[Solved] Counting Values in R Vector
I like findInterval for these sorts of tasks: x <- c(1,2,3,20,21,22,40,41,42,60,61,62,80,81,82) table(findInterval(x,c(0,20,40,60,80))) 1 2 3 4 5 3 3 3 3 3 4 solved Counting Values in R Vector
I like findInterval for these sorts of tasks: x <- c(1,2,3,20,21,22,40,41,42,60,61,62,80,81,82) table(findInterval(x,c(0,20,40,60,80))) 1 2 3 4 5 3 3 3 3 3 4 solved Counting Values in R Vector
So the way your question is posed is a bit sloppy, but an example of what you might be trying to do is to take the average of each 2000 x 22 array for each of the 3 of them. here is how this would be done: arr = array(1, dim=c(2000,22,3)) dim(arr) m = NULL … Read more
I am not sure if the code below is what you are after out1 <- ds %>% group_by(id) %>% summarise(n = n()) %>% mutate(ncomb2 = choose(n,2)) such that > out1 # A tibble: 3 x 3 id n ncomb2 <dbl> <int> <dbl> 1 1 4 6 2 2 4 6 3 3 4 6 solved … Read more
Let ll <- c(0.2, 0.2, 0.15, 0.15) ul <- c(0.5, 0.5, 0.4, 0.3) where ll and ul correspond to lower and upper limits for each of the random variables. Next, x <- runif(length(ll), 0, ul – ll) is a vector of uniform random draws from intervals [0,0.3], [0,0.3], [0,0.25], and [0,0.15]. The reason for this … Read more
If I understand correctly, the trick is that you want to fill downward except for the bottommost NAs. And the problem with tidyr‘s fill is that it goes all the way down. This isn’t a fully-tidyverse solution, but for this data: library(dplyr) library(tidyr) data <- tribble( ~Date, ~time_series_1, ~time_series_2, as.Date(“2019-01-01”), NA, 10, as.Date(“2019-02-01”), 5, NA, … Read more
Introduction Group_by and combinations are two powerful tools in the R programming language. They allow users to quickly and easily manipulate data and create meaningful insights. Group_by allows users to group data by one or more variables, while combinations allow users to create all possible combinations of a set of variables. In this article, we … Read more
Introduction If you are looking for a way to create four random defined percentages that sum to 1, then you have come to the right place. In this article, we will provide a step-by-step guide on how to generate four random defined percentages that sum to 1. We will also discuss the importance of using … Read more
Usually when questions like this are asked some effort needs to be shown. So please take consideration to state the exact problem with at least some effort on what you have attempted next time. To get you started here is an example using the XML package and applying XPath along with strsplit to grab the … Read more
You can run an R script in VBA by creating Windows Shell obejct and passing it a string that executes an R script Sub RunRscript() ‘runs an external R code through Shell ‘The location of the RScript is ‘C:\R_code’ ‘The script name is ‘hello.R’ Dim shell As Object Set shell = VBA.CreateObject(“WScript.Shell”) Dim waitTillComplete As … Read more
This is a really unusual thing to do. No, it’s worse than unusual. It’s probably just bad. Matrices and arrays are useful, appropriate and very fast if all of the data is of a single class. If you have mixed classes, then work with data frames instead. But here you go. In the first example … Read more
If your schedule is stored as a matrix or data frame, you can use the reshape2 package: # generate a fake schedule sched <- matrix(rbinom(25, 1, 1/2), 5, dimnames = list(1:5, c(“Mo”, “Tu”, “We”, “Th”, “Fr”))) library(reshape2) melt(sched) # Var1 Var2 value # 1 1 Mo 0 # 2 2 Mo 0 # 3 3 … Read more
We can use paste iris$Observation <- paste0(“Obs_”, seq_len(nrow(iris))) If we are using tidyverse, there is row_number() function library(dplyr) iris %>% mutate(Observation = paste0(“Obs_”, row_number())) solved How to add column in a data frame with values based on the length of data as as string [duplicate]
Warning, a very large table will be created! dt <- as.data.table(matrix(runif(1000*1000000),ncol=1000)) dt[,lapply(.SD,max)] 14 solved What is the substitute idiom for sapply in data.table? [duplicate]
This question is very vague and does not have any specific code to further understand what you mean (reproducible example). It seems to me that you need to understand the data using descriptive statistics first. Functions such as summary, head, names, levels, sapply and nlevels (by column) can help you to begin understanding what you’re … Read more
You will probably speed things up a lot if you get rid of the outer loop and construct an entire column of your matrix at once: nsim <- 1e5 nt <- 200 U <- matrix(NA, nrow=nsim, ncol=200, byrow=TRUE) V_last <- V <- rep(0.04, nsim) U[,1] <- 40 Inv.phi <- function(y,p) ifelse (y <= p, 0, … Read more