在1DNumPy数组中寻找奇点/局部极大值/极小值集（再一次）

Question

我想有一个函数，可以检测局部极大值/极小值在数组中的位置（即使存在一组局部极大值/极小值）。示例：

给定数组

test03 = np.array([2,2,10,4,4,4,5,6,7,2,6,5,5,7,7,1,1])

我希望输出如下：

^{pr2}$

从这个例子中可以看出，不仅检测到奇异值，而且还检测到局部极大值/极小值集。在

{我只知道一些极小值集的答案，但忽略了它们中的许多极值点。在

在提出这个问题之前，我自己编写了一个函数，它与我前面描述的完全相同（这个函数位于这个问题的末尾：local_min(a)）。通过我做的测试，它工作正常）。在

问题：然而，我也确信这不是使用Python的最佳方法。是否有我可以使用的内置函数、API、库等？还有其他功能建议吗？一行指令？全矢量解？在

def local_min(a):
    candidate_min=0
    for i in range(len(a)):

        # Controlling the first left element
        if i==0 and len(a)>=1:
            # If the first element is a singular local minima
            if a[0]<a[1]:
                print("local minima, i = 0")
            # If the element is a candidate to be part of a set of local minima
            elif a[0]==a[1]:
                candidate_min=1
        # Controlling the last right element
        if i == (len(a)-1) and len(a)>=1:
            if candidate_min > 0:
                if a[len(a)-1]==a[len(a)-2]:
                    print("set of " + str(candidate_min+1)+ " local minima => array["+str(i-candidate_min)+"]:array["+str(i)+"]")
            if a[len(a)-1]<a[len(a)-2]:
                print("local minima, i = " + str(len(a)-1))
        # Controlling the other values in the middle of the array
        if i>0 and i<len(a)-1 and len(a)>2:
            # If a singular local minima
            if (a[i]<a[i-1] and a[i]<a[i+1]):
                print("local minima, i = " + str(i))
                # print(str(a[i-1])+" > " + str(a[i]) + " < "+str(a[i+1])) #debug
            # If it was found a set of candidate local minima
            if candidate_min >0:
                # The candidate set IS a set of local minima
                if a[i] < a[i+1]:
                    print("set of " + str(candidate_min+1)+ " local minima => array["+str(i-candidate_min)+"]:array["+str(i)+"]")
                    candidate_min = 0
                # The candidate set IS NOT a set of local minima
                elif a[i] > a[i+1]:
                    candidate_min = 0
                # The set of local minima is growing
                elif a[i] == a[i+1]:
                    candidate_min = candidate_min + 1
                # It never should arrive in the last else
                else:
                    print("Something strange happen")
                    return -1
            # If there is a set of candidate local minima (first value found)
            if (a[i]<a[i-1] and a[i]==a[i+1]):
                candidate_min = candidate_min + 1

Note: I tried to enrich the code with some comments to let understand what I do. I know that the function that I propose is not clean and just prints the results that can be stored and returned at the end. It was written to give an example. The algorithm I propose should be O(n).

更新：

有人建议导入from scipy.signal import argrelextrema并使用如下函数：

def local_min_scipy(a):
    minima = argrelextrema(a, np.less_equal)[0]
    return minima

def local_max_scipy(a):
    minima = argrelextrema(a, np.greater_equal)[0]
    return minima

拥有这样的东西才是我真正想要的。然而，当局部极小/极大值集合有两个以上的值时，它就不能正常工作。例如：

test03 = np.array([2,2,10,4,4,4,5,6,7,2,6,5,5,7,7,1,1])

print(local_max_scipy(test03))

输出为：

[ 0  2  4  8 10 13 14 16]

当然在test03[4]我有一个最小值，而不是最大值。我该如何纠正这种行为？（我不知道这是不是另一个问题，或者这是问这个问题的正确地点。）

Answer 1

有多种方法可以解决这个问题。这里列出了一种方法。 您可以创建一个自定义函数，并在查找mimima时使用maximums来处理边缘情况。在

import numpy as np
a = np.array([2,2,10,4,4,4,5,6,7,2,6,5,5,7,7,1,1])

def local_min(a):
    temp_list = list(a)
    maxval = max(a) #use max while finding minima
    temp_list = temp_list + [maxval] #handles last value edge case.

    prev = maxval #prev stores last value seen
    loc = 0 #used to store starting index of minima
    count = 0 #use to count repeated values
    #match_start = False
    matches = []
    for i in range(0, len(temp_list)): #need to check all values including the padded value
        if prev == temp_list[i]:
            if count > 0: #only increment for minima candidates
                count += 1
        elif prev > temp_list[i]:
            count = 1
            loc = i
    #        match_start = True
        else: #prev < temp_list[i]
            if count > 0:
                matches.append((loc, count))
            count = 0
            loc = i
        prev = temp_list[i]
    return matches

result = local_min(a)

for match in result:
    print ("{} minima found starting at location {} and ending at location {}".format(
            match[1], 
            match[0],
            match[0] + match[1] -1))

