Displaying 20 results from an estimated 10000 matches similar to: "Split a data.frame"
2018 May 19
0
Split a data.frame
Hello,
Maybe something like the following.
splitDF <- function(data, col, s){
n <- nrow(data)
inx <- which(data[[col]] %in% s)
lapply(seq_along(inx), function(i){
k <- if(inx[i] < n) (inx[i] + 1):(inx[i + 1])
data[k, ]
})
}
splitDF(DF, "name", split_str)
Hope this helps,
Rui Barradas
On 5/19/2018 12:07 PM, Christofer Bogaso
2017 Aug 10
3
Zoo rolling window with increasing window size
Hi Joshua, thanks for your prompt reply. However as I said, sum()
function I used here just for demonstrating the problem, I have other
custom function to implement, not necessarily sum()
I am looking for a generic solution for above problem.
Any better idea? Thanks,
On Fri, Aug 11, 2017 at 12:04 AM, Joshua Ulrich <josh.m.ulrich at gmail.com> wrote:
> Use a `width` of integer index
2017 Aug 23
4
Getting all possible combinations
Hi again,
I am exploring if R can help me to get all possible combinations of
members in a group.
Let say I have a group with 5 members : A, B, C, D, E
Now I want to generate all possible unique combinations with all
possible lengths from that group e.g.
1st combination : A
2nd combination : B
.....
5th combination : E
6th combination : A, B
7th combination : B, C
....
last combination: A, B,
2017 Aug 10
0
Zoo rolling window with increasing window size
Replace "sum" with your custom function's name. I don't see any
reason why that wouldn't work, and the problem with my solution is not
clear in your response.
r <- rollapplyr(x, seq_along(x), yourCustomFunctionGoesHere)
On Thu, Aug 10, 2017 at 1:39 PM, Christofer Bogaso
<bogaso.christofer at gmail.com> wrote:
> Hi Joshua, thanks for your prompt reply. However
2017 Aug 23
3
Getting all possible combinations
ummm, Ista, it's 2^n.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Aug 23, 2017 at 8:52 AM, Ista Zahn <istazahn at gmail.com> wrote:
> On Wed, Aug 23, 2017 at 11:33 AM, Christofer Bogaso
>
2017 Aug 10
2
Zoo rolling window with increasing window size
Hi again,
I am wondering there is any function for 'zoo' time series, where I
can apply a user defined function rolling window basis, wherein window
size is ever increasing i.e. not fixed. For example, let say I have
below user defined function and a zoo time series :
> library(zoo)
> UDF = function(x) sum(x)
> TS = zoo(rnorm(10), seq(as.Date('2017-01-01'),
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 ==
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
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
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,
2011 Jan 11
5
A question on dummy variable
Dear all, I would like to ask one question related to statistics, for
specifically on defining dummy variables. As of now, I have come across 3
different kind of dummy variables (assuming I am working with Seasonal
dummy, and number of season is 4):
> dummy1 <- diag(4)
> for(i in 1:3) dummy1 <- rbind(dummy1, diag(4))
> dummy1 <- dummy1[,-4]
>
> dummy2 <- dummy1
>
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
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
2023 Oct 12
4
if-else that returns vector
Hi,
Following expression returns only the first element
ifelse(T, c(1,2,3), c(5,6))
However I am looking for some one-liner expression like above which
will return the entire vector.
Is there any way to achieve this?
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,
2013 Mar 28
4
How to replace '$' sign?
Hello again,
I want to remove "$" sign and replace with nothing in my text.
Therefore I used following code:
> gsub("$|,", "", "$232,685.35436")
[1] "$232685.35436"
However I could not remove '$' sign.
Can somebody help me why is it so?
Thanks and regards
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 =
2011 Nov 10
5
A question on Programming
Dear all. Let say I have a group of codes which will be used in many places
in my overall R-code files. These group of codes will be used within a
for-loop (with a big length, like 10000 times) and also many other places
outside of that for loop. As this group of codes are being used in many
places, I thought to put them within a user-defined function.
Here my question is, is there any speed
2011 Jul 22
2
Indexing problem with matrix
Dear all, assume I have a matrix with just 1 row. Now suppose I want to
fetch 1st few rows from that matrix, however resulting object becomes
vector. Here is 1 such example:
> matrix(1:5, 1)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 3 4 5
>
> matrix(1:5, 1)[,-1]
[1] 2 3 4 5
Can somebody point me how to keep resulting object as matrix with same row?
Ofcourse I