find_peaks 未识别数组开头的峰

Question

我正在尝试找到一种矢量化方法来找到数组中第一个位置，其中的值不高于前面 n 个数字的最大值。我想过用scipy.signal的find_peaks方法求局部最大值。我认为如果您将距离定义为 10 n 就是 10，那么它就是这样做的。但不幸的是，距离的条件必须在两个方向上都满足 - 先前和即将到来的数字。有没有其他方法或方法可以找到这样的东西？

示例：

arr1 = np.array([1.        , 0.73381293, 0.75649351, 0.77693474, 0.77884614,
       0.81055903, 0.81402439, 0.78798586, 0.78839588, 0.82967961,
       0.8448    , 0.83276451, 0.82539684, 0.81762916, 0.82722515,
       0.82101804, 0.82871127, 0.82825041, 0.82086957, 0.8347826 ,
       0.82666665, 0.82352942, 0.81270903, 0.81191224, 0.83180428,
       0.84975767, 0.84044236, 0.85057473, 0.8394649 , 0.80000001,
       0.83870965, 0.83962262, 0.85039371, 0.83359748, 0.84019768,
       0.83281732, 0.83660132])

from scipy.signal import find_peaks
peaks, _ = find_peaks(arr1, distance=10)

在这种情况下，它找到位置 10 和 27。但位置 0 也有 10 个不更高的后续元素。我怎样才能找到这些？

Answer 1

def rolling_window(a, window):
    shape = a.shape[:-1] + (a.shape[-1] - window + 1, window)
    strides = a.strides + (a.strides[-1],)
    return np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)

def get_peaks(arr, window):
    maxss = np.argmax(rolling_window(arr1, window), axis=1)
    return np.where(maxss == 0)[0]

>>> arr1 = np.array([1.        , 0.73381293, 0.75649351, 0.77693474, 0.77884614,
       0.81055903, 0.81402439, 0.78798586, 0.78839588, 0.82967961,
       0.8448    , 0.83276451, 0.82539684, 0.81762916, 0.82722515,
       0.82101804, 0.82871127, 0.82825041, 0.82086957, 0.8347826 ,
       0.82666665, 0.82352942, 0.81270903, 0.81191224, 0.83180428,
       0.84975767, 0.84044236, 0.85057473, 0.8394649 , 0.80000001,
       0.83870965, 0.83962262, 0.85039371, 0.83359748, 0.84019768,
       0.83281732, 0.83660132])
>>> get_peaks(arr1, 10)
array([ 0, 10, 27])

滚动学分 window 功能：Rolling window for 1D arrays in Numpy?

Answer 2

很遗憾，find_peaks() works by comparing neighboring values - so will not identify peaks that occur at the beginning or end of the array. One workaround is to use np.concatenate()要在首尾插入数组的最小值，然后peaks变量减1：

>>> import numpy as np
>>> peaks, _ = find_peaks(np.concatenate(([min(arr1)],arr1,[min(arr1)])), distance=10)
>>> peaks-1
array([ 0, 10, 27], dtype=int64)

Answer 3

我们可以使用1D sliding-windowed max filter from SciPy。此外，您似乎正在与 n 之前的元素进行比较。因为，第一个元素不会有任何前面的元素，我们需要让它忽略。

因此，我们会像这样实现 -

from scipy.ndimage.filters import maximum_filter1d

def peaks_previousN(a, n):
    W = (n-1)//2
    return np.flatnonzero(a[1:]>=maximum_filter1d(a, n, origin=W)[:-1])+1

具有给定样本数组的样本运行 -

In [107]: peaks_previousN(arr1, n=10)
Out[107]: array([25, 27])

find_peaks 未识别数组开头的峰

find_peaks does not identify a peak at the start of the array

python

numpy

vectorization

scipy