I have no experience with this but a Google search came up with
CRAN support.BWS
See https://cran.r-project.org/web/packages/support.BWS/index.html
On Wed, Apr 13, 2022 at 4:37 PM Dr Cazhaow Qazaz
wrote:
> Hi All,
>
> Any recommendations for a R package to perform MaxDiff analysis?
>
> Thank y
Hi All,
Any recommendations for a R package to perform MaxDiff analysis?
Thank you,
Cazhaow Qazaz
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Technically "opening" a file does not involve reading it into memory at all, so
any limits would arise from the OS.
As for "reading" the file, the amount of virtual memory [1] is the key factor,
but different file formats can require more or less overhead memory to parse
the data. Also, differe
Hello, I´d like to ask you if the max size of a file that R can open or
import to work depends of the memory RAM?? or has other restriction???
Thank you very much
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list -- To
> On 20 Jan 2018, at 07:53 , Suharto Anggono Suharto Anggono via R-help
> wrote:
>
> Extremes.Rd, that documents 'max' and 'pmax', has this in "Details" section,
> in the paragraph before the last.
> By definition the min/max of a numeric vector containing an NaN is NaN,
> except that the
Extremes.Rd, that documents 'max' and 'pmax', has this in "Details" section, in
the paragraph before the last.
By definition the min/max of a numeric vector containing an NaN is NaN, except
that the min/max of any vector containing an NA is NA even if it also contains
an NaN.
--
> Michal Burda
> on Mon, 15 Jan 2018 12:04:13 +0100 writes:
> Dear R users, is the following OK?
>> max(NA, NaN)
> [1] NA
>> max(NaN, NA)
> [1] NA
>> pmax(NaN, NA)
> [1] NA
>> pmax(NA, NaN)
> [1] NaN
> ...or is it a bug?
> Documentation
Dear R users,
is the following OK?
> max(NA, NaN)
[1] NA
> max(NaN, NA)
[1] NA
> pmax(NaN, NA)
[1] NA
> pmax(NA, NaN)
[1] NaN
...or is it a bug? Documentation says that NA has a higher priority over
NaN.
Best regards, Michal Burda
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___
On 5/5/2014 3:42 PM, William Dunlap wrote:
Have you looked at the width.cutoff and nlines arguments to deparse?
width.cutoff controls how long a line can be and it quits producing
output (saving time and space) after nlines of output are produced.
You want essentially the following
f0 <- func
Have you looked at the width.cutoff and nlines arguments to deparse?
width.cutoff controls how long a line can be and it quits producing
output (saving time and space) after nlines of output are produced.
You want essentially the following
f0 <- function(x, maxChar=20) deparse(x, width.cutoff=ma
Is there a standard or or a standard utility to limit the size of
deparse(substitute(.))?
Below please find an example of the problem plus one solution.
If another solution already exists, I might prefer to use it.
Thanks,
Spencer
##
## Problem
##
deparse.x0 <-
Homework? We don't do homework here...
-- Bert
On Wed, Aug 28, 2013 at 10:02 AM, Edoardo Baldoni wrote:
> Dear R-help,
>
> I would like to ask you how to find the critical points of a function of
> several variables such as the following one:
>
> x = seq(-10,10,0.2)
> y = seq(-10,10,0.2)
> z
Dear R-help,
I would like to ask you how to find the critical points of a function of
several variables such as the following one:
x = seq(-10,10,0.2)
y = seq(-10,10,0.2)
zfunc = function(x,y) x^2 + y^2
z = outer(x,y,zfunc)
persp(x,y,z,theta = 30,phi=1,ticktype='detailed')
Thanks
Edoardo
Technically this is correct for raw R functionality.
in practice various modules impose their own limits on variables so you have
to check.
For example the coxreg package truncates all variables to 16 characters.
for example the test below
> res
Call:
coxreg(formula = Surv(vtime, vstatus) ~
abcde
Dear all,
How can I calculate (in runtime) the max width and height I can use in order to
avoid the content in graphics window appears truncate?. In Windows I can avoid
it (using windows(width=my.width, height=my.height, rescale="fixed") because
the value "fixed" in rescale makes the scroll bar
HI,
Using
c11<- 0.01
c12<- 0.01
c1<- 0.10
c2<- 0.10
One possible problem is that:
dim(res5)
#[1] 513 20
res6<-aggregate(.~m1+n1+m+n,data=res5[,c(1:6,9:12,21:24)] ,max)
#Error in `[.data.frame`(res5, , c(1:6, 9:12, 21:24)) :
# undefined columns selected
A.K.
