>It is easy in a loop but that will take ages. Is there any vectorised
>apply-like solution to this?
If you showed the loop that takes ages, along with small inputs for
it (and an indication of how to expand those small inputs to big ones),
someone might be able to show you some code that does the
I don't know what you mean by "without having to use any special
interfaces", but "reference classes" will do what I think you want. E.g.,
the following makes a class called 'SortedNumeric' that only sorts the
vector when you want to get its value, not when you append values. It
stores the sorted
Does using zz<-bxp(boxplot(data,plot=FALSE)) do what you want? E.g.,
d <- transform(data.frame(t=1:15), x = sin(t)+log2(t), group =
paste("Group", t%/%4))
groupedX <- with(d, split(x, group))
zz <- bxp(boxplot(groupedX, plot=FALSE)) # bxp returns the x positions of
the boxes
points(col="blue", pc
"1X","5F8","F10"),
> > c("1X","2F6","3A15","F8","F5","F5"),
> > c("1X","F7","2A15","F9","F5")),
> >col.names=c("
It has been a while since I used Fortran formatted input, but the following,
without dots in the format, works:
> txt <- "1950. .614350 .026834 .087227 .006821 .180001 4.56E-2"
> print(read.fortran(textConnection(txt), c("f5", "6f8")), digits=10)
V1 V2 V3 V4 V5 V6
>If you want to drop levels, use droplevels() either on the factor or on
the >subset of your data frame. Example:
>droplevels(f[1]) #One element, only one level
Calling factor() on a factor, as the OP did, also drops any unused levels,
as the examples showed.
> str(factor(factor(letters)[11:13]))
You did not change df$quant - you made a new object called 'subdf'
containing a column called 'quant' that had only one level. Changing
subdf has no effect on df.
> df <- data.frame(quant=factor(letters))
> str(df)
'data.frame': 26 obs. of 1 variable:
$ quant: Factor w/ 26 levels "a","b","c",
Just to be clear, do you really want your 'condition' groups to be be
subsets
of one another? Most (all?) of the *ply functions assume you want
non-overlapping groups so they do a split-summarize-combine sequence.
You would have to replace the split part of that.
Bill Dunlap
TIBCO Software
wdunla
If you really want to return all the objects in a function, I think it
is better to return as.list(environment()), perhaps adding the
all.names=TRUE argument to capture names starting with a dot.
I only have done this while debugging a function and then I
find it is more convenient to return just e
The following usage of stopifnot seems reasonable to me and it
would be nice if the 2nd call caused the message 'is.data.frame(df) is not
TRUE'.
f <- function(df) {
stopifnot(is.data.frame(df), is.integer(df$ID))
range(df$ID)
}
f(data.frame(ID=4:7))
# [1] 4 7
f(4:7)
# Error in df$ID : $ op
ave(A, i, FUN=cummax) loops but is faster than your aggregate-based
solution. E.g.,
> i <- rep(1:1, sample(0:210, replace=TRUE, size=1))
> length(i)
[1] 1056119
> a <- sample(-50:50, replace=TRUE, size=length(i))
> system.time( vAve <- ave(a, i, FUN=cummax) )
user system elapsed
0.
ave() encapsulates the split/lapply/unsplit stuff so
transform(mydf, v1.mod = ave(v1, blocks, FUN=mynorm))
also gives what you got above.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, May 13, 2016 at 7:44 AM, Massimo Bressan <
massimo.bres...@arpa.veneto.it> wrote:
> yes, thanks
>
> yo
You can also use match(code, unique(code)), as in
transform(dd.2, codex2 = paste0("Person", match(code, unique(code
It is not guaranteed that x!=y implies digest(x)!=digest(y), but it is
extremely
unlikely to fail. This match idiom guarantees that.
Bill Dunlap
TIBCO Software
wdunlap tibco.c
Does this do what you want?
z <- data.frame(Name=c("One","Three","Twelve","Eleven"), Count=c(1,3,12,11))
with(z, barplot(Count, names=Name, horiz=TRUE))
with(z, barplot(Count, names=Name, horiz=TRUE, las=1))
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, May 12, 2016 at 10:19 AM, yoursurr
Why are you predicting from tr_m$finalModel instead of from tr_m?
