Displaying 20 results from an estimated 10000 matches similar to: "if-else that returns vector"
2023 Oct 13
1
if-else that returns vector
?ifelse
'ifelse' returns a value with the same shape as 'test' which is
filled with elements selected from either 'yes' or 'no' depending
on whether the element of 'test' is 'TRUE' or 'FALSE'.
This is actually rather startling, because elsewhere in the
S (R) language, operands are normally replicated to the length
of the longer.
2018 Mar 04
3
Change Function based on ifelse() condtion
Below is my full implementation (tried to make it simple as for demonstration)
Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) {
if (Apply_MC) {
return(mclapply(X, FUN, ...))
} else {
if (any(names(list(...)) == 'mc.cores')) {
myList = list(...)[!names(list(...)) %in% 'mc.cores']
}
return(lapply(X, FUN, myList))
}
}
Lapply_me(as.list(1:4), function(xx) {
if (xx ==
2018 Mar 04
2
Change Function based on ifelse() condtion
My modified function looks below :
Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) {
if (Apply_MC) {
return(mclapply(X, FUN, ...))
} else {
if (any(names(list(...)) == 'mc.cores')) {
myList = list(...)[!names(list(...)) %in% 'mc.cores']
}
return(lapply(X, FUN, myList))
}
}
Here, I am not passing ... anymore rather passing myList
On Sun, Mar 4, 2018 at 10:37 PM,
2018 Mar 04
2
Change Function based on ifelse() condtion
@Eric - with this approach I am getting below error :
Error in FUN(X[[i]], ...) : unused argument (list())
On Sun, Mar 4, 2018 at 10:18 PM, Eric Berger <ericjberger at gmail.com> wrote:
> Hi Christofer,
> You cannot assign to list(...). You can do the following
>
> myList <- list(...)[!names(list(...)) %in% 'mc.cores']
>
> HTH,
> Eric
>
> On Sun, Mar
2018 Mar 04
0
Change Function based on ifelse() condtion
The reason that it works for Apply_MC=TRUE is that in that case you call
mclapply(X,FUN,...) and
the mclapply() function strips off the mc.cores argument from the "..."
list before calling FUN, so FUN is being called with zero arguments,
exactly as it is declared.
A quick workaround is to change the line
Lapply_me(as.list(1:4), function(xx) {
to
Lapply_me(as.list(1:4),
2018 Mar 04
2
Change Function based on ifelse() condtion
Hi,
As an example, I want to create below kind of custom Function which
either be mclapply pr lapply
Lapply_me = function(X = X, FUN = FUN, ..., Apply_MC = FALSE) {
if (Apply_MC) {
return(mclapply(X, FUN, ...))
} else {
if (any(names(list(...)) == 'mc.cores')) {
list(...) = list(...)[!names(list(...)) %in% 'mc.cores']
}
return(lapply(X, FUN, ...))
}
}
However when Apply_MC =
2018 Mar 04
0
Change Function based on ifelse() condtion
That's fine. The issue is how you called Lapply_me(). What did you pass as
the argument to FUN?
And if you did not pass anything that how is FUN declared?
You have not shown that in your email.
On Sun, Mar 4, 2018 at 7:11 PM, Christofer Bogaso <
bogaso.christofer at gmail.com> wrote:
> My modified function looks below :
>
> Lapply_me = function(X = X, FUN = FUN, Apply_MC =
2018 Mar 04
0
Change Function based on ifelse() condtion
Hi Christofer,
Before you made the change that I suggested, your program was stopping at
the statement: list(...) = list(..) .etc
This means that it never tried to execute the statement:
return(lapply(X,FUN,...))
Now that you have made the change, it gets past the first statement and
tries to execute the statement: return(lapply(X,FUN,...)).
That attempt is generating the error message because
2018 Mar 04
0
Change Function based on ifelse() condtion
Hi Christofer,
You cannot assign to list(...). You can do the following
myList <- list(...)[!names(list(...)) %in% 'mc.cores']
HTH,
Eric
On Sun, Mar 4, 2018 at 6:38 PM, Christofer Bogaso <
bogaso.christofer at gmail.com> wrote:
> Hi,
>
> As an example, I want to create below kind of custom Function which
> either be mclapply pr lapply
>
> Lapply_me =
2013 Apr 24
2
Looking for a better code for my problem.
