我正在尝试创建一个仅在第n个（在示例中为第8个）位置提供值的掩码数组（或至少填充了NaN）。数组的长度应与原始数组的长度相同。
有没有一个比较荒谬的方法呢？
b = np.array([[i for i in 7*[np.nan] + [val]] for val in a[::8]]).flatten()[7:]
一种方法是使用切片分配：
>>> a
array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33,
34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50,
51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63])
>>> b = numpy.array([numpy.NaN] * len(a))
>>> b[::8] = a[::8]
>>> b
array([ 0., nan, nan, nan, nan, nan, nan, nan, 8., nan, nan,
nan, nan, nan, nan, nan, 16., nan, nan, nan, nan, nan,
nan, nan, 24., nan, nan, nan, nan, nan, nan, nan, 32.,
nan, nan, nan, nan, nan, nan, nan, 40., nan, nan, nan,
nan, nan, nan, nan, 48., nan, nan, nan, nan, nan, nan,
nan, 56., nan, nan, nan, nan, nan, nan, nan])
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