由于输入为data.table，因此下面是使用data.table的一种方法

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]

• 输出

> 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

或者，另一种选择是使用命名向量

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]

• 输出

> 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

或者创建1和0的计数进行相乘

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

这是一个tidyverse方式的解决方案。它使用一个data.frame，并在第一步中使其变长。然后使用case_when实现不同的条件。
pivot_wider把它带回了更广泛的格式。

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

使用dplyr::across()，你可以跨多列应用一个函数，它支持tidy selections，这样你就可以根据变量的名称或属性巧妙地选择变量。

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

一个狡猾的方法

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

使用矩阵乘法：

# like @akrun using a named vector for conversion, to avoid ifelse/case/switch:
keyval <- setNames(c(2, 2, 1, 1), c("0/0", "1/1", "1/0", "0/1"))

#convert and make the matrix, then multiply
matrix(keyval[ as.matrix(dt[, -1 ]) ] * dt[[ 1 ]], ncol = ncol(dt) - 1)
#        [,1]   [,2]   [,3]
# [1,]  0.660  0.330  0.660
# [2,]  0.340  0.680  0.340
# [3,]  0.300  0.600  0.600
# [4,] -0.440 -0.220 -0.440
# [5,]  0.464  0.464  0.232

使用更大的数据集进行基准测试：

library(dplyr)
library(tidyr)

#bigger data
n = 1000
set.seed(1); dt <- data.table(cbind(
  Score = runif(n),
  data.frame(matrix(sample(c("0/0", "0/1", "1/0", "1/1"), n * n, replace = TRUE), ncol = n))))

与 dplyr 相比，矩阵乘法应给予3- 7倍：

m <- microbenchmark::microbenchmark(
  matrix = {
    matrix(keyval[ as.matrix(dt[, -1 ]) ] * dt[[ 1 ]], ncol = ncol(dt) - 1)
  },
  dplyr1 = {
    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)
  },
  dplyr2 = {
    dt %>%
      mutate(across(-Score, ~ Score * (.x %in% c("0/0", "1/1") + 1)))
  })

print(m, unit = "relative")
# Unit: relative
#    expr      min       lq     mean   median       uq      max neval
#  matrix 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000   100
#  dplyr1 7.697692 8.468686 7.279598 8.071069 7.652855 3.423847   100
#  dplyr2 3.862794 3.708899 3.399736 3.560082 3.687698 2.096620   100