-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77843-4352
From: Jose Claudio Faria
Sent: Monday, September 24, 2018 2:01 PM
To: David L Carlson
Cc: Jeff Newmiller ; r-help@r-project.org
Subject: Re: [R] cut{base}:
;> Department of Anthropology
>> Texas A&M University
>> College Station, TX 77843-4352
>>
>> -Original Message-
>> From: Jeff Newmiller
>> Sent: Monday, September 24, 2018 10:41 AM
>> To: r-help@r-project.org; David L Carlson ; Jose
>>
ty
> College Station, TX 77843-4352
>
> -Original Message-
> From: Jeff Newmiller
> Sent: Monday, September 24, 2018 10:41 AM
> To: r-help@r-project.org; David L Carlson ; Jose
> Claudio Faria ; r-help@r-project.org
> Subject: Re: [R] cut{base}: is it a bug?
>
> &
t.org; David L Carlson ; Jose Claudio
Faria ; r-help@r-project.org
Subject: Re: [R] cut{base}: is it a bug?
"Subtracting a bit" only fixes the problem for the test data... it introduces a
bias in any continuous data you happen to throw at it. However, if you have
data with known r
,1.59)6
>[1.59,1.79)6
>[1.79,1.99)5
>[1.99,2.19]6
>
>
>David L Carlson
>Department of Anthropology
>Texas A&M University
>College Station, TX 77843-4352
>
>
>-Original Message-
>From: R-help On B
[1.79,1.99)5
[1.99,2.19]6
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77843-4352
-Original Message-
From: R-help On Behalf Of Jose Claudio Faria
Sent: Monday, September 24, 2018 9:32 AM
To: r-help@r-project.or
Dears members,
Is the below a bug of the cut {base} function?
dat <- c(
0.6, 0.6, 0.6, 0.7, 0.7, 0.7, 0.7, 0.7, #(8)
0.8, 0.8, 0.8, 0.9, 0.9, 0.9, 0.9,#(7)
1.0, 1.0, 1.0, 1.0, 1.1, 1.1, 1.1,#(7)
1.2, 1.2, 1.2, 1.2, 1.3, 1.3, 1.3,#(7)
1.4, 1.4, 1.4, 1.5, 1.5, 1.5,
Week",1:52,sep=" "))
Jim
On Wed, Jun 8, 2016 at 12:44 PM, TJUN KIAT TEO wrote:
>
> It does not seem to work for me. I will show you exactly my data format
>
> SMA$TIME_DATE
>
> "28/9/2014" "17/6/2014" "19/9/2014" "17/1/2015
"2014-06-16" "2014-06-16" "2014-06-17" "2014-06-23"
What I would like to do is the cut the first set of dates into the second set
of dates
> From: drjimle...@gmail.com
> Date: Fri, 3 Jun 2016 18:59:11 +1000
> Subject: Re: [R] Cut Dates into b
fiddle with as.numeric and what do you mean by
bin dates.
Regards
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of TJUN KIAT
> TEO
> Sent: Friday, June 3, 2016 10:01 AM
> To: r-help@r-project.org
> Subject: [R] Cut Dates into bins
Hi Tjun Kiat,
This seems to work:
daily_date<-as.Date(paste("2000-01",1:28,sep="-"),"%Y-%m-%d")
weekly_date<-as.Date(paste(c(1,8,15,22,28),"01/2000",sep="/"),
"%d/%m/%Y")
cut(daily_date,breaks=weekly_date,include.lowest=TRUE,
labels=paste("Week",1:4))
Jim
On Fri, Jun 3, 2016 at 6:00 PM, TJUN
dear all,
I have a set of data that can be defined as bimodal; would be possible
to define a cut-off to split it? my mathematical/statistical knowledge
is limited, my apologies, thus i am not sure what kind of area this
problem belongs to; at the moment I can fit -- thanks to previous help
from the
Thanks this was very helpful.
@Olivier Crouzet: Yes, round (x) would do the job but it was a principal
confusion ...
