ere the pairwise comparison is obvious, while the numeric
conversion isn't.
A simple one would be a list of string vectors of different lengths,
where you want to sort lexicographically.
Duncan
Have fun with the remainder of the advent!
Another Martin
From: R-help on behalf of Martin Møller Skar
h for S3 generics), but rather define a
class (e.g., that requires vectors person and value) and implement a
corresponding `xtfrm()` method.
Have fun with the remainder of the advent!
Another Martin
From: R-help on behalf of Martin Møller
Skarbiniks Pedersen
Date: Thursday, December 14, 2023
On Thu, 14 Dec 2023 at 12:02, Duncan Murdoch wrote:
>
> class(df$value) <- "sizeclass"
>
> `>.sizeclass` <- function(left, right) custom_sort(unclass(left),
> unclass(right)) == 1
>
> `==.sizeclass` <- function(left, right) custom_sort(unclass(left),
> unclass(right)) == 0
>
> `[.sizeclass` <- fu
> This sounds suspiciously like homework (which is off-topic... see the Posting
> Guide)
It is not homework.
Currently I am trying to solve this: https://adventofcode.com/2023/day/7
But it is something that has puzzled me for a long time.
In many programming languages, you can give a "less" func
On 14/12/2023 3:00 a.m., Martin Møller Skarbiniks Pedersen wrote:
Hi,
I need to sort a data.frame based on a custom sorting function.
It is easy in many languages but I can't find a way to do it in R.
In many cases I could just use an ordered factor but my data.frame
contains poker han
This sounds suspiciously like homework (which is off-topic... see the Posting
Guide), and you haven't indicated how you plan to encode your poker hands, and
most core features of other languages are possible in R so if you really
understand these other techniques and R then you should be able to
Make the grouping variable into a factor end define the level ordering.
?factor will tell you more.
> On 14.03.2022, at 17:39, Ebert,Timothy Aaron wrote:
>
> Set the factor levels rather than using the R default.
>
> -Original Message-
> From: R-help On Behalf Of Fernando Archuby
> Sen
Hi Bert,
Good reference and David Urbina's example showed that a simple swap
was position dependent. The reason I pursued this is that it seems
more efficient to sequentially apply the precedence rules to the
arbitrarily sorted elements of the vector than to go through the
directed graph approach.
Thanks Bert. Excellent reference, I learned a lot from it!
Just a note: I did use search engines for at least 2 days before posting. BUT
as often happens, I did not use the right keywords. I tried several variants of
"Convert ordered pairs to sorted", "Sort vector on paired comparisons" and
abo
If I understand correctly, the answer is a topological sort.
Here is an explanation
https://davidurbina.blog/on-partial-order-total-order-and-the-topological-sort/
This was found by a simple web search on
"Convert partial ordering to total ordering"
Btw. Please use search engines before posting
Hi Pedro,
This looks too simple to me, but it seems to work:
swap<-function(x,i1,i2) {
tmp<-x[i1]
x[i1]<-x[i2]
x[i2]<-tmp
return(x)
}
mpo<-function(x) {
L<-unique(as.vector(x))
for(i in 1:nrow(x)) {
i1<-which(L==x[i,1])
i2<-which(L==x[i,2])
if(i2 wrote:
>
> Dear All,
>
> This should be
This is called topological sorting in some circles. The function below
will give you one ordering that is consistent with the contraints but not
all possible orderings. I couldn't find such a function in core R so I
wrote one a while back based on Kahn's algorithm, as described in Wikipedia.
> S
Thanks for this.
Yes, this is checked before trying to process this.
Pedro
On 14/03/2019 14.09, Bert Gunter wrote:
This cannot be done unless transitivity is guaranteed. Is it?
S L
a b
b c
c a
Bert
On Thu, Mar 14, 2019, 4:30 AM Pedro Conte de Barros
mailto:pbar...@ualg.pt>> wrote:
Dea
This cannot be done unless transitivity is guaranteed. Is it?