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, May 9, 2016 at 11:46 AM, Muhammad Bilal <
muhammad2.bi...@live.uwe.ac.uk> wrote:
> Please find the sample dataset attached along with R code pasted below to
> reproduce the issu
I think as.POSIXct will just pass through a POSIXct object without any
changes. E.g.,
> dput(as.POSIXct( structure( list(quote(foo)),
class=c("POSIXct","POSIXt"
structure(list(foo), class = c("POSIXct", "POSIXt"))
If as.POSIXct( POSIXctObject, tz="ZONE") changed the time zone then a fair
You could install and load the 'lubridate' package, which has month()
and month<-() functions so you can do the following:
> z <- as.Date(c("2015-01-29", "2016-01-29", "2016-05-07", "2016-12-25"))
> z
[1] "2015-01-29" "2016-01-29" "2016-05-07" "2016-12-25"
> month(z) <- month(z) + 1
> z
[1] NA
No matter how expert you are at writing regular expressions,
it is important to list which sorts of strings you want matched
and which you do not want matched. Saying you want to match
"age" but not "age2" leads to lots of possibilities. Saying how
you want to categorize each string in a vector o
The following base R code does roughly what Bert suggests. It does
no checking that the data is in the format you describe. The
split-by-cumsum
trick is a handy idiom.
# lines <- readLines(yourFile), or, for this example:
lines <- c("#One","X Y Z","1 2 3","4 5 6","",
"#Two", "X Y
, 2016 at 3:54 AM, T.Riedle wrote:
> Something is wrong here. The formula
>
> pmin(psi/VaR,exp(((abs(B)/Bmax)^w2)*log((psi/VaR),2.718182)))
>
>
> provides a time series. Nevertheless, the returned values are incorrect
> and it produces NANs.
>
>
>
> ___
For starters, use 'pmin' (parallel min) instead of 'min'.
substitute(MIN(psi/K14,EXP(((ABS(H14)/peak)^omega)*LN(psi/K14))),
list(MIN=quote(pmin), K14=quote(VaR), ABS=quote(abs),
EXP=quote(exp), LN=quote(log), H14=quote(Bmax),
omega=quote(w2)))
# pmin(psi/VaR, exp(((abs(Bma
Your code looks fine to me. What did t2pd look like?
I tried reproducing the problem in R-3.2.4(Revised) and everything worked
(although the output of str() looked a bit different - perhaps you have an
old version of R)
> df <- data.frame(TE=1:10, TR=101:110,
Command=c("pd_local_abdomen_t2","kne
> dt1[ vapply(dt1, FUN=is.numeric, FUN.VALUE=NA) ]
a c
1 1 1.1
2 2 1.0
...
10 10 0.2
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Apr 29, 2016 at 9:19 AM, Carl Sutton via R-help <
r-help@r-project.org> wrote:
> Good morning RGuru's
> I have a data frame of 575 columns. I wan
I think you are seeing the Moire (interference) pattern arising from
the interaction of lty=4 (dotted/dashed lines) and closely spaced
parallel lines. Use lty=1 (solid lines) to see this.
[BTW, your script did not work because you changed the capitalization
of top and bottom halfway through it.]
table converts its non-factor arguments to factors using the exclude
argument that you supply. If you want the arguments to be handled
differently, then convert them to factors
yourself, in the way you want. E.g.,
> with(df, table(x=factor(x, exclude=1), y))
y
x 1 2 3
2 0 1 0
3 1 0 0
>
You could use transform() instead of [[<- to add columns to your data.frame
so the new columns get transformed they way they do when given to the
data.frame function itself. E.g.,
> dd <- data.frame(X=1:5, Y=11:15)
> str(transform(dd, Z=matrix(X+Y,ncol=1,dimnames=list(NULL, "NewZ"
'data.frame
> i <- seq_len(length(x)-1)
> split(x, cumsum(c(TRUE, (x[i]==0) != (x[i+1]==0
$`1`
[1] 0.144872972504 0.850797178400
$`2`
[1] 0 0 0
$`3`
[1] 0.199304859380 2.063609410700 0.939393760782 0.838781367540
$`4`
[1] 0 0 0 0 0
$`5`
[1] 0.374688091264 0.488423999452 0.783034615362 0.626990428900
0.
