import pandas as pd

# 示例DataFrame
df = pd.DataFrame({
    'fruit_id': [1, 2, 3, 1, 2],
    'quantity': [10, 15, 8, 12, 9]
})

# 键值映射字典
fruit_map = {
    1: 'Apple',
    2: 'Banana',
    3: 'Orange'
}

# 方法1: 使用map函数
df['fruit_name'] = df['fruit_id'].map(fruit_map)

# 方法2: 使用replace函数
df['fruit_name'] = df['fruit_id'].replace(fruit_map)

# 方法3: 使用字典推导式(适用于大型DataFrame)
fruit_series = pd.Series(fruit_map)
df['fruit_name'] = df['fruit_id'].map(fruit_series)

print(df)

# 示例数据框
df <- data.frame(
  fruit_id = c(1, 2, 3, 1, 2),
  quantity = c(10, 15, 8, 12, 9)
)

# 键值映射列表
fruit_map <- c("1" = "Apple", "2" = "Banana", "3" = "Orange")

# 方法1: 使用factor和levels
df$fruit_name <- factor(df$fruit_id, levels = names(fruit_map), labels = fruit_map)

# 方法2: 使用dplyr的recode
library(dplyr)
df <- df %>% mutate(fruit_name = recode(fruit_id, !!!fruit_map))

print(df)

-- 假设有一个fruit_mapping表存储映射关系
SELECT 
    t.fruit_id,
    fm.fruit_name,
    t.quantity
FROM 
    transactions t
JOIN 
    fruit_mapping fm ON t.fruit_id = fm.fruit_id;

# 使用map的fillna参数或设置默认值
df['fruit_name'] = df['fruit_id'].map(fruit_map).fillna('Unknown')

# 或者
default_value = 'Unknown'
df['fruit_name'] = df['fruit_id'].map(lambda x: fruit_map.get(x, default_value))

# 创建反向映射字典
reverse_map = {v: k for k, v in fruit_map.items()}
df['fruit_id'] = df['fruit_name'].map(reverse_map)