S L
a b
b c
c a
Bert
On Thu, Mar 14, 2019, 4:30 AM Pedro Conte de Barros wrote:
> Dear All,
>
> This should be a quite established algorithm, but I have been searching
> for a couple days already without finding any satisfact
Try this. Anything that appears only in Smaller is candidate for smallest.
Among those, order is arbitrary.
Anything that appears only in Larger is a candidate for largest. Among
those order is arbitrary.
Remove rows of matComp containing the already classified items. Repeat
with the smaller set
Thanks for all suggestions.
With my build (from the CRAN repo) I don't get ICU support, and setting
LC_COLLATE to "C" did not help.
> capabilities("ICU")
ICU
FALSE
> sessionInfo()
R version 3.3.2 (2016-10-31)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 14.04.5 LTS
locale:
Hello,
What is your sessionInfo()?
With me it works as expected:
> sort(c("-", "+"))
[1] "-" "+"
> sort(c("+", "-"))
[1] "-" "+"
> x5 <- c("+Aa","-Ab")
> sort(x5)
[1] "-Ab" "+Aa"
> sessionInfo()
R version 3.3.2 (2016-10-31)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64
On 08 Nov 2016, at 13:18 , Pascal A. Niklaus wrote:
> I just got caught by the way in character vectors are sorted.
>
> It seems that on my machine "sort" (and related functions like "order") only
> consider characters related to punctuation (at least here the "+" and "-")
> when there is no
I suspect it is a problem related to locales: R and the base Ubuntu might
be using different locales, hence the source of the different sorting.
Can't say if this is the problem in your case, but it might be.
Adrian
On Sat, Apr 9, 2016 at 12:18 AM, Maria Ninova wrote:
> Hello, I came across the
Wrong list. This is not the Ubuntu shell support mailing list. The fact that
you are using R to get at the operating system command line doesn't make this
an R question.
--
Sent from my phone. Please excuse my brevity.
On April 8, 2016 2:18:40 PM PDT, Maria Ninova wrote:
>Hello, I came acros
"%in%" is a wrapper for match(), which uses hashing I believe
(correction welcome!),and so is generally very fast.
See ?Rprof for profiling R code to get timings. (Haven't used it
myself, so not sure how useful it would be for this situation).
-- Bert
Bert Gunter
"The trouble with having an o
?order
is what I think you want. There is a slight wrinkle here, however. You
want to sort Cluster in *increasing* order and Byers_EMT in
*decreasing*, as I understand. order() will do both increasing, or
both decreasing, but not differently. So a simple but inefficient way
around this is to first
...
> mydf[2] # ???
B
1 4
2 5
3 6
A data frame is "really" a list of columns, so giving a single value
returns that column.
False. It returns a data frame consisting of a single column = a list
containing a single component.
mydf[[2]]
returns a single component/column.
While these differ
On Tue, Jan 26, 2016 at 4:24 PM, Robert Sherry wrote:
>
> Thank you for the response. As expected, the following expression worked:
> df[order(df$x),]
This says to sort the rows, and leave the columns alone.
Subsetting a 2-dimensional object is via
[rows, columns]
> I would expect the follo
Thank you for the response. As expected, the following expression worked:
df[order(df$x),]
I would expect the following expression to work also:
df[order(df$x)]
However it does not. That is, the comma is needed. Please tell me why
the comma is there.
