Dear all,
Thank you for your email and help. I solved the problem!
All the best!
Catalin
On Tue, 7 Jul 2020 at 20:43, Rasmus Liland wrote:
> On 2020-07-07 19:23 +0200, Thierry Onkelinx wrote:
> >
> > Don't use the cut() function.
>
> Ah, I see it now. Changing
>
> fill = cut(cor, zCu
On 2020-07-07 19:23 +0200, Thierry Onkelinx wrote:
>
> Don't use the cut() function.
Ah, I see it now. Changing
fill = cut(cor, zCuts)
to
fill = cor
did it, probably. Perhaps Catalin agrees.
__
R-help@r-project.org mailing list -
Don't use the cut() function.
ir. Thierry Onkelinx
Statisticus / Statistician
Vlaamse Overheid / Government of Flanders
INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE AND
FOREST
Team Biometrie & Kwaliteitszorg / Team Biometrics & Quality Assurance
thierry.onkel...@inbo.be
On 2020-07-07 12:44 +0200, Thierry Onkelinx via R-help wrote:
> Op di 7 jul. 2020 om 12:02 schreef Catalin Roibu :
> >
> > Dear R users,
> >
> > I want to create a plot for multiple
> > sites and to keep the same color
> > range scale (the correlation values
> > range from -0.5 to 0.7 for all d
Dear Catalin,
use scale_fill_gradient() and set fixed limits
ggplot(df1, aes(x=as.factor(spei), y=as.factor(month), fill = cut(cor,
zCuts))) +
geom_tile() +
scale_fill_gradient(limits = c(-0.7, 0.7))
Best regards,
ir. Thierry Onkelinx
Statisticus / Statistician
Vlaamse Overheid / Governmen
Dear R users,
I want to create a plot for multiple sites and to keep the same color range
scale (the correlation values range from -0.5 to 0.7 for all data, but I
have sites with different min and max).
I used this code:
cols<-c("#0288D1", "#039BE5", "#03A9F4","#29B6F6", "#4FC3F7", "#FFCDD2",
"#E
ty
Hamilton, Ontario, Canada
Web: socialsciences.mcmaster.ca/jfox/
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Knut
> Krueger
> Sent: Monday, November 19, 2018 9:42 AM
> To: r-help@r-project.org >> r-help mailing list
> Subject
Hello,
Try
i <- !(duplicated(Dup) | duplicated(Dup, fromLast = TRUE))
Dup[i]
or in one line, I post it like this to make it more clear.
Hope this helps,
Rui Barradas
Às 14:41 de 19/11/2018, Knut Krueger escreveu:
It should be simple but i do not find the right keywords:
Dup = c(1,2,3,4,
Hi
and maybe sloightly less complicated
setdiff(Dup,Dup[duplicated(Dup)])
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Knut Krueger
> Sent: Monday, November 19, 2018 3:42 PM
> To: r-help@r-project.org >> r-help mailing list
> Subject: [R] unique()
Hi
Dup %in% Dup[duplicated(Dup)]
Dup[!(Dup %in% Dup[duplicated(Dup)])]
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Knut Krueger
> Sent: Monday, November 19, 2018 3:42 PM
> To: r-help@r-project.org >> r-help mailing list
> Subject: [R] unique() d
It should be simple but i do not find the right keywords:
Dup = c(1,2,3,4,1,2,3,5)
I need 4,5 as result
unique(Dup) gives me [1] 4 1 2 3 5
duplicated(Dup) gives me
[1] FALSE FALSE FALSE FALSE TRUE TRUE TRUE FALSE
I need
[1] TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE
Kind regards Knut
library(data.table)
setDT(df)
setkeyv(df, c("Subject", "dates"))
unique(df) #gets what you want.
