python - 在 1D-NumPy 数组中查找单数/局部最大值/最小值集(再次)

Question

我想要一个可以检测局部最大值/最小值在数组中的位置的函数(即使有一组局部最大值/最小值)。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])

我想要这样的输出:

set of 2 local minima => array[0]:array[1]
set of 3 local minima => array[3]:array[5]
local minima, i = 9
set of 2 local minima => array[11]:array[12]
set of 2 local minima => array[15]:array[16]

从示例中可以看出，不仅检测到奇异值，还检测到局部最大值/最小值集。

我知道 this question有很多好的答案和想法，但没有一个能完成所描述的工作:其中一些只是忽略数组的极值点，并且都忽略局部最小值/最大值的集合。

在问这个问题之前，我自己写了一个函数，它完全按照我上面描述的(函数在这个问题的最后:local_min(a))。通过我做的测试，它工作正常)。

问题:但是，我也确信这不是使用 Python 的最佳方式。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？全矢量解决方案？

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

拥有这样的东西是我真正想要的。但是，当局部最小值/最大值集具有两个以上的值时，它无法正常工作。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))

输出是:

[ 0  2  4  8 10 13 14 16]

当然在 test03[4] 中我有一个最小值而不是最大值。如何解决此问题？ (我不知道这是另一个问题还是问这个问题的地方是否合适。)

Answer 1

全向量解决方案:

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)

结果:

[0, 1] [3, 4, 5] [9] [11, 12] [15, 16]

编辑

不幸的是，一旦它们至少有 3 个项目大，这也会检测到最大值，因为它们被视为平坦的局部最小值。这样一个 numpy 补丁会很丑。

为了解决这个问题，我提出了另外 2 个解决方案，先用 numpy，然后用 numba。

使用 np.diff 的 numpy :

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)    

兼容numba加速的直接解决方案:

#@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)