在数字列表中检测峰并记录其位置 [英] Detect peaks in list of numbers and record their positions

问题描述

我正在尝试创建一些代码，以返回数字数组的峰值"(或局部最大值)的位置和值.

I am trying to create some code that returns the positions and the values of the "peaks" (or local maxima) of a numeric array.

例如，列表 arr = [0、1、2、5、1、0] 在位置 3 处具有峰值，值为5 (因为 arr [3] 等于 5 ).

For example, the list arr = [0, 1, 2, 5, 1, 0] has a peak at position 3 with a value of 5 (since arr[3] equals 5).

数组的第一个和最后一个元素将不被视为峰(在数学函数的上下文中，您不知道其前后是什么，因此，您不知道它是峰还是峰不是).

The first and last elements of the array will not be considered as peaks (in the context of a mathematical function, you don't know what is after and before and therefore, you don't know if it is a peak or not).

def pick_peaks(arr):
    print(arr)
    posPeaks = {
        "pos": [],
        "peaks": [],
    }
    startFound = False
    n = 0
    while startFound == False:
        if arr[n] == arr[n+1]:
            n += 1
        else:
            startFound = True

    endFound = False
    m = len(arr) - 1
    while endFound == False:
        if arr[m] == arr[m-1]:
            m -= 1
        else:
            endFound = True

    for i in range(n+1, m):
        if arr[i] == arr[i-1]:
            None
        elif arr[i] >= arr[i-1] and arr[i] >= arr[i+1]:
            posPeaks["pos"].append(i)
            posPeaks["peaks"].append(arr[i])

    return posPeaks

我的问题是高原. [1,2,2,2,1] 有一个峰值，而 [1,2,2,2,3] 没有.当高原是峰值时，将记录高原的第一位置.

My issue is with plateaus. [1, 2, 2, 2, 1] has a peak while [1, 2, 2, 2, 3] does not. When a plateau is a peak, the first position of the plateau is recorded.

感谢您的帮助.

推荐答案

我建议您使用

I suggest you use groupby to group contiguous equal values, then for each group store the first position, example for [1, 2, 2, 2, 1] it creates the following list following list of tuples [(1, 0), (2, 1), (1, 4)], putting all together:

from itertools import groupby


def peaks(data):
    start = 0
    sequence = []
    for key, group in groupby(data):
        sequence.append((key, start))
        start += sum(1 for _ in group)

    for (b, bi), (m, mi), (a, ai) in zip(sequence, sequence[1:], sequence[2:]):
        if b < m and a < m:
            yield m, mi


print(list(peaks([0, 1, 2, 5, 1, 0])))
print(list(peaks([1, 2, 2, 2, 1])))
print(list(peaks([1, 2, 2, 2, 3])))

输出

[(5, 3)]
[(2, 1)]
[]

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