2015-10-06 21:57 GMT+02:00 Marc Schwartz :
>
> > On Oct 6, 2015, at 2:20 PM, Hermann Norpois wrote:
> >
> > Hello,
> >
> > why do I get NA for the following:
> >
> > cut (x, seq
Hi,
On Tue, 6 Oct 2015 21:20:13 +0200
Hermann Norpois wrote:
> Hello,
>
> why do I get NA for the following:
>
> cut (x, seq (0, max(x), by=1), label=FALSE)
> [1] 1322 1175 1155 1149 1295 1173 1289 1197 NA 1129
The NA comes from your max value and it's due to your seq(0, max(x),
by = 1) cr
> On Oct 6, 2015, at 2:20 PM, Hermann Norpois wrote:
>
> Hello,
>
> why do I get NA for the following:
>
> cut (x, seq (0, max(x), by=1), label=FALSE)
> [1] 1322 1175 1155 1149 1295 1173 1289 1197 NA 1129
>
> dput (x)
> c(1321.55376901374, 1174.35657200935, 1154.02042504008, 1148.6098192594
Because
> tail(seq(0, max(x), by=1))
[1] 1350 1351 1352 1353 1354 1355
> tail(seq(0, ceiling(max(x)), by=1))
[1] 1351 1352 1353 1354 1355 1356
and max(x)=1355.88836502166 is beyond the range
of the former.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Oct 6, 2015 at 12:20 PM, Herm
Hello,
why do I get NA for the following:
cut (x, seq (0, max(x), by=1), label=FALSE)
[1] 1322 1175 1155 1149 1295 1173 1289 1197 NA 1129
dput (x)
c(1321.55376901374, 1174.35657200935, 1154.02042504008, 1148.60981925942,
1294.6166388941, 1172.45806806869, 1288.31933914639, 1196.26080041462,
1
0, 0, 0, 0, 0, 0, 0, 0,
> > > 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
> > > 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), RO = c(0,
> > > 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
> > > 0, 0, 0, 0, 0, 0, 0, 0, 0
64175415, 34.4827562570572, 7.95454531908035,
> > 75, 0, 0, 0, 0, 0, 0, 5.26393800973892, 0, 0, 0, 0, 0, 0, 0,
> > 0, 0, 74.6153831481934, 84.6153914928436, 0, 5.09554147720337,
> > 0, 0, 0, 21.0884347558022, 18.4549376368523, 6.1224490404129,
> > 25.373136997222
224490404129,
> > 25.3731369972229, 2.12765969336033, 0, 84.3988716602325, 0, 0,
> > 0, 100), awc_class = c(106.228088378906, 78.2306137084961,
> 80.9311141967773,
> > 32.4921531677246, 54.8475151062012, 80.6665878295898, 116.331588745117,
> > 54.84751510620
2812, 453.118286132812,
> 95.2380981445312, 63), se025 = c(163.529998779297, 2.70000004768372,
> 157, 5.5, 6.3019073486, 36.024577637, 86, 5.0990463257,
> 6.4009536743, 6, 8.6980926514, 4, 6.3019073486, 5.8019073486,
> 8.8019073486, 2, 118.809997558594,
"2", "3", "4", "5", "6", "7",
"8", "11", "12", "13", "14", "15", "16", "17", "18", "19", "20",
"21", "22", "2
Hi Thierry and Petr,
I really appreciate the comments you already gave. Thank you very much for
that.
Below you can find a link to the data and the code. Hopefully this helps in
spotting the error.
I still think the issue is that the cut2 function only accepts numbers, and
not an "i" that refers
day, August 14, 2015 10:16 AM
> To: Michael Dewey
> Cc: r-help@r-project.org
> Subject: Re: [R] cut variable within a loop
>
> Hey Michael,
>
> Sorry for the late reply!
>
> Thanks for your comment, but for the cut2 command, this is not the
> case. If I enter for insta
Hey Michael,
Sorry for the late reply!
Thanks for your comment, but for the cut2 command, this is not the case. If
I enter for instance
Alldata$irri=cut2(irrigation,3)
Then I get 2 intervals from 0-3 and from 3-100.
Janka
2015-08-11 17:25 GMT+02:00 Michael Dewey :
> Dear Janka
> If you suppl
Dear Janka
If you supply a single number to the breaks parameter of cut I think it
is the number of intervals.
On 11/08/2015 13:57, Janka Vanschoenwinkel wrote:
Hi Thierry!
Thanks for your answer. I tried this, but I get this error:
"Error in cut.default(x, k2) : invalid number of intervals"
Hi Thierry!
