package. It might be suitable.
Peter Ehlers
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' with 'myradial.plot' and
save the edited version as 'mypolar.plot', say. Now
use mypolar.plot() with whatever grid.lwd value suits
your purpose.
Peter Ehlers
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, data = benthos,
FUN = sum)
Then just use benthos2 as the data argument to barchart().
Peter Ehlers
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On 2012-10-23 15:22, bwone wrote:
Hello all,
Is it possible to change the radii line type in radial plots? I wasn't able
to find anything online.
Do please be more specific: which package's radial plot function
are you using? Package plotrix has radial.plot() with an 'lty'
argument.
Peter
)
segments(0, 0, xpos, ypos, col = grid.col)
and add the argument 'lty = grid.lty' to each.
Then add 'grid.lty = 1' to the function arguments.
Save the modified function as myradial.plot and source()
it into your R session and then use it with whatever
grid.lty setting you prefer.
Peter Ehlers
, INDICES=df$ID, FUN=function(DF) DF[which.max(DF$week), ]))
ID week outcome
1 16 42
4 4 12 85
9 9 12 84
With the plyr package:
library(plyr)
ddply(df, .(ID), function(x) tail(x, 2))
or, slightly simpler:
ddply(df, .(ID), tail, 2)
Peter Ehlers
---
why is the suffixes argument ignored?
I mean, I expected that the second a to be a.y.
The 'suffixes' argument refers to _non-by_ names only (as per ?merge).
Peter Ehlers
(when I omit suffixes, the result is the same).
Thanks.
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On 2012-10-07 10:50, arun wrote:
Hi,
Though, this does give the result you wanted when the column names are the same.
y1-y
colnames(y1)-c(a,b)
merge(x,y1,by=a,all=TRUE,suffixes=c(,.y))
# a b b.y
#1 1 4a
#2 2 5b
#3 3 6 NA
A.K.
Yes, because 'b' is _not_ a 'by'-name.
Peter Ehlers
On 2012-10-07 14:44, Sam Steingold wrote:
* Peter Ehlers ruy...@hpnytnel.pn [2012-10-07 10:03:42 -0700]:
On 2012-10-07 08:34, Sam Steingold wrote:
I know it does not look very good - using the same column names to mean
different things in different data frames, but here you go:
--8
:
count - sum(with(x, A*B 5))
Peter Ehlers
Hope this helps,
Rui Barradas
Em 16-09-2012 11:41, SirRon escreveu:
Hello,
I'm working with a dataset that has 2 columns and 1000 entries. Column 1 has
either value 0 or 1, column 2 has values between 0 and 10. I would like to
count how often Column 1
(1,1, xlab=expression(phantom()^32m*K) )
Error: unexpected symbol in plot(1,1, xlab=expression(phantom()^32m
plot(1,1, xlab=expression(phantom()^32m*K) ) # succeeds
I think what PD was trying to say is that a preferred solution would be:
plot(1,1, xlab=expression(phantom()^{32*m}*K) )
Peter
) as
you have done in your post. And what makes you think that there
_should_ be a function mesh()? The answer to that question may well
provide a sufficient clue to its whereabouts. Finally, searching for
functions is relatively simple with the findFn() function in the sos
package.
Peter Ehlers
of
the legend region.
barplot(x,beside=TRUE,
ylim=c(0,90))
abline(h=c(seq(10,90,10)))
box()
barplot(x,beside=TRUE,
xlab=Ailment,
ylab=Percent,
legend.text=TRUE,
args.legend=list(topright,title=Treatment,bg=white),
add=TRUE)
Peter Ehlers
$datamean))
Peter Ehlers
Thank you for your help.
Best regards,
Dominic
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On 2012-08-31 16:03, Bert Gunter wrote:
?which ##
as in ix - which(x==values)
-- Bert
Or maybe ?identify.
Peter Ehlers
On Fri, Aug 31, 2012 at 2:09 PM, Michael comtech@gmail.com wrote:
Hi all,
I am using locator to select the points from a scatter plot...
This is all fine
I think that what the OP is looking for comes under the heading of
inverse regression or the calibration problem. One reference
with a simple explanation including confidence intervals is Applied
regression analysis by Draper and Smith. (It's in section 3.2 in
my 3rd edition).
Peter Ehlers
- data.frame(g=gl(5, 10), x=rnorm(50), y=rnorm(50))
by(d[,2:3], d$g, cor, method=spearman)
It may seem a bit of overkill, but the plyr package is handy and
gives a nice output:
library(plyr)
ddply(d, .(g), summarize, correlation = cor(x, y, method = 's'))
Peter Ehlers
want to keep (e.g. the
names), but otherwise works on your example.
