如何使用函数使用R计算相关系数？

Question

你好！ 我正在尝试编写一个函数来推导出 Pearson 相关系数的公式。 我编写了以下代码，但是当我尝试传递值时，它返回空输出。 请指出我的错误，我一无所知！ 非常感激。

correlation = function(X, Y, n = length(X)){
sum_X = 0
sum_Y = 0
sum_XY = 0
squareSum_X = 0
squareSum_Y = 0
i = 0
while (i < n ) { 
    # sum of elements of array X. 
    sum_X = sum_X + X[i] 

    # sum of elements of array Y. 
    sum_Y = sum_Y + Y[i] 

    # sum of X[i] * Y[i]. 
    sum_XY = sum_XY + X[i] * Y[i] 

    # sum of square of array elements. 
    squareSum_X = squareSum_X + X[i] * X[i] 
    squareSum_Y = squareSum_Y + Y[i] * Y[i] 

    i =+ 1
}
# combine all into a final formula
final = (n * sum_XY - (sum_X * sum_Y))/ (sqrt((n * squareSum_X - sum_X * sum_X)* (n * squareSum_Y - 
sum_Y * sum_Y))) 
return (final)
}

Answer 1

R 是 1 索引语言。 从i = 1开始并更改为while(i <= n) （并按照注释中的说明修复迭代计数器： i = i + 1 。然后您的函数可以正常工作。

n <- 100
x <- rnorm(n)
y <- rnorm(n)

round(correlation(x, y), 4) == round(cor(x, y), 4) # TRUE

但是请注意，R 也非常适合矢量化操作，您可以完全跳过显式循环。 像这样的事情是提高效率的一步：

correlation2 <- function(X, Y){
  n <- length(X)
  sum_X <- sum(X)
  sum_Y <- sum(Y)
  sum_XY <- sum(X * Y)
  squareSum_X <- sum(X * X)
  squareSum_Y <- sum(Y * Y)
  final <- (n * sum_XY - (sum_X * sum_Y)) / (sqrt((n * squareSum_X - sum_X * sum_X)* (n * squareSum_Y - sum_Y * sum_Y))) 
  return (final)
}

round(correlation2(x, y), 4) == round(cor(x, y), 4) # TRUE

或者甚至只是：

correlation3 <- function(X, Y){
  n = length(X)
  sum_x = sum(X)
  sum_y = sum(Y)
  (n * sum(X * Y) - sum_x * sum_y) / 
    (sqrt((n * sum(x^2) - sum_x^2) * (n * sum(Y^2) - sum_y^2)))
}