R-Help,
I discovered a mis-feature is ghostscript, which is used by the bitmap
function. It seems that specifying file names in the form ~/abc.png
rather than /home/directory/abc.png causes my GS to crash when I open
the bitmap device on my Linux box.
The easiest solution would seem to be to
R-help,
Given a simple linear model, say lm(x ~ y + z), I would like to remove
model terms that are factors with only one level. Thus, if 'z' were a
factor with only one level, lm(x ~ y + z) becomes lm(x ~ y + 1).
Likewise, if both 'y' and 'z' are one-level factors, then the resulting
calculation
Hello,
I've decided to take the leap and try my hand at the lattice package,
though I am getting stuck at what one might consider a trivial problem,
plotting text at a point in a graph. Apologies in advance if (that) I'm
missing something extremely basic.
Consider in base graphics:
plot(1:10)
Hello,
I'm looking to calculate a 95% confidence interval about my estimate for
a sample's weighted mean, where the calculated confidence interval would
equal the t-test confidence interval of the sample in the case when all
of the weights are equal.
My initial thought was to simply implement a
Hello,
I have a system command that occasionally fails and writes output to
standard error, which R will print to the screen when ignore.stderr =
FALSE.
For example:
system(BadCommand)
sh: line 1: BadCommand: command not found
I would like to know if the above command fails, and can presumably
Moreno,
As much of my processor time is often spent doing basic linear algebra
operations (matrix inversion, quadratic programming, etc), I recently
recompiled R using a BLAS implementation (ATLAS) tuned for parallel
processing. The speed improvement for linear algebra operations was
significant
I was going to suggest sqlUpdate in RODBC, but it looks like that
function also uses the UPDATE command repeated nrow times.
A second strategy that I generally prefer because it does not require
RODBC (as much) and better supports transaction control is to first
create a temporary table with the
Hadley,
There are several reasons that running one large load and one large
update would be significantly faster than thousands of individual
updates.
First, the time it takes to execute a query does not grow linearly with
the size of a query. That is, the statement: SELECT TOP 100 * FROM
table
Not sure I completely followed (especially what a Knockback is), but
I'll give this a try.
If you want to precisely define the convolution of multiple attacks, you
would want to use a language that utilizes symbolic integration, which R
currently does not support (e.g. see Mathematica). That
Actually, if you're going to go out transform the data, why not use a
Fourier transform and take advantage of the fact that F[f*g] = F[f]F[g]?
That's the easiest way in my mind to do multiple convolutions of fitted
data. Taylor series are neither good at modeling Schwartz functions (a
rapidly
Also check out the GDD package created by Simon Urbanek if bitmap does
not fit your needs. On some systems, bitmap is slow or produces an
inferior quality plot, in part because anti-aliasing is not supported
(at least on my system). GDD, however, produces excellent anti-aliased
graphs using the GD
code for ideas.
HTH,
Robert
-Original Message-
From: Laurent Gautier [mailto:[EMAIL PROTECTED]
Sent: Thursday, March 02, 2006 3:20 AM
To: McGehee, Robert
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] prepared query with RODBC ?
Well, I may not have been clear enough. My experience
R-help(ers),
Does anyone know of an R function available for calculating a confidence
interval for a Spearman correlation? If no such resource is available,
is using the confidence interval from a Pearson correlation a reasonable
proxy if the vectors come from normal distributions (i.e. likely
I may be misunderstanding you, but why can't you execute a prepared
query the same in RODBC as you would directly on your SQL server? In
Microsoft SQL server, for instance, I would just set up an ADO
application and set the Prepared and CommandText properties before
running the query.
Here is an
You could also use a function like this to transform a number x into your
beautiful notation, where x is the number and digits is the number of digits
you would like to see after the decimal. Then you could use the axis and labels
syntax suggested by Iain, as such:
sciNotation - function(x,
I would compare the Shannon entropy of your test vector with the entropy
of your expected probability distribution to see if they are close. That
is, if you're binary probability distribution is half 1 and half -1,
then if your string is long you would expect about half the numbers in
your vector
Hello,
Why not just copy and paste the examples from the help pages? No need to
type anything.
However, if you'd like to run the entire help section for a function,
check out the example function, i.e.:
example(plot)
Robert
-Original Message-
From: Mark Leeds [mailto:[EMAIL PROTECTED]
I would throw a tolower() around s1 and s2 so that 'canada' matches with
'CANADA', and perhaps consider using a Levenshtein distance rather than
the longest common subsequence.
An algorithm for Levenshtein distance can be found here (courtesy of
Stephen Upton)
Jan T. Kim wrote:
There is a draft R Coding Convention available at
http://www.maths.lth.se/help/R/RCC/
which may be useful for finding a style that is good because it is
widely used and therefore familiar to a large number of readers.--
However, as the author Henrik Bengtsson
Hello,
If we define IRR implicitly such that NPV(C, IRR) = 0, then we can just
write an IRR function that finds the zeros of the NPV function. Here are
two such functions:
NPV - function(C, r) {
sum(C / (1 + r) ^ (seq(along = C) - 1))
}
IRR - function(C) {
uniroot(NPV, c(0, 1), C =
It sounds like you want `try` with the argument `silent = TRUE`. This
will allow you to keep running your program without errors. If you want
to check if the line had an error, you can error control by seeing if
the class of the resulting object is try-error. For example, let's say
I wanted to
I'm trying to understand why I can rbind but not cbind dataframes to
NULLs.
For 'cbind' ('rbind'), vectors of zero length (including 'NULL')
are ignored unless the result would have zero rows (columns), for
S compatibility. (Zero-extent matrices do not occur in S3 and are
Spencer,
On the first iteration of your simulation, all of the Qp.t + z.t 0, so
you're adding a vector of rep(4.5, 2) to a random distribution with
mean -0.1. So one would expect on iteration 2, your mean would have
dropped by about 0.1 (which it does). This process continues until about
the
Hello all,
I'm trying to setup a simple construct that will email me all errors,
warnings, and output of a sample R script. My initial thought was to use
sink() and tryCatch around the script such that at the end of the script
any output would be mailed to me, i.e.:
sinkFile - tempfile()
/messages to the same file to preserve the order of
the messages/output, if this is possible.
Robert
-Original Message-
From: Thomas Lumley [mailto:[EMAIL PROTECTED]
Sent: Wednesday, July 27, 2005 10:37 AM
To: McGehee, Robert
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Redirecting Messages
setMethod([,foo,function(x,i, j,
.,drop=TRUE)callGeneric([EMAIL PROTECTED],i,j,drop=drop) )
^^^
You have a typo. Use ... instead of . (note ?[).
Best,
Robert
Robert McGehee
Quantitative Analyst
Geode Capital Management, LLC
53 State Street, 5th Floor | Boston,
a - data.frame(a = 1:2, b = c(a, b), c = I(c(a, b)))
a - a[ , !sapply(a, class) %in% factor]
-Original Message-
From: E. Michael Foster [mailto:[EMAIL PROTECTED]
Sent: Friday, June 17, 2005 10:58 AM
To: r-help@stat.math.ethz.ch
Subject: [R] drop elements of vector by class
i'm trying
I'm seeing some inconsistent behavior when re-assigning values in a data
frame. The first assignment turns all of the 0s in my data frame to 2s,
the second fails to do so.
df1 - data.frame(a = c(NA, 0, 3, 4))
df2 - data.frame(a = c(NA, 0, 0, 4))
df1[df1 == 0] - 2 ## Works
df2[df2 == 0] - 2
Jim,
I had some of the same difficulties. The NIR data frame consists of a
column of y variables and a matrix of X variables (and until looking at
this dataset, I had not realized that data frames could hold matrices).
So, after consulting the R-help sages, I turned by data into an
identical
Because rank and order are (supposed to be) inverses of each other.
For example:
a - c(3, 1, NA)
a[order(a[rank(a)])]
[1] 3 1 NA
a[rank(a[order(a)])]
[1] 3 1 NA
BUT
a[order(a[rank(a, na.last = FALSE)])]
[1] 1 NA 3
a[rank(a[order(a)], na.last = FALSE)]
[1] 1 NA 3
-Original
This will do it.
a - array(1:12, c(4, 3))
d - function(a,b,c) {a+b+c}
apply(a, 1, function(x) do.call(d, sapply(x, list)))
[1] 15 18 21 24
-Original Message-
From: BJ [mailto:[EMAIL PROTECTED]
Sent: Wednesday, May 18, 2005 9:10 AM
To: r-help@stat.math.ethz.ch
Subject: [R] applying a
You could try temporarily switching the list to an array, then just run
an apply on rows and columns:
apply(array(do.call(cbind, z), c(dim(z[[1]]), length(z))), c(1, 2),
mean)
Best,
Robert
-Original Message-
From: David Kane [mailto:[EMAIL PROTECTED]
Sent: Tuesday, May 10, 2005 10:04
Hello,
For reasons I don't understand, data() imports CSV (Comma-Separated
Values) as if they were delimited by semicolons instead of commas. (Are
semicolon-separated Comma-Separated-Value files common somewhere?) Given
that this is the case, if I choose to put comma-delimited CSV files in
my
Hello,
I'm playing around with the PLS package and found a data set (NIR) whose
structure I don't understand. Forgive me if this is a stupid question,
as I feel like it must be since I am less experienced with aspects of
modeling.
My problem, the pls NIR data frame does not seem to be a typical
Sure. Here are three ways. Using pmax() is probably the most elegant.
a - function(t) {
pmax(t + 1, 0)
}
OR
a - function(t) {
sapply(t + 1, max, 0)
}
OR
a - function(t) {
t - t + 1
t[which(t 0)] - 0
t
}
-Original Message-
From: Jorge Ahumada
This short function capitalizes the first letter of each word in a
character string, which is what I think you want.
capitalize - function(x) {
x - strsplit(x, )
for (i in seq(along = x)) {
substr(x[[i]], 1, 1) - toupper(substr(x[[i]], 1, 1))
}
sapply(x, function(z)
I doubt it's possible to use Recall inside a sapply like this. sapply
(and lapply) make a copy of the function (FUN - match.fun(FUN)) before
passing it inside an internal function (rval - .Internal(lapply(X,
FUN)). So, while I'm not precisely sure how Recall is coded up, I bet
that once it is
Hello all,
When I view or print the below plot on my Linux machine under R 2.0.1 I
see a nice thick solid and dashed line with a legend. However, while the
lines are distinguishable, the legend is not. That is, the short (solid)
line next to line1 and the short (dashed) line next to line2 seem to
Use substitute() instead of expression(); choose to use either MYPLOT or
myplot because they are different variables; and use parentheses around
your function arguments instead of braces.
-Original Message-
From: Ronny Klein [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 09, 2005 12:22
This isn't an environment problem. Assigning something to a get call
doesn't make any sense. Use assign.
a - 5
get(a) - 10
Error: couldn't find function get-
And from the ?assign help page, you can pick what environment you want
to make the assignment. Just pick the parent environment.
x - .001
x
[1] 0.001
sub(+0, , x)
[1] .001
Of course, that means you'll be storing this number as a character, but
if you're looking to format how the number is printed, that's probably
what you want.
-Original Message-
From: Laura Holt [mailto:[EMAIL PROTECTED]
Sent: Monday,
Not without extending character to another class. If you try to define
this + method using S4, you get a proper error indicating that
addition over characters is not allowed.
setMethod(+, signature(character, character),
function(e1,e2) paste(e1,e2, sep = ))
Error in
?coef should do the trick.
-Original Message-
From: mingan [mailto:[EMAIL PROTECTED]
Sent: Thursday, February 24, 2005 5:12 PM
To: r-help@stat.math.ethz.ch
Subject: [R] help: how to get coefficients from the result
I am using timereg and Survreg packages
below is the output, but
Hi all,
I develop and print from both Windows and Linux, and am seeing some
printing inconsistencies first described about a year and a half ago by
Andy Liaw (see below). Specifically, the line widths on my windows plots
are about 5 times smaller than that on Linux, and my windows printouts
do not
Why not just save your current global environment and reload it after
running the offending code? See ?save.image to save, then rm(list =
ls()) to clear the global environment.
If a program tries to get or assign something to a global environment,
then either you rewrite the program to do
Perhaps an easier way would be to throw away the offending text at the
end of the strings, rather than matching all possible numeric
formulations at the beginning of the string, that is:
sub(\\.*[[:alpha:]]+$, , x)
Easier to read, if nothing else, and it allows for 2e-7 as a valid
number. This
I was unaware until recently that partial matching was used to index
data frames and lists. This is now causing a great deal of problems in
my code as I sometimes index a list without knowing what elements it
contains, expecting a NULL if the column does not exist. However, if
partial matching is
.
Thanks,
Robert
PS. None of this applies to partial matching of function arguments, as
this is certainly widely used.
-Original Message-
From: Huntsinger, Reid [mailto:[EMAIL PROTECTED]
Sent: Thursday, January 27, 2005 10:15 AM
To: 'McGehee, Robert'; r-help@stat.math.ethz.ch
Subject: RE: [R
get() doesn't take a list. You need to wrap it in an lapply so that it
grabs the dataframes individually and then wraps it into a list for
do.call().
The easy way, however, that doesn't involve do.call or get is to simply
construct the proper text rbind query and then run it with eval.
Certainly this _is_ possible. However, there are no built-in functions to do
just this (that I know of). Here's a function that attaches like you say (but
still makes a local copy)
attachslot - function(x) {
xname - substitute(x)
sl - names(getSlots(class(x)))
slotnames - paste(sl,
The method below looks for arguments of sec and numeric. Sounds like
you want to pass in sec and missing (the class of missing data) and
get a default behavior. So in addition to the method you already wrote,
try this one as well:
setMethod(signif, signature(x = sec, digits = missing),
Hello,
?as.character says that the as.character function is a generic with
usage: as.character(x, ...). So, I want to create an S4 object with an
as.character method following the above usage, but I get the below error
telling me that ... isn't in the generic for as.character.
setClass(tmp,
the point entirely. Any clarification is greatly appreciated.
Thanks,
Robert
-Original Message-
From: Prof Brian Ripley [mailto:[EMAIL PROTECTED]
Sent: Thursday, January 13, 2005 5:58 PM
To: McGehee, Robert
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] as.character methods
You seem
I would suggest reading the posting guide,
(http://www.r-project.org/posting-guide.html) and give a reproducible
example, with the error message that you received. As is, I have no idea
what you are doing here, and certainly cannot run this code. You use
... as an argument to your functions (why I
Adam, I'm about to encounter a very similar problem, so would be curious
if you find a good solution, although your thoughts with reg.finalizer
look promising. The thoughts I had sketched out for tackling this
problem:
1) Have all ancillary resources freed when the package is detached. The
new()
Is it correct (by its lack of mention in the R-Language Definition
Manual) that it is impossible to create a user-defined unary operator?
Ex: (This doesn't work, but it's an example of what I'm looking for)
%PLUSONE% - function(x) x + 1
%PLUSONE% 2
[1] 3
And if the above is impossible, am I
It sounds like what you want is a rudimentary spell-checker whose word
is the input name, and whose dictionary is an array of your database
names. Spell checking rules are designed to find missing repeats,
transposed letters, extra letters... precisely the reasons you're not
matching your names to
R-help,
I'm the primary developer for an increasingly large R package with over
three thousand lines of code. Unfortunately, do the complexity of the
code, I sometimes am required to change several interoperating parts of
the package before testing for bugs and performance. And sometimes
unnoticed
R users, I am having a problem with the modulus operator for large
numbers as follows,
a - 2
n - 561
## n is the first Carmichael number, so by Fermat's Little Theorem the
below should equal zero.
(a^(n-1) - 1) %% n
[1] 2.193172e+152
## Seems that R and Fermat disagree
## Also,
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