我想知道如何在ggplot上添加回归线方程和R^2。我的代码是:

library(ggplot2)

df <- data.frame(x = c(1:100))
df$y <- 2 + 3 * df$x + rnorm(100, sd = 40)
p <- ggplot(data = df, aes(x = x, y = y)) +
            geom_smooth(method = "lm", se=FALSE, color="black", formula = y ~ x) +
            geom_point()
p

任何帮助都将不胜感激。


当前回答

这里有一个解决方案

# GET EQUATION AND R-SQUARED AS STRING
# SOURCE: https://groups.google.com/forum/#!topic/ggplot2/1TgH-kG5XMA

lm_eqn <- function(df){
    m <- lm(y ~ x, df);
    eq <- substitute(italic(y) == a + b %.% italic(x)*","~~italic(r)^2~"="~r2, 
         list(a = format(unname(coef(m)[1]), digits = 2),
              b = format(unname(coef(m)[2]), digits = 2),
             r2 = format(summary(m)$r.squared, digits = 3)))
    as.character(as.expression(eq));
}

p1 <- p + geom_text(x = 25, y = 300, label = lm_eqn(df), parse = TRUE)

编辑。我找到了这个代码的来源。下面是ggplot2谷歌组中原始帖子的链接

其他回答

我修改了stat_smooth和相关函数的源代码中的几行代码,以创建一个添加了拟合方程和R平方值的新函数。这也适用于小面图!

library(devtools)
source_gist("524eade46135f6348140")
df = data.frame(x = c(1:100))
df$y = 2 + 5 * df$x + rnorm(100, sd = 40)
df$class = rep(1:2,50)
ggplot(data = df, aes(x = x, y = y, label=y)) +
  stat_smooth_func(geom="text",method="lm",hjust=0,parse=TRUE) +
  geom_smooth(method="lm",se=FALSE) +
  geom_point() + facet_wrap(~class)

我使用了@Ramnath回答中的代码来格式化方程。stat_smooth_func函数不是很健壮,但是使用它应该不难。

https://gist.github.com/kdauria/524eade46135f6348140。如果出现错误,请尝试更新ggplot2。

另一种选择是创建一个自定义函数,使用dplyr和broom库生成方程:

get_formula <- function(model) {
  
  broom::tidy(model)[, 1:2] %>%
    mutate(sign = ifelse(sign(estimate) == 1, ' + ', ' - ')) %>% #coeff signs
    mutate_if(is.numeric, ~ abs(round(., 2))) %>% #for improving formatting
    mutate(a = ifelse(term == '(Intercept)', paste0('y ~ ', estimate), paste0(sign, estimate, ' * ', term))) %>%
    summarise(formula = paste(a, collapse = '')) %>%
    as.character
  
}

lm(y ~ x, data = df) -> model
get_formula(model)
#"y ~ 6.22 + 3.16 * x"

scales::percent(summary(model)$r.squared, accuracy = 0.01) -> r_squared

现在我们需要将文本添加到情节中:

p + 
  geom_text(x = 20, y = 300,
            label = get_formula(model),
            color = 'red') +
  geom_text(x = 20, y = 285,
            label = r_squared,
            color = 'blue')

我修改了Ramnath的帖子,a)使其更通用,以便它接受线性模型作为参数,而不是数据帧;b)更适当地显示负号。

lm_eqn = function(m) {

  l <- list(a = format(coef(m)[1], digits = 2),
      b = format(abs(coef(m)[2]), digits = 2),
      r2 = format(summary(m)$r.squared, digits = 3));

  if (coef(m)[2] >= 0)  {
    eq <- substitute(italic(y) == a + b %.% italic(x)*","~~italic(r)^2~"="~r2,l)
  } else {
    eq <- substitute(italic(y) == a - b %.% italic(x)*","~~italic(r)^2~"="~r2,l)    
  }

  as.character(as.expression(eq));                 
}

用法将更改为:

p1 = p + geom_text(aes(x = 25, y = 300, label = lm_eqn(lm(y ~ x, df))), parse = TRUE)

使用ggpubr:

library(ggpubr)

# reproducible data
set.seed(1)
df <- data.frame(x = c(1:100))
df$y <- 2 + 3 * df$x + rnorm(100, sd = 40)

# By default showing Pearson R
ggscatter(df, x = "x", y = "y", add = "reg.line") +
  stat_cor(label.y = 300) +
  stat_regline_equation(label.y = 280)

# Use R2 instead of R
ggscatter(df, x = "x", y = "y", add = "reg.line") +
  stat_cor(label.y = 300, 
           aes(label = paste(..rr.label.., ..p.label.., sep = "~`,`~"))) +
  stat_regline_equation(label.y = 280)

## compare R2 with accepted answer
# m <- lm(y ~ x, df)
# round(summary(m)$r.squared, 2)
# [1] 0.85

受到这个答案中提供的方程风格的启发,一个更通用的方法(多个预测器+ latex输出作为选项)可以是:

print_equation= function(model, latex= FALSE, ...){
    dots <- list(...)
    cc= model$coefficients
    var_sign= as.character(sign(cc[-1]))%>%gsub("1","",.)%>%gsub("-"," - ",.)
    var_sign[var_sign==""]= ' + '

    f_args_abs= f_args= dots
    f_args$x= cc
    f_args_abs$x= abs(cc)
    cc_= do.call(format, args= f_args)
    cc_abs= do.call(format, args= f_args_abs)
    pred_vars=
        cc_abs%>%
        paste(., x_vars, sep= star)%>%
        paste(var_sign,.)%>%paste(., collapse= "")

    if(latex){
        star= " \\cdot "
        y_var= strsplit(as.character(model$call$formula), "~")[[2]]%>%
            paste0("\\hat{",.,"_{i}}")
        x_vars= names(cc_)[-1]%>%paste0(.,"_{i}")
    }else{
        star= " * "
        y_var= strsplit(as.character(model$call$formula), "~")[[2]]        
        x_vars= names(cc_)[-1]
    }

    equ= paste(y_var,"=",cc_[1],pred_vars)
    if(latex){
        equ= paste0(equ," + \\hat{\\varepsilon_{i}} \\quad where \\quad \\varepsilon \\sim \\mathcal{N}(0,",
                    summary(MetamodelKdifEryth)$sigma,")")%>%paste0("$",.,"$")
    }
    cat(equ)
}

model参数需要一个lm对象,latex参数是一个布尔值,要求一个简单的字符或一个乳胶格式的方程,而…参数将其值传递给format函数。

我还添加了一个选项来输出它为latex,这样你就可以在rmarkdown中使用这个函数:


```{r echo=FALSE, results='asis'}
print_equation(model = lm_mod, latex = TRUE)
```

现在使用它:

df <- data.frame(x = c(1:100))
df$y <- 2 + 3 * df$x + rnorm(100, sd = 40)
df$z <- 8 + 3 * df$x + rnorm(100, sd = 40)
lm_mod= lm(y~x+z, data = df)

print_equation(model = lm_mod, latex = FALSE)

这段代码产生: Y = 11.3382963933174 + 2.5893419 * x + 0.1002227 * z

如果我们要求一个乳胶方程,将参数四舍五入为3位:

print_equation(model = lm_mod, latex = TRUE, digits= 3)

这个收益率: