R replace each row based on the string match and column value-CodePudding

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?

CodePudding user response：

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

CodePudding user response：

With dplyr::across(), you can apply a function across multiple columns. It supports tidy selections so that you can cleverly select variables based on their names or properties.

library(dplyr)

dt %>%
  mutate(across(-Score, ~ ifelse(.x %in% c("0/0", "1/1"), Score * 2, Score)))

#     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

A tricky way

dt %>%
  mutate(across(-Score, ~ Score * (.x %in% c("0/0", "1/1")   1)))

CodePudding user response：

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