On Mon, Nov 14, 2016 at 11:38 PM, Jim Lemon wrote:
> Hi Farnoosh,
> Try this:
>
> for(id in unique(df$Subject)) {
> whichsub<-df$Subject==id
> if(exists("newdf"))
> newdf<-rbind(newdf,df[whichsu
Hi Farnoosh,
Try this:
for(id in unique(df$Subject)) {
whichsub<-df$Subject==id
if(exists("newdf"))
newdf<-rbind(newdf,df[whichsub,][which(!duplicated(df$dates[whichsub])),])
else newdf<-df[whichsub,][which(!duplicated(df$dates[whichsub])),]
}
Jim
On Tue, Nov 15, 2016 at 9:38 AM, Farnoosh
Hi Farnoosh,
you can use unique in the R-base or distinct from the dplyr library.
Best
Ulrik
On Tue, 15 Nov 2016 at 06:59 Farnoosh Sheikhi via R-help <
r-help@r-project.org> wrote:
> Hi,
> I have a data set like below:
> Subject<- c("2", "2", "2", "3", "3", "3", "4", "4", "5", "5", "5",
> "5")d
Hi,
I have a data set like below:
Subject<- c("2", "2", "2", "3", "3", "3", "4", "4", "5", "5", "5",
"5")dates<-c("2011-01-01", "2011-01-01", "2011-01-03" ,"2011-01-04",
"2011-01-05", "2011-01-06" ,"2011-01-07", "2011-01-07", "2011-01-09"
,"2011-01-10" ,"2011-01-11" ,"2011-01-11")deps<-
Hi Edward,
I'm not really sure that this is what you want as I can't figure out
what the "earn" factor is, but:
epdat[order(epdat$Var2,epdat$Freq,decreasing=TRUE),]
Jim
On Sat, Apr 23, 2016 at 4:08 AM, Patzelt, Edward wrote:
> Hi R-Help,
>
> data at bottom
>
> I've been struggling with a probl
Hi R-Help,
data at bottom
I've been struggling with a problem where I need to order based on 1) the
Frequency "Freq" and 2) keeping each group of 3 of the same type together
"Var2" but I want across all groups it to go "high to low" based on the
earn factor.
Thank you!
structure(list(Var1 = str
cumsum(c(x[1],pmax(0,diff(x*x
Am 23.11.2015 um 15:59 schrieb PIKAL Petr:
> Dear all
>
> I have a vector ones and zeroes like that
> x<-c(rep(0,5), rep(1,5), rep(0,10), rep(1,8))
>
> and I need to get result like that
> x.i<-c(rep(0,5), rep(1,5), rep(0,10), rep(2,8))
>
> It means I need an u
> f <- function(x)cumsum(c(x[1]==1, x[-length(x)]==0 & x[-1]==1)) * (x==1)
> f(x)
[1] 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2
> f(rev(x))
[1] 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 0 0 0 0 0
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Nov 23, 2015 at 6:59 AM, PIK
[mailto:jholt...@gmail.com]
> *Sent:* Monday, November 23, 2015 4:14 PM
> *To:* PIKAL Petr
> *Cc:* r-help@r-project.org
> *Subject:* Re: [R] unique identifier for number sequence
>
>
>
> Here is one way of doing it:
>
>
>
> > x<-c(rep(0,5), rep(1,5), rep(0,10)
Hi
Cool, thanks. I knew I am missing some obvious way.
Cheers
Petr
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Monday, November 23, 2015 4:14 PM
To: PIKAL Petr
Cc: r-help@r-project.org
Subject: Re: [R] unique identifier for number sequence
Here is one way of doing it:
> x<-c(r
Here is one way of doing it:
> x<-c(rep(0,5), rep(1,5), rep(0,10), rep(1,8))
> x
[1] 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
>
> # mark changes from 0->1 and create increments
> indx <- cumsum(c(FALSE, diff(x) == 1))
>
> # keep just matches with '1'
> x.i <- ifelse(x == 1, indx, 0
Dear all
I have a vector ones and zeroes like that
x<-c(rep(0,5), rep(1,5), rep(0,10), rep(1,8))
and I need to get result like that
x.i<-c(rep(0,5), rep(1,5), rep(0,10), rep(2,8))
It means I need an unique identifier for each sequence of ones.