Thanks for your answer. I tried this, but I get this error:
"Error in cut.default(x, k2) : invalid number of intervals"
Which is strange because I am not specifying intervals, but the number at
where the sample has to be cut?
Greetings from Belgium! :-)
2015-08-11 14:52 GMT+02:00 T
You'll need to send a reproducible example of the code. We can't run the
code that you send. Hence it is hard to help you. See e.g.
http://adv-r.had.co.nz/Reproducibility.html
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kw
Dear Janka,
You loop goes for 0 to 100. It should probably go from 1:99
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belg
Dear list members,
I have a loop where I want to do several calculations for different samples
and save the results for each sample. These samples are for each loop
different. I want to use the "i" in the loop to cut the samples.
So for instance:
- In loop 1 (i=1), I have a sample from 0-1 an
On Apr 3, 2015, at 5:09 AM, Wing Keong Lew wrote:
> Hi,
>
> Is it a requirement to provide the break intervals of the cut function in
> ascending order?
Apparently not. I get teh sam splits even with random permutations. It is
apparently a "requirement" to make sure you labels match the sorte
Hi,
Is it a requirement to provide the break intervals of the cut function in
ascending order? The help documentation for cut didn't specify this but the
labels returned are reversed if I indicate the break intervals in a descending
order. Here is an example
tbl<-data.frame(x=c(0:10))
tbl$asce
Witold,
Here's one way:
Vec <- rnorm(30)
Vec.cut <- cut(Vec, breaks=c(quantile(Vec, probs = seq(0, 1, by = 0.20))),
labels=c("0-20","20-40","40-60","60-80","80-100"),
include.lowest=TRUE)
table(Vec.cut)
or determine the breaks automatically:
cut.size <- function(x, size) {
cut.prob <
Here's one way:
Vec <- rnorm(30)
Vec.cut <- cut(Vec, breaks=c(quantile(Vec, probs = seq(0, 1, by = 0.20))),
labels=c("0-20","20-40","40-60","60-80","80-100"), include.lowest=TRUE)
table(Vec.cut)
or determine the breaks automatically:
cut.size <- function(x, size) {
cut.prob <- size/leng
Greg,
Good catch. My recollection was that the vector would be broken up into
'breaks' groups of equal size, however it is range(x) that is split into
'breaks' intervals, each of which is equal width.
Thanks,
Marc
On Jul 18, 2013, at 10:55 AM, Greg Snow <538...@gmail.com> wrote:
> Marc,
>
my fault. qcut() is a python function in pandas ;-)
what i meant is cut2() in hmisc.
sorry for messing up.
On Jul 18, 2013 11:51 AM, "Greg Snow" <538...@gmail.com> wrote:
> Wensui,
>
> ?qcut on my machine gives an error and ??qcut does not find anything in
> the installed packages. Which packa
Marc,
Your method works fine when the data is perfectly uniform, but try it with
"Vec <- rnorm(30)" and you will see that there are more observations in the
middle groups and fewer in the tail groups. Something like quantile needs
to be used to find the unequally spaced breaks that will give equa
Wensui,
?qcut on my machine gives an error and ??qcut does not find anything in
the installed packages. Which package is qcut in?
On Wed, Jul 17, 2013 at 4:43 PM, Wensui Liu wrote:
> ?qcut
> On Jul 17, 2013 5:45 PM, "Witold E Wolski" wrote:
>
> > I would like to "cut" a vector into groups o
On Jul 17, 2013, at 4:43 PM, Witold E Wolski wrote:
> I would like to "cut" a vector into groups of equal nr of elements.
> looking for a function on the lines of cut but where I can specify
> the size of the groups instead of the nr of groups.
In addition to the other options, if the 'breaks'
?qcut
On Jul 17, 2013 5:45 PM, "Witold E Wolski" wrote:
> I would like to "cut" a vector into groups of equal nr of elements.
> looking for a function on the lines of cut but where I can specify
> the size of the groups instead of the nr of groups.
>
>
>
>
> --
> Witold Eryk Wolski
>
> __
:04 PM
Subject: Re: [R] cut into groups of equal nr of elements...
