python-3.x - 在数字序列的顶部或底部找到点(Find the point on the top or bottom of a sequence of numbers)

Question

I have a problem like this and i would like to write a snippet of code to solve this problem.

(我有这样的问题，我想写一段代码来解决这个问题。)

1. Sequences like: [1,2,3,2], [1,3,2], [1,3,2,1] -> i want to output 3 (maximum) because the sequence increases to 3 and then decreases again

   (像[1,2,3,2], [1,3,2], [1,3,2,1]之类的序列->我想输出3（最大），因为序列增加到3，然后又减少)

2. Sequences like [3,2,1,2], [3,1,2], [3,1,2,3] -> i want to output 1 (minimum) because the sequence decreases to 1 and then increases again

   (像[3,2,1,2], [3,1,2], [3,1,2,3]之类的序列->我想输出1（最小），因为该序列递减为1，然后又增加)

Any idea on how to do this automatically?

(关于如何自动执行此操作的任何想法？)

  ask by Tuyen Vo Quang translate from so

Answer 1

Try getting local maximas and/or local minimas:

(尝试获取局部最大值和/或局部最小值：)

import numpy as np
from scipy.signal import argrelextrema

a = np.array([3,2,1,3])

res=a[np.hstack([argrelextrema(a, np.greater),argrelextrema(a, np.less)]).ravel()]

This will return both local maximas and minimas.

(这将同时返回局部最大值和最小值。)

You can mark them somehow separately, if it's better for your use case.

(如果对您的用例更好，则可以单独标记它们。)

From your question I assumed it can be just one extremum.

(根据您的问题，我认为这可能只是一个极值。)

Also - depending on your data you might consider using np.less_equal or np.greater_equal instead of np.less or np.greater respectively.

(另外-根据您的数据，您可能会考虑分别使用np.less_equal或np.greater_equal而不是np.less或np.greater 。)