It probably can be done by rle, cumsum and some fid
You messed up the quote marks and the "library" function is not capitalized.
You defined your search list by the name "list", which is also the name of a
commonly used base function in R. Also, the vector you gave to GetAllSubsets
had several misleading invisible conversions to character because
Hello list
I would like to know how can i detect dataframe columns that have as
unique values elements of a list.
For example, at a given dataframe to look for columns that unique values
are elements of a list like this one
dataframe<-data.frame(
x = c("yes, "1, "no, "no),
y = c("black,
sorry.. don't know unique().. such a great function
From: Bert Gunter
Cc: "r-help@r-project.org"
Sent: Tuesday, January 28, 2014 2:21 PM
Subject: Re: [R] unique rows
Inline.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650)
Hi,
use ?unique
unique(dat)
A.K.
Hi, I wanted to remove redundant rows (with same entry in columns) in a data
frame. For example, with this data frame:
> dat<-cbind(x=c('a','a','b','b','c','c'),y=c('x','x','d','s','g','g'))
> dat
x y
[1,] "a" "x"
[2,] "a" "x"
[3,] "b" "d"
[4,] "
Inline.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
H. Gilbert Welch
On Tue, Jan 28, 2014 at 2:06 PM, array chip wrote:
> Hi, I wanted to remove redundant rows (with sa
Hi, I wanted to remove redundant rows (with same entry in columns) in a data
frame. For example, with this data frame:
> dat<-cbind(x=c('a','a','b','b','c','c'),y=c('x','x','d','s','g','g'))
> dat
x y
[1,] "a" "x"
[2,] "a" "x"
[3,] "b" "d"
[4,] "b" "s"
[5,] "c" "g"
[6,] "c" "g"
after re
We answered this on StackOverflow already. Excel was doing
case-insensitive duplicate matching.
http://stackoverflow.com/questions/20759346/counting-unique-values-in-r-and-excel/20759523#20759523
Barry
On Tue, Dec 24, 2013 at 5:43 PM, David Winsemius wrote:
>
> On Dec 24, 2013, at 1:08 AM, Kous
On Dec 24, 2013, at 1:08 AM, Koushik Saha wrote:
> i have a wired problem. i want to count the unique entry in a certain
> column.Here i have attached my csv file.
Files named with extension .csv do not typically make it through the R-help
mail server.
>
> i am doing this to get the unique en
On 13-12-24 4:08 AM, Koushik Saha wrote:
i have a wired problem. i want to count the unique entry in a certain
column.Here i have attached my csv file.
i am doing this to get the unique entries in the column.
dat<-read.csv("C:/Project/Gawk-scripts/Book1.csv")
names(dat)<-c("user_name")
unique(d
i have a wired problem. i want to count the unique entry in a certain
column.Here i have attached my csv file.
i am doing this to get the unique entries in the column.
dat<-read.csv("C:/Project/Gawk-scripts/Book1.csv")
names(dat)<-c("user_name")
unique(dat$user_name)
results says i have 170 uniq
roject.org
Subject: Re: [R] Unique in discerning missing values NA
Hello,
Your data example is difficult to read into an R session. Next time, post the
output of ?dput. Like this:
dput(menPatients[1:40, 1]) # post the output of this
The help page for unique says that "Missing values are re
Hello,
Your data example is difficult to read into an R session. Next time,
post the output of ?dput. Like this:
dput(menPatients[1:40, 1]) # post the output of this
The help page for unique says that "Missing values are regarded as
equal" so you should expect one NA to still be present i
Hi,
testUnique <- unique(testData[!is.na(testData)])
or
testUnique <- unique(na.omit(testData))
And probably some other solutions.
