Re: [R] any and all

Hi Avi, As Duncan already mentioned, a reproducible example would be helpful to assist you better. Having said that, I think you misunderstand how `dplyr::filter` works: it performs row-wise filtering, so the filtering expression shall return a logical vector of the same length as the

Re: [R] Functional Programming Problem Using purr and R's data.table shift function

y downside to it? Or is the side note meant to tell me to drop the last: "%>% `[`"? Thank you, == Michael Lachanski PhD Student in Demography and Sociology MA Candidate in Statistics University of Pennsylvania mikel...@sas.upenn.edu <mailto:mikel...@sas.upenn.edu> On

Re: [R] Functional Programming Problem Using purr and R's data.table shift function

Hi Michael, Note that you have to be very careful when using by-reference operations in data.table (see `?data.table::set`), especially in a functional programming approach. In your function, you avoid this problem by calling `data.table(A)` which makes a copy of A even if it is already a

Re: [R] intersection in data frame

Or if your data is really large, you can try data.table::dcast(). > library(data.table) > dcast(ID ~ station, data = as.data.table(df1)) ID xy xz 1: 12 15 20 2: 13 16 19 (Note: instead of `as.data.table()`, you can use `setDT` or create your object as a data.table in the first place.) On

Re: [R] understanding as.list(substitute(...()))

Hi Tim, I have also asked a similar question a couple of months ago, and someone else did the same recently, maybe on r-devel. We received no "official" response, but Deepayan Sarkar (R Core Team member) claimed that: " There is no documented reason for this to work (AFAIK), so again, I

Re: [R] row combining 2972 files

On 3/18/20 9:02 PM, Bert Gunter wrote: Untested in the absence of example data, but I think combined <- do.call(rbind, lapply(ls2972, function(x)get(x)[[2]])) Or if you have largish data, use rbindlist() from the data.table package: combined <- data.table::rbindlist( lapply(ls2972,

Re: [R] "chi-square" | "chi-squared" | "chi squared" | "chi square" ?

Dear Martin, Others struggle with this inconsistency as well; I found this discussion useful: https://math.stackexchange.com/questions/1098138/chi-square-or-chi-squared Denes On 10/18/19 2:51 PM, Martin Maechler wrote: As it's Friday .. and I also really want to clean up help files and

Re: [R] How to create a new column based on the values from multiple columns which are matching a particular string?

Hi Bert, see inline. On 7/30/19 1:12 AM, Bert Gunter wrote: While Eric's solution is correct( mod "corner" cases like all NA's in a row), it can be made considerably more efficient. One minor improvement can be made by using the idiom any(x == "A") instead of matching via %in% for the simple

Re: [R] need help in if else condition

On 7/10/19 5:54 PM, Richard O'Keefe wrote: Expectation: ifelse will use the same "repeat vectors to match the longest" rule that other vectorised functions do. So a <- 1:5 b <- c(2,3) ifelse(a < 3, 1, b) => ifelse(T T F F F <<5>>, 1 <<1>>, 2 3 <<2>>) => ifelse(T T F F F <<5>>, 1 1 1 1 1

Re: [R] missRanger package

Hi Rebecca, I think it was me how suggested you the missRanger package, so this is actually a follow-up of you previous question about censored imputation of missing values (as far as I can remember). The missRanger package uses predictive mean matching, so take a look at ?missRanger::pmm

Re: [R] randomForrest-imputation

Hi, The missRanger package performs predictive mean matching which should generate positive values if the non-missing values are positive. Regards, Denes On 11/09/2018 01:11 PM, Rebecca Bingert wrote: Hi! How can I generate only positive data with randomForrest-imputation? I'm working

Re: [R] Finding unique terms

It works perfectly as I want! Thanks a lot. On Fri, Oct 12, 2018 at 6:29 AM Dénes Tóth wrote: On 10/12/2018 12:12 AM, roslinazairimah zakaria wrote: Dear r-users, I have this data: structure(list(STUDENT_ID = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L), .Label = c("AA1

Re: [R] Finding unique terms

On 10/12/2018 12:12 AM, roslinazairimah zakaria wrote: Dear r-users, I have this data: structure(list(STUDENT_ID = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L), .Label = c("AA15285", "AA15286"), class = "factor"), COURSE_CODE = structure(c(1L, 2L, 5L, 6L, 7L, 8L, 2L, 3L,