___
16001 5500519
> #20 ok 2010-01-15 06:19:59 40193.51 64 17016001 5500519
>
> #If you need a rolling max,
> library(zoo)
> rollapply(A$lig,2,max,align="right",fill=NA)
> #[1] NA 64 64 44 32 22 13 4 8 8 0 45 64 64 64 64 64 64 51 64
>
>
> A.K.
>
>
>
y(zoo)
rollapply(A$lig,2,max,align="right",fill=NA)
#[1] NA 64 64 44 32 22 13 4 8 8 0 45 64 64 64 64 64 64 51 64
A.K.
- Original Message -
From: zuzana zajkova
To: Rui Barradas
Cc: r-help@r-project.org
Sent: Tuesday, February 12, 2013 6:53 PM
Subject: Re: [R] M
Hi,
sorry for not useable data...
When I tried to use the dput function to the dataframe let's call it A, had
some problems with the time and clock variables... however I attach an
other dataframe B, which is created by merging A with other dataframe. The
B dataframe has a little different struct
Hello,
Your data example is a mess. Can't you please use ?dput to post it?
Supposing your data is named 'dat', use
dput(head(dat, 20)) # Paste the output of this in a post
Hope this helps,
Rui Barradas
Em 12-02-2013 15:30, zuzana zajkova escreveu:
Hello,
I would like to ask you for help
Hello,
I would like to ask you for help. I have quite a big dataframe (119 313
rows), this is a part of it:
"jul" "ind" "time" "secs" "geo" "act" "lig" "date.x" "clock" "h" "m" "s"
"d" "mo" "y" "dtime" "land" "date.y" "sriseIC" "ssetIC" "dssetIC" "dsriseIC"
"1207" 14628 5500519 2010-01-19 15:14:5
Hi,
Try this:
summary(x,digits=max(5))
# Min. 1st Qu. Median Mean 3rd Qu. Max.
# 1.0 6901.5 13802.0 13802.0 20702.0 27603.0
A.K.
- Original Message -
From: Sam Steingold
To: r-help@r-project.org
Cc:
Sent: Friday, September 28, 2012 12:14 PM
Subject: [R] max & sum
On 28/09/2012 12:14 PM, Sam Steingold wrote:
why does summary report max 27600 and not 27603?
> x <- c(27603, 1)
> max(x)
[1] 27603
> summary(x)
Min. 1st Qu. MedianMean 3rd Qu.Max.
16902 13800 13800 20700 27600
Because you asked for 3 digit accuracy. See ?summ
why does summary report max 27600 and not 27603?
> x <- c(27603, 1)
> max(x)
[1] 27603
> summary(x)
Min. 1st Qu. MedianMean 3rd Qu.Max.
16902 13800 13800 20700 27600
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.chi
Hi
>
> On 2012-05-15 08:36, Melissa Rosenkranz wrote:
> > Here is an R problem I am struggling with:
> > My dataset is organized like this...
> >
> > subject sessionvariable_x variable_y
> > 01 11
> > 01 1
On 2012-05-15 08:36, Melissa Rosenkranz wrote:
Here is an R problem I am struggling with:
My dataset is organized like this...
subject sessionvariable_x variable_y
01 11
01 12
01 1
Here is an R problem I am struggling with:
My dataset is organized like this...
subject sessionvariable_x variable_y
01 11
01 12
01 13
01 2
On 01/11/2012 12:01 PM, Rui Esteves wrote:
> Is there any constant that represents the maximum value of an integer?
Yes, there is (assuming you refer to the 'integer' type). See ?.Machine.
> .Machine$integer.max
[1] 2147483647
> as.integer(2147483647)
[1] 2147483647
> as.integer(2147483648)
[1]
rmx wrote
>
> Hi.
> Is there any constant that represents the maximum value of an integer?
> If I need to setup by myself what is the maximum value?
>
?.Machine
i.e.
.Machine$integer.max
Berend
--
View this message in context:
http://r.789695.n4.nabble.com/Max-value-of-an-integer-tp428495
Hi.
Is there any constant that represents the maximum value of an integer?
If I need to setup by myself what is the maximum value?
Best,
Rui
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PLEASE do read the posting guid
Thanks for these various tips.
Sarah, this is not a howework, but a simplified dataset speecificly for
this question.
Laura
.