Regards,
Pascal
2013/7/5 Pancho Mulongeni
> Hi,
> I am trying to remove duplicate Patient numbers in a clinical record, I
> used unique
> menPatients[1:40,1]
>
Hi,
I am trying to remove duplicate Patient numbers in a clinical record, I used
unique
menPatients[1:40,1]
[1] abr1160(C)/001 ABR1363(A)/001 ABR1363(A)/001 ABR1363(A)/001 abr1772(B)/001
[6] AFR0003/001AFR0003/001afr0290(C)/001 afr1861(B)/001 Aga0007/001
[11] AGA1548(A)/001 AGA1548(A)
You could give 1-nearest neighbor classification a try. For example,
a <- data.frame(person=1:10, ht=rnorm(10, mean=5, sd=1),
wt=rnorm(10, mean=180, sd=30), bp=rnorm(10, mean=120, sd=10))
meas.err <- data.frame(ht=rnorm(10, sd=0.1),
wt=rnorm(10, sd=3), bp=rnorm(10, sd=1))
b <- (a[, -1]
Dear list,
I've searched the archives and tried some code, however would appreciate some
input - even a pointer in the direction of the correct function to use.
Given N samples each of which is measured for characteristics x1, x2, x3,... (m
6) where each characteristic is a roughly normally di
i meant "uncountry". i am creating one now/ dput().
thanks.
~Nicole Ford
Ph.D. student
Graduate Assistant/ Instructor
University of South Florida
Government and International Affairs
office: SOC 012M
On Mar 28, 2013, at 11:55 AM, Sarah Goslee wrote:
> Hi,
>
> I'm confused.
>
> On Thu, M
Now R isn't running on my computer AT ALL. i deleted it and d/l'd again,
trying to get it to work, it keeps crashing. so perhaps this is a hardware
issue. the absolute worst timing.
~Nicole Ford
Ph.D. Student
Instructor: Empirical Political Analysis
Department of Government and International Affa
Hi,
I'm confused.
On Thu, Mar 28, 2013 at 10:49 AM, Nicole Ford wrote:
> i am using mac OSX 10.7.5, running R version 2.15.2 (2012-10-26) -- "Trick or
> Treat"
>
> when i do:
> uncountry <- unique(wvsAB[,7])
> wvsAB$numcountry <- match(wvsAB$country, uncountry)
> "unstate" isn't attaching.
Wha
i am using mac OSX 10.7.5, running R version 2.15.2 (2012-10-26) -- "Trick or
Treat"
when i do:
uncountry <- unique(wvsAB[,7])
wvsAB$numcountry <- match(wvsAB$country, uncountry)
"unstate" isn't attaching.
> library(base)
> uncountry <- unique(wvsAB[,7])
> wvsAB$numcountry <- match(wvsAB$country
If I understand your problem correctly (as Milan has pointed out, sample data
and code would help enormously) this should get you where you want:
unique( shopdata$name[ shopdata$employee > 10 ] )
If not, something is wrong from the outset (or with my understanding, but then
... see above)!
R
Le mardi 16 octobre 2012 à 14:45 +0200, paladini a écrit :
> Hello everybody,
> I've got a problem concerning the function unique. I have got a
> data.frame "shopdata" with 1000 shop which were evaluated at different
> points in time.
>
> With function subset I chose those shops with more then 1
Hello everybody,
I've got a problem concerning the function unique. I have got a
data.frame "shopdata" with 1000 shop which were evaluated at different
points in time.
With function subset I chose those shops with more then 10 employee and
store it in data.frame "bigshopdata" with 700 shops.
On 9/6/2012 7:24 PM, Bert Gunter wrote:
... and if it Duncan's suggestion won't do, maybe approaching it via
clustering might be useful.
But do note that, as stated, the problem is not well defined, because
transitivity fails: consider
v <- c(1,2,3,4,5,10)
with a tolerance of <=2. Then 1 is the
Thanks to both of you for suggestions.
