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R 为什么协方差2()给出了错误的答案?_R_Statistics - Fatal编程技术网

R 为什么协方差2()给出了错误的答案?

R 为什么协方差2()给出了错误的答案?,r,statistics,R,Statistics,为什么协方差2给出了错误的输出?根据输出应相同。在协方差2()中,您需要除以长度(x),而不是长度(x)-1 sMean = function(x) { sum = 0; for (i in 1:length(x)) { sum = sum + x[i] } return(sum/length(x)) } covariance = function(x,y) { #Formula is E((x-E[x])(y-E[y])) => E[xy] - E[x]E[

为什么协方差2给出了错误的输出?根据输出应相同。

协方差2()
中,您需要除以
长度(x)
,而不是
长度(x)-1

sMean = function(x) {
  sum = 0;
  for (i in 1:length(x)) {
    sum = sum + x[i]
  }
  return(sum/length(x))
}

covariance = function(x,y) {
  #Formula is E((x-E[x])(y-E[y])) => E[xy] - E[x]E[y]
  meanX = sMean(x)
  meanY = sMean(y)
  cov = 0;
  for (i in 1:length(x)) {
    cov = cov + ((x[i] - meanX) * (y[i] - meanY))
  }
  cov = cov/(length(x)-1)
  return (cov)
}

covariance2 = function(x,y) {
  #Formula is E((x-E[x])(y-E[y])) => E[xy] - E[x]E[y]
  meanX = (sMean(x) * length(x)) /(length(x)-1)
  meanY = (sMean(y) * length(x)) /(length(x)-1)
  meanXY = (sMean(x*y) * length(x)) /(length(x)-1)
  return (meanXY - meanX*meanY)
}

#Output
  #> cov(arr,arr2)
  #[1] 16.75
  #> covariance(arr,arr2)
  #[1] 16.75
  #> covariance2(arr,arr2) #Why this function give wrong output?
  #[1] -9.5
协方差2=函数(x,y){
我想
length(x-1)
应该是
length(x)-1
开始吧?@konvas,非常感谢你的回答,修复了那部分,它仍然不起作用。更新了问题。请检查。你为什么要除以
length(x)-1而不是
length(x)
?也不需要函数
sMean
,您可以使用
R
附带的函数
mean()
编写
mean(x)
covariance2 = function(x,y) {
    meanX <- sMean(x)
    meanY <- sMean(y)
    meanXY <- sMean(x*y)
    return((meanXY - meanX * meanY) * length(x) / (length(x) - 1))
}