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16 de ene. de 2019 · 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.
A newer solution is to use case_match from dplyr. library(dplyr) dat %>% mutate(var = case_match(var, "Candy" ~ var, .default ~ "Not Candy"))
18 de mar. de 2022 · In this example, I’ll show how to replace particular values in a data frame variable by using the mutate and replace functions in R. More precisely, the following R code replaces each 2 in the column x1: data_new <- data %>% # Replacing values. mutate ( x1 = replace ( x1, x1 == 2, 99))
31 de may. de 2024 · Use mutate() method from dplyr package to replace the R DataFrame column value. The following example replaces all instances of the street with st on the address column. library("dplyr") # Replace on selected column df <- df %>% mutate(address = str_replace(address, "St", "Street")) df
df <-tibble (x = c ("a", "b"), y = c (1, 1), z = c (-1, 1)) # Find all rows where EVERY numeric variable is greater than zero df %>% filter (if_all (where (is.numeric), ~.x > 0)) #> # A tibble: 1 × 3 #> x y z #> <chr> <dbl> <dbl> #> 1 b 1 1 # Find all rows where ANY numeric variable is greater than zero df %>% filter (if_any (where (is.numeric ...
dplyr::na_if() to replace specified values with NAs; dplyr::coalesce() to replaces NAs with values from other vectors.
27 de jun. de 2022 · You can use the following methods to replace a string in a specific column of a data frame using functions from the dplyr package: Method 1: Replace One String with New String. library(dplyr) library(stringr) df %>%. mutate(across('column_name', str_replace, 'old_value', 'new_value')) Method 2: Replace Multiple Strings with New String.