Duncan Murdoch
It seems that class listof also works:
class(x) - listof
x
Peter Ehlers
x
$setosa
x[, Sepal.Length]
n missing uniqueMean .05 .10 .25 .50 .75
50 0 15
that the page is clear:
... NA index picks an unknown element and so *returns* NA
in the corresponding element of ...
(my emphasis)
Isn't that exactly what occurs?
Peter Ehlers
Thanks again,
Mauricio
[[alternative HTML version deleted
,...)
},
key = list(corner = c(1,0),
text = list(lab = levels(iris[[Species]])),
points = list(pch = 15:17, col = 2:4)))
If you use panel.abline for regression lines, be sure to
specify the regression formula with the reg= argument.
Peter Ehlers
)
Peter Ehlers
I will try to dig into the code to see how this new parameter can be
accomodated into the code. I must confess that I am far from a pro at
this sort of stuff. I will appreciate it if I can get some help.
However, I will also give it a try myself. Will come back later if I run
with
your favourite decompression utility.
Peter Ehlers
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and provide commented, minimal, self
)
statement and use
with(survived, myballoonplot())
Peter Ehlers
I will appreciate any help that I can get.
Thanks,
Ravi
[[alternative HTML version deleted]]
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Here's another pretty straightforward solution, using the plyr pkg:
DF - data.frame(id, month, distance, bearing)
# variables as defined in the OP
require(plyr)
DF1-ddply(DF, .(id,month), summarize,
maxdist = max(distance),
maxbearing = bearing[which.max(distance)])
Peter
. the four suspect cases
came out just as they should). But what my mailer provides as your
data may not be what you really have.
Oh, and get a bandage for that head bruise.
Peter Ehlers
WWHHH Why why why why why why why? Why?
(Sorry, I've been trying to figure this out for hours
,
my_flag = ( (PM.EXP 0) (PM.DIST.TOT != 1) ) * 1 )
It looks as though your PM.DIST.TOT variable is meant to be
integer. If so, you might want to ensure that it is that type.
Otherwise, you might want to use Michael's suggestion of using
abs(... - 1) 1e-05.
Peter Ehlers
On 2012-08-24 14:56, Jennifer
Have you checked help(SSbiexp) ?
Peter Ehlers
On 2012-08-23 04:54, vincent guyader wrote:
Hi everyone,
I'm trying to perform a bi exponential Fit with the package NLS. the
plinear algorithm seems to be a good choice
see:
p-3000
q-1000
a--0.03
b--0.02
t-seq(0:144);t
y-p*exp(a*t) + q*exp(b*t
You can set grid parameters in the grid.pars component of
trellis.par.get().
x - 1:4; y - c(1,3,2,4)
xyplot(y ~ x, type = l, lwd = 20) # default linejoin
## set linejoin to 'mitre'
trellis.par.set(grid.pars = list(linejoin = mitre))
xyplot(y ~ x, type = l, lwd = 20)
Peter Ehlers
On 2012
I should add that you can also use the par.settings()
mechanism to set your linejoin specification on the fly:
x - 1:4; y - c(1,3,2,4)
xyplot(y ~ x, type = l, lwd = 20,
par.settings = list(grid.pars = list(linejoin = mitre)))
Peter Ehlers
On 2012-08-22 12:28, Peter Ehlers wrote:
You can
the quote symbols with the word.
Peter Ehlers
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R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained
On 2012-08-21 11:02, m p wrote:
Hello,
I have a problem writing a variable to an existing file.
Below is a part of my script and how it fails.
I can't find create.var.ncdf in help
You probably want 'create.ncdf'.
Peter Ehlers
Thanks for any help.
Mark
nc - open.ncdf(ncname, readunlim=FALSE
.))
or
text(x=1, y=11, expression(This is a * test^1 * of the
Emergency Broadcast System.))
If one is going to use plotmath expressions frequently, then these
two handy little symbols are worth knowing.
Peter Ehlers
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
of an outcome that is 5 units *less*
(or lower) than the EV:
pbinom(52, 112, .512)
# 0.1799121
Add the two probabilities to get:
0.2161936 + 0.1799121
# 0.3961057
and that's what binom.test() reports.
For details, have a look at the binom.test code.
Peter Ehlers
I'd be happy to know what I
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