R中使用lapply调用依赖全局常量的函数出错，求解决方案

问题背景 #

尝试在R中使用lapply将依赖全局向量BASELINE_CLASSIFICATION_THRESHOLDS的value_to_classification函数应用到数据框列，实现数值到Mild/Moderate/Severe/Extreme四个等级的转换。原代码如下：

BASELINE_CLASSIFICATION_THRESHOLDS  <- c(0, 3.5, 6.5, 10.0000001)

value_to_classification <- function(x){

  if((x >= BASELINE_CLASSIFICATION_THRESHOLDS[1]) && (x < BASELINE_CLASSIFICATION_THRESHOLDS[2])){
    classification <- "Mild"
  } else if((x >= BASELINE_CLASSIFICATION_THRESHOLDS[2]) && (x < BASELINE_CLASSIFICATION_THRESHOLDS[3])){
    classification <- "Moderate"
  } else if((x >= BASELINE_CLASSIFICATION_THRESHOLDS[3]) && (x < round(BASELINE_CLASSIFICATION_THRESHOLDS[4])){
    classification <- "Severe"
  } else {
    classification <- "Extreme"
  }
  return(classification)
}

df <- data.frame(x = runif(10, min = 0, max = 10),
                 y = runif(10, min = 0, max = 10),
                 z = runif(10, min = 0, max = 10))

执行lapply(df["x"], value_to_classification)时出现错误警告，仅返回一个"Mild"结果；但lapply(df[["x"]], value_to_classification)可正常运行。需要实现类似df[c("x1", "x2")] <- lapply(df[c("x", "y")], value_to_classification)的批量列转换。

错误原因 #

• 数据类型差异：
  df["x"]返回的是单列数据框，lapply处理数据框时会将整列（向量）作为单个元素传入函数；df[["x"]]返回的是向量，lapply会将向量的每个元素逐个传入函数。原函数是为单个数值设计的，当传入整列向量时，if语句的条件会返回长度大于1的逻辑向量，而if仅接受长度为1的逻辑值，触发警告后仅取第一个元素的判断结果，导致只返回"Mild"。
• 语法错误：
  第三个else if的条件表达式缺少闭合括号，正确写法应为：

  else if((x >= BASELINE_CLASSIFICATION_THRESHOLDS[3]) && (x < round(BASELINE_CLASSIFICATION_THRESHOLDS[4]))){
  

修复方案 #

方案1：使用向量化函数（推荐） #

利用R的向量化特性，用cut函数实现分类，代码更简洁高效：

BASELINE_CLASSIFICATION_THRESHOLDS <- c(0, 3.5, 6.5, 10.0000001)
classification_labels <- c("Mild", "Moderate", "Severe", "Extreme")

value_to_classification <- function(x) {
  # include.lowest = TRUE 包含左边界，right = FALSE 区间为左闭右开
  cut(x, breaks = BASELINE_CLASSIFICATION_THRESHOLDS, 
      labels = classification_labels, 
      include.lowest = TRUE, 
      right = FALSE)
}

# 批量处理多列并赋值
df[c("x1", "x2")] <- lapply(df[c("x", "y")], value_to_classification)

方案2：修改原函数为向量化版本 #

用嵌套ifelse实现向量化判断，兼容向量输入：

BASELINE_CLASSIFICATION_THRESHOLDS  <- c(0, 3.5, 6.5, 10.0000001)

value_to_classification <- function(x){
  ifelse(x >= BASELINE_CLASSIFICATION_THRESHOLDS[1] & x < BASELINE_CLASSIFICATION_THRESHOLDS[2], "Mild",
         ifelse(x >= BASELINE_CLASSIFICATION_THRESHOLDS[2] & x < BASELINE_CLASSIFICATION_THRESHOLDS[3], "Moderate",
                ifelse(x >= BASELINE_CLASSIFICATION_THRESHOLDS[3] & x < round(BASELINE_CLASSIFICATION_THRESHOLDS[4]), "Severe",
                       "Extreme")))
}

# 批量处理多列
df[c("x1", "x2")] <- lapply(df[c("x", "y")], value_to_classification)

方案3：用Vectorize包装原函数 #

如果不想修改原函数逻辑，可通过Vectorize将其转为支持向量输入的函数：

# 先修复原函数的语法错误
value_to_classification <- function(x){
  if((x >= BASELINE_CLASSIFICATION_THRESHOLDS[1]) && (x < BASELINE_CLASSIFICATION_THRESHOLDS[2])){
    classification <- "Mild"
  } else if((x >= BASELINE_CLASSIFICATION_THRESHOLDS[2]) && (x < BASELINE_CLASSIFICATION_THRESHOLDS[3])){
    classification <- "Moderate"
  } else if((x >= BASELINE_CLASSIFICATION_THRESHOLDS[3]) && (x < round(BASELINE_CLASSIFICATION_THRESHOLDS[4]))){
    classification <- "Severe"
  } else {
    classification <- "Extreme"
  }
  return(classification)
}

# 转为向量化函数
vec_classify <- Vectorize(value_to_classification)

# 批量处理
df[c("x1", "x2")] <- lapply(df[c("x", "y")], vec_classify)

内容的提问来源于stack exchange，提问作者Thomas Philips