Displaying 20 results from an estimated 10000 matches similar to: "Memory allocation problem (again!)"
2013 Jan 24
2
Question on matrix calculation
Hello again,
Ley say I have 1 matrix and 1 data frame:
> mat <- matrix(1:15, 5)
> match_df <- data.frame(Seq = 1:5, criteria = sample(letters[1:5], 5, replace = T))
> mat
[,1] [,2] [,3]
[1,] 1 6 11
[2,] 2 7 12
[3,] 3 8 13
[4,] 4 9 14
[5,] 5 10 15
> match_df
Seq criteria
1 1 c
2 2 e
3 3 c
4 4 c
5
2010 Mar 30
2
Problem with expand.grid() function
Hi, good morning,
I got following error which looks strange to me while executing this code :
> temp <- expand.grid(rep(list(c(1,0)),40))
Error in rep.int(rep.int(seq_len(nx), rep.int(rep.fac, nx)), orep) :
invalid 'times' value
In addition: Warning message:
In rep.int(rep.int(seq_len(nx), rep.int(rep.fac, nx)), orep) :
NAs introduced by coercion
However if I put a small
2013 Apr 17
2
On matrix calculation
Hello again,
Let say I have a matrix:
Mat <- matrix(1:12, 4, 3)
And a vector:
Vec <- 5:8
Now I want to do following:
Each element of row-i in 'Mat' will be divided by i-th element of Vec
Is there any direct way to doing that?
Thanks for your help
2011 Oct 01
1
error using ddply to generate means
Dear list,
I encounter an error when I try to use ddply to generate means as follows:
fun3<-structure(list(sector = structure(list(gics_sector_name = c("Financials",
"Financials", "Materials", "Materials")), .Names = "gics_sector_name",
row.names = structure(c("UBSN VX Equity",
"LLOY LN Equity", "AI FP Equity",
2013 Apr 29
1
Need help on matrix calculation
Hello again,
Let say I have 1 matrix:
Mat <- matrix(1:12, 4, 3)
rownames(Mat) <- letters[1:4]
Now I want to subscript of Mat in following way:
Subscript_Vec <- c("a", "e", "b", "c")
However when I want to use this vector, I am geting following error:
Mat[Subscript_Vec, ]
Error: subscript out of bounds
Basically I want to get my final matrix
2012 Dec 01
4
Getting all possible contingency tables
Hello all,
Let say I have 2-way contingency table:
Tab <- matrix(c(8, 10, 12, 6), nr = 2)
and the Chi-squared test could not reject the independence:
> chisq.test(Tab)
Pearson's Chi-squared test with Yates' continuity correction
data: Tab
X-squared = 1.0125, df = 1, p-value = 0.3143
However I want to get all possible contingency tables under this
independence
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,
2012 Jan 06
1
ggplot using scale_x_date gives Error in seq.int(r1$year, to$year, by)
Dear all,
ggplot gives me an error when trying to plot time series data using a
date variable as the x axis.
g<-structure(list(Date = c("2011-12-23", "2011-12-30", "2012-01-06",
"2011-12-23", "2011-12-30", "2012-01-06", "2011-12-23", "2011-12-30",
"2012-01-06"), variable = structure(c(1L, 1L, 1L, 2L, 2L,
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
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
>
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,
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
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
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
2010 Sep 14
5
Problem with cat()
Dear all, I have a problem with the cat() function. Let say I have following:
fn1 <- function(n = 5){
mat <- matrix(rnorm(5*5), 5, 5)
cat(as.character(mat))
return(n)
}
However when I run above function I get this:
> fn1()
-0.601930631438248 -1.16950049447942 0.469257329394626
-1.39766868242906 -1.02580943892082 1.4067931110327 -1.07245318857022
-0.0205043699310245 0.234628727206755