Re: [R] Erase content of dataframe in a single stroke

Hi Luigi, Actually I doubt that the original problem you try to solve requires the initialization of an empty data.frame with a particular structure. However, if you think you really need this step, I would write a function for it and also consider edge cases. getSkeleton <- function(x,

Re: [R] Creatng new variable based upon conditions

On 07/26/2018 08:58 PM, JEFFERY REICHMAN wrote: Given something like ... x <- c(3,2,4,3,5,4,3,2,4,5) y <- c("A","B","B","A","A","A","A","B","A","B") xy <- data.frame(x,y) xy$w <- ifelse(xy$y=="A",xy$w[,x]*10,xy$w[,x]*15 ) You should learn the basics about how to extract or replace part of

Re: [R] initiate elements in a dataframe with lists

On 07/25/2018 10:23 AM, Ivan Calandra wrote: Just for my understanding: Is a data.frame with list columns still a data.frame? Isn't it then a list? A data.frame is a list of equally sized vectors - that is, each vector must be of the same length. It is not required that the vector is an

Re: [R] Regexp bug or misunderstanding

Hi Martin, I assume you want to check whether a particular character string contains a digit. In this case you should use the following pattern: "[[:digit:]]" instead of "[:digit:]". From ?regex: "A character class is a list of characters enclosed between [ and ] which matches any single

Re: [R] Package 'data.table' in version R-3.5.0 not successfully being installed

You might find this discussion useful, too: https://github.com/Rdatatable/data.table/issues/2797 On 04/26/2018 11:01 PM, Henrik Bengtsson wrote: If you're installing packages to the default location in your home account and you didn't remove those library folders, you still have you R 3.4

Re: [R] Portable R in zip file for Windows

Hi Juan, you might find this useful: https://sourceforge.net/projects/rportable/ Cheers, Denes On 01/26/2018 11:57 AM, Juan Manuel Truppia wrote: Pretty good question Gabor. I can execute R once it is installed (if someone with rights installs it before) but not the installer. I can download

Re: [R] Faster Subsetting

Hi Harold, Generally: you can not beat data.table, unless you can represent your data in a matrix (or array or vector). For some specific cases, Hervé's suggestion might be also competitive. Your problem is that you did not put any effort to read at least part of the very extensive

Re: [R] Treating a vector of characters as object names to create list

On 09/05/2016 12:07 AM, Bert Gunter wrote: Time for an R tutorial or two to learn how to use the "apply" family in R. I think what you want is: merged_list <- lapply(merging, get) Or even: named_merged_list <- mget(merging) Anyway, probably you could arrive to a list of parameters

Re: [R] Matrix

On 07/17/2016 01:39 AM, Duncan Murdoch wrote: On 16/07/2016 6:25 PM, Ashta wrote: > Hi all, > > I have a large square matrix (60 x 60) and found it hard to > visualize. Is it possible to change it as shown below? > > Sample example (3 x 3) > > A B C > A 3 4 5 > B 4

Re: [R] Find mean of values in three-dimensional array

On 06/15/2016 09:05 PM, peter dalgaard wrote: > >> On 15 Jun 2016, at 19:37 , Nick Tulli wrote: >> >> Hey R-Help, >> >> I've got a three dimensional array which I pulled from a netcdf file. >> The data in array are the humidity values of locations in the United >>

Re: [R] Reshaping an array - how does it work in R

Hi Martin, On 03/22/2016 10:20 AM, Martin Maechler wrote: >>>>>Dénes Tóth<toth.de...@ttk.mta.hu> >>>>> on Fri, 18 Mar 2016 22:56:23 +0100 writes: > Hi Roy, > R (usually) makes a copy if the dimensionality of an array is modified,

Re: [R] Reshaping an array - how does it work in R

Hi Roy, R (usually) makes a copy if the dimensionality of an array is modified, even if you use this syntax: x <- array(1:24, c(2, 3, 4)) dim(x) <- c(6, 4) See also ?tracemem, ?data.table::address, ?pryr::address and other tools to trace if an internal copy is done. Workaround: use

Re: [R] How to extract same columns from identical dataframes in a list?

Hi, Although you did not provide any reproducible example, it seems you store the same type of values in your data.frames. If this is true, it is much more efficient to store your data in an array: mylist <- list(a = data.frame(week1 = rnorm(24), week2 = rnorm(24)), b =

Re: [R] does save.image() also save the random state?

