如何在R语言中编写if-else函数生成指定规则？

方案1：枚举所有情况的if-else判断 #

这是最直观的写法，枚举所有8种输入组合，逻辑清晰不易出错：

def rule(a, b, c):
    if a == 0 and b == 0 and c == 0:
        return 0
    elif a == 0 and b == 0 and c == 1:
        return 1
    elif a == 0 and b == 1 and c == 0:
        return 1
    elif a == 0 and b == 1 and c == 1:
        return 1
    elif a == 1 and b == 0 and c == 0:
        return 1
    elif a == 1 and b == 0 and c == 1:
        return 0
    elif a == 1 and b == 1 and c == 0:
        return 0
    elif a == 1 and b == 1 and c == 1:
        return 0

方案2：简化逻辑表达式 #

通过分析输出规律，可将条件合并简化，减少代码量：
输出为1的场景仅两种：

• 第一个参数a为0，且不是三个参数全为0；
• 第一个参数a为1，且b和c都为0。

对应代码：

def rule(a, b, c):
    if (a == 0 and not (b == 0 and c == 0)) or (a == 1 and b == 0 and c == 0):
        return 1
    return 0

也可以用布尔运算进一步精简（0和1可直接参与逻辑运算）：

def rule(a, b, c):
    return 1 if (not a and (b or c)) or (a and not b and not c) else 0

方案3：查表法 #

将三个二进制输入转换为十进制索引，通过预定义的查找表直接取值，代码最简洁，修改规则只需调整查找表：

def rule(a, b, c):
    # 索引计算：4*a + 2*b + c，对应二进制a b c的十进制值
    lookup_table = [0, 1, 1, 1, 1, 0, 0, 0]
    return lookup_table[4*a + 2*b + c]

内容的提问来源于stack exchange，提问作者Bayes Rule