2011/11/14 Dennis Murphy
> Groupwise data summarization is a very common task, and it is worth
> learning the various ways to do it in R. Josh showed you one way to
> u
Groupwise data summarization is a very common task, and it is worth
learning the various ways to do it in R. Josh showed you one way to
use aggregate() from the base package and Michael showed you one way
of using the plyr package to do the same; another way would be
ddply(df, .(Patient, Region),
I took a stab at this using ddply() from the plyr package. How's this
look to you?
x<- textConnection("Col Patient Region Score Time
11 X19 28
21 X20 126
31 X22 100
41 X25 191
52 Y121
62 Y
Hi Laura,
You were close. Just use range() instead of min/max:
## your data (read in and then pasted the output of dput() to make it easy)
dat <- structure(list(Patient = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 6L,
Hi Laura,
This looks suspiciously like homework. Nonetheless, you may wish to
check out ?cbind.
Sarah
On Mon, Nov 14, 2011 at 11:10 AM, B Laura wrote:
> dear R-team
>
> I need to find the min, max values for each patient from dataset and keep
> the output of it as a dataframe with the following
dear R-team
I need to find the min, max values for each patient from dataset and keep
the output of it as a dataframe with the following columns
- Patient nr
- Region (remains same per patient)
- Min score
- Max score
Patient Region Score Time
11 X19 28
21
On Jul 16, 2011, at 9:55 AM, Thomas Chesney wrote:
I know there's a really easy way to do this but I just can't track
it down. I experimented with various apply functions but couldn't
get it quite right.
I have a matrix like this:
1, 16
1, 23
1, 21
1, 6
1, 25
2, 4
2, 17
2, 45
2, 11
2, 20
I know there's a really easy way to do this but I just can't track it down. I
experimented with various apply functions but couldn't get it quite right.
I have a matrix like this:
1, 16
1, 23
1, 21
1, 6
1, 25
2, 4
2, 17
2, 45
2, 11
2, 20
and I'd like to find the max value in Column 2 (or the in
On 13.07.2011 14:42, Rilana Prenger wrote:
Dear all,
I have a question regarding the output of the coxph function. What does the
'max possible' exactly mean in the output below? Many thanks.
coef exp(coef) se(coef) robust se z Pr(>|z|)
smocc_zyban -0.43840.6451
Dear all,
I have a question regarding the output of the coxph function. What does the
'max possible' exactly mean in the output below? Many thanks.
coef exp(coef) se(coef) robust se z Pr(>|z|)
smocc_zyban -0.43840.6451 0.86670.9473 -0.4630.644
self
Thanks
I eventually tracked down the problem to something unrelated to this
question (one out of the millions of character strings happened to be
"NA" by chance, which of course was parsed as a missing value, breaking
the code a long way downstream.)
Richard
On 28/09/2010 04:01, Michael
You have provided no information as to what you mean by "my analysis
fails". Exactly what error message are you getting, what operation
system do you have, how much memory do you have, how much are you
using for all the other objects in your address space, etc..
Information like this would hel
Hello Richard,
Since no one else has answered yet I'll venture a guess.
The following works on my little macbook...
x <- as.factor(sapply(letters[1:26], function(x) paste(rep(x, 10),
collapse="")))
So each of the 26 factor levels in x has a string representation of
100,000 chars. So I'm *g
Hi
Is there a maximum length for the character string representing a level
of a factor? I have a set of several million variables, each a factor
of length 19. Each factor level is a character string which in some
cases can be many thousands of characters long. I am trying to find out
why my
It is easy to store a list of that size:
> x <- list(1:1e6, 1:1e6, 1:1e6)
> object.size(x)
12000112 bytes
> str(x)
List of 3
$ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...
$ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...
$ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...
Now it really depends on how you a
Hi,
Is it possible for me to store a list of vectors of 1 million entries?
like,
cc<-list(c(1,2,1million),c(1,2,1million))
also
what is the length of the longest string in R? I keep getting info from a
socket and keep pasting on a string...when will this start becoming a pr
On 20-May-10 19:29:31, Jonathan wrote:
> Hi all,
> I'm hoping this question has a simple answer, but I can't find it
> through searching or trying commands.
>
> I have a list of numeric vectors called 'husk'. I'd just like to
> treat the set of all numbers from all vectors in the list as if
> it w
Hello Jonathan,
Look at ?sapply Does something like this do what you want?
max(sapply(yourlist, max))
Josh
On Thu, May 20, 2010 at 12:29 PM, Jonathan wrote:
> Hi all,
> I'm hoping this question has a simple answer, but I can't find it through
> searching or trying commands.
>
> I have a li
You have a list that you want to treat as a vector, so ?unlist it.
Jonathan wrote:
Hi all,
I'm hoping this question has a simple answer, but I can't find it through
searching or trying commands.
