R语言绘制logistic回归图时mutate生成prob列报错如何解决

• 核心报错来自mutate()中ifelse()的语法错误：原代码中ifelse(outcome == "pos", 0",","1")存在引号错位问题，0后的引号、包裹逗号的引号都是无效写法，导致R解析出了多余的第4个参数，触发unused argument报错。
• 额外逻辑问题：你的y轴是「糖尿病阳性概率」，如果要对应outcome为pos的样本，ifelse的返回值写反了：outcome等于pos时应该返回1，否则返回0，否则拟合出的概率曲线是反向的。

假设你的原始数据集名为df，修正后代码如下：

df %>%
  mutate(prob = ifelse(outcome == "pos", 1, 0)) %>%
  ggplot(aes(plasma_glucose, prob)) +
  geom_point(alpha = 0.3) + # 若点无重叠可保留原来的alpha=1
  geom_smooth(method = "glm", method.args = list(family = "binomial")) +
  labs(
    title = "Logistic Regression Model", 
    x = "Plasma Glucose Concentration",
    y = "Probability of being diabete-pos"
  )

你也可以不用提前生成0/1变量，直接在ggplot映射中处理，代码更简洁：

df %>%
  ggplot(aes(plasma_glucose, as.numeric(outcome == "pos"))) +
  geom_point(alpha = 0.3) +
  geom_smooth(method = "glm", method.args = list(family = "binomial")) +
  labs(
    title = "Logistic Regression Model", 
    x = "Plasma Glucose Concentration",
    y = "Probability of being diabete-pos"
  )

内容的提问来源于stack exchange，提问作者Div