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
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)$r.squared, accuracy = 0.01) -> r_squared
现在我们需要将文本添加到情节中:
p +
geom_text(x = 20, y = 300,
label = get_formula(model),
color = 'red') +
geom_text(x = 20, y = 285,
label = r_squared,
color = 'blue')
其他回答
我修改了stat_smooth和相关函数的源代码中的几行代码，以创建一个添加了拟合方程和R平方值的新函数。这也适用于小面图!
library(devtools)
source_gist("524eade46135f6348140")
df = data.frame(x = c(1:100))
df$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。
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:
我的包ggpmisc中的统计数据stat_poly_eq()使它可以根据线性模型拟合添加文本标签。
“ggpmisc”(>= 0.4.0)和“ggplot2”(>= 3.3.0)的答案已于2022-06-02更新。 在示例中，我使用stat_poly_line()而不是stat_smooth()，因为它的方法和公式的默认值与stat_poly_eq()相同。在所有代码示例中，我都省略了stat_poly_line()的附加参数，因为它们与添加标签的问题无关。
library(ggplot2)
library(ggpmisc)
#> Loading required package: ggpp
#>
#> Attaching package: 'ggpp'
#> The following object is masked from 'package:ggplot2':
#>
#> annotate
# artificial data
df <- data.frame(x = c(1:100))
df$y <- 2 + 3 * df$x + rnorm(100, sd = 40)
df$yy <- 2 + 3 * df$x + 0.1 * df$x^2 + rnorm(100, sd = 40)
# using default formula, label and methods
ggplot(data = df, aes(x = x, y = y)) +
stat_poly_line() +
stat_poly_eq() +
geom_point()
# assembling a single label with equation and R2
ggplot(data = df, aes(x = x, y = y)) +
stat_poly_line() +
stat_poly_eq(aes(label = paste(after_stat(eq.label),
after_stat(rr.label), sep = "*\", \"*"))) +
geom_point()
# adding separate labels with equation and R2
ggplot(data = df, aes(x = x, y = y)) +
stat_poly_line() +
stat_poly_eq(aes(label = after_stat(eq.label))) +
stat_poly_eq(label.y = 0.9) +
geom_point()
# regression through the origin
ggplot(data = df, aes(x = x, y = y)) +
stat_poly_line(formula = y ~ x + 0) +
stat_poly_eq(formula = y ~ x + 0, aes(label = after_stat(eq.label))) +
geom_point()
# fitting a polynomial
ggplot(data = df, aes(x = x, y = yy)) +
stat_poly_line(formula = y ~ poly(x, 2, raw = TRUE)) +
stat_poly_eq(formula = y ~ poly(x, 2, raw = TRUE),
aes(label = after_stat(eq.label))) +
geom_point()
# adding a hat as asked by @MYaseen208 and @elarry
ggplot(data = df, aes(x = x, y = y)) +
stat_poly_line() +
stat_poly_eq(eq.with.lhs = "italic(hat(y))~`=`~",
aes(label = paste(after_stat(eq.label),
after_stat(rr.label), sep = "*\", \"*"))) +
geom_point()
# variable substitution as asked by @shabbychef
# same labels in equation and axes
ggplot(data = df, aes(x = x, y = y)) +
stat_poly_line() +
stat_poly_eq(eq.with.lhs = "italic(h)~`=`~",
eq.x.rhs = "~italic(z)",
aes(label = after_stat(eq.label))) +
labs(x = expression(italic(z)), y = expression(italic(h))) +
geom_point()
# grouping as asked by @helen.h
dfg <- data.frame(x = c(1:100))
dfg$y <- 20 * c(0, 1) + 3 * df$x + rnorm(100, sd = 40)
dfg$group <- factor(rep(c("A", "B"), 50))
ggplot(data = dfg, aes(x = x, y = y, colour = group)) +
stat_poly_line() +
stat_poly_eq(aes(label = paste(after_stat(eq.label),
after_stat(rr.label), sep = "*\", \"*"))) +
geom_point()
ggplot(data = dfg, aes(x = x, y = y, linetype = group, grp.label = group)) +
stat_poly_line() +
stat_poly_eq(aes(label = paste(after_stat(grp.label), "*\": \"*",
after_stat(eq.label), "*\", \"*",
after_stat(rr.label), sep = ""))) +
geom_point()
# a single fit with grouped data as asked by @Herman
ggplot(data = dfg, aes(x = x, y = y)) +
stat_poly_line() +
stat_poly_eq(aes(label = paste(after_stat(eq.label),
after_stat(rr.label), sep = "*\", \"*"))) +
geom_point(aes(colour = group))
# facets
ggplot(data = dfg, aes(x = x, y = y)) +
stat_poly_line() +
stat_poly_eq(aes(label = paste(after_stat(eq.label),
after_stat(rr.label), sep = "*\", \"*"))) +
geom_point() +
facet_wrap(~group)
由reprex包于2022-06-02创建(v2.0.1)
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)$r.squared, accuracy = 0.01) -> r_squared
现在我们需要将文本添加到情节中:
p +
geom_text(x = 20, y = 300,
label = get_formula(model),
color = 'red') +
geom_text(x = 20, y = 285,
label = r_squared,
color = 'blue')
真的很喜欢@Ramnath的解决方案。为了允许使用自定义回归公式(而不是固定为y和x作为字面变量名)，并将p值添加到打印输出中(正如@Jerry T评论的那样)，下面是mod:
lm_eqn <- function(df, y, x){
formula = as.formula(sprintf('%s ~ %s', y, x))
m <- lm(formula, data=df);
# formating the values into a summary string to print out
# ~ give some space, but equal size and comma need to be quoted
eq <- substitute(italic(target) == a + b %.% italic(input)*","~~italic(r)^2~"="~r2*","~~p~"="~italic(pvalue),
list(target = y,
input = x,
a = format(as.vector(coef(m)[1]), digits = 2),
b = format(as.vector(coef(m)[2]), digits = 2),
r2 = format(summary(m)$r.squared, digits = 3),
# getting the pvalue is painful
pvalue = format(summary(m)$coefficients[2,'Pr(>|t|)'], digits=1)
)
)
as.character(as.expression(eq));
}
geom_point() +
ggrepel::geom_text_repel(label=rownames(mtcars)) +
geom_text(x=3,y=300,label=lm_eqn(mtcars, 'hp','wt'),color='red',parse=T) +
geom_smooth(method='lm')
不幸的是，这对facet_wrap或facet_grid不起作用。