Thanks
Bob
On 1/26/2016 8:19 A
> On 23.01.2016 01:21, Robert Sherry wrote:
> > In R, I run the following commands:
> > df = data.frame( x=runif(10), y=runif(10) )
> > df2 = df[order(x),]
>
>
> You use another x from your workspace, you actually want to
>
>
> df2 = df[order(df[,"x"]),]
or
df[order(df$x),]
And
On 23.01.2016 01:21, Robert Sherry wrote:
In R, I run the following commands:
df = data.frame( x=runif(10), y=runif(10) )
df2 = df[order(x),]
You use another x from your workspace, you actually want to
df2 = df[order(df[,"x"]),]
Best,
Uwe Ligges
The first, as I would expect
Thank you all again. It works with :
library("gtools")
dat[mixedorder(A),]
considering :
A = c("A1","A10","A11","A2")
B = c(1,2,3,4)
dat = data.frame(A,B
On Sun, Jun 21, 2015 at 10:52 AM, Bert Gunter
wrote:
> Diego:
>
> Nonsense! Look at the results of your code -- you have failed to order
Diego:
Nonsense! Look at the results of your code -- you have failed to order
the results as had been requested by the OP. It's also unnecessarily
complicated. The following suffices (where I have used regular
expressions rather substring() to get the numeric part of the strings
-- **assuming** th
Bogdan,
Follow my suggestion.
letter <- substring(A, 1, 1)
number <- substring(A, 2, nchar(A))
new.data <- paste0(letter, formatC(as.numeric(number), width = 2, flag =
"0"))
Em 20/06/2015 21:21, "Bogdan Tanasa" escreveu:
> Dear all,
>
> I am looking for a suggestion please regarding sorting a d
thank you all, it is working fine. happy weekend ;) !
On Sat, Jun 20, 2015 at 6:15 PM, David Winsemius
wrote:
>
> On Jun 20, 2015, at 5:18 PM, Bogdan Tanasa wrote:
>
> > Dear all,
> >
> > I am looking for a suggestion please regarding sorting a dataframe with
> > alphanumerical components :
> >
On Jun 20, 2015, at 5:18 PM, Bogdan Tanasa wrote:
> Dear all,
>
> I am looking for a suggestion please regarding sorting a dataframe with
> alphanumerical components :
>
> let's assume we have :
>
> A = c("A1","A10","A11","A2")
> B = c(1,2,3,4)
>
> C = data.frame(A,B)
>
> how could I sort C
Re:
Try with sort() function
sort(x, decreasing = FALSE, ...) As you can see, u can set the argument
decreasing=TRUE, then, it'll sort by decreasing order.
For more details, call:
help(sort) under R console
Message: 23
Date: Thu, 9 Apr 2015 03:11:37 -0300
From: Luis Fernando Garc?a
To: r-hel
> On 09 Apr 2015, at 08:11 , Luis Fernando García wrote:
>
> Dear R experts,
>
> I am newbie on the R use, now I want to sort my data by mean in a
> descending and not acending order (like I have done so far). If any of you
> could help me with this, it would be really appreciated!
How about
Use order(), as in
sortListByMean <- function(List) {
List[order(vapply(List, mean, 0))]
}
sortedL <- sortListByMean(l)
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Mar 3, 2015 at 11:01 AM, wrote:
> Hello R-helpers,
>
> I have a list of 999 dataframes and I would like to s
Here is an implementation.
> t <-
> data.frame(x=c(1,1,1,1,1,2,2,2,2,2),y=c("a","a","a","b","b","a","a","b","b","b"))
> t
x y
1 1 a
2 1 a
3 1 a
4 1 b
5 1 b
6 2 a
7 2 a
8 2 b
9 2 b
10 2 b
> assignSeq
function(test)
{
temp <- test[order(test$x),]
InC <- numeric(length(test))
inD <-
Hi,
Maybe a beginning of solution with this?
test <-
data.frame(x=c(1,1,1,1,1,1,2,2,2,2,2,2),y=c("a","a","a","b","b","b","a","a","b","b","b","a"))
test[order(test$x),]
out <- split(test,test$x)
for (i in 1:length(out)) {
foo <- unique(out[[i]][,2])
out[[i]][,2] <- rep(foo,(nrow(out[[
Thanks for the quick response.
My work-around was suggested as a quick fix for right-censored data, not as a
general sort method for censored data.
My concern was that sort does not work on right-censored data as described in
the xtfrm documentation.
Mik Bickis
> On Feb 13, 2015, at 05:53 A
Your work around is not as "easy" looking to me.
Survival times come in multiple flavors: left censored, right censored, interval censored,
left-truncated and right censored, and multi-state. Can you give me guidance on how each
of these should sort? If a sort method is added to the package i
On 12/02/2015 16:26, Professor Bickis wrote:
It seems that Surv objects do not sort correctly. This seems to be a bug.
Anyone else found this?
This is presumably about Surv() from package survival, not mentioned.
There was a bug, corrected in R-devel (and I will port to R-patched
before 3.
On 02/12/2014 12:28, Pierrick Bruneau wrote:
I'll try to be more specific:
- ?order points to a common man page for both order() and sort.list()
- sort.list() has a "method" argument, i.e. a sorting algorithm. As far as
I can read, I'll rather say the default is "shell"
Not quite: the page says
On 02/12/2014, 7:28 AM, Pierrick Bruneau wrote:
> I'll try to be more specific:
> - ?order points to a common man page for both order() and sort.list()
> - sort.list() has a "method" argument, i.e. a sorting algorithm. As far as
> I can read, I'll rather say the default is "shell"
> - order() seems
I'll try to be more specific:
- ?order points to a common man page for both order() and sort.list()
- sort.list() has a "method" argument, i.e. a sorting algorithm. As far as
I can read, I'll rather say the default is "shell"
- order() seems not to support this argument. My questions were thus 1)
w
Hi
Can you be more specific? How do you want parametrize order?
There is no sort.list argument in order.
For sort list, the default method is "radix", AFAIK.
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of
> Pierrick Bruneau
> Sent: Tu
Thank you.
Original Message
Subject: Re: Sorting within a dataframe
From: "jfzeac [via R]" < ml-node+s789695n4699000...@n4.nabble.com >
Date: Thu, October 30, 2014 8:57 am
To: dodie < do...@statcourse.com >
Hi, a = data[order(data$x),] If you reply to this email, yo
Hi,
Try:
res1 <- aggregate(PROFIT~DATE, data=dat, FUN=sum)
res1[rev(order(as.Date(res1$DATE, format="%d/%m/%Y"))),]
# DATE PROFIT
#2 02/07/2014 -1350
#1 01/07/2014 9400
#4 30/06/2014 11325
#3 27/06/2014 6850
#if you don't wanted another column in the original dataset with `sum`
re
Hi A.K
I modified and got the results
thanks A.K
library(XML)
URL <- "
http://money.securebank.in/index.php?option=com_dashboard&view=history&Itemid=56&startdate=01/01/2014&enddate=02/07/2014&exchange=MCX&sid=1
"
doc <- htmlParse(URL)
tableNodes <- getNodeSet(doc, "//table")
l=length(tableNodes)
Warning in install.packages :
package âdplyrâ is not available (for R version 2.15.3)
Is there any alternate way to sorting data.frame datewise in a
descending order?.(not using dplyr)
On Thu, Jul 3, 2014 at 12:48 PM, Velappan Periasamy
wrote:
> Hi A.K
> I modified and got the result
Hi A.K
-
library(XML)
URL <- "
http://money.securebank.in/index.php?option=com_dashboard&view=history&Itemid=56&startdate=01/01/2013&enddate=6/9/2014&exchange=MCX&sid=1";
doc <- htmlParse(URL)
tableNodes <- getNodeSet(doc, "//table")
dat1 <- readHTMLTable(tableNodes[[
Hi veepsirtt,
If `dat` is the dataset
library(dplyr)
dat %>%
group_by(DATE) %>%
summarize(PROFIT=sum(PROFIT)) %>%
arrange(desc(as.Date(DATE,format="%d/%m/%Y")))
Source: local data frame [4 x 2]
DATE PROFIT
1 02/07/2014 -1350
2 01/07/2014 9400
3 30/06/2014 11325
4 27/06/2014
If you already have the order stored in a list or so:
For example:
dat1 <- as.data.frame(mat,stringsAsFactors=FALSE)
lst1 <- list(c("OF","ON"), c("US","UK", "WW","BO","BR","CA"),
c("P2","P3","P1"),c("S2","S1","S3"))
dat1[] <- lapply(seq_along(lst1),function(i)
factor(dat1[,i],levels=lst1[[i]]))
Hi,
Not sure whether this helps:
dat1 <- as.data.frame(mat,stringsAsFactors=FALSE)
dat1$c4 <- factor(dat1$c4,levels=c("OF","ON"))
dat1$c1 <- factor(dat1$c1,levels=c("US","UK","WW","BO","BR","CA"))
dat1$c2 <- factor(dat1$c2, levels=c("P2","P3","P1"))
dat1$c3 <- factor(dat1$c3, levels=c("S2","S1"
Hi,
Hope this is what you meant..
