similar to: [.data.frame speedup

On Wed, 13 Dec 2023 09:04:18 +0100 Hilmar Berger via R-devel <r-devel at r-project.org> wrote: > Still, I feel that default partial matching cripples the functionality > of data.frame for larger tables. Changing the default now would require a long deprecation cycle to give everyone who uses `[.data.frame` and relies on partial matching (whether they know it or not) enough time to

Hi Hilmar and Ivan, I have used your code examples to write a blog post about this topic, which has figures that show the asymptotic time complexity of the various approaches, https://tdhock.github.io/blog/2023/df-partial-match/ The asymptotic complexity of partial matching appears to be quadratic O(N^2) whereas the other approaches are asymptotically faster: linear O(N) or log-linear O(N log N).

Hi all, I was looking at the data frame subscription operator (attached in the end of this e-mail) and got puzzled by the following line: class(x) <- attr(x, "row.names") <- NULL This appears to set the class and row.names attributes of the incoming data frame to NULL. So far I was not able to figure out why this is necessary - could anyone help ? The reason I am

Dear Ivan, thanks a lot, that is helpful. Still, I feel that default partial matching cripples the functionality of data.frame for larger tables. Thanks again and best regards Hilmar On 12.12.23 13:55, Ivan Krylov wrote: > ? Mon, 11 Dec 2023 21:11:48 +0100 > Hilmar Berger via R-devel <r-devel at r-project.org> ?????: > >> What was unexpected is that in this case was that

Full_Name: Steven McKinney Version: 2.9.0 OS: Mac OS X 10.5.6 Submission from: (NULL) (142.103.207.10) A corrupt data frame can be constructed as follows: foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4]

Hello, I would like to use mapply to avoid using a loop but for some reason, I can't seem to get it to work. I've included copies of my code below. The first set of code uses a loop (and it works fine), and the second set of code attempts to use mapply but I get a "subscript out of bounds" error. Any guidance would be greatly appreciated. Xj, Yj, and Wj are also lists, and s2,

They differ in whether the row names are "automatic": > .row_names_info(a1) [1] -3 > .row_names_info(a2) [1] 3 Best, -Deepayan On Tue, 14 Nov 2023 at 08:23, Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > What is going on here? In the lines ending in #### the inputs and outputs > are identical yet one gives a warning and the other does not. > >

In that case identical should be FALSE but it is TRUE identical(a1, a2) ## [1] TRUE On Tue, Nov 14, 2023 at 8:58?AM Deepayan Sarkar <deepayan.sarkar at gmail.com> wrote: > > They differ in whether the row names are "automatic": > > > .row_names_info(a1) > [1] -3 > > .row_names_info(a2) > [1] 3 > > Best, > -Deepayan > > On Tue, 14 Nov

Conversion of a data frame to a matrix using as.matrix() when a column of the data frame is POSIXt and all other columns are numeric has changed in R 1.9.0 from R 1.8.1. The new behavior issues a warning message and converts to a character matrix. In R 1.8.1, such an object was converted to a numeric matrix. Here is an example. #### R 1.9.0 #### > foo <- data.frame(

Also why should that difference result in different behavior? On Tue, Nov 14, 2023 at 9:38?AM Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > In that case identical should be FALSE but it is TRUE > > identical(a1, a2) > ## [1] TRUE > > > On Tue, Nov 14, 2023 at 8:58?AM Deepayan Sarkar > <deepayan.sarkar at gmail.com> wrote: > > > >

Converting a data.frame with a nested data.frame to a matrix fails: x <- structure(list(a = data.frame(letters)), class = "data.frame", row.names = .set_row_names(26)) as.matrix(x) #> Error in ncol(xj) : object 'xj' not found The offending code is here, in the definition of as.matrix.data.frame (source/base/all.R): for (j in pseq) {

I got the error. I haven't been able to get a stand along case so that I can show it here. But could somebody give some clue on what could cause this error? Since I never defined xj[i], I don't understand where this error come from. Error in xj[i] : invalid subscript type 'list'

On Tue, 14 Nov 2023 at 09:41, Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > Also why should that difference result in different behavior? That's justifiable, I think; consider: > d1 = data.frame(a = 1:4) > d2 = d3 = data.frame(b = 1:2) > row.names(d3) = c("a", "b") > data.frame(d1, d2) a b 1 1 1 2 2 2 3 3 1 4 4 2 > data.frame(d1,

I would like to suggest a modification to the make.names() function. The current implementation has two problems: 1. It doesn't check if a name matches an R keyword (like "function"). 2. The uniqueness algorithm is not invariant to concatenation. In other words, make.names(c("a","a","a"),unique=T) !=

I would like to suggest a modification to the make.names() function. The current implementation has two problems: 1. It doesn't check if a name matches an R keyword (like "function"). 2. The uniqueness algorithm is not invariant to concatenation. In other words, make.names(c("a","a","a"),unique=T) !=

Dear readers of the list, I have a problem a comparison of two data from a vector. The comparison yields FALSE but should be TRUE. I have checked for mode(), length() and attributes(). See the following code (R2.10.0): ----------------------------------------------- # data vector of 66 double data X =

>>>>> Jan van der Laan <rhelp at eoos.dds.nl> >>>>> on Fri, 6 Oct 2017 12:13:39 +0200 writes: > It is actually model.matrix that crashes, not glm. Same > crash occurs with e.g. lm. > model.matrix(dob_mon ~ dob_day*dob_mon, data = tab) > also crashes R. Yes, segmentation fault. It only happens when these are *logical*

Dear R-ians, I'm looking for a computational simplified formula to calculate a measure for heterogeneity (let's say H ): H = sqrt [ (Si (Sj (Xi - Xj)?? ) ) /n ] where: sqrt = square root Si = summation over i (= 0 to n) Sj = summation over j (= 0 to n) Xi = element of X with index i Xj = element of X with index j I can simplify the formula to: H = sqrt [ ( 2 * n * Si (Xi) - 2 Si (Sj

Hi, I am tring to write a loop to compute this, ========================== x1=c( rep(-1,4), rep(1,4) ) x2=c( rep(c(-1,-1,1,1),2) ) x3=c( rep(c(-1,1),4) ) x1*x2 x1*x3 x2*x3 ======================== suppose i have x1,x2,x3 i want to compute their ' two factor interactions', x1x2,x1x3 and x2x3, I wrote ======================== for(i in 1:2){ for( j in i+1:3){ xij=c()

I have the 'breaks' for the histogram ('hist') but I want weight the cells instead of using actual observations. I thought that using freq=FALSE implied that the numbers in 'x' were weights but this turned out to be wrong. Any help and/or comment is very much appreciated. Regards, M?rten M?rten Bjellerup Doctoral Student in Economics School of Management and Economics