python - find positive/negative/positive pattern in numpy array

Question

Does numpy have an easy way to determine if an array has a pattern that looks like this: (positives numbers followed by negative numbers followed by positive numbers)?

Example:

myArray = np.array([5, 5, 3, 6, -2, -5, 4, 9])
myArray2 = np.array([5, 5, 3, 6, 8, 4, -5, -8])

if foundPatern(myArray):  # True
if foundPatern(myArray2): # False

A pure numpy solution: check = np.all((x[np.hstack((True,np.diff(x>=0)))]>=0)==[True,False,True]) where x is your array. I keep only the value where the sign change and check if the sign is indeed Positive Negative Positive. — obchardon
– obchardon, Commented Jun 9, 2021 at 14:48

grey_ranger · Accepted Answer · 2021-06-10 05:56:09Z

1

There exists a numpy built-in to convert each element of an array to 1, 0, -1 based on the sign. You could easily use itertools.groupby for this task after converting the sign:

from itertools import groupby
import numpy as np


def foundPattern(lst):
    g = list(groupby(np.sign(lst)))
    return (len(g)==3 and g[0][0]==1.0)

This works as follows:

The items of your array are grouped based on their sign (+ or -)
If exactly 3 groups exist and the first one is a group of positives, you return True; in all other cases you return False

Note that you have not specified how edge cases are to be treated (e.g., 0 in the array) so I will not attempt to handle them in any way. Be mindful however that they do exist. Another example of an edge case could be NaN.

edited Jun 10, 2021 at 5:56

grey_ranger

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answered Jun 9, 2021 at 14:33

Ma0

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4 Comments

Farhan Ahmad Over a year ago

This is perfect! Didn't know of itertools until now. Thanks for the detailed explanation! :)

Ma0 Over a year ago

@FarhanAhmad then you better have a look because itertools is an extremely useful library. Cheers.

grey_ranger Over a year ago

There is a numpy built-in for this: np.sign. You can modify the first line of the function to: g = list(groupby(np.sign(lst)))

Ma0 Over a year ago

@SNygard thats even better! Feel free to edit, I am away from the pc atm

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