This is not a solution but your lsfit attempt
#Error in lsfit(A, b) : only 3 cases, but 4 variables
lsfit(A,b)
gave that error because lsfit adds a column of 1 to
its first argument unless you use intercept=FALSE.
Then it will give you an answer (but I think it converts
your sparse matrix int
(quantile(x,.25),0) - round(quantile(x,0.75),0)"
>> >>
>> >> no matter what the argument of your function, x. Hence
>> >>
>> >> return(paste(...)) will return this exact character string and never
>> >> evaluates x.
>> >>
>>
uot; :(. thanks for the reply
> through
>
> On Tue, Apr 19, 2016 at 10:34 AM, William Dunlap
> wrote:
>
>> > That didn't work Jim!
>>
>> It always helps to say how the suggestion did not work. Jim's
>> function had a typo in it - was that
> That didn't work Jim!
It always helps to say how the suggestion did not work. Jim's
function had a typo in it - was that the problem? Or did you not
change the call to ddply to use that function. Here is something
that might "work" for you:
library(plyr)
data <- data.frame(groupColumn=rep
Since these are color strings, you can use functions in the grDevices
package (other others) to manipulate them. E.g., you can convert them
to various color spaces and perhaps use the mean in one of those
spaces as your 'average color'.
> myColors <- c(One="#FF7C00",Two="#00BF40",Three="#00
Since you only have 3 predictors, each categorical with a small number of
categories, you can use expand.grid to make a data.frame containing all
possible combinations and give that the predict method for your model to
get all possible predictions.
Something like the following untested code.
n
%=>% would have precendence ('order of operations') problems also.
A + B %=>% C
is equivalent to
A + ( B %=>% C)
and I don't think that is what you want.
as.list(quote(A + B %=>% C)) shows the first branch in the parse tree. The
following function, str.language, shows the entire parse tr
Use all.equal instead of identical if you want to gloss over
integer/numeric class differences and minor floating point differences (and
a host of others).
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Apr 11, 2016 at 5:25 PM, Paulson, Ariel wrote:
> Hi Jeff,
>
>
> We are splitting hairs
not like anything short of a year. This seems like I'm
> overlooking something obvious, but I can't get this for the life of me...
>
> Thanks for your time,
> r
>
>
> On Wed, Mar 30, 2016 at 1:03 PM, William Dunlap wrote:
>
>> You said you specified frequency=
You said you specified frequency=96 when you constructed the time
series, but when I do that the decomposition looks reasonable:
> time <- seq(0,9,by=1/96) # 15-minute intervals, assuming time unit is day
> measurement <- sqrt(time) + 1/(1.2+sin(time*2*pi)) +
rnorm(length(time),0,.3)
> plot(decomp
Using the print.level=2 argument to nlm can help track this down.
Also, set options(warn=1) before calling nlm so warnings get printed
as soon as they are generated.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Mar 28, 2016 at 3:32 PM, Rolf Turner
wrote:
> On 29/03/16 09:19, heba eldeeb
; "matrix"
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Mar 24, 2016 at 4:54 AM, Martin Maechler wrote:
> >>>>> William Dunlap via R-help
> >>>>> on Wed, 23 Mar 2016 13:56:35 -0700 writes:
>
> > I don't know what is in R&
done in the multivariate cases.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Mar 23, 2016 at 1:41 PM, Liviu Andronic
wrote:
> On Wed, Mar 23, 2016 at 9:29 PM, William Dunlap wrote:
> > I think the worst aspect of this restriction in poly() is that when
> > you use poly
I think the worst aspect of this restriction in poly() is that when
you use poly in the formula of a model-fitting function you cannot
have any missing values in the data, even if you supply
na.action=na.exclude.
> d <- transform(data.frame(y=c(-1,1:10)), x=log(y))
Warning message:
In log(y)
One technique for dealing with this is called 'multiple imputation'.
Google for 'multiple imputation in R' to find R packages that implement
it (e.g., the 'mi' package).
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Mar 15, 2016 at 8:14 AM, Lorenzo Isella
wrote:
> Dear All,
> A situation
I would suggesting using "" instead of NULL for rz, throughout this code.
(I would also suggest making sure the code can be copied into R without
causing a syntax error before posting the request for help.)
