如何使用 DRY 原则减少这段代码？

Question

我在我的 R 代码中实践了 DRY 原则，但我已经到了无法减少代码行数的地步，我发现它非常重复，我需要你的帮助。

这是一个可重现的例子：

library(tidyverse)
set.seed(2023)

# first, I generate the data
data <- data.frame(var1_01 = c(sample(0:1, 7, replace=T)),
                   var1_02 = c(sample(0:1, 7, replace=T)),
                   var1_03 = c(sample(0:1, 7, replace=T)),
                   var1_04 = c(sample(0:1, 7, replace=T)),
                   var1_05 = c(sample(0:1, 7, replace=T)),
                   var1_06 = c(sample(0:1, 7, replace=T)),
                   var1_07 = c(sample(0:1, 7, replace=T)),
                   var1_08 = c(sample(0:1, 7, replace=T)),
                   var1_09 = c(sample(0:1, 7, replace=T)),
                   var1_10 = c(sample(0:1, 7, replace=T)),
                   var2_01 = c(sample(30:100, 7, replace=T)),
                   var2_02 = c(sample(30:100, 7, replace=T)),
                   var2_03 = c(sample(30:100, 7, replace=T)),
                   var2_04 = c(sample(30:100, 7, replace=T)),
                   var2_05 = c(sample(30:100, 7, replace=T)),
                   var2_06 = c(sample(30:100, 7, replace=T)),
                   var2_07 = c(sample(30:100, 7, replace=T)),
                   var2_08 = c(sample(30:100, 7, replace=T)),
                   var2_09 = c(sample(30:100, 7, replace=T)),
                   var2_10 = c(sample(30:100, 7, replace=T)))

这是我要减少的代码：

data %<>%
  mutate(var3_01 = case_when(var1_01 == 1 ~ var2_01 + 0,
                              TRUE ~ 0),
         var3_02 = case_when(var1_02 == 1 ~ var2_02 + 0,
                              TRUE ~ 0),
         var3_03 = case_when(var1_03 == 1 ~ var2_03 + 0,
                              TRUE ~ 0),
         var3_04 = case_when(var1_04 == 1 ~ var2_04 + 0,
                              TRUE ~ 0),
         var3_05 = case_when(var1_05 == 1 ~ var2_05 + 0,
                              TRUE ~ 0),
         var3_06 = case_when(var1_06 == 1 ~ var2_06 + 0,
                              TRUE ~ 0),
         var3_07 = case_when(var1_07 == 1 ~ var2_07 + 0,
                              TRUE ~ 0),
         var3_08 = case_when(var1_08 == 1 ~ var2_08 + 0,
                              TRUE ~ 0),
         var3_09 = case_when(var1_09 == 1 ~ var2_09 + 0,
                              TRUE ~ 0),
         var3_10 = case_when(var1_10 == 1 ~ var2_10 + 0,
                              TRUE ~ 0))

目标是如果 var1_* == 1，它为每一行取 var2_* 的值，但是我无法在较短的版本中复制此代码（tidyverse o R Base 版本，这无关紧要).我试过这个：

numbers <- c(paste0("0", 1:5))

data %<>% 
  mutate(across(starts_with("var1_"), ~ifelse(isTRUE(.x==1), .x:=data[, 6:10], 0), .names="var3_{numbers}"))

...但是这段代码不会生成与扩展版本相同的结果

感谢任何建议！

Answer 1

如果你想维护宽数据格式，我会使用矩阵。像这样的东西：

set.seed(2023)
DF <- do.call(data.frame, 
              c(setNames(replicate(10, sample(0:1, 7, replace=T), simplify = FALSE),
                         sprintf("var1_%02d", 1:10)),
                setNames(replicate(10, sample(30:100, 7, replace=T), simplify = FALSE),
                         sprintf("var2_%02d", 1:10))))
foo <- function(a, b) {
  a <- as.matrix(a)
  b <- as.matrix(b)
  b[a == 0] <- 0
  colnames(b) <- gsub("var2", "var3", colnames(b))
  as.data.frame(b)
}

DF <- cbind(DF, foo(DF[, grepl("var1", names(DF))],    
                    DF[, grepl("var2", names(DF))]))

all.equal(data, DF)
#[1] TRUE

Answer 2

留在 tidyverse 中

可以用

across

，用

get

在

case_when

内使用，免去重复

cols = names(data)[1:10]

data |> 
  mutate(across({cols}, \(x){
    partName <- gsub("var\\d", "", cur_column())
    case_when(x == 1 ~ get(paste0("var2", partName)), TRUE ~ 0)
  }, .names = "{paste0('var3_', c(paste0('0', seq(cols)[-length(cols)]), length(cols)))}"))

  var1_01 var1_02 var1_03 var1_04 var1_05 var1_06 var1_07 var1_08 var1_09 var1_10 var2_01 var2_02 var2_03 var2_04
1       0       0       1       1       1       0       0       1       1       1      31      74      42      60
2       0       1       0       0       1       0       1       0       1       1      92      63      57      98
3       1       1       0       1       0       0       0       1       1       0      53      89      64      42
4       0       1       0       0       0       1       0       1       1       1      55      37      41      97
5       0       0       0       0       1       1       0       0       0       1      47      87      56      60
6       0       0       1       0       1       0       0       0       0       1      99      73      79      31
7       1       0       0       1       0       0       0       1       1       0      61      44      52      90
  var2_05 var2_06 var2_07 var2_08 var2_09 var2_10 var3_01 var3_02 var3_03 var3_04 var3_05 var3_06 var3_07 var3_08
1      60      55      57      67      97      40       0       0      42      60      60       0       0      67
2      97      78      74      30      90      49       0      63       0       0      97       0      74       0
3      77      43      52      84      43      78      53      89       0      42       0       0       0      84
4      95      94      65      86      32      82       0      37       0       0       0      94       0      86
5      47      65     100      70      91      40       0       0       0       0      47      65       0       0
6      93      77      92      57      76      93       0       0      79       0      93       0       0       0
7      46     100      74      35      38      56      61       0       0      90       0       0       0      35
  var3_09 var3_10
1      97      40
2      90      49
3      43       0
4      32      82
5       0      40
6       0      93
7      38       0

Answer 3

您可以使用

across(...)

选择变量并重命名，然后将其传递给

ifelse()

。

data %>%
  mutate(as_tibble(
    ifelse(as.matrix(across(contains("var1"), .names = "{sub('var1', 'var3', .col)}")) == 1,
           as.matrix(across(contains("var2"))), 0)
  ))