Question

给出这样的一系列峰：

peaks = [0, 5, 0, 3, 2, 0, 1, 7, 0]

如何创建一个指示最近峰值的步骤数组，如下所示：

steps = [0, 5, 5, 3, 3, 3, 3, 7, 7]

要求：

这将用于对大型3D图像（1000 ** 3）进行图像分析，因此需要快速进行，这意味着无需进行循环或列表解析...仅适用于numpy向量化。
我在上面给出的示例是一个线性列表，但这需要同样适用于ND图像。这意味着沿单个轴进行操作，但允许同时使用多个轴。

注意

我最近asked a question原来是骗子（用scipy.maximum.accumulate很容易解决），但是我的问题还包含一个可选的“如果...会更好”扭曲，如上所述。事实证明，我实际上也需要第二种行为，所以我只重新发布了这一部分。

Answer 1

这是处理ND并可以检测“ ..., 0, 4, 4, 4, 3, ...”而不是“ ..., 0, 4, 4, 4, 7, ...”的“宽峰”的解决方案。

import numpy as np
import operator as op

def keep_peaks(A, axis=-1):
    B = np.swapaxes(A, axis, -1)
    # take differences between consecutive elements along axis
    # pad with -1 at the start and the end
    # the most efficient way is to allocate first, because otherwise
    # padding would involve reallocation and a copy
    # note that in order to avoid that copy we use np.subtract and its
    # out kwd
    updown = np.empty((*B.shape[:-1], B.shape[-1]+1), B.dtype)
    updown[..., 0], updown[..., -1] = -1, -1
    np.subtract(B[..., 1:], B[..., :-1], out=updown[..., 1:-1])
    # extract indices where the there is a change along axis
    chnidx = np.where(updown)
    # get the values of the changes
    chng = updown[chnidx]
    # find indices of indices 1) where we go up and 2) the next change is
    # down (note how the padded -1's at the end are useful here)
    # also include the beginning of each 1D subarray
    pkidx, = np.where((chng[:-1] > 0) & (chng[1:] < 0) | (chnidx[-1][:-1] == 0))
    # use indices of indices to retain only peak indices
    pkidx = (*map(op.itemgetter(pkidx), chnidx),)
    # construct array of changes of the result along axis
    # these will be zero everywhere
    out = np.zeros_like(A)
    aux = out.swapaxes(axis, -1)
    # except where there is a new peak
    # at these positions we need to put the differences of peak levels
    aux[(*map(op.itemgetter(slice(1, None)), pkidx),)] = np.diff(B[pkidx])
    # we could ravel the array and do the cumsum on that, but raveling
    # a potentially noncontiguous array is expensive
    # instead we keep the shape, at the cost of having to replace the
    # value at the beginning of each 2D subarray (we do not need the
    # "line-jump" difference but the plain 1st value there)
    aux[..., 0] = B[..., 0]
    # finally, use cumsum to go from differences to plain values
    return out.cumsum(axis=axis)

peaks = [0, 5, 0, 3, 2, 0, 1, 7, 0]

print(peaks)
print(keep_peaks(peaks))

# show off axis kwd and broad peak detection
peaks3d = np.kron(np.random.randint(0, 10, (3, 6, 3)), np.ones((1, 2, 1), int))

print(peaks3d.swapaxes(1, 2))
print(keep_peaks(peaks3d, 1).swapaxes(1, 2))

样品运行：

[0, 5, 0, 3, 2, 0, 1, 7, 0]
[0 5 5 3 3 3 3 7 7]
[[[5 5 3 3 1 1 4 4 9 9 7 7]
  [2 2 9 9 3 3 4 4 3 3 7 7]
  [9 9 0 0 2 2 5 5 7 7 9 9]]

 [[1 1 3 3 9 9 3 3 7 7 0 0]
  [1 1 1 1 4 4 5 5 0 0 3 3]
  [5 5 5 5 8 8 1 1 2 2 7 7]]

 [[6 6 3 3 8 8 2 2 3 3 2 2]
  [6 6 9 9 3 3 9 9 3 3 9 9]
  [1 1 5 5 7 7 2 2 7 7 1 1]]]
[[[5 5 5 5 5 5 5 5 9 9 9 9]
  [2 2 9 9 9 9 4 4 4 4 7 7]
  [9 9 9 9 9 9 9 9 9 9 9 9]]

 [[1 1 1 1 9 9 9 9 7 7 7 7]
  [1 1 1 1 1 1 5 5 5 5 3 3]
  [5 5 5 5 8 8 8 8 8 8 7 7]]

 [[6 6 6 6 8 8 8 8 3 3 3 3]
  [6 6 9 9 9 9 9 9 9 9 9 9]
  [1 1 1 1 7 7 7 7 7 7 7 7]]]

将一系列峰转换为代表最新峰值的一系列步骤

1 个答案: