R语言在散点图中添加lm线性回归公式的问题

1. 简单的线性回归

函数自带的例子（R 中键入?lm），lm(y ~ x)回归y=kx + b， lm( y ~ x -1 )省略b，不对截距进行估计：

require(graphics)

## Annette Dobson (1990) "An Introduction to Generalized Linear Models".
## Page 9: Plant Weight Data.
ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group <- gl(2, 10, 20, labels = c("Ctl","Trt"))
weight <- c(ctl, trt)
lm.D9 <- lm(weight ~ group)
lm.D90 <- lm(weight ~ group - 1) # omitting intercept

anova(lm.D9)
summary(lm.D90)

opar <- par(mfrow = c(2,2), oma = c(0, 0, 1.1, 0))
plot(lm.D9, las = 1)      # Residuals, Fitted, ...
par(opar)

使用R中自带的mtcars数据，可以得到截距和斜率，也可以得到解释率R-square：

require(ggplot2)
library(dplyr) #加载dplyr包
library(ggpmisc) #加载ggpmisc包
library(ggpubr)
require(gridExtra)
model=lm(mtcars$wt ~ mtcars$mpg)
model
## 输出：
Call:
lm(formula = mtcars$wt ~ mtcars$mpg)

Coefficients:
(Intercept)   mtcars$mpg  
    6.047       -0.141
    ```
```handlebars
summary(model)

## 输出：
Call:
lm(formula = mtcars$wt ~ mtcars$mpg)

Residuals:
 Min     1Q Median     3Q    Max 
-0.652 -0.349 -0.138  0.319  1.368 

Coefficients:
          Estimate Std. Error t value Pr(>|t|)    
(Intercept)   6.0473     0.3087   19.59  < 2e-16 ***
mtcars$mpg   -0.1409     0.0147   -9.56  1.3e-10 ***
---
Signif. codes:  0 ‘***' 0.001 ‘**' 0.01 ‘*' 0.05 ‘.' 0.1 ‘ ' 1

Residual standard error: 0.494 on 30 degrees of freedom
Multiple R-squared:  0.753,	Adjusted R-squared:  0.745 
F-statistic: 91.4 on 1 and 30 DF,  p-value: 1.29e-10

提取回归R-square值：

通过summary提取：
## 上面的例子


## mtcars例子
model=lm(mtcars$wt ~ mtcars$mpg)
res=summary(model)
str(res) 
## 提取各个值：
res$r.squared
res$coefficients
res$adj.r.squared  ## df 矫正后的结果
res$coefficients[1,1]
res$coefficients[2,1]

使用默认的plot绘制回归散点：

plot(mtcars$mpg, mtcars$wt, pch=20,cex=2)
abline(model,col="red",lwd=2)

计算Confidence interval（95%）：

test=mtcars[c("mpg","wt")]
head(test)
colnames(test)=c("x","y")
model = lm(y ~ x, test)

test$predicted = predict(
object = model,
newdata = test)

test$CI = predict(
object = model,
newdata = test,
se.fit = TRUE
)$se.fit * qt(1 - (1-0.95)/2, nrow(test))

test$predicted = predict(
object = model,
newdata = test)

test$CI_u=test$predicted+test$CI
test$CI_l=test$predicted-test$CI
plot(mtcars$mpg, mtcars$wt, pch=20,cex=1) ##  have replicated x values
abline(model,col="red",lwd=2)
lines(x=test$x,y=test$CI_u,col="blue")
lines(x=test$x,y=test$CI_l,col="blue")

上面的图蓝线有点奇怪，简单绘制最初的plot：

plot(mtcars$mpg, mtcars$wt, pch=20,cex=1,type="b") ##  have replicated x values

实际上面的计算方法没问题，但是数据不合适，因为数据x含有重复值，所以要考虑这个。

2. 使用ggplot2展示

ggplot2例子：

p <- ggplot(df, aes(x=yreal, y=ypred)) +
geom_point(color = "grey20",size = 1, alpha = 0.8)
#回归线
#添加回归曲线
p2 <- p + geom_smooth(formula = y ~ x, color = "red",
                    fill = "blue", method = "lm",se = T, level=0.95) +
theme_bw() +
stat_poly_eq(
  aes(label = paste(..eq.label.., ..adj.rr.label.., sep = '~~~')),
  formula = y ~ x,  parse = TRUE,color="blue",
  size = 5, #公式字体大小
  label.x = 0.05,  #位置 ，0-1之间的比例
  label.y = 0.95) + 
labs(title="test",x="Real Value (Huang Huaihai 1777)" , y="Predicted Value (Correlation: 0.5029)")
p2

ggplot版本的手动计算：

require(ggplot2)
library(dplyr) #加载dplyr包
library(ggpmisc) #加载ggpmisc包
library(ggpubr)
require(gridExtra)
ggplot(data=df, aes(x=yreal, y=ypred)) +
geom_smooth(formula = y ~ x, color = "blue",
            fill = "grey10", method = "lm")  +
geom_point() +
stat_regline_equation(label.x=0.1, label.y=-1.5) +
stat_cor(aes(label=..rr.label..), label.x=0.1, label.y=-2)

test=df
head(test)
colnames(test)=c("x","y")
model = lm(y ~ x, test)
test$predicted = predict(
object = model,
newdata = test)

test$CI = predict(
object = model,
newdata = test,
se.fit = TRUE
)$se.fit * qt(1 - (1-0.95)/2, nrow(test))

ggplot(test) +
aes(x = x, y = y) +
geom_point(size = 1,colour="grey40") +
geom_smooth(formula =y ~ x,method = "lm",  fullrange = TRUE, color = "black") +
geom_line(aes(y = predicted + CI), color = "blue") + # upper
geom_line(aes(y = predicted - CI), color = "red") + # lower
theme_classic()

参考：https://stackoverflow.com/questions/23519224/extract-r-square-value-with-r-in-linear-models （提取R2）https://blog.csdn.net/LeaningR/article/details/118971000 （提取R2等）https://stackoverflow.com/questions/45742987/how-is-level-used-to-generate-the-confidence-interval-in-geom-smooth （添加lm线）https://zhuanlan.zhihu.com/p/131604431 （知乎）

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