这两个协方差函数之间的区别是什么？

Question

我想重写这个协方差函数：

 def cov1(a, b):
    a_mean = np.mean(a)
    b_mean = np.mean(b)
    sum = 0
    for i in range(0, a.size):
        sum = ((a[i] - a_mean) * (b[i] - b_mean)) + sum
    return sum/(len(a)-1)

我尝试重新编写求和部分，使其成为np.sum：

def cov(a, b):
    a_mean = np.mean(a)
    b_mean = np.mean(b)
    for i in range(0, a.size):
        summation = np.sum((a[i] - a_mean) * (b[i] - b_mean))
    return summation/(len(a)-1)

但当我使用两个数组时：

a = np.arange(1,11,1)
b = np.arange(10,21,1)

我尝试了两个不同的函数，得到了不同的答案。功能cov1是正确的：

print(cov1(a,b))
print(cov(a,b))

9.166666666666666
2.0

为什么会这样？如何修复函数cov（a，b）使其与cov1（a，b）相同

Answer 1

首先，我认为两个函数都应该确保输入的长度相同

第二，你的功能可能是

def cov(a, b):
    a_mean = np.mean(a)
    b_mean = np.mean(b)
    return ((a - a_mean) * (b - b_mean)).sum() / (len(a) - 1)

第三，numpy有cov功能，试试看

np.cov(a, b)

这将返回变量的协方差矩阵。在您的例子中，您可以只使用np.cov（a，b）[0，0]

Answer 2

您忘记在cov中定义summation，并且忘记将summation添加到新的总和中。 请尝试以下方法：

def cov(a, b):
    a_mean = np.mean(a)
    b_mean = np.mean(b)
    # Added summation and assigned 0 to it, like for sum in cov1
    summation = 0
    for i in range(0, a.size):
        # Added + summation here, just like in cov1
        summation = np.sum((a[i] - a_mean) * (b[i] - b_mean)) + summation 
    return summation/(len(a)-1)