tidyverse

I want to recode a bunch of variables with as few function calls as possible. I have one data.frame where I want to recode a number of variables. I create a named list of all variable names and the recoding arguments I want to execute. Here I have no problem using map and dpylr . However, when it comes to recoding I find it much easier using recode from the car package, instead of dpylr 's own recoding function. A side question is whether there is a nice way of doing the same thing with dplyr::recode . As a next step I break the data.frame down into a nested tibble. Here I want to do specific

While working with R I encountered a strange problem: I am processing date in the follwing manner: Reading data from a database into a dataframe, filling missing values, grouping and nesting the data to a combined primary key, creating a timeseries and forecastting it for every group, ungroup and clean the data, write it back into the DB. Somehting like this: https://cran.rstudio.com/web/packages/sweep/vignettes/SW01_Forecasting_Time_Series_Groups.html For small data sets this works like a charm, but with lager ones (over about 100000 entries) I do get the "R Session Aborted" screen from R

I am having serious troubles using the tidyverse package that I cannot debug. As an example, "mutate" does not work properly even on past project I have already produced. This all started when I installed the following package: library(pdftools) library(tm) library(stringi) library(tidyverse) (or library(dplyr) library(tidyr)) library(purrr) ) And it still remains when I do a rm(list=ls()) . The only thing I haven't tried so forth is deinstalling R/RStudio and reinstalling it. I use RStudio version 1.0.153 and R version 3.4.1. I actually tried to reproduce the bug on other computers and this

The question replace NA in a dplyr chain results into the solution dt %.% group_by(a) %.% mutate(b = ifelse(is.na(b), mean(b, na.rm = T), b)) with dplyr. I want to impute all colums with dplyr chain. There is no single column to group by, rather I want all numeric columns to have all NAs replaced by the means such as column means. What is the most elegant way to replace all NAs with column means with tidyverse/dp? We can use mutate_all with ifelse dt %>% group_by(a) %>% mutate_all(funs(ifelse(is.na(.), mean(., na.rm = TRUE), .))) If we want a compact option, then use the na.aggregate from zoo

I am writing my function and want to use dplyr's filter() function to select rows of my data frame that satisfy a condition. This is my code: library(tidyverse) df <-data.frame(x = sample(1:100, 50), y = rnorm(50), z = sample(1:100,50), w = sample(1:100, 50), p = sample(1:100,50)) new <- function(ang,brad,drau){ df%>%filter(!!drau %in% 1:50)%>%select(ang,brad) -> A return(A) } brand <- c("z","w","p") lapply(1:3, function(i) new(ang = "x", brad = "y", drau = brand[i]))%>%bind_rows() Anytime I run this function, it looks like filter doesn't select any rows that satisfy the condition. How can I

I would like to find and extract the longest word of a string, if possible using a tidyverse package. library(tidyverse) tbl <- tibble(a=c("ab cde", "bcde f", "cde fg"), b=c("cde", "bcde", "cde")) tbl # A tibble: 3 x 1 a <chr> 1 ab cde 2 bcde f 3 cde fg The result I am looking for is: # A tibble: 3 x 2 a b <chr> <chr> 1 ab cde cde 2 bcde f bcde 3 cde fg cde The closest post to the question I have found is this: longest word in a string . Does anyone have an idea for an even simpler way? Solution using base R: # Using OPs provided data tbl$b <- sapply(strsplit(tbl$a, " "), function(x) x[which