On 02/05/2016 05:25 PM, Duncan Murdoch wrote: On 05/02/2016 11:14 AM, Jinsong Zhao wrote: Dear there, Here is a snipped code, > rm(list = ls()) > x <- 123 > save.image("abc.RData") > rm(list = ls()) > load("abc.RData") > sample(10) [1] 3 7 4 6 10 2 5 9 8 1 > rm(list

Re: [R] Efficient way to create new column based on comparison with another dataframe

Hi, I have not followed this thread from the beginning, but have you tried the foverlaps() function from the data.table package? Something along the lines of: --- # create the tables (use as.data.table() or setDT() if you # start with a data.frame) mapfile <- data.table(Name = c("S1", "S2",

Re: [R] Lists heading in an array of lists in R

Hi, Provide a list of a list in the second assignment: -- TunePar <- matrix(list(NULL), 2, 2) TunePar[2,1] <- list(list(G = 2)) TunePar[2,1] TunePar[2,1][[1]]$G TunePar[[2]]$G --- The point is that "[" returns the list element of the same level as the original object (TunePar in the present

Re: [R] Transfer a 3-dimensional array to a matrix in R

Hi, Bill was faster than me in suggesting aperm() instead of apply(), however, his solution is still suboptimal. Try to avoid array(), and set the dimensions directly if possible. fn1 <- function(x) { apply(x, 3, t) } fn2 <- function(x) { array(aperm(x, c(2, 1, 3)),

Re: [R] Plotting EEG signals as a "head" using R

Hi, the eegkit package (https://cran.r-project.org/web/packages/eegkit/index.html) might help you if you happen to work with a standard electrode cap. Best, Denes On 10/19/2015 02:41 PM, Charles Novaes de Santana wrote: Dear all, I have .csv file with the evoked potential of different

Re: [R] fast way to create composite matrix based on mixed indices?

Hi Matt, you could use matrix indexing. Here is a possible solution, which could be optimized further (probably). # The old matrix (old.mat <- matrix(1:30,nrow=3,byrow=TRUE)) # matrix of indices index <- matrix(c(1,1,1,4, 1,3,5,10, 2,2,1,3,

Re: [R] Multiple if function

On 09/16/2015 04:41 PM, Bert Gunter wrote: Yes! Chuck's use of mapply is exactly the split/combine strategy I was looking for. In retrospect, exactly how one should think about it. Many thanks to all for a constructive discussion . -- Bert Bert Gunter Use mapply like this on large

Re: [R] Extracting elements out of list in list in list

Hi, Here is a solution which is restricted to lists with identically shaped branches (like your example). The idea is to transform the list to an array and make use of the fact that unlist(x, use.names=FALSE) is much much faster for large lists than unlist(x). # function which transforms a

Re: [R] Extract values from multiple lists

Dear Jeff, On 12/17/2014 01:46 AM, Jeff Newmiller wrote: You are chasing ghosts of performance past, Denes. In terms of memory efficiency, yes. In terms of CPU time, there can be significant difference, see below. The data.frame function causes no problems, and if it is used then the OP

Re: [R] Extract values from multiple lists

On 12/16/2014 06:06 PM, SH wrote: Dear List, I hope this posting is not redundant. I have several list outputs with the same components. I ran a function with three different scenarios below (e.g., scen1, scen2, and scen3,...,scenN). I would like to extract the same components and group

Re: [R] Equivalent to matlab .* operator in R

Hi, It is better to use sweep() for these kinds of problems, see ?sweep y - matrix(cbind(c(0, 0.5, 1),c(0, 0.5, 1)),ncol=2) z - matrix(c(12, -6),ncol=2) sweep(y, 2, z, *) Best, Denes On 11/19/2014 03:50 PM, Berend Hasselman wrote: On 19-11-2014, at 15:22, Ruima E. ruimax...@gmail.com

Re: [R] Equivalent to matlab .* operator in R

Hi, just for the records, your original code seems incorrect, see inline. On 11/19/2014 03:22 PM, Ruima E. wrote: Hi, I have this: y = matrix(cbind(c(0, 0.5, 1),c(0, 0.5, 1)),ncol=2) z = matrix(c(12, -6),ncol=2) In matlab I would do this y .* x Here you wrote 'x' which I guess refers to

Re: [R] arrow egdes and lty

Hi, Dear R developers, I want to draw an arrow in a figure with lty=2. The lty argument also affects the edge of the arrow, which is unfortunate. Feature or bug? Is there some way to draw an arrow with intact edge, still with lty=2? AFAIK there is no such option in the arrow function,

