在1D-NumPy数组中查找局部最大值/最小值的奇异点/集(再次) [英] Finding singulars/sets of local maxima/minima in a 1D-NumPy array (once again)

问题描述

我想拥有一个可以检测局部最大值/最小值在数组中的位置的函数(即使存在一组局部最大值/最小值).示例:

I would like to have a function that can detect where the local maxima/minima are in an array (even if there is a set of local maxima/minima). Example:

给出数组

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

我想要一个输出，如:

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]

从示例中可以看到，不仅检测到奇异值，还检测到局部最大值/最小值集.

As you can see from the example, not only are the singular values detected but, also, sets of local maxima/minima.

我在这个问题中知道有很多很好的答案和想法，但都没有完成所描述的工作:其中一些只是忽略了数组的极点，而全部都忽略了局部极小值/极大值的集合.

I know in this question there are a lot of good answers and ideas, but none of them do the job described: some of them simply ignore the extreme points of the array and all ignore the sets of local minima/maxima.

在问问题之前，我自己编写了一个函数，该函数的功能与我上面所述的完全相同(该函数位于问题的末尾:local_min(a).经过我的测试，它可以正常工作).

Before asking the question, I wrote a function by myself that does exactly what I described above (the function is at the end of this question: local_min(a). With the test I did, it works properly).

问题:但是，我也确信这不是使用Python的最佳方法.我可以使用内置的函数，API，库等吗?还有其他功能建议吗?单行指令?完整的矢量解决方案?

Question: However, I am also sure that is NOT the best way to work with Python. Are there builtin functions, APIs, libraries, etc. that I can use? Any other function suggestion? A one-line instruction? A full vectored solution?

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

注意:我试图通过添加一些注释来丰富代码，以使我理解自己的工作.我知道我建议的功能是 不干净，只是打印可以存储和返回的结果 在最后.它被写为一个例子.我建议的算法应该是O(n).

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并使用如下功能:

Somebody was suggesting to import from scipy.signal import argrelextrema and use the function like:

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，它将无法正常工作.例如:

To have something like that is what I am really looking for. However, it doesn't work properly when the sets of local minima/maxima have more than two values. For example:

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]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在哪里问这个问题.)

Of course in test03[4] I have a minimum and not a maximum. How do I fix this behavior? (I don't know if this is another question or if this is the right place where to ask it.)

推荐答案

完整的矢量解决方案:

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:

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

编辑

不幸的是，由于它们被视为平坦的局部极小值，因此一旦它们至少有3个项目，它也会立即检测出最大值.这样的numpy补丁会很丑.

Unfortunately, This detects also maxima as soon as they are at least 3 items large, since they are seen as flat local minima. A numpy patch will be ugly this way.

为解决此问题，我提出了另外两个解决方案，分别是numpy和numba.

To solve this problem I propose 2 other solutions, with numpy, then with 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)    

与numba加速兼容的直接解决方案:

A direct solution compatible with numba acceleration :

#@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)  

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