with(dat1,mean(V21[(V2==1|V2==0) & V24<25],na.rm=TRUE))
#[1] 2.8125
sapply(0:1,function(i) with(dat1,mean(V21[V2==i & V24<25],na.rm=TRUE)))
#[1] 2.75 3.00
". who are in condition 1 or 0 (V2) and then vice
versa ..."
dat1[,21]
# [1] NA 3.40 3.00 3
Hi,
It's not clear what you really wanted. May be this helps:
dat1<- read.table(text="
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21
V22 V23 V24 V25 V26
1 1 1 3 1 NA NA 1 NA NA 1 2 7 5 3 2 1.0 15 17 5 1
NA 1 2 19 1 1
2
Dear Sir,
Thanks a lot for your valuable input and guidance.
Regards
Katherine
--- On Mon, 15/4/13, Jeff Newmiller wrote:
From: Jeff Newmiller
Subject: Re: [R] Sorting data.frame and again sorting within data.frame
To: "David Winsemius" , "Katherine Gobin"
Cc: r-hel
On Apr 15, 2013, at 9:33 AM, Jeff Newmiller wrote:
> Yes, that would be because she converted to Date on the fly in her example,
> and so apparently did not need this reminder.
I apologize, Iobviously missed that. So the answer was simply to put a minus
sign in front of the as.Date() expressi
Yes, that would be because she converted to Date on the fly in her example, and
so apparently did not need this reminder.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#
On Apr 14, 2013, at 11:01 PM, Katherine Gobin wrote:
> Dear R forum,
>
> I have a data.frame as defied below -
>
> df = data.frame(names = c("C", "A", "A", "B", "C", "B", "A", "B", "C"), dates
> = c("4/15/2013", "4/13/2013", "4/15/2013", "4/13/2013", "4/13/2013",
> "4/15/2013", "4/14/2013",
/2013 29
#5 C 4/13/2013 11
A.K.
- Original Message -
From: arun
To: Katherine Gobin
Cc: R help
Sent: Monday, April 15, 2013 8:57 AM
Subject: Re: [R] Sorting data.frame and again sorting within data.frame
library(plyr)
arrange(df,names,desc(dates))
# names dates values
#1
library(plyr)
arrange(df,names,desc(dates))
# names dates values
#1 A 4/15/2013 31
#2 A 4/14/2013 102
#3 A 4/13/2013 31
#4 B 4/15/2013 34
#5 B 4/14/2013 47
#6 B 4/13/2013 17
#7 C 4/15/2013 10
#8 C 4/14/2013 29
#9 C 4/13/2013
The examples in ?order show a method that could be applied if you avoid the
decreasing argument and instead convert the Date to numeric for purposes of
sorting.
---
Jeff NewmillerThe . ..
Dear LondonPhd,
assuming that you have assigned 'mod' to your VAR() call, you can run the
following:
lapply(coef(mod), function(x) x[sort(rownames(x)), ])
In general, the coef-method will retrieve the estimated coefficients and you
can then do the reordering to your liking.
Best,
Bernhard
--
Hi,
after all, brute force seems the way to go.
