我有一个对数日志图,我通过使用得到了回归线:
geom_smooth(formula = y ~ x, method='lm')
但是现在我想获得这一行的等式(例如 y = a * x ^( - b))并打印出来。我设法在lin-lin情节中得到它但在这种情况下不是。 这是代码:
mydataS<-data.frame(DurPeak_h[],IntPeak[],IntPeakxDurPeak[],ID[]) #df peak
names(mydataS)<-c("x","y","ID","IDEVENT")
plotID<-ggplot(mydataS, aes(x=x, y=y, label=IDEVENT)) +
geom_text(check_overlap = TRUE, hjust = 0, nudge_x = 0.02)+
geom_point(colour="black", size = 2) + geom_point(aes(colour = ID)) +
geom_quantile(quantiles = qs, colour="green")+
scale_colour_gradient(low = "white", high="red") +
scale_x_log10(limits = c(min(DurEnd_h),max(DurEnd_h))) +
scale_y_log10(limits = c(min(IntEnd),max(IntEnd))) +
geom_smooth(formula = y ~ x, method='lm')
ggsave(height=7,"plot.pdf")
答案 0 :(得分:0)
mydataS<-data.frame(DurPeak_h[],IntPeak[],IntPeakxDurPeak[],ID[])
names(mydataS)<-c("x","y","ID","IDEVENT")
model <- lm(y~x, header = T)
summary(model)
使用截距值为“b”,系数为“a”
答案 1 :(得分:0)
是否采用了解决方法:使用nls
计算两个参数 a 和 b ,正好:
nlsPeak <- coef(nls(y ~ a*(x)^b, data = mydataS, start = list(a=30, b=-0.1)))
然后使用annotate
绘制线条(参见一些示例here),最后使用函数打印等式:
power_eqn = function(ds){
m = nls(y ~ a*x^b, start = list(a=30, b=-0.1), data = ds);
eq <- substitute(italic(y) == a ~italic(x)^b,
list(a = format(coef(m)[1], digits = 4),
b = format(coef(m)[2], digits = 2)))
as.character(as.expression(eq));
}
如下所示:
annotate("text",x = 3, y = 180,label = power_eqn(mydataS), parse=TRUE, col="black") +
希望它有所帮助!