R在图表本身上打印线性回归方程

Question

我们如何在图上打印线的等式？

我有2个自变量，想要这样的等式：

y=mx1+bx2+c

where x1=cost, x2 =targeting

我可以绘制最佳拟合线，但如何在图上打印等式？

也许我不能在一个等式中打印出2个自变量，但我怎么做呢 至少y=mx1+c？

这是我的代码：

fit=lm(Signups ~ cost + targeting)
plot(cost, Signups, xlab="cost", ylab="Signups", main="Signups")
abline(lm(Signups ~ cost))

Answer 1

我试着稍微自动化输出：

fit <- lm(mpg ~ cyl + hp, data = mtcars)
summary(fit)
##Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 36.90833    2.19080  16.847  < 2e-16 ***
## cyl         -2.26469    0.57589  -3.933  0.00048 ***
## hp          -0.01912    0.01500  -1.275  0.21253 


plot(mpg ~ cyl, data = mtcars, xlab = "Cylinders", ylab = "Miles per gallon")
abline(coef(fit)[1:2])

## rounded coefficients for better output
cf <- round(coef(fit), 2) 

## sign check to avoid having plus followed by minus for negative coefficients
eq <- paste0("mpg = ", cf[1],
             ifelse(sign(cf[2])==1, " + ", " - "), abs(cf[2]), " cyl ",
             ifelse(sign(cf[3])==1, " + ", " - "), abs(cf[3]), " hp")

## printing of the equation
mtext(eq, 3, line=-2)

希望它有所帮助，

亚历

Answer 2

您使用?text。此外，您不应该使用abline(lm(Signups ~ cost))，因为这是一个不同的模型（请参阅我在CV上的答案：Is there a difference between 'controling for' and 'ignoring' other variables in multiple regression）。无论如何，请考虑：

set.seed(1)
Signups   <- rnorm(20)
cost      <- rnorm(20)
targeting <- rnorm(20)
fit       <- lm(Signups ~ cost + targeting)

summary(fit)
# ...
# Coefficients:
#             Estimate Std. Error t value Pr(>|t|)
# (Intercept)   0.1494     0.2072   0.721    0.481
# cost         -0.1516     0.2504  -0.605    0.553
# targeting     0.2894     0.2695   1.074    0.298
# ...

windows();{
  plot(cost, Signups, xlab="cost", ylab="Signups", main="Signups")
  abline(coef(fit)[1:2])
  text(-2, -2, adj=c(0,0), labels="Signups = .15 -.15cost + .29targeting")
}

Answer 3

这是使用 tidyverse 包的解决方案。

关键是broom包，它简化了提取模型数据的过程。例如：

fit1 <- lm(mpg ~ cyl, data = mtcars)
summary(fit1)

fit1 %>%
    tidy() %>%
    select(estimate, term)

结果

# A tibble: 2 x 2
  estimate term       
     <dbl> <chr>      
1    37.9  (Intercept)
2    -2.88 cyl 

我编写了一个函数来使用 dplyr 提取和格式化信息：

get_formula <- function(object) {
    object %>% 
        tidy() %>% 
        mutate(
            term = if_else(term == "(Intercept)", "", term),
            sign = case_when(
                term == "" ~ "",
                estimate < 0 ~ "-",
                estimate >= 0 ~ "+"
            ),
            estimate = as.character(round(abs(estimate), digits = 2)),
            term = if_else(term == "", paste(sign, estimate), paste(sign, estimate, term))
        ) %>%
        summarize(terms = paste(term, collapse = " ")) %>%
        pull(terms)
}

get_formula(fit1)

结果

[1] " 37.88 - 2.88 cyl"

然后使用 ggplot2 绘制线条并添加标题

mtcars %>%
    ggplot(mapping = aes(x = cyl, y = mpg)) +
    geom_point() +
    geom_smooth(formula = y ~ x, method = "lm", se = FALSE) +
    labs(
        x = "Cylinders", y = "Miles per Gallon", 
        caption = paste("mpg =", get_formula(fit1))
    )

Plot using geom_smooth()

这种绘制线条的方法实际上只对可视化两个变量之间的关系有意义。正如@Glen_b 在评论中指出的那样，我们从建模 mpg 作为 cyl (-2.88) 的函数得到的斜率与我们从建模 mpg 得到的斜率不匹配cyl 和其他变量的函数（-1.29）。例如：

fit2 <- lm(mpg ~ cyl + disp + wt + hp, data = mtcars)
summary(fit2)

fit2 %>%
    tidy() %>%
    select(estimate, term)

结果

# A tibble: 5 x 2
  estimate term       
     <dbl> <chr>      
1  40.8    (Intercept)
2  -1.29   cyl        
3   0.0116 disp       
4  -3.85   wt         
5  -0.0205 hp 

也就是说，如果您想为包含未出现在图中的变量的模型准确绘制回归线，请改用 geom_abline() 并使用 broom 获取斜率和截距包函数。据我所知，geom_smooth() 公式不能引用尚未映射为美学的变量。

mtcars %>%
    ggplot(mapping = aes(x = cyl, y = mpg)) +
    geom_point() +
    geom_abline(
        slope = fit2 %>% tidy() %>% filter(term == "cyl") %>% pull(estimate),
        intercept = fit2 %>% tidy() %>% filter(term == "(Intercept)") %>% pull(estimate),
        color = "blue"
    ) +
    labs(
        x = "Cylinders", y = "Miles per Gallon", 
        caption = paste("mpg =", get_formula(fit2))
    )

Plot using geom_abline()