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Mar 11, 2016 at 9:10 AM, Adams, Jean
How much do you care about dealing with misformatted date strings, like
"111-Oct"
or "12-Mai"? Flagging those may be more important than milliseconds of CPU
time.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Mar 9, 2016 at 5:24 PM, Dalthorp, Daniel wrote:
> How about the following, whi
If your start/end pairs are not overlapping you can use findInterval() to
do this
pretty quickly. E.g.,
isInABound <- function (x, low, high)
{
stopifnot(length(low) == length(high))
bounds <- rep(low, each = 2)
bounds[seq(2, length(bounds), by = 2)] <- high
stopifnot(!is.unsorted(
> mod_c <- aov(dv ~ myfactor_c + Error(subject/myfactor_c), data=mydata_c)
>
> summary.lm(mod_c)
> Error in if (p == 0) { : argument is of length zero>
You called the lm method for summary() on an object of class c("aovlist",
"listof"). You should not expect a method for one class to work on an
You can add the argument na.print=" " to print() to make
the missing values print as " " instead of as NA. Some,
but not all print methods support this. E.g.,
> print(c(1,2,NA,4,5,NA), na.print="-")
[1] 1 2 - 4 5 -
> print(matrix(c(1,2,NA,4,5,NA),2), na.print="-")
[,1] [,2] [,3]
[1,]
You can do things like
while ( !is.nan( r <- randomFunction(x) )) {}
# r will be a non-NaN value of randomFunction(x) now
or
for(i in seq_len(1000)) {
if (!is.nan( r <- randomFunction(x))) {
break
}
if (i == 1000) {
stop("no good values of randomFuncti
uot;", c("$VAR", "$20/oz."))
[1] "VAR" "/oz."
Modern Linuxen's tools like sed do not seem to have this requirement.
% echo '$VAR' '$20/oz.' | sed -e 's/$[0-9]*//g'
VAR /oz.
% echo '$VAR' '$2
s NOT generally necessary to wrap it in brackets to remove the special
> meaning unless it would otherwise be at the end of the pattern string.
> --
> Sent from my phone. Please excuse my brevity.
>
> On February 10, 2016 10:10:40 PM PST, William Dunlap via R-help <
> r-help@r-proje
> y
[1] "$1,000.00 " "$1,000.00 " "$1,000.00 " "$2,600.00 " "$2,600.00 "
> gsub("$", "", y)
[1] "$1,000.00 " "$1,000.00 " "$1,000.00 " "$2,600.00 " "$2,600.00 “ #
no change. Why?
"$" as a regular expression means "end of string", which has zero length -
replacing "end
of string" with n
Most of the mass of that distribution is within 3e-100 of 2.
You have to be pretty lucky to have a point in sequence
land there. (You will get at most one point there because
the difference between 2 and its nearest neightbors is on
the order of 1e-16.)
seq(-2,4,len=101), as used by default in cu
You could try pulling some of the repeated subscripting operations,
especially the insertions, out of the loop. E.g.,
values <- observations[,"values"];
date <- observations[,"date"] ;
groups$average <- vapply(seq_len(NROW(groups)), function(i)
mean(values[date >= groups[i, "start"] &
I tried the following but it does not work:
data <- lapply(
paste(("C:/Research3/simulation1/second_gen/pheno_
1000ind_4000m_add_h70_prog_",[1:2],"_",[2:3],".csv",sep=''),
read.csv, header=TRUE, sep=',' )
names(data) <- paste("d", LETTERS[1:3], sep='')
I tried that and R comp
If 'tab' is a data.frame then new.tab <- tab[[mvar]] is a column from that
data.frame, not a data.frame with one column. new.tab <- tab[ , mvar,
drop=FALSE ] will give you a data.frame that you can add to with either of
nvar <- "newName"
new.tab[ , nvar] <- newColumn
new.tab[[nvar]] <-
nks for the discussion,
>
> Stefano
>
>
>
>
>
>
>
> 2016-02-04 18:19 GMT+01:00 William Dunlap :
>
>> The "missingness" of an argument gets passed down through nested function
>> calls. E.g.,
>> fOuter <- function(x) c(outerMissing=missin
The "missingness" of an argument gets passed down through nested function
calls. E.g.,
fOuter <- function(x) c(outerMissing=missing(x), innerMissing=fInner(x))
fInner <- function(x) missing(x)
fInner()
#[1] TRUE
fOuter()
#outerMissing innerMissing
# TRUE TRUE
It is only
Perhaps you can try clustering the output of the Hough transform.