I will use a simplified example to illustrate what I want (dump of dat4 is
below):
suppose dat4:
ID rrt Mnd Result
1 0.45 00.1
1 0.48 00.3
1 1.24 00.5
2 0.45 30.2
2 0.48 30.6
2 1.22 30.4
I want to ge
old 0.3 0.6 1.20 1.80
#4 1.2225 old 0.5 0.4 0.45 0.50
A.K.
- Original Message -
From: Bart Joosen
To: Dennis Murphy ; r-help@r-project.org
Cc:
Sent: Friday, January 25, 2013 1:48 AM
Subject: Re: [R] sorting/grouping/classification problem?
Nice suggestion for the extra "Time" column
y possible
combination of shuffling within certain limits (eg max 10% or so), calculate r2
for each combination and maximize?
Seems so brute force and low elegant?
Bart
> Date: Fri, 25 Jan 2013 10:01:44 -0800
> From: smartpink...@yahoo.com
> Subject: Re: [R] sorting/grouping/classific
clear now.
Thank you all for your suggestions
Bart
> Date: Thu, 24 Jan 2013 15:01:40 -0800
> Subject: Re: [R] sorting/grouping/classification problem?
> From: djmu...@gmail.com
> To: bartjoo...@hotmail.com
>
> Hi:
>
> Here's a potential workaround:
>
> # Add
dat3 is the dataframe where there are some rrt values merged, which is actually
the problem: how on Earth discide which rows van be merged
Thanks for your input!
Bart
-Original Message-
From: arun
Sent: 24 Jan 2013 20:18:28 GMT
To: Bart Joosen
Cc: R help
Subject: Re: [R] sorting
HI,
If I understand your question, "dat3" is not you wanted.
Is it something like this you wanted?
library(reshape2)
dcast(dat,rrt~Mnd,value.var="Result")
# rrt 0 3 6 9
#1 0.35 NA NA 0.05 NA
#2 0.36 NA NA NA 0.06
#3 0.44 NA NA 0.40 NA
#4 0.45 0.1 0.2 NA 0.60
#5 0.46 N
Spring 0.1927715
> #16 Spring 0.1927715
>
> The process you describe does not get me there
>
> Any other recommendations?
>
> -Original Message-
> From: arun [mailto:smartpink...@yahoo.com]
> Sent: Thursday, October 11, 2012 10:33 AM
> To: ROLL Josh F
> Cc: R help
>
ly, the Obs column has only 4 values. Do you want to
get the means???
A.K.
- Original Message -
From: ROLL Josh F
To: 'arun'
Cc: R help
Sent: Thursday, October 11, 2012 1:42 PM
Subject: RE: [R] Sorting a data frame by specifying a vector
Sorry if I wasn't clear but th
Winter 0.9318599
> #10 Winter 0.9318599
> #14 Winter 0.9318599
> #4 Spring 0.1927715
> #8 Spring 0.1927715
> #12 Spring 0.1927715
> #16 Spring 0.1927715
>
> Any other thoughts?
>
> JR
>
>
> -Original Message-
> From: Bert Gunter [mailto:gunter.ber
tober 11, 2012 10:33 AM
To: ROLL Josh F
Cc: R help
Subject: Re: [R] Sorting a data frame by specifying a vector
Hi,
In your dataset, it seems like it is already ordered in the way you wanted to.
df.. <- data.frame(Season=rep(c("Summer","Fall","Winter","Spring&
Hi,
In your dataset, it seems like it is already ordered in the way you wanted to.
df.. <- data.frame(Season=rep(c("Summer","Fall","Winter","Spring"),4),Obs=
runif(length(rep(c("Summer","Fall","Winter","Spring"),4
#Suppose the order you want is:
vec2<-c("Summer","Winter","Fall","Spring")
df1
?order
df[order(yourcolumn, ]
-- Bert
On Thu, Oct 11, 2012 at 10:08 AM, LCOG1 wrote:
> Hello all,
>I cannot seem to figure out this seemingly simple procedure.