I settled for the round() approach.
much obliged,
Michael Bach
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-g
... and if it Duncan's suggestion won't do, maybe approaching it via
clustering might be useful.
But do note that, as stated, the problem is not well defined, because
transitivity fails: consider
v <- c(1,2,3,4,5,10)
with a tolerance of <=2. Then 1 is the same as 2 and 3, 2 and 3 are
the same as
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Michael Bach
> Sent: Thursday, September 06, 2012 12:48 PM
> To: r-h...@stat.math.ethz.ch
> Subject: [R] unique with tolerance
>
> Dear R Users a
On 06/09/2012 6:48 AM, Michael Bach wrote:
Dear R Users and Developers,
I am trying to do the equivalent of
v <- c(1,2,3,3,2,1,)
vu <- unique(v)
for a vector such as
v2 <- c(1.02, 2.03, 1.00, 3.04, 3.06)
vut <- ...
As indicated in the subject, we need approximately unique values with a defin
Dear R Users and Developers,
I am trying to do the equivalent of
v <- c(1,2,3,3,2,1,)
vu <- unique(v)
for a vector such as
v2 <- c(1.02, 2.03, 1.00, 3.04, 3.06)
vut <- ...
As indicated in the subject, we need approximately unique values with a defined
tolerance, i.e. for the v2 vector the resu
Bert,
Is it important that you end up with a data frame? If not, it would be
very easy to generate a list with the unique values for each column. For
example:
df <- data.frame(v1 = sample(5, 20, T), v2 = sample(7, 20, T),
v3 = sample(9, 20, T), v4 = sample(11, 20, T))
lapply(df, uniq
Hi,
I was wondering what the best way is to create a new dataframe based on an
existing dataframe with only the unique available levels for each column (22
columns in total) in it.
If some columns have less unique values than others, then those columns can
be filled with blanks for the remaining
x2),fun(x2))
#[1] TRUE
A.K.
- Original Message -
From: Peter Ehlers
To: arun
Cc: Nico902 ; R help
Sent: Monday, July 9, 2012 4:14 PM
Subject: Re: [R] unique vs duplicate problem
On 2012-07-09 11:07, arun wrote:
> Hi,
> Try this:
> #Duplicated:
> x<-c(1:3,3)
> x==x[
eter Ehlers
- Original Message -
From: Nico902
To: r-help@r-project.org
Cc:
Sent: Monday, July 9, 2012 12:42 PM
Subject: [R] unique vs duplicate problem
Hi,
Let say I have a numeric vector: x <- c(1, 2, 3, 3).
I want on one hand numbers which are not duplicated ie "1,2" and d
Hi,
Try this:
#Duplicated:
x<-c(1:3,3)
x==x[duplicated(x)]
#[1] FALSE FALSE TRUE TRUE
#Unique:
x[!x==x[duplicated(x)]]
#[1] 1 2
A.K.
- Original Message -
From: Nico902
To: r-help@r-project.org
Cc:
Sent: Monday, July 9, 2012 12:42 PM
Subject: [R] unique vs duplicate prob
excellent!!! thanks a lot!!
--
View this message in context:
http://r.789695.n4.nabble.com/unique-vs-duplicate-problem-tp4635868p4635874.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.eth
Hello,
Maybe this function.
fun <- function(x) x %in% x[duplicated(x)]
x <- c(1, 2, 3, 3)
fun(x)
Hope this helps,
Rui Barradas
Em 09-07-2012 17:42, Nico902 escreveu:
Hi,
Let say I have a numeric vector: x <- c(1, 2, 3, 3).
I want on one hand numbers which are not duplicated ie "1,2" a
Here is one way of doing it -- you can create your own functions:
> x <- c(1, 2, 3, 3)
>
> allDup <-
+ function (value)
+ {
+ duplicated(value) | duplicated(value, fromLast = TRUE)
+ }
>
> duped <- unique(x[allDup(x)])
> duped
[1] 3
>
> setdiff(unique(x), duped)
[1] 1 2
>
>
On Mon, Jul 9, 2
Hi,
Let say I have a numeric vector: x <- c(1, 2, 3, 3).