HI,
Not sure whether this is what you wanted.
vec1<- 1:7
fun1<- function(x,nr) {((x-1)%/%nr)+1}
fun1(vec1,2)
#[1] 1 1 2 2 3 3 4
fun1(vec1,3)
#[1] 1 1 1 2 2 2 3
split(vec1,fun1(vec1,2))
A.K.
- Original Messag
nesday, July 17, 2013 5:43 PM
Subject: [R] cut into groups of equal nr of elements...
I would like to "cut" a vector into groups of equal nr of elements.
looking for a function on the lines of cut but where I can specify
the size of the groups instead of the nr of groups.
--
Wi
You could use the quantile function to choose the cut points and pass that
to cut.
Or you could sort the vector and just take the first n elements as the 1st
group, etc.
On Wed, Jul 17, 2013 at 3:43 PM, Witold E Wolski wrote:
> I would like to "cut" a vector into groups of equal nr of elements
I would like to "cut" a vector into groups of equal nr of elements.
looking for a function on the lines of cut but where I can specify
the size of the groups instead of the nr of groups.
--
Witold Eryk Wolski
__
R-help@r-project.org mailing list
http
There is a discussion of the ways to do this
task (which can be trickier than it seems) at:
http://www.portfolioprobe.com/2012/12/24/miles-of-iles/
Pat
On 26/02/2013 17:39, Martin Batholdy wrote:
Hi,
I would like to cut a vector of values in parts.
Each part should have an equal number of el
ollege Station, TX 77843-4352
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Rui Barradas
> Sent: Tuesday, February 26, 2013 1:59 PM
> To: Martin Batholdy
> Cc: r-help@r-project.org
> Subject: Re: [R] cut
Hello,
Try the following.
x <- rnorm(500)^2
split(x, cut(x, quantile(x, probs = seq(0, 1, by = 0.2
Hope this helps,
Rui Barradas
Em 26-02-2013 17:39, Martin Batholdy escreveu:
Hi,
I would like to cut a vector of values in parts.
Each part should have an equal number of elements.
for e
?sort
?cut
On Tue, Feb 26, 2013 at 12:39 PM, Martin Batholdy
wrote:
> Hi,
>
> I would like to cut a vector of values in parts.
> Each part should have an equal number of elements.
>
> for example:
>
> x <- (rnorm(500)^2)
>
> now I want 5 vectors each with 100 elements.
> The first vector should i
Hi,
I would like to cut a vector of values in parts.
Each part should have an equal number of elements.
for example:
x <- (rnorm(500)^2)
now I want 5 vectors each with 100 elements.
The first vector should include the 100 lowest values of x and so on
(so that the fifth vector contains the 100 h
At Tue, 1 Jan 2013 02:00:14 +,
Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
> Although David's solution (putting the right parenthesis, which I had missed)
> has resolved the issue, I would like to try yours as well.
>
> Could you please clarify the six elements: c(-1e-8, 0, 0, 0, 0, 1e8)?
It's a v
From: Neal H. Walfield [n...@walfield.org]
Sent: Monday, December 31, 2012 5:42 PM
To: Muhuri, Pradip (SAMHSA/CBHSQ)
Cc: R help
Subject: Re: [R] cut ()
At Mon, 31 Dec 2012 22:25:25 +,
Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
> The issue is that, for Utah, I am getting an instead of (42,48.7
help'
Subject: RE: [R] cut ()
A misplaced right parenthesis caused the problem:
p1_st_data$ob_mrj_cat <- cut (p1_st_data$obt_mrj_p, quantile
(p1_st_data$obt_mrj_p, (0:5/5), include.lowest=TRUE))
Should be
p1_st_data$ob_mrj_cat <- cut (p1_st_data$obt_mrj_p, quantile
(p1_st_data$o
radip (SAMHSA/CBHSQ)
> Sent: Monday, December 31, 2012 4:25 PM
> To: R help
> Subject: [R] cut ()
>
> Hello List,
>
>
> My goal is to create a 5 category variable (p1_st_data$ob_mrj_cat),
> based on the p1_st_data$obt_mrj_p variable, using the following code
At Mon, 31 Dec 2012 22:25:25 +,
Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
> The issue is that, for Utah, I am getting an instead of (42,48.7] in the
> ob_mrj_cat column.
The problem is likely due to comparisons of floating point numbers.