>
> I want to sort a data frame by a specified character vector.
>
> So for :
>
> df.. <- data.frame(Season=rep(c("Summer","Fall","
Please show the data set...
--
View this message in context:
http://r.789695.n4.nabble.com/sorting-dataframe-tp840507p4644857.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
h
HI,
I guess this might be what you want.
dat1<-structure(c(0.585528817843856, 0.709466017509524, 0, -0.453497173462763,
0.605887455840393, -1.81795596770373, 0.630098551068391,
-0.276184105225216,
-0.284159743943371, -0.919322002474128, -0.116247806352002,
1.81731204370422,
0.370627864257954, 0.520
HI,
Not sure I understand the last part.
Try this:
dat1<-structure(c(0.585528817843856, 0.709466017509524, 0, -0.453497173462763,
0.605887455840393, -1.81795596770373, 0.630098551068391,
-0.276184105225216,
-0.284159743943371, -0.919322002474128, -0.116247806352002,
1.81731204370422,
0.370627864257
Met [mailto:nicome...@gmail.com]
> *Sent:* Wednesday, August 29, 2012 4:56 PM
> *To:* PIKAL Petr
>
> *Subject:* Re: [R] Sorting of columns of a matrix
>
> ** **
>
> Thanks a lot.
>
> ** **
>
> But I also want to attach row names (if avai
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Nico Met
> Sent: Wednesday, August 29, 2012 4:25 PM
> To: Berend Hasselman
> Cc: r-help@r-project.org
> Subject: Re: [R] Sorting of columns of a matrix
Hi
From: Nico Met [mailto:nicome...@gmail.com]
Sent: Wednesday, August 29, 2012 5:27 PM
To: PIKAL Petr
Cc: r-help
Subject: Re: [R] Sorting of columns of a matrix
Yes, each column shall have different sequence of row names
And what? I do not understand? You still did not describe how do you
Met [mailto:nicome...@gmail.com]
Sent: Wednesday, August 29, 2012 4:56 PM
To: PIKAL Petr
Subject: Re: [R] Sorting of columns of a matrix
Thanks a lot.
But I also want to attach row names (if available) associated with each column.
In that case does your second code works in the same way?
thanks
Please find the require info:
set.seed(12345)
X<-matrix(rnorm(5*10),nrow=5)
dim(X)
X
[,1] [,2][,3] [,4] [,5] [,6]
[,7] [,8] [,9] [,10]
[1,] 1.2774431 -1.2427735 0.81933548 -2.1098586 -1.6726799 -2.2994684
-0.28823228 0.192930
On 29-08-2012, at 16:08, Nico Met wrote:
> Hello john,
>
> thanks for the suggestion. Please find an example:
>
>> X<-matrix(rnorm(5*10),nrow=5)
>
>> dim(X)
> [1] 5 10
>> X
> [,1] [,2][,3] [,4] [,5] [,6]
> [,7] [,8] [,9] [,10]
> [1,
Hello john,
thanks for the suggestion. Please find an example:
> X<-matrix(rnorm(5*10),nrow=5)
> dim(X)
[1] 5 10
> X
[,1] [,2][,3] [,4] [,5] [,6]
[,7] [,8] [,9] [,10]
[1,] 1.2774431 -1.2427735 0.81933548 -2.1098586 -1.6726799 -2.2
Perhaps you could supply the matrix using dput() ? It is a handy way to supply
sample data. Just do dput(mydata), copy the results and paste into email.
At the moment your matrix is almost unreadable.
