case_when() 给出的具有不同多个条件的 R mutate()

Question

我正在尝试使用 mutate() 和 case_when() 在 R 中创建具有值的 table，因为 mutate() 的计算可能因 [= 的多个条件而异23=]()。但是 case_when() 的多个条件的数量也可能因我使用的模型而异。下面是一个可重现的例子（来源：https://doseresponse.github.io/medrc/articles/medrc.html）：

library(medrc)
library(dplyr)
library(tidyr)

data(spinach)
spinach$CURVE <- as.factor(spinach$CURVE)

#I can have a model using LL.4() or LL.3() function (as in 'fct' bellow)
sm1 <- metadrm(SLOPE ~ DOSE, 
               data=spinach,
               fct=LL.3(),
               ind=CURVE,
               cid2=HERBICIDE,
               struct="UN")

#or in other situations
sm1 <- metadrm(SLOPE ~ DOSE, 
               data=spinach,
               fct=LL.4(),
               ind=CURVE,
               cid2=HERBICIDE,
               struct="UN")

#Extracting the coefficients from the model
minor_coef_table <- as.data.frame(sm1$estimates$ind)
minor_coef_table <- cbind(minor_coef_table, sm1$estimates$coefficient)
minor_coef_table <- cbind(minor_coef_table, sm1$estimates$estimate)
colnames(minor_coef_table) <- c("curves","minor_coef", "minor_estimates")

#I need to create a "table" ('pdata') with a calculated column "SLOPE_per_CURVE" in a way that it is independent of the number of levels of CURVE in the data (spinach, in the example).
pdata <- spinach %>%
  group_by(CURVE, HERBICIDE) %>%
  expand(DOSE=exp(seq(-5, 5, length=50)))

#One of the conditions is the fct used in the model [LL.4() or LL.3()]
#Other conditions are the curve IDs
ncurves <- length(levels(spinach$CURVE))

#This is an IDEA of what I need, but it is not working as the column "SLOPE_per_CURVE is not created
pdata <- pdata %>% mutate(
  SLOPE_per_CURVE =
    for(i in 1:ncurves){
    case_when(
      sm1$fct$name == "LL.4" & CURVE == levels(pdata$CURVE)[i] ~ minor_coef_table[(ncurves+i),3]+((minor_coef_table[(2*ncurves+i),3]-minor_coef_table[(ncurves+i),3])/((1+ exp(minor_coef_table[i,3] * (log(DOSE) - log(minor_coef_table[(3*ncurves+i),3])))))),
      sm1$fct$name == "LL.3" & CURVE == levels(pdata$CURVE)[i] ~ ((minor_coef_table[(ncurves+i),3])/((1+ exp(minor_coef_table[i,3] * (log(DOSE) - log(minor_coef_table[(2*ncurves+i),3]))))))
    )
  }
)

