当累积概率为 50% 时计算“x”

如何解决当累积概率为 50% 时计算“x”

我有不同年龄患特定疾病的累积概率（这是已发布数据的一个例子）。我想获得累积概率为 50% 的年龄。我怎样才能在 R 中完成这项工作？

data<-structure(list(Age = structure(1:65,.Label = c("cumF0","cumF1","cumF2","cumF3","cumF4","cumF5","cumF6","cumF7","cumF8","cumF9","cumF10","cumF11","cumF12","cumF13","cumF14","cumF15","cumF16","cumF17","cumF18","cumF19","cumF20","cumF21","cumF22","cumF23","cumF24","cumF25","cumF26","cumF27","cumF28","cumF29","cumF30","cumF31","cumF32","cumF33","cumF34","cumF35","cumF36","cumF37","cumF38","cumF39","cumF40","cumF41","cumF42","cumF43","cumF44","cumF45","cumF46","cumF47","cumF48","cumF49","cumF50","cumF51","cumF52","cumF53","cumF54","cumF55","cumF56","cumF57","cumF58","cumF59","cumF60","cumF61","cumF62","cumF63","cumF64"
),class = "factor"),normCumFreq = c(0.0175245968789,0.0578992981113,0.115553204108,0.173581449516,0.222035397344,0.270097945681,0.315896329923,0.354579579166,0.387149573944,0.415871590616,0.440978280114,0.461444925286,0.47917749443,0.493999004038,0.505508619684,0.51512487903,0.52411789833,0.532735008033,0.540812972269,0.548113686003,0.554798158422,0.561563099934,0.567823106219,0.573724015785,0.579703589596,0.585608672506,0.591383628199,0.597301585054,0.603266669771,0.609342088192,0.61559812477,0.622057526279,0.628818973321,0.635910562207,0.643342620371,0.650990220485,0.659028041826,0.667365640833,0.676046511395,0.684819235513,0.693721906451,0.702852737329,0.712051335692,0.721215208117,0.730622008883,0.740269523173,0.750050662044,0.760004826004,0.770091216483,0.780442405891,0.790907696368,0.8018259855,0.813018530957,0.8246166078,0.836600903053,0.848877171158,0.861608224775,0.874792841362,0.888399062405,0.902531605004,0.917202798245,0.932317029944,0.948013615225,0.96439871036,0.981432598965),AgeNum = c(0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64)),row.names = c(NA,65L
),class = "data.frame")

谢谢。

解决方法

使用 approx() 将年龄线性内插为累积频率的函数，然后预测 cumfreq==0.5 的值。

a50 <- with(data,approx(NormCumFreq,AgeNum,xout=0.5))$y

这基本上等同于@neilfws 的回答。当您的值间隔密集时，线性插值应该可以正常工作：如果您有较稀疏的 (x,y) 对并且想要使用平滑插值函数，您可以使用 interpSpline() 和 backSpline() 的组合（内置）splines 包以拟合样条曲线，然后将其反转。

library(ggplot2); theme_set(theme_bw())
gg0 <- ggplot(data,aes(AgeNum,NormCumFreq)) + geom_step() +
    geom_hline(yintercept=0.5,lty=2) +
    geom_vline(xintercept=a50,lty=2)
print(gg0)

,

你可以试试approxfun。

af1 <- approxfun(df1$NormCumFreq,df1$AgeNum)

af1(0.5)
[1] 13.52139

library(ggplot2)
data %>% 
  ggplot(aes(AgeNum,NormCumFreq)) + 
  geom_point() + 
  geom_vline(xintercept = 13.52139) + 
  geom_hline(yintercept = 0.5)