您最好的选择是使用 numpy 的掩码数组提供的方法,其中之一是在存在掩码项时计算协方差矩阵:
>>> import numpy as np
>>> mask_value = -9999
>>> a = np.array([1, 2, mask_value, 4])
>>> b = np.array([1, mask_value, 3, 4])
>>> c = np.vstack((a,b))
>>>
>>> masked_a, masked_b, masked_c = [np.ma.array(x, mask=x==mask_value) for x in (a,b,c)] # note: testing for equality is a bad idea if you're working with floats. I'm not, these are integers, so it's okay.
>>>
>>> result = np.ma.cov(masked_c)
>>> result
masked_array(data =
[[2.333333333333333 4.444444444444445]
[4.444444444444445 2.333333333333333]],
mask =
[[False False]
[False False]],
fill_value = 1e+20)
>>> np.cov([1,2,4]) # autocovariance when just one element is masked is the same as the previous result[0,0]
array(2.333333333333333)
结果因您的调用方式而异np.ma.cov
:
>>> np.ma.cov(masked_a, masked_b)
masked_array(data =
[[4.5 4.5]
[4.5 4.5]],
mask =
[[False False]
[False False]],
fill_value = 1e+20)
>>> np.cov([1,4]) # result of the autocovariance when 2 of the 4 values are masked
array(4.5)
原因是后一种方法结合了 2 个变量的掩码,如下所示:
>>> mask2 = masked_c.mask.any(axis=0)
>>> all_masked_c = np.ma.array(c, mask=np.vstack((mask2, mask2)))
>>> all_masked_c
masked_array(data =
[[1 -- -- 4]
[1 -- -- 4]],
mask =
[[False True True False]
[False True True False]],
fill_value = 999999)
>>> np.ma.cov(all_masked_c) # same function call as the first approach, but with a different mask!
masked_array(data =
[[4.5 4.5]
[4.5 4.5]],
mask =
[[False False]
[False False]],
fill_value = 1e+20)
因此,请使用np.ma.cov
但请注意当存在不重叠的掩码值时您希望如何解释数据。