Replace each row based on the string match and column value

Question

I have a dataset like this

dt <- data.table(Score = c(0.33,0.34,00.3, -0.22, 0.232), 
                 Id2 = c("0/0","0/1","1/0","0/0","0/0"), 
                 Kps = c("0/1","0/0","1/1","0/1","0/0"), 
                 Inr = c("0/0","0/1","1/1","0/0","0/1"))

I need to replace the values of each row based on the Score column as like this

• If "0/0" or "1/1" then Score * 2
• If "1/0" or "0/1" then Score

Usually, it can be done by using the base function like this

dt$Id2 <- dt$Score * 2

But here I have to consider each row and I have around 1000 columns so it can be only done with loop

The expected output

Score  Id2    Kps    Inr 
0.330  0.66   0.330  0.66
0.340  0.340  0.68  0.340
0.300  0.300  0.6   0.6
-0.220 -0.44 -0.22 -0.44
0.232  0.464 0.464  0.232

Any suggestions?

Answer 1

As the input is data.table, here is one approach with data.table

library(data.table)
 dt[, (names(dt)[-1]) := lapply(.SD, \(x)
    fcase(x %chin% c("0/0", "1/1"), Score *2,
    x %chin% c("1/0", "0/1"), Score)), .SDcols = -1]

-output

> dt
    Score    Id2    Kps    Inr
1:  0.330  0.660  0.330  0.660
2:  0.340  0.340  0.680  0.340
3:  0.300  0.300  0.600  0.600
4: -0.220 -0.440 -0.220 -0.440
5:  0.232  0.464  0.464  0.232

---

Or another option is to make use of named vector

keyval <- setNames(c(2, 2, 1, 1), c("0/0", "1/1", "1/0", "0/1"))
dt[, (names(dt)[-1]) := lapply(.SD, \(x) Score *keyval[x]), .SDcols = -1]

-output

> dt
    Score    Id2    Kps    Inr
1:  0.330  0.660  0.330  0.660
2:  0.340  0.340  0.680  0.340
3:  0.300  0.300  0.600  0.600
4: -0.220 -0.440 -0.220 -0.440
5:  0.232  0.464  0.464  0.232

---

Or create a count of 1s and 0s to multiply

library(stringr)
dt[, (names(dt)[-1]) := lapply(.SD, \(x) Score * 1 + 
   (str_count(x, "0")!= 1)) , .SDcols = -1]
> dt
    Score   Id2    Kps   Inr
1:  0.330 1.330  0.330 1.330
2:  0.340 0.340  1.340 0.340
3:  0.300 0.300  1.300 1.300
4: -0.220 0.780 -0.220 0.780
5:  0.232 1.232  1.232 0.232

Answer 2

Here is a tidyverse -way solution. It uses a data.frame and makes it longer in a first step. Then with case_when the different conditions were implemented.

pivot_wider brought it back to a wider format.

library(tidyverse)

dt<- data.frame(Score = c(0.33,0.34,00.3, -0.22, 0.232), 
                Id2=c("0/0","0/1","1/0","0/0","0/0"), 
                Kps=c("0/1","0/0","1/1","0/1","0/0"), 
                Inr=c("0/0","0/1","1/1","0/0","0/1"))


dt |> 
  pivot_longer(-Score) |> 
  mutate(value = case_when(
    value == '0/0' | value == "1/1" ~ Score *2,
    value == '1/0' | value == "0/1" ~ Score 
  )) |> 
  pivot_wider(names_from = name, values_from = value)
#> # A tibble: 5 × 4
#>    Score    Id2    Kps    Inr
#>    <dbl>  <dbl>  <dbl>  <dbl>
#> 1  0.33   0.66   0.33   0.66 
#> 2  0.34   0.34   0.68   0.34 
#> 3  0.3    0.3    0.6    0.6  
#> 4 -0.22  -0.44  -0.22  -0.44 
#> 5  0.232  0.464  0.464  0.232