Filling a row from a certain column under certain condition in r-CodePudding

I want to fill a row with "-3" from some specific columns (B and E in this example) to the end when these columns contain "-3" in that row. I figured out a solution, but it is extremely slow in my original dataset (2435 x 431 cells) and with 15 columns to check for values == "-3".

In this example, the rows to fill with "-3" are 4 and 10 from column "B" and 3 from column "D". Note that 4 and 10 also contain values == "-3" in column "E" but they were already filled when iterated over column "B"

library(tidyverse)

values <- as.character(-3:3)

set.seed(123)

data <- tibble(
  A = sample(values, 10, replace = T),
  B = sample(values, 10, replace = T),
  C = sample(values, 10, replace = T),
  D = sample(values, 10, replace = T),
  E = sample(values, 10, replace = T),
  F = sample(values, 10, replace = T)
)

fill_minus_three <- function(x){
  for (i in 1:length(x)){
    if ((names(x)[i] %in% c("B", "E")) && x[i] == "-3"){
      x[i:length(x)] <- "-3"
      break
    }
  }
  return(x)
}

t(apply(data, 1, fill_minus_three)) %>% 
  as_tibble()

#> # A tibble: 10 x 6
#> A     B     C     D     E     F    
#> <chr> <chr> <chr> <chr> <chr> <chr>
#> 1  3    0     0    -1     1     1    
#> 2  3    2    -3     0     3    -1   
#> 3 -1    2    -3     2    -3    -3   
#> 4  2   -3    -3    -3    -3    -3   
#> 5 -1   -2    -1    -1    -2    -2   
#> 6 -2   -1    -2     3     3     1    
#> 7 -2    1     3     1    -1     1    
#> 8  2   -1    -2     0     0     0    
#> 9 -1   -1    -3     3     1     3    
#> 10 1   -3    -3    -3    -3    -3

In adittion, I would like to use map_* family since the rest of the scripts follow the tidyverse approach (however, this is optional).

CodePudding user response：

Si entenc bé, you're trying to change values in multiple columns according to the values in columns B and E.

No need for for loops or map/apply functions, you can just use mutate and pair it with across:

library(dplyr)

data |> 
  mutate(across(C:F, ~ if_else(B == "-3", "-3", .x)),
         F = if_else(E == "-3", "-3", F))

Output

#> # A tibble: 10 × 6
#>    A     B     C     D     E     F    
#>    <chr> <chr> <chr> <chr> <chr> <chr>
#>  1 3     0     0     -1    1     1    
#>  2 3     2     -3    0     3     -1   
#>  3 -1    2     -3    2     -3    -3   
#>  4 2     -3    -3    -3    -3    -3   
#>  5 -1    -2    -1    -1    -2    -2   
#>  6 -2    -1    -2    3     3     1    
#>  7 -2    1     3     1     -1    1    
#>  8 2     -1    -2    0     0     0    
#>  9 -1    -1    -3    3     1     3    
#> 10 1     -3    -3    -3    -3    -3

^{Created on 2022-06-02 by the reprex package (v2.0.1)}

CodePudding user response：

I give a thought and I realized that I do not need to iterate over all columns, just the ones I need to check for value == "-3". So I changed the function a bit, I works much faster (0.3 s in the entire dataset while previous took two minutes). Yet, it is not tidy-friendly. :/

positions <- which(names(data) %in% c("B", "E"))

fill_minus_three <- function(x, positions){
  for (i in positions){
    if (x[i] == "-3"){
      x[i:length(x)] <- "-3"
      break
    }
  }
  return(x)
}

t(apply(data, 1, function(x) fill_minus_three(x, positions))) %>% 
  as_tibble()