如果这对你有用，请告诉我。这个想法很简单，你需要在列表中迭代一次，并在看到它们时继续存储最小值。通过在每一端填充最大值来处理边。（或填充最后一个结尾，并使用最大值进行初始比较）

Answer 2

全矢量解决方案：

test03 = np.array([2,2,10,4,4,4,5,6,7,2,6,5,5,7,7,1,1])  # Size 17
extended = np.empty(len(test03)+2)  # Rooms to manage edges, size 19
extended[1:-1] = test03
extended[0] = extended[-1] = np.inf

flag_left = extended[:-1] <= extended[1:]  # Less than successor, size 18
flag_right = extended[1:] <= extended[:-1]  # Less than predecessor, size 18

flagmini = flag_left[1:] & flag_right[:-1]  # Local minimum, size 17
mini = np.where(flagmini)[0]  # Indices of minimums
spl = np.where(np.diff(mini)>1)[0]+1  # Places to split
result = np.split(mini, spl)

result：

^{pr2}$

编辑

不幸的是，这也检测到了最大值，因为它们至少有3个项目大，因为它们被视为平坦的局部极小值。这样的话，裸体补丁会很难看。在

为了解决这个问题，我提出了另外两种解决方案，用numpy，然后用numba。在

使用np.diff的数字：

import numpy as np
test03=np.array([12,13,12,4,4,4,5,6,7,2,6,5,5,7,7,17,17])
extended=np.full(len(test03)+2,np.inf)
extended[1:-1]=test03

slope = np.sign(np.diff(extended))  # 1 if ascending,0 if flat, -1 if descending
not_flat,= slope.nonzero() # Indices where data is not flat.   
local_min_inds, = np.where(np.diff(slope[not_flat])==2) 

#local_min_inds contains indices in not_flat of beginning of local mins. 
#Indices of End of local mins are shift by +1:   
start = not_flat[local_min_inds]
stop =  not_flat[local_min_inds+1]-1

print(*zip(start,stop))
#(0, 1) (3, 5) (9, 9) (11, 12) (15, 16)    

与numba加速度兼容的直接解决方案：

#@numba.njit
def localmins(a):
    begin= np.empty(a.size//2+1,np.int32)
    end  = np.empty(a.size//2+1,np.int32)
    i=k=0
    begin[k]=0
    search_end=True
    while i<a.size-1:
         if a[i]>a[i+1]:
                begin[k]=i+1
                search_end=True
         if search_end and a[i]<a[i+1]:   
                end[k]=i
                k+=1
                search_end=False
        i+=1
    if search_end and i>0  : # Final plate if exists 
        end[k]=i
        k+=1 
    return begin[:k],end[:k]

    print(*zip(*localmins(test03)))
    #(0, 1) (3, 5) (9, 9) (11, 12) (15, 16)  

Answer 3

下面是一个基于将数组重新划分为iterable窗口的答案：

import numpy as np
from numpy.lib.stride_tricks import as_strided

def windowstride(a, window):
    return as_strided(a, shape=(a.size - window + 1, window), strides=2*a.strides)

def local_min(a, maxwindow=None, doends=True):
    if doends: a = np.pad(a.astype(float), 1, 'constant', constant_values=np.inf)
    if maxwindow is None: maxwindow = a.size - 1

    mins = []
    for i in range(3, maxwindow + 1):
        for j,w in enumerate(windowstride(a, i)):
            if (w[0] > w[1]) and (w[-2] < w[-1]):
                if (w[1:-1]==w[1]).all():
                    mins.append((j, j + i - 2))

    mins.sort()
    return mins

测试一下：

^{pr2}$

输出：

[(0, 2), (3, 6), (9, 10), (11, 13), (15, 17)]

不是最有效的算法，但至少它很短。我很确定它是O(n^2)，因为有大约1/2*(n^2 + n)个窗口需要迭代。这只是部分矢量化，所以可能有一种方法可以改进它。在

编辑

为了澄清，输出是包含局部最小值运行的切片的索引。事实上，他们超过了运行的结束是故意的（有人只是试图在编辑中“修复”这个问题）。可以使用输出迭代输入数组中最小值的片段，如下所示：

for s in local_mins(test03):
    print(test03[slice(*s)])

输出：

[2 2]
[4 4 4]
[2]
[5 5]
[1 1]