table(data.o$Var.1[1:6]) #scheme 3
>> >
>> > A B C D
>> > 2 1 2 1
>> >
>> > > table(data.o$Var.1[1:7]) # scheme1
>> > A B C D
>> > 2 1 2 2
>> >
>> > > table(data.o$Var.1[1:8]) # no such scheme, so scheme 1 is chos
Var.1[1:8]) # no such scheme, so scheme 1 is chosen one
> > A B C D
> > 2 1 2 3
> >
> > #Now you need to select values based on scheme 1.
> > # 3A - 3B - 2C - 2D
> >
> > sss <- split(Order, Order$Var.1)
> > selection <- c(3,3,2,2)
> > re
on scheme 1.
> # 3A - 3B - 2C - 2D
>
> sss <- split(Order, Order$Var.1)
> selection <- c(3,3,2,2)
> result <- vector("list", 4)
>
> #I would use loop
>
> for(i in 1:4) {
> result[[i]] <- sss[[i]][1:selection[i],]
> }
>
> Maybe someone com
ult[[i]] <- sss[[i]][1:selection[i],]
}
Maybe someone come with other ingenious solution.
Cheers
Petr
From: Silvano Cesar da Costa
Sent: Monday, August 23, 2021 7:54 PM
To: PIKAL Petr
Cc: r-help@r-project.org
Subject: Re: [R] Selecting elements
Hi,
I apologize for the confusion. I will t
gt; $D
> [1] 39 77 114 141 166 189 209 223 229 232
>
> Highest value is in D so either 3A - 3B - 2C - 2D or 3A - 3B - 2C - 2D
> should be appropriate. And here I am again lost as both sets are same.
> Maybe you need to reconsider your statements.
>
> Cheers
> Petr
>
>
alue is in D so either 3A - 3B - 2C - 2D or 3A - 3B - 2C - 2D should
be appropriate. And here I am again lost as both sets are same. Maybe you need
to reconsider your statements.
Cheers
Petr
From: Silvano Cesar da Costa
Sent: Friday, August 20, 2021 9:28 PM
To: PIKAL Petr
Cc: r-help@r-project.
Well, I don't think language is as much of a problem as your failure to compose
your messages using plain text format. Your examples are all mushed together
since the mailing list removes formatting. See what we see below for example.
On August 20, 2021 12:27:50 PM PDT, Silvano Cesar da Costa
Hi, thanks you for the answer.
Sorry English is not my native language.
But you got it right.
> As C is first and fourth biggest value, you follow third option and
select 3 highest A, 3B 2C and 2D?
I must select the 10 (not 15) highest values, but which follow a certain
order:
3A - 3B - 2C - 2D
Agreed. Need the rest of a complete example.
On August 19, 2021 11:27:59 PM PDT, PIKAL Petr wrote:
>Hallo
>
>I am confused, maybe others know what do you want but could you be more
>specific?
>
>Let say you have such data
>set.seed(123)
>Var.1 = rep(LETTERS[1:4], 10)
>Var.2 = sample(1:40, replac
Hallo
I am confused, maybe others know what do you want but could you be more
specific?
Let say you have such data
set.seed(123)
Var.1 = rep(LETTERS[1:4], 10)
Var.2 = sample(1:40, replace=FALSE)
data = data.frame(Var.1, Var.2)
What should be the desired outcome?
You can sort
data <- data[order
On Feb 4, 2014, at 11:54 AM, Francesca Pancotto wrote:
> Hello A. k.
> thanks for the suggestion.
>
> I tried this but it does not work. I probably use it in the wrong way.
> This is what it tells me,
>
>
> do.call(rbind,lapply(bank.list,function(x) x[x[,"p_made"]==406,]))
>
> Errore in mat
On 02/05/2014 06:54 AM, Francesca Pancotto wrote:
Hello A. k.
thanks for the suggestion.
I tried this but it does not work. I probably use it in the wrong way.
This is what it tells me,
do.call(rbind,lapply(bank.list,function(x) x[x[,"p_made"]==406,]))
Errore in match.names(clabs, names(xi)
Hi,
Looks like the colnames of list elements are not the same.
For e.g.
lst1 <- list(structure(list(bankname = structure(c(1L, 1L, 1L, 1L, 1L,
1L), .Label = "CIB", class = "factor"), date = structure(c(1L,
2L, 3L, 1L, 2L, 3L), .Label = c("10/02/06", "10/23/06", "11/22/06"
), class = "factor"), px
Hello A. k.
thanks for the suggestion.
I tried this but it does not work. I probably use it in the wrong way.
This is what it tells me,
do.call(rbind,lapply(bank.list,function(x) x[x[,"p_made"]==406,]))
Errore in match.names(clabs, names(xi)) :
names do not match previous names
What am I
Hi,
Try:
If `lst1` is the list:
do.call(rbind,lapply(lst1,function(x) x[x[,"p_made"]==406,]))
A.K.
On Tuesday, February 4, 2014 8:53 AM, Francesca
wrote:
Dear Contributors
sorry but the message was sent involuntary.
I am asking some advice on how to solve the following problem.
I have a list
Dear Contributors
sorry but the message was sent involuntary.
I am asking some advice on how to solve the following problem.
I have a list composed of 78 elements, each of which is a matrix of factors
and numbers, similar to the following
bank_name date px_last_CIB Q.Yp_made p_for
1
Yes the lapply function sorted it out. Thanks for the advice.
--
View this message in context:
http://r.789695.n4.nabble.com/Selecting-elements-from-all-items-in-a-list-tp4387045p4387505.html
Sent from the R help mailing list archive at Nabble.com.