I want on one hand numbers which are not duplicated ie "1,2" and duplicated
"3".
so I did:
>duplicated(x)
FALSE FALSE FALSE TRUE
> unique(x)
1 2 3
which is not what I want. Is there a function in R to have the following
result:
>dupli
It's a bit of a hack, but I think you can try something like this:
x <- c(1,2,3,4,5,2,5)
duplicated(x) | duplicated(x, fromLast=T)
Michael
On Tue, Mar 27, 2012 at 7:02 PM, z2.0 wrote:
> I'm sure there's a better way to do this using plyr. I just can't nail the
> right series of commands.
>
> I
I'm sure there's a better way to do this using plyr. I just can't nail the
right series of commands.
I've got a vector of strings. I want to remove all where the string's count
within the vector is > 1.
Right now I'm using:
Where x2 is the original data.frame and my character strings live i
Thanks.
--
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http://r.789695.n4.nabble.com/Unique-in-DataFrame-tp4488943p4489554.html
Sent from the R help mailing list archive at Nabble.com.
__
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This may be what you want:
> x<-data.frame(city="Barcelona",sales=253639)
> x<-rbind(x,data.frame(city="Madrid",sales=223455))
> x<-rbind(x,data.frame(city="Lisbon",sales=273633))
> x<-rbind(x,data.frame(city="Madrid",sales=266535))
> x<-rbind(x,data.frame(city="Barcelona",sales=258369))
> x<-rbin
Hello,
I have little doubt, and I do not think that the way I solve the problem
is the best way to do it.
The following is a small dataset
x<-data.frame(city="Barcelona",sales=253639)
x<-rbind(x,data.frame(city="Madrid",sales=223455))
x<-rbind(x,data.frame(city="Lisbon",sales=273633))
x<-rb
On Wed, 21 Dec 2011, Keith Jewell wrote:
Thanks Uwe,
I was happy that my 2 lines gave what the OP asked for, albeit in a
different order.
My puzzlement arose from Jeff Newmillers comment:
You could read the help for expand.grid very carefully for the answer to
this question.
... which I reas
Thanks Uwe,
I was happy that my 2 lines gave what the OP asked for, albeit in a
different order.
My puzzlement arose from Jeff Newmillers comment:
> You could read the help for expand.grid very carefully for the answer to
> this question.
... which I reas as saying that ?expand.grid would lead t
On 21.12.2011 14:39, Keith Jewell wrote:
OK, someone point it out to me; my wife tells me I can't see what's in front
of me :-}
I read ?expand.grid carefully, went to ?combn and ?choose but still couldn't
see an easy way to get what the OP asked for. The neatest I can get (which
isn't very nea
OK, someone point it out to me; my wife tells me I can't see what's in front
of me :-}
I read ?expand.grid carefully, went to ?combn and ?choose but still couldn't
see an easy way to get what the OP asked for. The neatest I can get (which
isn't very neat!) is:
> myVec <- c(1,2,3)
> eg <- expan
You could read the help for expand.grid very carefully for the answer to this
question.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
On Dec 21, 2011, at 08:59 , Antje Niederlein wrote:
> Hi there,
>
> I have a vector and would like to create a data frame, which contains
> all unique combination of two elements, regardless of order.
>
> myVec <- c(1,2,3)
>
> what expand.grid does:
>
> 1,1
> 1,2
> 1,3
> 2,1
> 2,2
> 2,3
> 3,1
Hi there,
I have a vector and would like to create a data frame, which contains
all unique combination of two elements, regardless of order.
myVec <- c(1,2,3)
what expand.grid does:
1,1
1,2
1,3
2,1
2,2
2,3
3,1
3,2
3,3
what I would like to have
1,1
1,2
1,3
2,2
2,3
3,3
Can anybody help?
