similar to: Calling a column from a matrix using a variable

Hi everyone, now I am trying to finish writing the code (I had asked for assistance on subtracting arrays) This is what I what I am running in R: > source("/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Example/NSEr.R") NSEr <- function (obs, sim) { {jjh <- (as.vector(obs) - sim)^2 Xjjhs <- apply(Xjjh, 2, sum) Yii <- (obs - mean(obs))^2 Yiis <- apply(Yii, 2,

Hello here. I'm struggling to understand R's subsetting behavior in couple of edge cases - subsetting row in a single column matrix and subsetting column in a single row matrix. I've read R's docs several times and haven't found answer. Consider following example: a = matrix(1:2, nrow = 2, dimnames = list(c("row1", "row2"), c("col1"))) a[1, ] # 1

Hi Rui. Thanks for answer, I'm aware of drop = FALSE option. Unfortunately it doesn't resolve the issue - I'm expecting to get a vector, not a matrix . ??, 21 ????. 2018 ?. ? 20:54, Rui Barradas <ruipbarradas at sapo.pt>: > Hello, > > Use drop = FALSE. > > a[1, , drop = FALSE] > # col1 > #row1 1 > > > Hope this helps, > > Rui Barradas

here's my question: suppose I have a matrix: mt<-matrix(1:12,ncol=6) now I have a vector vt<-c(1,2,2,2,1,2) which means I want to get: the 1st row for column1; the 2nd row for column2; the 2nd row for column3; the 2nd row for column4; ... that what I want is this vector: 1,4,6,8,9,12 Does anyone know how to do this fast? I know I can use for-loop to travel all columns,but

Hello, Right! I copied from the OP's question without thinking about it. Corrected would be bootMedianSE <- function(data, indices){ d <- data[indices, ] fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) ypred <- predict(fit) y <- d$crp median((y - ypred)^2) } Sorry, rui Barradas On 5/22/2018 11:32 AM, Daniel Nordlund wrote: > On 5/22/2018

I forgot, you should also set.seed() before calling boot() to make the results reproducible. Rui Barradas On 5/22/2018 10:00 AM, Rui Barradas wrote: > Hello, > > If you want to bootstrap a statistic, I suggest you use base package boot. > You would need the data in a data.frame, see how you could do it. > > > library(boot) > > bootMedianSE <- function(data,

?s 22:54 de 11/06/2023, javad bayat escreveu: > Dear Rui; > Many thanks for your email. I used one of your codes, > "data2$LU[which(data2$Layer == "Level 12")] <- "Park"", and it works > correctly for me. > Actually I need to expand the codes so as to consider all "Levels" in the > "Layer" column. There are more than hundred

Dear R-experts, I am trying to bootstrap (and average) the median squared error evaluation metric for a robust regression. I can't get it. What is going wrong ? Here is the reproducible example. ############################# install.packages( "quantreg" ) library(quantreg) crp <-c(12,14,13,24,25,34,45,56,25,34,47,44,35,24,53,44,55,46,36,67) bmi

Hello, I'm very new to R, and have managed to plot xy graphs, but can't seem to plot a matrix properly. The first column is my time points, and the following columns 2-4 are the replicates of my experiment. I've tried using row.names=1 to make the first column the value for that row (whereas before I had 1-31 sequence) and then my 1st column) but I can't work out how to tell it

Hello, If you want to bootstrap a statistic, I suggest you use base package boot. You would need the data in a data.frame, see how you could do it. library(boot) bootMedianSE <- function(data, indices){ d <- data[indices, ] fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) ypred <- predict(fit) y <- d$crp median(y - ypred)^2 } dat <-

Dear R users, I have a matrix composed of lists: m <- matrix( list(), nrow=1, ncol=3 ) m[[ 1, 1 ]] <- list("A", "B") m[[ 1, 2 ]] <- list("A", "C") m[[ 1, 3 ]] <- list("A", "B") and want to get the sub-matrix where cells contain "B". But m[ , "B" %in% m[ 1, ], drop=F ] as well as m[ , "B"

Hi all, I have a matrix with 10000 rows and 10 columns. The last columns contains another identifiers but the values are not uniques so that I want to generate another matrix with rows with unique values in the last column. If I did tmp<-unique(my_mat$col10) this will give me 8560 unique entries so the ideal matrix will be 8560X10 columns now then. I tried sub_mat<-my_mat[tmp,] but

hello i have a very long column of numbers. i want R to make a new column every time the value changes from zero. example for the column: 90.1194354 87.94788274 80.34744843 64.06080347 30.40173724 0 0 0 0 0 16.28664495 23.88707926 29.31596091 48.85993485 13.02931596 0 0 0 7.600434311 20.62975027 29.31596091 32.5732899 for this example i want to get 3 columns. thanks! [[alternative HTML version

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On 5/22/2018 2:32 AM, Rui Barradas wrote: > bootMedianSE <- function(data, indices){ > ???? d <- data[indices, ] > ???? fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) > ???? ypred <- predict(fit) > ???? y <- d$crp > ???? median(y - ypred)^2 > } since the OP is looking for the "median squared error", shouldn't the final line of the

Dear Rui; Many thanks for your email. I used one of your codes, "data2$LU[which(data2$Layer == "Level 12")] <- "Park"", and it works correctly for me. Actually I need to expand the codes so as to consider all "Levels" in the "Layer" column. There are more than hundred levels in the Layer column. If I use your provided code, I have to write it

Hello, Sorry, but I think my first answer is wrong. You probably want something along the lines of sp <- split(priceStore_Grps, priceStore_Grps$StorePC) res <- lapply(seq_along(sp), function(i){ sp[[i]]$StoreID <- paste("Store", i, sep = "_") sp[[i]] }) res <- do.call(rbind, res) row.names(res) <- NULL Hope this helps, Rui Barradas On 5/26/2018

Hello, Please always cc the list. As for the question, yes, it does. If you want to remove just the ones with exactly 73.1 use the pattern grep("^73\\.1$", etc) Explanation: Beginning of string: ^ End of string: $ Escape special characters: \\ (needed because the period is a special character.) Hope this helps, Rui Barradas On 5/22/2018 12:50 PM, Ahmed Serag wrote: > Thank

Hello, See if this is it: priceStore_Grps$StoreID <- paste("Store", seq_len(nrow(priceStore_Grps)), sep = "_") Hope this helps, Rui Barradas On 5/26/2018 2:03 PM, Jeff Reichman wrote: > ALCON > > > > I'm trying to figure out how to rename groups in a data frame after groups > by selected variabels. I am using the dplyr library to group my

ALCON I'm trying to figure out how to rename groups in a data frame after groups by selected variabels. I am using the dplyr library to group my data by 3 variables as follows # group by lat (StoreX)/long (StoreY) priceStore <- LapTopSales[,c(4,5,15,16)] priceStore <- priceStore[complete.cases(priceStore), ] # keep only non NA records priceStore_Grps <- priceStore %>%