PET::hough() will compute it, given a matrix like
gplots::hist2d(1/temp[,1],temp[,2])$hData. I do not have much experience
here.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Feb 2, 2016 at 3:35 AM, PIKAL Petr wrote:
> De
In addition to the other fine answers, you might find it convenient
to represent the points as complex numbers and use the Arg function
to get the angle (and abs() or Mod() the distance).
> z <- complex(real=0.8660254, imaginary=0.5)
> Arg(z) / base::pi * 180
[1] 30
> Arg(-z) / base::pi *
Did you somehow get a copy of the Splus par function into your
R workspace?
Use the conflicts() function to see if you have something masking par.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Jan 27, 2016 at 1:56 AM, Kim Pilegaard wrote:
> I have a function that calls par, it has work
You could start by sorting the components, by lambda (size) or by mu (mean)
since, if you don't supply starting values, the order of the components is
random. You could use the following to sort normalmixEM's output:
sort.mixEM <- function (x, decreasing = FALSE, ..., by = "lambda")
{
stopifn
> 28 PiB. Storing such a large matrix even on file is not possible.
The ads for Amazon Red Shift say it is possible. E.g.,
Amazon Redshift is a fast, fully managed, petabyte-scale data
warehouse that makes it simple and cost-effective to analyze
all your data using your existing business i
Using column names where you used column numbers would work:
example <- data.frame(
check.names = FALSE,
Nuclei = c(133L, 96L, 62L, 60L),
`Positive Nuclei` = c(96L, 70L, 52L, 50L),
Slide = factor(c("A1", "A1", "A2", "A2"), levels = c("A1", "A2")))
aggregate(example["Nuclei"], by=ex
Note that the expression
x[1] <- 10
is equivalent not to
`[<-`(x, 1, value=10)
but to
x <- `[<-`(x, 1, value=10)
so there is no conflict between your two expressions.
Saying
c(1,2,3) <- `[<-`(c(1,2,3), 1, value=10)
is not allowed because there is no name to assign something to.
There
Is the following function, myFormula(), what you are looking for?
myFormula <- function(stem, env = parent.frame()) {
eval(bquote(.(as.name(paste0("Disch.",stem))) ~
.(as.name(paste0("Base.",stem))),
list(stem=as.name("rolling"))), envir=env)
}
str(myFormula("myStem"))
#Class "formula" languag
> getwd()
[1] "/Users/johnbeyer"
> dir
function (path = ".", pattern = NULL, all.files = FALSE, full.names =
FALSE,
recursive = FALSE, ignore.case = FALSE, include.dirs = FALSE,
Note what a difference the parentheses after the function name make. With
parentheses,
often with ar
helper_fun))
> lapply(df2, levels)
$x1
[1] "[0,2]" "(2,3]" "(3,4]" "(4,5]" "(5,6]" "(6,7]" "(7,14]"
$x2
[1] "[1,4]" "(4,5]" "(5,6]" "(6,7]" "(7,8]" "(8,9]&
Don't use vapply() here - use lapply() instead and then leave cut's output
alone.
vapply() will combine its outputs to create a character matrix and
data.frame will pull apart the character matrix into its columns. Skipping
the matrix intermediary solves
lots of issues.
Bill Dunlap
TIBCO Softwar
I tend to use bquote, as in
x_label <- bquote(bold(species) ~ (italic(N1)))
plot(1:10,main=bquote("This is the expression for" ~ .(x_label) * "!"))
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Jan 11, 2016 at 7:59 AM, Evan Cooch wrote:
> Suppose I've specified that the xlab for a
The following shows how to get different colors for most features of a
scatterplot:
plot(1:11,log(1:11),ann=FALSE,axes=FALSE,col="pink",pch=16)
box(col="gray")
title(xlab="X Axis Label", col.lab="light blue")
title(ylab="Y Axis Label", col.lab="light green")
axis(side=1, at=c(2,3,5,7,11), lab=as.e
.table(text=txt, header=TRUE))
'data.frame': 2 obs. of 2 variables:
$ ColA: int 101 201
$ ColB: int 102 202
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Jan 6, 2016 at 1:26 PM, Andy Schneider
wrote:
>
> Hi Bill -
>
> Thanks so much! This was actually
You may have read in your data incorrectly - a column you expected to be
numeric was not recognized as such so it was read in a character and then
converted to a 'factor'.