_
The error I am referring to is in unique.c in Base R, it cannot
accomodate greater than 2^29 values, even though it appears the
overflow protection should be 2^30. The only relevance of the arules
package is I was using it while I discovered this issue.
Thanks,
Patrick
2011/10/6 Uwe Ligges :
>
I see. For now: Yes, you need to change and recompile.
I will take a look what was actually changed and will run some test cases.
Best,
Uwe
On 06.10.2011 16:50, Patrick McCann wrote:
The error I am referring to is in unique.c in Base R, it cannot
accomodate greater than 2^29 values, even thou
On 05.10.2011 22:15, Patrick McCann wrote:
Hi,
I am trying to read in a rather large list of transactions using the
arules library.
You mean the arules package?
It seems in the coerce method into the dgCmatrix, it
somewhere calls unique. Unique.c throws an error when n> 536870912;
howev
Hi,
I am trying to read in a rather large list of transactions using the
arules library. It seems in the coerce method into the dgCmatrix, it
somewhere calls unique. Unique.c throws an error when n > 536870912;
however, when 4*n was modified to 2*n in 2004, the overflow protection
should have cha
Hi
I am using the 64 bit version. To check that i went in the bin folder and
executed "file r ". It gave the following output
ELF 64-bit LSB executable, AMD x86-64, version 1 (SYSV), for GNU/Linux
2.6.9, dynamically linked (uses shared libs), for GNU/Linux 2.6.9, not
stripped
The error i got when
Could it be that you are running on a 32-bit version of R? 536870912
* 4 = 2GB if those were integers which would use up all of memory.
You never did show what your error message was or what system you were
using.
On Wed, Jan 5, 2011 at 12:08 AM, Indrajeet Singh wrote:
> Hi
> I am using R with i
Hi
I am using R with igraph to analyze an edgelist that is greater than the said
amount. Does anyone know a way around this?
Thanks
Inder
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PLEASE do read the posting guide
Hi,
On Nov 29, 2010, at 7:40 PM, wrote:
How do I create a second dataframe that shows unique column values
from first dataframe and the number count of rows in first dataframe
where column value appears?
For example, first dataframe is this:
x <- matrix(c(101:104,101:104,105:106,1:10),
How do I create a second dataframe that shows unique column values from first
dataframe and the number count of rows in first dataframe where column value
appears?
For example, first dataframe is this:
> x <- matrix(c(101:104,101:104,105:106,1:10), nrow=10, ncol=2)
> x
[,1] [,2]
[1,] 1
Oops, yeah I didn't see that.
Thanks,
Andrew
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I already gave you three examples of how this works. Your last request
can be done in exactly the same way. Give it a try and see what
happens (use example data of course!). As a last resort you could read
the documentation:
?Comparison
?Extract
-Ista
On Wed, Sep 22, 2010 at 2:22 PM, AndrewPage
Hi Andrew,
Perhaps you did not notice my previous email. The answer is still the
same (see below):
On Wed, Sep 22, 2010 at 1:48 PM, AndrewPage wrote:
>
> How about this:
>
>
> s = c("aa", "bb", "cc", "", "aa", "dd", "", "aa")
>
> n = c(2, 3, 5, 6, 7, 8, 9, 3)
>
> b = c(TRUE, FALSE, TRUE, TRUE, FA
Thanks-- that works for what I'm trying to do. I was also wondering, in the
data frame example you gave, if I just wanted to get rid of rows where the
"a" value is 5, how would I do that?
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How about this:
s = c("aa", "bb", "cc", "", "aa", "dd", "", "aa")
n = c(2, 3, 5, 6, 7, 8, 9, 3)
b = c(TRUE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, FALSE)
df = data.frame(n, s, b) # df is a data frame
I want to display df with no value in s occurring more than once. Also, I
want to
Hi Andrew,
You can use duplicated() to index the rows you wish to keep, like this:
test.dat <- data.frame(a=c(1,1:5,5:10), b=1:12, c=letters[1:12]) #make up data
duplicated(test.dat$a) # see what duplicated() function does
!duplicated(test.dat$a) # see how we can invert using the ! function
so t
I just figured that out, but the real data I'm using is a data frame for
sure, so I'll find another example.