Try moving your lower and upper bounds out a tiny bit. Whe
Hello List,
My goal is to create a 5 category variable (p1_st_data$ob_mrj_cat), based on
the p1_st_data$obt_mrj_p variable, using the following code for 50 States and
District of Columbia (N=51).
p1_st_data$ob_mrj_cat <- cut (p1_st_data$obt_mrj_p, quantile
(p1_st_data$obt_mrj_p, (0:5/5), incl
Please use dput while supplying data.
Best Regards,
Bhupendrasinh Thakre
Sent from my iPhone
On Oct 25, 2012, at 5:31 AM, Soheila Khodakarim wrote:
> var1
>
> var2
>
> var3
>
> var4
>
> var5
>
> var6
>
> var7
>
> var8
>
> var9
>
> var10
>
> gold
>
> 2
>
> 3
>
> 1
>
> 2
>
> 4
>
var1
var2
var3
var4
var5
var6
var7
var8
var9
var10
gold
2
3
1
2
4
0
1
4
4
3
2
2
4
2
4
3
4
2
4
4
4
2
3
3
0
0
4
1
0
2
4
4
2
1
4
0
3
2
0
0
2
4
4
2
3
4
0
2
2
0
0
0
3
4
2
2
2
3
2
2
0
0
0
2
4
2
2
4
1
1
2
0
0
3
3
3
2
3
x 0 0
>
> Thank you very much!
>
> Regards,
> Yan
>
> -Original Message-
> From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
> Sent: Tuesday, October 18, 2011 5:23 AM
> To: Li, Yan
> Cc: r-help@r-project.org
> Subject: Re: [
1 x 0
0
Thank you very much!
Regards,
Yan
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Tuesday, October 18, 2011 5:23 AM
To: Li, Yan
Cc: r-help@r-project.org
Subject: Re: [R] cut data into sevral group and ass
On 17.10.2011 20:53, Li, Yan wrote:
Hi All,
I have some data from which I set four points to be breaks. Based on these
points, I cut the dataset into four groups and assign a number to it:
<=331.04 assign 0
331.04<=476.07 assign data-331.04/(476.07-331.04)
476.07<=608.66 assign 1
608.66<=
Hi All,
I have some data from which I set four points to be breaks. Based on these
points, I cut the dataset into four groups and assign a number to it:
<=331.04 assign 0
>331.04 <=476.07 assign data-331.04/(476.07-331.04)
>476.07<=608.66 assign 1
>608.66 <=791.5 assign 791.5- data/(791.5-608.
On 04/20/2011 12:32 AM, Santosh wrote:
Dear Rxperts,
Below is a small sample of values (cut short due to space considerations
while posting).. I was wondering if it is possible to construct boundaries
(or intervals) based on the distribution of points. Is it anything similar
to boundary detecti
On Apr 19, 2011, at 10:32 AM, Santosh wrote:
Dear Rxperts,
Below is a small sample of values (cut short due to space
considerations
while posting).. I was wondering if it is possible to construct
boundaries
(or intervals) based on the distribution of points. Is it anything
similar
to b
Dear Rxperts,
Below is a small sample of values (cut short due to space considerations
while posting).. I was wondering if it is possible to construct boundaries
(or intervals) based on the distribution of points. Is it anything similar
to boundary detection of distributions?
x1 <-
c(0.00,0.25,0.
У Срд, 09/03/2011 у 23:29 -0800, Dennis Murphy піша:
> Hi:
>
> Here's one approach, although I imagine there are more efficient ways.
>
> # A function to strip spaces and return the first three non-blank elements
> of a string
> keyset <- function(x) substr(gsub(' ', '', x)[1], 1, 3)
Hello to ev
Try this:
cut(DF$A, c(-Inf, 0, 4, 8, 12, 14, 16, 20))
On Tue, Oct 26, 2010 at 7:01 PM, LCOG1 wrote:
>
> Hi everyone,
>
> I have been using R too long to have to ask this but here i am. How do i
> create a separate bin for the 0 value? So for:
>
> #Create data frame
> DF<-data.frame(A=0:20)
Hi everyone,
I have been using R too long to have to ask this but here i am. How do i
create a separate bin for the 0 value? So for:
#Create data frame
DF<-data.frame(A=0:20)
#Create label vector
labs<-1:6
#Create buckets and label
DF$Cut<-cut(DF$A,c(0,4,8,12,14,16,20),labels=labs,include.l
Thanks all! That did it.