FAQ 7.10 tells how to work around the problem
https://cran.r-project.org/doc/FAQ/R-FAQ.html#How-do-I-convert-factors-to-numer
You did not show the structure of your datasets (with, e.g.,
dump(c("datafile1","datafile2"),file=stdout())) nor what your call to
merge() was. However, it may be that you did not use the by.x and by.y
arguments to merge() to specify which columns to match.
txt1 <- "date1 xva
rame, stratified by ARM. Using a suggestion provided by both William
> Dunlap and Rolf Turner, I have written the code below which fails with the
> error:
> Error in match.fun(FUN) : 'sqrt(var(x))' is not a function, character or
> symbol
>
>
> code:
> mySD <-
What do you hope to do with this data while it is in R?
E.g., do you want to plot it, fit a model to it, to select a few
rows or columns from it, sort it, summarize lots of small subsets
of it, or something else?
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Dec 28, 2015 at 1:39 PM, Mark F
by(dataFrame, groupId, FUN) applies FUN a bunch of data.frames (row subsets
of the dataFrame input). mean() returns NA for data.frames. You could use
FUN=colMeans if you wanted column means or FUN=function(x)mean(colMeans(x))
or FUN=function(x)mean(unlist(x)) if you wanted some version of a grand
Do you know what sort of S+ objects do not survive the
dump-from-S+-source-into-R procedure unscathed? E.g., do you have S4
classes that don't make it or certain classes
of data objects?
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, Dec 26, 2015 at 3:17 AM, Kim Pilegaard wrote:
> I have
Since you only want the diagonal of the correlation matrix, the following
will probably
do the job using less memory. The mapply versions works on the data.frames
you supplied, but will not work on matrices - be careful not to conflate
the two classes of data objects.
> vapply(colnames(df1), fu
What answer do you want for the following data?
x <- c(2,2,3,4,4,4,4,5,5,5,3,1,1,0,0,0,1,1,1)
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Dec 22, 2015 at 11:34 PM, Makram Belhaj Fraj <
belhajfraj.mak...@gmail.com> wrote:
> Dear colleagues
> i need your generous help to solve the fo
Have you considered thin plate splines, which are a natural
extension of one-dimensional splines to higher dimensional
spaces, but are not piecewise polynomials? The Tps function
in the fields package will fit a thin plate spline to irregularly
spaced data.
Bill Dunlap
TIBCO Software
wdunlap tib
You can fit a spline to the points and evaluate the derivative
of the fitted spline where you want. The built-in splines package
has functions for this, as do other packages on CRAN.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Dec 18, 2015 at 1:09 AM, BARLAS Marios 247554
wrote:
> Hi
is there a way to achieve that in
> IronPython, i am open for both options :)
>
> Thanks again for reply.
>
> On Wed, Dec 16, 2015 at 9:26 PM, William Dunlap wrote:
>>
>> Your code works in TERR under Spotfire, but since Spotfire deals with
>> rectangular data
mple 1: f(x) = x^2 , xmin=-4, xmax=4
> ### ===
>
>
>
> showIntegral.med(f=function(x)x^2, xmin=-4, xmax=4, n=16)
>
>
>
> with many thanks
> steve
>
> On Sat, Nov 28, 2015 at 1:11 PM, William Dunlap wrote:
>>
>
any help from TIBCommunity, but my code
> worked in RStudio.
>
> On Tue, Dec 15, 2015 at 4:41 PM, William Dunlap wrote:
>>
>> It looks like you are calling TERR from Spotfire. The Spotfire/TERR
>> interface
>> can only pass TERR data.frames (eq. to Spotfire tables) b
It looks like you are calling TERR from Spotfire. The Spotfire/TERR interface
can only pass TERR data.frames (eq. to Spotfire tables) back to Spotfire and
XMLInternalDocuments cannot be columns of data.frames (in neither TERR nor
R).