__
On Feb 14, 2012, at 8:44 AM, geotheory wrote:
Basic question (if you know the answer)... I am dealing with a list
of
commonly-formatted sub-lists, for example:
l <- list("")
l[[1]] <- c("A1","A2","A3")
l[[2]] <- c("B1","B2","B3")
l[[3]] <- c("C1","B2","B3")
Lets say I need to extract every
How about:
l <- list("")
l[[1]] <- c("A1","A2","A3")
l[[2]] <- c("B1","B2","B3")
l[[3]] <- c("C1","B2","B3")
lapply(l, "[[", 2)
or
sapply(l, "[[", 2)
I hope it helps.
Best,
Dimitris
On 2/14/2012 2:44 PM, geotheory wrote:
Basic question (if you know the answer)... I am dealing with a li
Try this:
!list %in% c("aa", "bb")
On Fri, Jan 14, 2011 at 10:19 AM, A M Lavezzi wrote:
> Hi everybody,
>
> I have the following problem. I have a vector containing character
> elements,
> such as:
>
> list = c("aa","bb","cc","dd","ee")
>
> I want to create an index which identifies the element
it works!
thank you so much
Mario
On Fri, Jan 14, 2011 at 1:29 PM, Henrique Dallazuanna wrote:
> Try this:
>
> !list %in% c("aa", "bb")
>
> On Fri, Jan 14, 2011 at 10:19 AM, A M Lavezzi wrote:
>
>> Hi everybody,
>>
>> I have the following problem. I have a vector containing character
>> element
Try this:
!list %in% c("aa", "bb")
On Fri, Jan 14, 2011 at 10:19 AM, A M Lavezzi wrote:
> Hi everybody,
>
> I have the following problem. I have a vector containing character
> elements,
> such as:
>
> list = c("aa","bb","cc","dd","ee")
>
> I want to create an index which identifies the element
aa[order(aa)] is the same as sort, although sort is much quicker and
has lower memory requirements:
> aa <- rnorm(1000)
> all(aa[order(aa)] == sort(aa))
[1] TRUE
> system.time(sort(aa))
user system elapsed
7.270.088.25
> system.time(aa[order(aa)])
user system elapsed
29.58
Yes Johannes - That helped, thank you.
On Sat, Mar 28, 2009 at 11:50 PM, Johannes Huesing wrote:
> Tal Galili [Sat, Mar 28, 2009 at 06:48:36PM CET]:
> > Hello people.
> >
> > I wish to reorder a simple vector of numbers by another vector of the
> order
> > (and then do the same, but with a da
Tal Galili [Sat, Mar 28, 2009 at 06:48:36PM CET]:
> Hello people.
>
> I wish to reorder a simple vector of numbers by another vector of the order
> (and then do the same, but with a data frame rows)
>
> I try this (which doesn't work) :
> > aa <- c(3, 1 ,2 )
> > aa[aa]
> [1] 2 3 1
To my mind, i
Thanks Jorge ,
I mistakingly (and foolishly) confused a vector of ranking, to a vector of
ordering that ranked vector...
Patrick Burns explained to me that I was looking for:
aa[order(aa)]
df[order(df[,1]), ]
Sorry,
Tal
On Sat, Mar 28, 2009 at 9:08 PM, Jorge Ivan Velez
wrote:
>
> Dear Tal,
Dear Tal,
It works as it should. In this code:
# Data
aa <- c(3, 1 ,2 )
aa
[1] 3 1 2
aa[aa]
[1] 2 3 1
you are telling R to do the following: take the vector aa and select the
elements aa (in R language that is aa[aa]). If you look carefully, the third
element of aa is 2, the first is 3 and the
Thanks Patrick,
Sorry to have missed that!
Tal
On Sat, Mar 28, 2009 at 9:01 PM, Patrick Burns wrote:
> I presume you are looking for:
>
> aa[order(aa)]
>
> and
>
> df[order(df[,1]), ]
>
>
> Patrick Burns
> patr...@burns-stat.com
> +44 (0)20 8525 0696
> http://www.burns-stat.com
> (home of
> I wish to reorder a simple vector of numbers by another vector of the order
> (and then do the same, but with a data frame rows)
>
> I try this (which doesn't work) :
> > aa <- c(3, 1 ,2 )
> > aa[aa]
> [1] 2 3 1
>
> The same won't work if I try to order a data frame:
> > data.frame(matrix(c(3,1
Thanks - that is exactly what I was looking for
Rainer
On Fri, Mar 14, 2008 at 1:10 PM, Henrique Dallazuanna <[EMAIL PROTECTED]> wrote:
> Try this:
>
> x[x %in% y]
>
>
>
> On 14/03/2008, Rainer M Krug <[EMAIL PROTECTED]> wrote:
> > Hi
> >
> > Consider the following code
> >
> > > x <- re
Use %in%:
x [ x %in% y ]
G.
On Fri, Mar 14, 2008 at 12:37:45PM +0200, Rainer M Krug wrote:
> Hi
>
> Consider the following code
>
> > x <- rep(1:13, 13)
>
> > y <- 1:3
>
> I want to select all elements in x which are equal to 1, 2 or 3.
>
> I know that I could use
>
> > sel <- x==y[1] | x=
Try this:
x[x %in% y]
On 14/03/2008, Rainer M Krug <[EMAIL PROTECTED]> wrote:
> Hi
>
> Consider the following code
>
> > x <- rep(1:13, 13)
>
> > y <- 1:3
>
> I want to select all elements in x which are equal to 1, 2 or 3.
>
> I know that I could use
>
> > sel <- x==y[1] | x==y[2] | x==y[3
32 matches
Mail list logo