I have a list of numeric vectors called 'husk'. I'd just like to treat the
set of all numbers fr
I think you want unlist with recursive = TRUE, see ?unlist. Bryan
*
Bryan Hanson
Acting Chair
Professor of Chemistry & Biochemistry
DePauw University, Greencastle IN USA
On 5/20/10 3:29 PM, "Jonathan" wrote:
> Hi all,
>I'm hoping this question has a simple answer, but I can't
Hi all,
I'm hoping this question has a simple answer, but I can't find it through
searching or trying commands.
I have a list of numeric vectors called 'husk'. I'd just like to treat the
set of all numbers from all vectors in the list as if it were one large
vector, because I'd like to extract
Try this:
> library(zoo)
> set.seed(1)
> z <- zoo(rnorm(100), as.Date(0) + 1:100)
> mx <- z[which.max(z)]; mx
1970-03-03
2.401618
> coredata(mx)
[1] 2.401618
> time(mx)
[1] "1970-03-03"
On Sat, Feb 13, 2010 at 3:39 PM, Edouard Tallent wrote:
> hi everyone.
> i am dealing on a zoo-class object.
hi everyone.
i am dealing on a zoo-class object.i am aware of the 'min' and 'max' function
to get the minimum and the maximum values.getting these maximum and minimum
values is ont thing.but, how to get the (zoo) dates these values occur ?
in fact, in analysing a time series i am interested in th
Belated answer:
A few remarks regarding your questions:
Your running max problem could be solved in the following way:
(which is a soution based o Duncan Murdoch's suggestion,
but a little bit more general.
foldOrbit<-function(x,fun){
res<-numeric(length(x))
res[1]<-x[1]
On 01/07/2009 9:10 PM, Carl Witthoft wrote:
> More generally, you can always write a loop. They aren't necesssrily
fast
> or elegant, but they're pretty general. For example, to calculate
the max
> of the previous 50 observations (or fewer near the start of a
vector), you
> could do
>
Hi
what about do inside some function a subset of your whole data frame
fff <- function( data, rows) {
data.1 <- data[1:rows,]
get all necessary stuf on data.1
return what you want
}
You can put a dimension check if you want the function to be more robust
Regards
Petr
r-help-boun...@r-pr
> More generally, you can always write a loop. They aren't necesssrily
fast
> or elegant, but they're pretty general. For example, to calculate
the max
> of the previous 50 observations (or fewer near the start of a
vector), you
> could do
>
> x <- ... some vector ...
>
> result <- numeric(le
For another generic approach, you might be interested in the Reduce
function,
rolling <- function( x, window=seq_along(x), f=max){
Reduce(f, x[window])
}
x= c(1:10, 2:10, 15, 1)
rolling(x)
#15
rolling(x, 1:10)
#10
rolling(x, 1:12)
#10
Of course this is only part of the solution to the
On Wed, Jul 1, 2009 at 12:54 PM, Duncan Murdoch wrote:
> On 01/07/2009 1:26 PM, Mark Knecht wrote:
>>
>> On Wed, Jul 1, 2009 at 9:39 AM, Duncan Murdoch
>> wrote:
>>>
>>> On 01/07/2009 11:49 AM, Mark Knecht wrote:
Hi,
I have a data.frame that is date ordered by row number - earliest
On 01/07/2009 1:26 PM, Mark Knecht wrote:
On Wed, Jul 1, 2009 at 9:39 AM, Duncan Murdoch wrote:
On 01/07/2009 11:49 AM, Mark Knecht wrote:
Hi,
I have a data.frame that is date ordered by row number - earliest
date first and most current last. I want to create a couple of new
columns that show
On Wed, Jul 1, 2009 at 9:39 AM, Duncan Murdoch wrote:
> On 01/07/2009 11:49 AM, Mark Knecht wrote:
>>
>> Hi,
>> I have a data.frame that is date ordered by row number - earliest
>> date first and most current last. I want to create a couple of new
>> columns that show the max and min values from
On 01/07/2009 11:49 AM, Mark Knecht wrote:
Hi,
I have a data.frame that is date ordered by row number - earliest
date first and most current last. I want to create a couple of new
columns that show the max and min values from other columns *so far* -
not for the whole data.frame.
It seems
Hi,
I have a data.frame that is date ordered by row number - earliest
date first and most current last. I want to create a couple of new
columns that show the max and min values from other columns *so far* -
not for the whole data.frame.
It seems this sort of question is really coming from m
quite understand it. What does the
option do and why did I initially run into an error ?
>From man R
--max-nsize
Set max number of cons cells to N
Why should there be a maximum?