You should contact TIBCO support and/or participate in the foru
Hadley wrote
> * all.equal is suboptimal because it returns TRUE or a text string
That feature works ok with stopifnot():
> stopifnot(all.equal("one", 1))
Error: all.equal("one", 1) are not all TRUE
and I suppose stopifnot could be enhanced to print the text strings that
all.equal() returns so
> as Bill or Jeff explained, "the empty set is always true"
My wording was that any(logical(0)) is FALSE because "there are no
TRUEs in logical(0)".
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, Dec 12, 2015 at 1:54 AM, Martin Maechler
wrote:
>> Henrik Bengtsson
>> on Fri,
In addition to the other fine replies, you should also know that
kmeans's results
depend on the relative scales of the data columns (since it is based
on distances
between points). Your x and y have quite different scales so the distance is
essentially determined only by the differences in the var
stats::sigma was added to R recently. It is is R-devel now, I don't know
about yesterday's R-3.2.3.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Dec 11, 2015 at 11:05 AM, Duncan Murdoch
wrote:
> On 11/12/2015 1:44 PM, Rich Shepard wrote:
>>
>> Trying to update package pbkrtest faile
The reason is probably that
any(logical())
and
any(!logical())
return FALSE (there are no TRUEs in logical(0)).
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Dec 11, 2015 at 5:38 AM, Dario Beraldi wrote:
> Hi All,
>
> I'd like to understand the reason why stopifnot(logical(0) == x)
You can use the approx() function (in that stats package) to put
5 equally spaced times between your high and low water times.
E.g., in the following 'tmp' will be your interpolated times, in seconds
since 1970, which I clumsily convert to POSIX times (I can never
remember how to deal with time zon
by() calls FUN with a data.frame as the argument. summary(), sum(), etc.
have methods that work on data.frames but sd() and mean() do not.
aggregate() calls its FUN with each column of a data.frame as the argument.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Dec 8, 2015 at 3:08 PM, Dim
Change the order of adding the themes from
theme(axis.ticks=element_blank()) + theme_bw()
to
theme_bw() + theme(axis.ticks=element_blank())
because theme_bw() adds axis.ticks = element_line(colour = "black").
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Dec 7, 2015 at 7:10 AM, Brian
I cannot see either your data or your picture. Does the following dataset
along with your ggplot command give a similar picture?
TS <- data.frame(Well=paste0("Well",rep(1:21,each=3)),
LOCATION=paste0("Loc",rep(LETTERS[1:7],each=9)),
Date=as.POSIXct(paste(sep="-", 2015, 10, rep(c(11,18,25),21))),
With a data.frame sorted by id, with ties broken by date, as in
your example, you can select rows that are either the start
of a new id group or the start of run of consecutive dates with:
> w <- c(TRUE, diff(uci$date)>1) | c(TRUE, diff(uci$id)!=0)
> which(w)
[1] 1 4 5 7
> uci[w,]
id date
Every package has in it, after it is installed, a character object
called ".packageName" containing its name. It is not exported from
the package. Functions in your package can refer to it as just
.packageName.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Dec 3, 2015 at 5:28 PM, Murat T
In your
> ordinalmodel <- polr(eduattain ~ dadedu, data = workdataset, weights =
> "SPFWT0", Hess = TRUE)
take the quotation marks off of SPFWT0. Like the subset argument, weights
is a literal expression, evaluated in the context of the data argument, not a
character string naming a column in the
The short answer is to provide an argument that your function does not
pass through substitute(), so standard evaluation takes place. E.g., change
SmallFn <- function(v,d) {
column <- deparse(substitute(v))
d[,column]
}
to
SmallFn <- function(v, d, column=deparse(substi
Your right <- (1:n)*dx mean that your leftmost rectangle's left edge
is at 0, but you want it to be at -4. You should turn this into a function
so you don't have to remember how the variables in your code depend
on one another. E.g.,
showIntegral <- function (f, xmin, xmax, n = 16)
{
curve(
rpois(100, 5) gives a different set of random numbers each time it is
called, so if you want repeatable results compute it once and use its
value in the calls to plot. E.g.,
r <- rpois(100, 5)
plot(table(r), type="h", col="red", lwd=10, main="hello")
Bill Dunlap
TIBCO Software
wdunlap tibco.
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