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On Sep 22, 2010, at 12:35 PM, AndrewPage wrote:
I understand how duplicated and unique work for a list where all
parts of a
given row are duplicated, or how to find duplicated values if I'm just
looking at that first column, but in this case the rows for 1954
and 1955
are not completely
I understand how duplicated and unique work for a list where all parts of a
given row are duplicated, or how to find duplicated values if I'm just
looking at that first column, but in this case the rows for 1954 and 1955
are not completely the same; only quarter 1 is duplicated, so I'm not sure
h
Hi,
Take a look at ?duplicated and ?unique
HTH,
Ivan
Le 9/22/2010 16:55, AndrewPage a écrit :
> Hi all,
>
> I'm looking at a large data set, and I'm interested in removing rows where
> only one variable is duplicated. Here's an example:
>
>> presidents
> Qtr1 Qtr2 Qtr3 Qtr4
> 1945 NA
Hi all,
I'm looking at a large data set, and I'm interested in removing rows where
only one variable is duplicated. Here's an example:
> presidents
Qtr1 Qtr2 Qtr3 Qtr4
1945 NA 87 82 75
1946 63 50 43 32
1947 35 60 54 55
1948 36 39 NA NA
1949 69 57 57
Hi Ralf,
Perhaps the following is what you are looking for:
d <- data.frame(timestamp=c(3,3,3,5,8), mylabel=c("a","a","a","b","c"))
d
d[!duplicated(d$timestamp),]
HTH,
Jorge
On Fri, Jul 30, 2010 at 12:18 AM, Ralf B <> wrote:
> I have to deal with data frames that contain multiple entries of t
I have to deal with data frames that contain multiple entries of the
same (based on an identifying collumn 'id'). The second collumn is
mostly corresponding to the the id collumn which means that double
entries can be eliminated with ?unique.
a <- unique(data.frame(timestamp=c(3,3,3,5,8), mylabel=
Hi:
Try this:
> for (i in 1:2) {
+ x=c(1,2,3,4)
+ y=c(10,20,30,40)
+ G <- paste("name", i, sep="")
+ assign(G, data.frame(x,y))
+ }
> name1
x y
1 1 10
2 2 20
3 3 30
4 4 40
> name2
x y
1 1 10
2 2 20
3 3 30
4 4 40
HTH,
Dennis
On Wed, Jun 23, 2010 at 3:56 PM, Douglas
\Why not put all the data frames in a list?
Kjetil
On Wed, Jun 23, 2010 at 6:56 PM, Douglas M. Hultstrand
wrote:
> Hello,
>
> I am trying to create a data frame with a unique name, based on indexing of
> for loop. I was wondering if there is a way to do this, I keep running into
> errors when I
Hello,
I am trying to create a data frame with a unique name, based on indexing
of for loop. I was wondering if there is a way to do this, I keep
running into errors when I try to do this. Below is a brief example, I
am trying to get two data frames (name1 and name2).
Any suggestions are a
f Dundee
> Mackenzie Building
> Kirsty Semple Way
> Dundee DD2 4BF
> Scotland UK
> PH: 01382-420119
>
>
> -Original Message-
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: 26 February 2010 1:49 PM
> To: Henrique Dallazuanna
> Cc: Todd DeWees;
David Winsemius [mailto:dwinsem...@comcast.net]
Sent: 26 February 2010 1:49 PM
To: Henrique Dallazuanna
Cc: Todd DeWees; r-help@r-project.org
Subject: Re: [R] Unique Values of a Matrix
On Feb 26, 2010, at 8:17 AM, Henrique Dallazuanna wrote:
> Try this:
>
> unique(c(X))
I did but it retu
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