On Thu, Jul 15, 2010 at 11:07 AM, Nikhil Kaza wrote:
> Building on Erik's solution and because it would easier to do date
> arithmetic..
>
> d1 <- as.character(date)
> d1 <- ifelse(nchar(d1)<4, paste(0,d1,sep=""),d1)
> d2 <- as.Date(date, "%m%d")
>
>
> On Jul 15, 2010, a
Building on Erik's solution and because it would easier to do date
arithmetic..
d1 <- as.character(date)
d1 <- ifelse(nchar(d1)<4, paste(0,d1,sep=""),d1)
d2 <- as.Date(date, "%m%d")
On Jul 15, 2010, at 1:21 PM, btc1 wrote:
Hello, I have a vector, "dates", as a series of 3 digit elements,
Try this:
gsub("^(.)", "\\1/", dates)
On Thu, Jul 15, 2010 at 2:21 PM, btc1 wrote:
>
> Hello, I have a vector, "dates", as a series of 3 digit elements, i.e. >
> date
> [1] 528 528 528 528 528 528 528 528 528 528 528 528 708 708 708 708 708
> 708
> [19] 708 708 708 708 529 529 529 529 529 5
gsub("(.*)(.{2})","\\1/\\2",dates)
-
A R learner.
--
View this message in context:
http://r.789695.n4.nabble.com/Cut-a-within-elements-by-length-not-value-of-vectors-tp2290440p2290471.html
Sent from the R help mailing list archive at Nabble.com.
btc1 wrote:
Hello, I have a vector, "dates", as a series of 3 digit elements, i.e. > date
[1] 528 528 528 528 528 528 528 528 528 528 528 528 708 708 708 708 708
708
[19] 708 708 708 708 529 529 529 529 529 529 529 529 529 529 529 529 529
529
[37] 529 624
I need to convert them into julian
uot; "12/08"
"4/23" "15/06" "3/21"
Martyn
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of btc1
Sent: 15 July 2010 18:22
To: r-help@r-project.org
Subject: [R] Cut a within elements by length,
Hello, I have a vector, "dates", as a series of 3 digit elements, i.e. > date
[1] 528 528 528 528 528 528 528 528 528 528 528 528 708 708 708 708 708
708
[19] 708 708 708 708 529 529 529 529 529 529 529 529 529 529 529 529 529
529
[37] 529 624
I need to convert them into julian, but have to i
Hi R users!
Is it possible to cut the edges from a surface graph without to cut the axes
using persp function?
I am using the following code:
op <- par(bg = "white")
persp(phi1, phi2, z,main="Bullwhip generated with AR(2) demand when L=1",
xlab ="phi1" , ylab ="phi2", zlab ="Bullwhip", t
Hi
I am a bit curious why one shall do such a twisted construction. Accessing
list is basically the same as accessing corresponding matrix row, you only
need to remember drop=FALSE option.
Regards
Petr
r-help-boun...@r-project.org napsal dne 17.03.2010 00:58:54:
> Here is a way of creating a
Here is a way of creating a list of the matrices:
> x <- matrix(1:(12*30), nrow=30)
> # create a list of single row matrices
> x.l <- lapply(seq(nrow(x)), function(a) x[a,, drop=FALSE])
>
> x.l
[[1]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,]1 31 61 91 121
Hey dear users
I'm trying to kind of split my matrix which looks as follows:
dim(out)
[1] 30 12
What I finally want is each line as it's own matrix which I can handle
then separately.
Like, say:
out1<- [1,]
out2<-[2,]
..
Would you do that with a for() loop or does exist an other approp
Luke,
Thanks for your previous advice on using misc3d; I've been able to get a good
start with it. Most of my colleagues use Tecplot for graphing and are
wondering if I can make "cut-aways" using R. An example is attached. The idea
is show a 3D isosurface of one property together with a 2D c
On Dec 1, 2009, at 3:28 PM, Gabor Grothendieck wrote:
> Try this:
>
> > library(gsubfn)
> > strapply(testvec, "[-+.0-9]+", as.numeric, simplify = ~
> colMeans(cbind(...)))
