如何在R中绘制与平滑曲线相切的直线？

问题描述

我正在尝试从原点（或Y轴上的任何点）绘制一条与黄土曲线相切的直线（一条仅接触该曲线的直线）。

The black line is the smoothed curve of the points,and I wish to draw that yellow line.

我将发布代码以获取下面的平滑曲线。

'''

library(quantmod)
library(plotly)
library(PerformanceAnalytics)
library(timetk)
library(tidyverse)

ticker = c('AMZN','AAPL','NFLX','XOM','T')
price_data = getSymbols(ticker,from = '2014-01-01',to = '2018-05-31')

prices = do.call(cbind,lapply(ticker,function(x) Ad(get(x))))

rets = Return.calculate(prices,method = 'log') %>%
  na.omit()
    
num_port = 1000
all_wts = matrix(nrow = num_port,ncol = length(ticker))
port_returns = vector('numeric',length = num_port)
port_risk = vector('numeric',length = num_port)
port_sr = vector('numeric',length = num_port)

for (i in seq_along(port_returns)) {
  
  wts = runif(n = length(ticker))
  wts = wts/sum(wts)
  
  all_wts[i,] = wts
  
  port = Return.portfolio(R = rets,weights = wts,verbose = TRUE)
  
  a = StdDev.annualized(port$returns)[1]
  b = SharpeRatio.annualized(port$returns,Rf = 0)[1]
  c = a*b
  
  port_returns[i] = c
  port_risk[i] = a
  port_sr[i] = b

}

all_wts = tk_tbl(all_wts)
colnames(all_wts) = colnames(rets)

pf_val = tibble(ret = port_returns,risk = port_risk,sr = port_sr)
pf_val = tk_tbl(cbind(all_wts,pf_val))

min_var = pf_val[which.min(pf_val$risk),]
max_sr = pf_val[which.max(pf_val$sr),]

library(tidyverse)

pf_line2 = pf_val[which((pf_val$ret %in% d$ret)),]


p3 = ggplot(aes(x = risk,y = ret,color = sr),data = pf_line2) +
  geom_point() + theme_classic() +
  scale_y_continuous(labels = scales::percent) +
  scale_x_continuous(labels = scales::percent) +
  labs(x = 'risk',y = 'return') +
  geom_smooth(method = 'loess',col='black',level=0.5,alpha=0.1)
  

ggplotly(p3)

'''

解决方法

我尝试运行您的数据，但在代码中未定义d，因此无法重现您的示例。

如果知道这些尺寸，则可以使用geom_abline添加具有自己的截距和坡度的线。

ggplot(aes(x = wt,y = mpg),data = mtcars) +
geom_point()+
labs(x = 'wt',y = 'mpg') +
geom_abline(intercept = 10,slope = 6)+
geom_smooth(method = 'loess',col='black',level=0.5,alpha=0.1)

这给了这样的东西。 line with geom_smooth。

我不是一个完整的专家，但是我希望这可以在某种程度上帮助我们。