以块的形式迭代一个 numpy 数组

Question

如何使用numpy.where() 或numpy.select() 将list_ 排列成数字组。我想将数据分成 3 组，然后计算这些函数的标准偏差std()。所以程序将采用457.334015,424.440002,394.795990 来计算标准dev 的第一个值，并将采用424.440002,394.795990, 408.903992 并计算标准偏差等。它将继续这样下去，直到它到达@ 的末尾987654327@。我希望在计算第二个块 424.440002,394.795990, 408.903992 之前删除第一个块 457.334015,424.440002,394.795990。我想从内存中删除块，所以我没有内存错误。使用 numpy 并且不使用 for 循环是否可以实现这一点。

number = 3
list_= np.array([457.334015,424.440002,394.795990,408.903992,398.821014,402.152008,435.790985,423.204987,411.574005,
404.424988,399.519989,377.181000,375.467010,386.944000,383.614990,375.071991,359.511993,328.865997,
320.510010,330.079010,336.187012,352.940002,365.026001,361.562012,362.299011,378.549011,390.414001,
400.869995,394.773010,382.556000])

Answer 1

来自this post：

from numpy.lib.stride_tricks import as_strided
def strided_app(a, L, S ):  # Window len = L, Stride len/stepsize = S
    nrows = ((a.size-L)//S)+1
    n = a.strides[0]
    return np.lib.stride_tricks.as_strided(a, shape=(nrows,L), strides=(S*n,n))
list_= np.array([457.334015,424.440002,394.795990,408.903992,398.821014,402.152008,435.790985,423.204987,411.574005,
404.424988,399.519989,377.181000,375.467010,386.944000,383.614990,375.071991,359.511993,328.865997,
320.510010,330.079010,336.187012,352.940002,365.026001,361.562012,362.299011,378.549011,390.414001,
400.869995,394.773010,382.556000])
np.std(strided_app(list_, 3, 1), axis=1)

但是，此代码不会从数组中删除任何元素。另外，请记住，这里使用的函数带有来自 numpy 文档的warning！