#The bellow code gives what I need and is an example of the final desired result, but it is not independent of the number of levels in CURVE as I have to write every condition. 
pdata <- pdata %>% mutate(
  SLOPE_per_CURVE =
    case_when(
      sm1$fct$name == "LL.4" & CURVE == levels(pdata$CURVE)[1] ~ minor_coef_table[6,3]+((minor_coef_table[11,3]-minor_coef_table[6,3])/((1+ exp(minor_coef_table[1,3] * (log(DOSE) - log(minor_coef_table[16,3])))))),
      sm1$fct$name == "LL.4" & CURVE == levels(pdata$CURVE)[2] ~ minor_coef_table[7,3]+((minor_coef_table[12,3]-minor_coef_table[7,3])/((1+ exp(minor_coef_table[2,3] * (log(DOSE) - log(minor_coef_table[17,3])))))),
      sm1$fct$name == "LL.4" & CURVE == levels(pdata$CURVE)[3] ~ minor_coef_table[8,3]+((minor_coef_table[13,3]-minor_coef_table[8,3])/((1+ exp(minor_coef_table[3,3] * (log(DOSE) - log(minor_coef_table[18,3])))))),
      sm1$fct$name == "LL.4" & CURVE == levels(pdata$CURVE)[4] ~ minor_coef_table[9,3]+((minor_coef_table[14,3]-minor_coef_table[9,3])/((1+ exp(minor_coef_table[4,3] * (log(DOSE) - log(minor_coef_table[19,3])))))),
      sm1$fct$name == "LL.4" & CURVE == levels(pdata$CURVE)[5] ~ minor_coef_table[10,3]+((minor_coef_table[15,3]-minor_coef_table[10,3])/((1+ exp(minor_coef_table[5,3] * (log(DOSE) - log(minor_coef_table[20,3])))))),
      sm1$fct$name == "LL.3" & CURVE == levels(pdata$CURVE)[1] ~ ((minor_coef_table[6,3])/((1+ exp(minor_coef_table[1,3] * (log(DOSE) - log(minor_coef_table[11,3])))))),
      sm1$fct$name == "LL.3" & CURVE == levels(pdata$CURVE)[2] ~ ((minor_coef_table[7,3])/((1+ exp(minor_coef_table[2,3] * (log(DOSE) - log(minor_coef_table[12,3])))))),
      sm1$fct$name == "LL.3" & CURVE == levels(pdata$CURVE)[3] ~ ((minor_coef_table[8,3])/((1+ exp(minor_coef_table[3,3] * (log(DOSE) - log(minor_coef_table[13,3])))))),
      sm1$fct$name == "LL.3" & CURVE == levels(pdata$CURVE)[4] ~ ((minor_coef_table[9,3])/((1+ exp(minor_coef_table[4,3] * (log(DOSE) - log(minor_coef_table[14,3])))))),
      sm1$fct$name == "LL.3" & CURVE == levels(pdata$CURVE)[5] ~ ((minor_coef_table[10,3])/((1+ exp(minor_coef_table[5,3] * (log(DOSE) - log(minor_coef_table[15,3]))))))
    )
)

编辑：

在M.Viking回答后，这是解决方案：

for(i in 1:ncurves){
pdata <- coalesce(pdata %>% mutate(
  SLOPE_per_CURVE =
      case_when(
        sm1$fct$name == "LL.4" & CURVE == levels(pdata$CURVE)[i] ~ minor_coef_table[(ncurves+i),3]+((minor_coef_table[(2*ncurves+i),3]-minor_coef_table[(ncurves+i),3])/((1+ exp(minor_coef_table[i,3] * (log(DOSE) - log(minor_coef_table[(3*ncurves+i),3])))))),
        sm1$fct$name == "LL.3" & CURVE == levels(pdata$CURVE)[i] ~ ((minor_coef_table[(ncurves+i),3])/((1+ exp(minor_coef_table[i,3] * (log(DOSE) - log(minor_coef_table[(2*ncurves+i),3]))))))
      )
    ), pdata)
  }

Answer 1

我用 iris 数据集

制作了一个 for 循环 mutate case_when 的玩具示例

第一个复杂化是在一个简单的 for (j in 1:10){out<-j} 循环中，输出数据被 j 的每个后续迭代覆盖，最后，只有 10th 的结果运行被保留。

然后我了解了coalesce()函数；它结合（合并？联合？加入？）等长的稀疏数据（类似于 Microsoft Excel 的特殊粘贴“跳过空白”转换）。

在下面的示例代码中，我们取 iris 数据集，并从 1 循环到 10，如果 iris$Sepal.Length（又名 iris[,1]）等于循环迭代 j ，我们将结果变量（斜率）更改为 100+j.

iris4 <- tibble(iris)

for (j in 1:10) {
  iris4 <- coalesce(iris4 %>% mutate(Slope = case_when(Sepal.Length == j ~ 100+j)),iris4)}

table(iris4$Slope)

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