> [1] -5.8500 -2.9800 -2.8160 -2.7120 -2.6325 -2.5680
Very, nice. Also tried on some other valid ("200,2") and
invalid )
Try this:
> library(gsubfn)
> strapply(testvec, "[-+.0-9]+", as.numeric, simplify = ~
colMeans(cbind(...)))
[1] -5.8500 -2.9800 -2.8160 -2.7120 -2.6325 -2.5680
On Tue, Dec 1, 2009 at 3:14 PM, David Winsemius wrote:
> I'm sitting here chuckling. Your solution is just so "pure".
>
> I would offer
I'm sitting here chuckling. Your solution is just so "pure".
I would offer an enhancement. When I tested with my cuts that had "-"
before the digits, you solution dropped them, so my suggestion for the
pattern would be: "[-[:digit:].]+"
I will admit that I thought it might fail with posit
Perhaps this shoul work too:
sapply(strsplit(gsub("^\\W|\\W$", "", testvec), ","),
function(x)sum(as.numeric(x))/2)
On Tue, Dec 1, 2009 at 5:41 PM, David Winsemius wrote:
> Starting with the head of a 499 element matrix whose column names are now
> the labels trom a cut() operation, I needed to
You also might want to look at
demo("gsubfn-cut")
On Tue, Dec 1, 2009 at 2:41 PM, David Winsemius wrote:
> Starting with the head of a 499 element matrix whose column names are now
> the labels trom a cut() operation, I needed to get to a vector of midpoints
> to serve as the basis for plotting
Starting with the head of a 499 element matrix whose column names are
now the labels trom a cut() operation, I needed to get to a vector of
midpoints to serve as the basis for plotting a calibration curve
( exp(linear predictor) vs. :
> dput(head(dimnames(mtcal)[2][[1]])) # was starting po
Try e.g. function recode from package car.
2009/9/22 Chris Hane :
> Hello R-users,
> I have a data frame with a factor of ages in 5 year increments, and various
> count data for each age group. I only have this summary information in R at
> the moment.
>
> I want to create a new factor that aggreg
Hello R-users,
I have a data frame with a factor of ages in 5 year increments, and various
count data for each age group. I only have this summary information in R at
the moment.
I want to create a new factor that aggregates the age factors if the
existing factors have insufficient counts. Then I
Shawn Rutledge wrote:
With floating point numbers I'm seeing 'cut' putting values in the wrong
bands. An example below places 0.3 in (0.3,0.6] i.e. 0.3 > 0.3.
x = 1:5*.1
x
[1] 0.1 0.2 0.3 0.4 0.5
cut(x, br=c(0,.3,.6))
[1] (0,0.3] (0,0.3] (0.3,0.6] (0.3,0.6] (0.3,0.6]
Leve
With floating point numbers I'm seeing 'cut' putting values in the wrong
bands. An example below places 0.3 in (0.3,0.6] i.e. 0.3 > 0.3.
> x = 1:5*.1
> x
[1] 0.1 0.2 0.3 0.4 0.5
> cut(x, br=c(0,.3,.6))
[1] (0,0.3] (0,0.3] (0.3,0.6] (0.3,0.6] (0.3,0.6]
Levels: (0,0.3] (0.3,0.6]
I'm sure this i
Thanks for the quick and nice reply.
I must apologise twice:
1-I should have reported I am using R 2.5.1 on Windows;
2- It was a spelling error on my part (rigth instead of right).
Pedro
At 11:27 2007/10/19, you wrote:
>Works for me with current R-2.6.0 (but you haven't told us about
>your R ver
Works for me with current R-2.6.0 (but you haven't told us about your R
version!):
> cut(1:5, 1:5)
[1] (1,2] (2,3] (3,4] (4,5]
Levels: (1,2] (2,3] (3,4] (4,5]
> cut(1:5, 1:5, right=FALSE)
[1] [1,2) [2,3) [3,4) [4,5)
Levels: [1,2) [2,3) [3,4) [4,5)
Uwe Ligges
Pedro de Barros wrote:
> Dear
Dear All,
I am trying to use cut() to produce intervals open on the right, but
it seems to ignore the argument right=F, contrary to what is
indicated on the help for cut.
Can anyone help?
Thanks,
Pedro
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