so I managed to calculate the count_new by checking the first appearance of a user_id and then merging with initial data adding a column that tell if a user is new by date and id then I counted the new by date: library(dplyr) firstshow<-Orders %>% group_by(user_id) %>% arrange(date) %>% slice(1L) %>% mutate(new = “new”) newdata<-merge.data.frame(Orders,firstshow,by=c(“date”,”user_id”),all … Read more
We can use duplicated dat[!(duplicated(dat$ID)|duplicated(dat$ID, fromLast = TRUE)),] 0 solved Keep only non-duplicate rows based on a Column Value [duplicate]
Try this: patients[patients$time > ‘2017-01-02 11:41:53’ & patients$time < ‘2017-08-07 09:06:07’,] 3 solved Fetch the time between two timestamps in R [duplicate]
I think what you are looking for (if you want to use dplyr) is a combination of the functions group_byand mutate. library(dplyr) city <- c(“a”, “a”, “b”, “b”, “c”) temp <- 1:5 df <- data.frame(city, temp) df %>% group_by(city) %>% mutate(mean(temp)) Which would output: city temp mean(temp) (fctr) (int) (dbl) 1 a 1 1.5 2 … Read more
You just want to group on-the-fly, which means passing an ad hoc variable to the grouper. In data.table, this would be done by using rep in by: library(data.table) set.seed(01394) DT <- data.table(VAR = rnorm(2520)) DT[ , mean(VAR), by = rep(1:18, each = 140)] # rep V1 # 1: 1 0.029683200 # 2: 2 0.150411987 # … Read more
The following will convert all non-zero numeric values to 1: df.richness %>% mutate_if(is.numeric, ~1 * (. != 0)) while df.richness %>% mutate_if(is.numeric, ~1 * (. > 0)) will do that with those greater than zero. solved Replace values greater than zero with 1 in r
Here you go. I acknowledge you specified a loop in your question, but in R I avoid loops wherever possible. This is better. This uses plyr::join_all to join all your data frames by Item and LC, then dplyr::mutate to do the calculations. Note you can put multiple mutations in one mutate() function: library(plyr) library(dplyr) library(tidyr) … Read more
Something like this (assuming your data is named df) # get the column names that follow this pattern cols = grepl(“[0-9]+_parameter”, names(df)) # see if any of those columns have a 97 any(df[cols] == 97) A base R way to filter for rows that have a 97 in any of those columns is this: df[rowSums(df[cols] … Read more
Promoting my comment to an answer as the OP appears to agree (and removing a spurious ) in the process): computeY <- function(x) { sqrt(1-(x-1)^2) } which at the prompt works as well: R> computeY <- function(x) { sqrt(1-(x-1)^2) } R> sapply((1:5)/10, computeY) [1] 0.435890 0.600000 0.714143 0.800000 0.866025 R> Disclaimer: No pipes were hurt … 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
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
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]
The best way to do this would be to use the glue package. This allows you to easily assign variable names based on strings: library(tidyverse) library(glue) > df <- data.frame( sample_id = c(‘SB013’, ‘SB014’, ‘SB013’, ‘SB014’, ‘SB015’, ‘SB016’, ‘SB016’), IN = c(1,2,3,4,5,6,7), OUT = c(rep(‘out’,7))) > df sample_id IN OUT 1 SB013 1 out 2 … Read more
1) eval/parse Create a cmd string, parse it and evaluate it: f2 <- function(DF, x, env = parent.frame()){ cmd <- sprintf(“mutate(%s, %s = mean(v1))”, deparse(substitute(DF)), x) eval(parse(text = cmd), env) } f2(DF, “v1_name”) giving v1 v1_mean 1 1 2 2 2 2 3 3 2 … etc … 2) eval/as.call Another way is to construct … Read more