Thanks for your time
Saptarshi Guha
__
R-help@r-project.org mail
Perform rollapply over the index rather than the series itself: The
result, sin.mx is a zoo series with three columns: the original
series, the series of maxima and their index into the original series.
library(zoo)
library(chron)
t1 <- chron("1/1/2006", "00:00:00")
t2 <- chron("1/31/2006", "23:
library(zoo)
library(chron)
t1 <- chron("1/1/2006", "00:00:00")
t2 <- chron("1/31/2006", "23:45:00")
deltat <- times("00:15:00")
tt <- seq(t1, t2, by = times("00:15:00"))
d <- sample(33:700, 2976, replace=TRUE)
sin.zoo <- zoo(d,tt)
#there are ninety six reading in a day
d.max <- rollapply(sin.zoo,
On Tue, 5 Aug 2008, rostam shahname wrote:
Hi R users,
I am trying to create a matrix, but R has problem with the size of dim,
wondering if there is anything that I can do?
No. See ?"Memory-limts", and consider a sparse matrix (e.g. package
Matrix).
Had this worked 'a' would have used 8*14
Are you aware that that matrix will have 147456^2 elements each of
size 8 bytes ("double") resulting in R trying to allocate
(147456^2)*8/1024^3 = 162 GB of RAM?
If you are aware of this and still trying to allocate a large matrix,
it is unfortunately too large due to "technical" limitations in R.
Hi R users,
I am trying to create a matrix, but R has problem with the size of dim,
wondering if there is anything that I can do?
> a <- diag(147456)
Error in array(0, c(n, p)) : 'dim' specifies too large an array
Thanks for your help
Rostam
[[alternative HTML version deleted]]
Hello everyone,
I have a vector of p-values and I am trying to find the max for the vector and
add it to a list.I am using a loop to loop through 50 times.I have used the
following code but it does not pick out the max and add to the list.Any help
would be appreciated.
w<-c(v[[1]][3],v[[2]][3]
I believe the original poster was looking for runs of consecutive
values. Here's a generalization of Tony's solution:
findlong = function(seq){
rr = rle(seq)
lens = rr$length
lens[rr$value == FALSE] = 0
ll = which.max(lens)
start = cumsum(c(1,rr$length))[ll]
list(start=st
If the increases or decreases could be any size,
rle(sign(diff(x))) could do it:
> x <- c(1, 2, 3, 4, 4, 4, 5, 6, 5, 4, 3, 2, 1, 1, 1, 1, 1)
> r <- rle(sign(diff(x)))
> r
Run Length Encoding
lengths: int [1:5] 3 2 2 5 4
values : num [1:5] 1 0 1 -1 0
> i1 <- which(r$lengths==max(r$lengths[r$v
Hi Erik,
If you look at first 4 numbers, you will se that there was one increase
between first and second number (1 and 2), immediatly after that increase,
there is an increase between second and third number (2 and 3) and finaly
third consecutive increse between third and fourth number (3 and 4).
rle(diff(sq)) could be helpful here,
best, Ingmar
On May 13, 2008, at 11:19 PM, Marko Milicic wrote:
Hi all R helpers,
I'm trying to comeup with nice and elegant way of "detecting"
consecutive
increases/decreases in the sequence of numbers. I'm trying with
combination
of which() and diff(
Hi all R helpers,
I'm trying to comeup with nice and elegant way of "detecting" consecutive
increases/decreases in the sequence of numbers. I'm trying with combination
of which() and diff() functions but unsuccesifuly.
For example:
sq <- c(1, 2, 3, 4, 4, 4, 5, 6, 5, 4, 3, 2, 1, 1, 1, 1, 1);
I'd
areless posting as well :^)
Cheers,
Giovanni
-Messaggio originale-
Da: Prof Brian Ripley [mailto:[EMAIL PROTECTED]
Inviato: martedì 27 novembre 2007 18:54
A: Millo Giovanni
Cc: r-help@r-project.org
Oggetto: Re: [R] max() and min() functions not found
You have a rogue package loaded in
You have a rogue package loaded into your session, one that was installed
for R < 2.6.0. The R posting guide asked for
For questions about unexpected behavior or a possible bug, you should,
at a minimum, copy and paste the output from sessionInfo() into your
message.
and that would hav
I guess you have some library in the search path that includes a base
package of an outdated R installation.
Uwe Ligges
Millo Giovanni wrote:
> Dear List,
>
> I just installed R 2.6.1 (on Win2K) and I get a strange error in
> functions min() and max():
>
>> min(1:3)
> Errore in .Internal(min
Dear List,
I just installed R 2.6.1 (on Win2K) and I get a strange error in
functions min() and max():
> min(1:3)
Errore in .Internal(min(..., na.rm = na.rm)) :
nessuna funzione interna "min"
which, as you may have guessed, means 'no internal function "min" '. The
same happens for max().
May
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