Hello again,
Let say I have following data:
Dat <- structure(list(AA = structure(c(3L, 1L, 2L, 1L, 2L, 3L, 3L, 2L,
3L, 1L, 1L, 3L, 3L, 2L, 2L, 3L, 2L, 1L, 1L, 1L), .Label = c("A",
"B", "C"), class = "factor"), BB = structure(c(2L, 3L, 2L, 2L,
2L, 3L, 2L, 2L, 2L, 1L, 1L, 2L, 3L, 1L, 3L, 2L, 1L, 2L, 2L, 3L
), .Label = c("a", "b",
2013 Mar 23
4
Converting a character vector to numeric
Hello again,
Let say I have following vector:
Vec <- c("0.0365780769", "(1.09738648244378)", "(0.812507787221523)",
"0.5778069963", "(0.452456601362355)", "-1.8900812605", "-1.8716093762",
"0.0055217041", "-0.4769192333", "-2.4133018880")
Now I want to convert this vector to numeric vector. I
2018 Mar 04
2
Change Function based on ifelse() condtion
Hi again,
I am looking for some way to alternately use 2 related functions,
based on some ifelse() condition.
For example, I have 2 functions mclapply() and lapply()
However, mclapply() function has one extra parameter 'mc.cores' which
lapply doesnt not have.
I know when mc.cores = 1, these 2 functions are essentially same,
however I am looking for more general way to control them
2013 Mar 22
2
A question on function return
Hello again,
Let say I have following user defined function:
fn <- function(x, y) {
Vec1 <- letters[1:6]
Vec2 <- 1:5
return(list(ifelse(x > 0, Vec1, NA), ifelse(y > 0, Vec2, NA)))
}
Now I have following calculation:
> fn(-3, -3)
[[1]]
[1] NA
[[2]]
[1] NA
> fn(3, -3)
[[1]]
[1] "a"
[[2]]
[1] NA
Here I can not understand why in the second case, I get
2012 Dec 31
3
Question on creating Date variable
Hello all,
Let say I have following (numeric) vector:
> x
[1] 11.00 11.25 11.35 12.01 11.14 13.00 13.25 13.35 14.01 13.14 14.50
14.75 14.85 15.51 14.64
Now, I want to create a 'Date' variable (i.e. I should be able to do all
calculations pertaining to date/time and also time-series plotting etc.)
like
2012-12-31 11:00:00 AM, 2012-12-31 11:25:00 AM, 2012-12-31 11:35:00 AM,
2018 Mar 04
0
Change Function based on ifelse() condtion
On 04/03/2018 10:39 AM, Christofer Bogaso wrote:
> Hi again,
>
> I am looking for some way to alternately use 2 related functions,
> based on some ifelse() condition.
>
> For example, I have 2 functions mclapply() and lapply()
>
> However, mclapply() function has one extra parameter 'mc.cores' which
> lapply doesnt not have.
>
> I know when mc.cores =
2013 Jan 15
2
Need some help on Text manipulation.
Dear all,
Let say I have following data-frame:
Dat <- structure(list(dat = c(-0.387795842956327, -0.23270882099043,
-0.89528973290562, 0.95857175595512, 1.61680582493783, -1.17738110289352,
0.210601060411423, -0.827369747447338, -0.36896112964414, 0.440288648776096,
1.28018410608809, -0.897113649961341, 0.342216546981718, -1.17288066266219,
-1.57994101992621, -0.913655547602414,
2010 Jul 10
7
Need help on date calculation
Hi all, please see my code:
> library(zoo)
> a <- as.yearmon("March-2010", "%B-%Y")
> b <- as.yearmon("May-2010", "%B-%Y")
>
> nn <- (b-a)*12 # number of months in between them
> nn
[1] 2
> as.integer(nn)
[1] 1
What is the correct way to find the number of months between "a" and "b",
still
2012 Dec 14
5
A question on list and lapply
Dear all, let say I have following list:
Dat <- vector("list", length = 26)
names(Dat) <- LETTERS
My_Function <- function(x) return(rnorm(5))
Dat1 <- lapply(Dat, My_Function)
However I want to apply my function 'My_Function' for all elements of
'Dat' except the elements having 'names(Dat) == "P"'. Here I have
specified the name
2017 Aug 02
4
Extracting numeric part from a string
Hi again,
I am struggling to extract the number part from below string :
"\"cm_ffm\":\"563.77\""
Basically, I need to extract 563.77 from above. The underlying number
can be a whole number, and there could be comma separator as well.
So far I tried below :
> library(stringr)
> str_extract("\"cm_ffm\":\"563.77\"",
2012 Mar 16
4
How to start R in maximized size???
Dear all, when I start R, I want that the console window should be in
the Maximized size automatically. Can somebody help me how to achieve
that?
Thanks and regards,