John Kane
Kingston ON Canada
> -Original Message-
> From: nicome...@gmail.com
> Sen
Hi Tony,
Try this:
dataT<-data.frame(Var1=rep(c("Sole","Lack","ABD","Zad"),rep(5,4)),
Var2=rnorm(20,0.5),Var3=runif(20,0.4))
dataT1<-dataT[with(dataT,order(Var1,Var2,Var3)),]
dataT1
Var1 Var2 Var3
12 ABD -0.19842353 0.4333720
13 ABD 0.14050814 0.9194297
11 ABD 1.07544531 0.453
If you want to sort by the columns with the names 'var1' and 'var2',
you would do:
newdata <- data[order(data$var1, data$var2), ]
On Wed, May 30, 2012 at 6:07 PM, tony.anderson wrote:
> I am a novice user of R and am stumbling on how to order a dataset produced
> during my session.
>
> I have
NICE DDE
It solves my problem !
Awesome stuff
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On 20-Feb-2012 Petr Savicky wrote:
> On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
>> I did, but this does not give the answer to my question...
>> Anybody knows how to tweack the behaviour of sort or how to do ?
>
> Hi.
> Try this
>
> Sys.setlocale("LC_COLLATE", "C")
>
> This c
On 2012-2-20 23:15, Rui Barradas wrote:
Could it be OS related?
Yes, it seems. I tried it on my local windows xp and redhat linux
server, and got different results. Hope it will be fixed in the future
versions. Maybe we should keep alert to check whether the results are
consistent when trans
Sorry, just made a mistake. This is the result from windows xp.
> sort(c("X.","X0B"))
[1] "X." "X0B"
> sort(c("X.Z","X0B.Z"))
[1] "X.Z" "X0B.Z"
> R.version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major
On Mon, Feb 20, 2012 at 04:56:21PM +0100, Petr Savicky wrote:
> On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
> > I did, but this does not give the answer to my question...
> > Anybody knows how to tweack the behaviour of sort or how to do ?
>
> Hi.
>
> Try this
>
> Sys.setlocale
It seems OS-dependent. I got different results when trying it on windows
xp and Redhat linux.
> R.version
_
platform x86_64-unknown-linux-gnu
arch x86_64
os linux-gnu
system x86_64, linux-gnu
status
major 2
minor 9.1
year
Ok I have :
R) str(R.Version())
List of 13
$ platform : chr "x86_64-unknown-linux-gnu"
$ arch : chr "x86_64"
$ os: chr "linux-gnu"
$ system: chr "x86_64, linux-gnu"
$ status: chr ""
$ major : chr "2"
$ minor : chr "12.2"
$ year
Hello,
statquant2 wrote
>
> Ok so it changed from 2.12.2 to 2.14.1 ??
> Can somebody tell me how to modify my sort or whatever to get the save
> resilt that I would get in 2.14.1 ?
>
> Cheers
>
I don't know about 2.12.2 but for 2.12.0 I get:
> R.version
_
On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
> I did, but this does not give the answer to my question...
> Anybody knows how to tweack the behaviour of sort or how to do ?
Hi.
Try this
Sys.setlocale("LC_COLLATE", "C")
This comes from ?locale and reads there
Sys.setloca
I don't *think* it's version specific, but rather it depends on your
(still unstated) locale, as the documentation goes to great lengths to
point out. Change that and you might see different behaviors.
Michael
On Mon, Feb 20, 2012 at 8:55 AM, statquant2 wrote:
> I did, but this does not give the
I did, but this does not give the answer to my question...
Anybody knows how to tweack the behaviour of sort or how to do ?
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_
Ok so it changed from 2.12.2 to 2.14.1 ??
Can somebody tell me how to modify my sort or whatever to get the save
resilt that I would get in 2.14.1 ?
Cheers
--
View this message in context:
http://r.789695.n4.nabble.com/Sorting-strings-tp4403696p4403858.html
Sent from the R help mailing list arch
See ?Comparison, which holds some warnings about what to expect when
sorting strings.
Am 20.02.2012 11:51, schrieb Petr Savicky:
On Mon, Feb 20, 2012 at 02:18:42AM -0800, statquant2 wrote:
Hi all, I am having difficulties to understand how R sort strings:
If I do
R) sort(c("X.","X0B"))
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