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library(corrplot) library(tidyverse) M <- cor(mtcars)
attach(mtcars) a <- sum( (mpg - mean(mpg)) * (disp-mean(disp)) ) b <- sqrt( sum( (mpg-mean(mpg))^2 ) * sum( (disp-mean(disp))^2 ) ) r <- a/b r cor(mpg,disp) #cor.mtest(count,conf.level=0.95) detach(mtcars)
-0.847551379262479 -0.847551379262479
r /sqrt( (1-r^2)/( length(mpg) -2 ) ) cor.test(mpg,disp)
-8.74715153409391 Pearson's product-moment correlation data: mpg and disp t = -8.7472, df = 30, p-value = 9.38e-10 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: -0.9233594 -0.7081376 sample estimates: cor -0.8475514
由于p=9.38e-10<0.05, 于是在\alpha =0.05的水准上拒绝H_0, 接受H_1, 可认为mpg与disp呈现负的线性相关。
with(mtcars,plot(mpg,disp,type="n")) with(mtcars,points(mpg,disp,col="red")) fit = lm(disp~mpg,mtcars) abline(fit,lwd=2)
keep <- rowSums(expr_lnRNA > 0.5)>=2 expr_lnRNA <- expr_lnRNA[keep,] #lnRNA_count<-lnRNA_count[!apply(lnRNA_count, 1,mean)==0,] cor_result <- (function(m6a_count,lnRNA_count){ cor_result <- data.frame() for (mRNA in rownames(m6a_count)){ #exp_cor<-rbind(m6a_count[1,],lnRNA_count) y <- as.numeric(m6a_count[mRNA,]) for (lnRNA in rownames(lnRNA_count)){ x <- as.numeric(lnRNA_count[lnRNA,]) test <- cor.test(x,y,type="pearson") cor_result <- rbind(mRNA=mRNA,lnRNA=lnRNA,cor_result,data.frame(pvalue=test$p.value,estimate=test$estimate)) } } return(cor_result) })(lnRAN_mRNA(fpkm_obj)@mRNA[gene,],expr_lnRNA)
R语言实现
多因素回归分析时,为探讨影响因素对结局事件的影响大小,可以利用森林图更直观的将回归结果可视化
使用Nomogram对结局事件的发生风险进行预测,列线图将复杂的回归方程,转变为了可视化的图形,使预测模型的结果更具有可读性
library(survival) library(rms) dd<-datadist(lung) options(datadist="dd") f <- cph(Surv(time,status)~age+sex,data=lung,x=TRUE,y=TRUE,surv=TRUE) survival <- Survival(f) survival1 <- function(x)survival(365,x) survival2 <- function(x)survival(730,x) nom <- nomogram(f,fun = list(survival1,survival2), fun.at = c(0.05,seq(0.1,0.9,by=0.05),0.95)) plot(nom)