Python 为什么np.var（x）和np.cov（x，y）给了我不同的值？

我是Python新手，想知道为什么np.var（x）给出的答案与np.cov（x，y）输出中的cov（x，x）值不同。它们不应该是一样的吗？我理解这和偏见或ddof有关，与正常化有关，但我不确定这意味着什么，也找不到任何具体回答我问题的资源。希望有人能帮忙

在numpy中，cov默认为“增量自由度”1，而var默认为ddof 0。从注释到numpy.var

Notes
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The variance is the average of the squared deviations from the mean,
i.e.,  ``var = mean(abs(x - x.mean())**2)``.

The mean is normally calculated as ``x.sum() / N``, where ``N = len(x)``.
If, however, `ddof` is specified, the divisor ``N - ddof`` is used
instead.  In standard statistical practice, ``ddof=1`` provides an
unbiased estimator of the variance of a hypothetical infinite population.
``ddof=0`` provides a maximum likelihood estimate of the variance for
normally distributed variables.

因此，您可以通过以下方式让他们同意：

In [69]: cov(x,x)#defaulting to ddof=1
Out[69]: 
array([[ 0.5,  0.5],
       [ 0.5,  0.5]])

In [70]: x.var(ddof=1)
Out[70]: 0.5

In [71]: cov(x,x,ddof=0)
Out[71]: 
array([[ 0.25,  0.25],
       [ 0.25,  0.25]])

In [72]: x.var()#defaulting to ddof=0
Out[72]: 0.25

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参考统计手册