Re: [R] R-help Digest, Vol 104, Issue 19

On Oct 21, 2011, at 09:01 , Martin Maechler wrote: ARE == Alex Ruiz Euler rruizeu...@ucsd.edu on Wed, 19 Oct 2011 14:05:16 -0700 writes: ARE Motion supported. Very. ARE On Wed, 19 Oct 2011 15:40:14 +0200 ARE peter dalgaard pda...@gmail.com wrote: Argh! Someone please

Re: [R] Structural equation modelling in R compared with Amos

Dear Ravi, I would also suggest you to have a look at 'lavaan' (www.lavaan.org). It has an extremely straightforward yet flexible model syntax and very good documentation. Although its statistical capabilities are not comparable to those of Mplus (which is the most powerful SEM-software I

Re: [R] Repeat a loop until...

If you want to generate truncated distributions, package 'tmvtnorm' can help you out. Regards, Denes Dear all, I know there have been various questions posted over the years about loops but I'm afraid that I'm still stuck. I am using Windows XP and R 2.9.2. I am generating some data

Re: [R] plots of correlation matrices

Hi, One way to do that is this (avoiding the use of a for loop): l.txt- id category attribute1 attribute2 attribute3 attribute4 661 SCHS 43.2 0 56.5 1 12202 SCHS 161.7 5.7 155 16 1182 SCHS 21.4 0 29 0 1356 SSS 8.8182 0.1818 10.6667 0.6667 1864 SCHS 443.7273 9.9091 537 46 12360 SOA

Re: [R] high and lowest with names

which.max is even faster: dims - c(1000,1000) tt - array(rnorm(prod(dims)),dims) # which system.time( replicate(100, which(tt==max(tt), arr.ind=TRUE)) ) # which.max ( arrayInd) system.time( replicate(100, arrayInd(which.max(tt), dims)) ) Best, Denes But it's simpler and probably faster to use

Re: [R] correlation matrix

And you might also consider packages like corrplot, corrgram etc. for other plotting options of a correlation matrix. They can be more informative than simply invoking image(heat) What a pleasant post to respond to - with self-contained code. :)

Re: [R] calc.relimp pmvd for US R-user

Dear All: I am calculating the relative importance of a regressor in a linear model. Does anyone know how I can obtain/install the 'pmvd' computation type? I am a US user. Regards, Y Hi, Do you have any specific reason to use the pmvd method, especially in the light of

Re: [R] studentized and standarized residuals

Dear Jen, Actually you can check out what R does by looking at the source. # first type the name of the function rstandard function (model, ...) UseMethod(rstandard) environment: namespace:stats # ?methods will list you the corresponding functions methods(rstandard) [1] rstandard.glm

Re: [R] How to count numbers of a vector and use them as index values?

See also ?tabulate. tabulate(x,8) Hi Paul, I would use something like this: x - c(2,2,3,3,4,6) table(x) x 2 3 4 6 2 2 1 1 x - factor(x, levels=1:8) table(x) x 1 2 3 4 5 6 7 8 0 2 2 1 0 1 0 0 Sarah On Sun, Jul 31, 2011 at 5:41 PM, Paul Menzel

Re: [R] finding a faster way to run lm on rows of predictor matrix

Hi, you can solve the task by simple matrix algebra. Note that I find it really inconvenient to have a matrix with variables in rows and cases in columns, so I transposed your predictors matrix. # your data regress.y = rnorm(150) predictors = matrix(rnorm(6000*150), ncol=150, nrow=6000) tpreds

Re: [R] Looping through data sets to change column from character to numeric

The problem is that you can not assign a variable to itself. rm(list=ls()) df1 - data.frame(ResultValue=as.character(1:5)) df2 - data.frame(ResultValue=as.character(1:10)) frames = ls() for (frame in frames){ temp - get(frame) temp[,ResultValue] = as.numeric(temp[,ResultValue])

Re: [R] Looping through data sets to change column from character to numeric

Sorry, I was wrong. Of course you can assign a variable to itself, but it does not make much sense... What you misunderstood was that in the assignment you assign the data frame (e.g. df1) to itself. You do not modify the frame object which remains a character string. The problem is that you

Re: [R] Finding/identifying a value within a factor

Hi, you provided a character vector as an example. I guess you meant something like: x - factor(c(1,2,3,4,1)) # You can identify those elements with an by ?grep or ?grepl: indices - grep(,as.character(x)) # You can transform those elements by ?as.numeric as.numeric(x[indices]) HTH, Denes