Counting elements of an array in Python

Question

I've been searching the answer for an exercise. This exercise gives us an array with numbers between -1 and 1. We are asked to count how many times the signal array changes sign. For example : you have this array : [0.7,0.5,0.6,-0.1,0.9,0.6,0.4,-0.2,-0.3,0.2,0.3,-0.3,-0.9,-0.7,0.1,0.2] It changes signs 7 times This is the code I have made that should get the "count" of how many times it changes signs. For some reason, it doesn't work, the count at the end is equal to 0 and it should be 7. Can you help me please ? Thank you very much.

--THE CODE--

import numpy as np

def zeroCrossing(signal):

    j=0
    ls=len(signal)
    print(ls)
    count=0
    x=0
    y=0
    for j in range(0,ls):
        if signal[j]>0 and x==1:
            j=j+1
            y=0
        elif signal[j]>0 and x==0:
            count==count+1
            x=1
            j=j+1
        elif signal[j]<0 and y==1:
            j=j+1
            x=0
        elif signal[j]<0 and y==0:
            count==count+1
            y=1
            j=j+1
    
    return count

print(zeroCrossing(np.array([0.7,0.5,0.6,-0.1,0.9,0.6,0.4,-0.2,-0.3,0.2,0.3,-0.3,-0.9,-0.7,0.1,0.2])))

It seems that you count the initial numbers as a "change" in sign? — Alexander Kalian, Dec 06 '20 at 19:27
In your own words, what do you expect `count==count+1` to do? — Karl Knechtel, Dec 06 '20 at 19:31
`arr = np.array([0.7,0.5,0.6,-0.1,0.9,0.6,0.4,-0.2,-0.3,0.2,0.3,-0.3,-0.9,-0.7,0.1,0.2]); print(sum(np.where(np.sign(arr) != np.sign(np.roll(arr, -1)), 1, 0)))` would do it, but I don't get where the 7 comes from — roganjosh, Dec 06 '20 at 19:42
Does this answer your question? [Python - counting sign changes](https://stackoverflow.com/questions/2936834/python-counting-sign-changes) — Chris, Dec 06 '20 at 19:43

Alexander Kalian · Answer 1 · 2020-12-06T20:02:08.333

You can achieve what you want simply by iterating through the array as follows:

sign_changes = 0
last_number = None
for n in my_array:
    if n < 0: # checks if n is negative
        if last_number != "negative": # checks if the last number was not negative
            sign_changes += 1
        last_number = "negative"
    else: # takes n as positive (including 0)
        if last_number != "positive": # checks if the last number was not positive
            sign_changes += 1
        last_number = "positive"

Edit:

This code will count 7 in the example array given by the OP, as the OP expects, which counts the first number as a "sign change". To avoid this (and instead count 6 sign changes), small tweaks to the code can be made. The most simple is to change the last_number != "positive" and last_number != "negative" conditions to last_number == "negative" and last_number == "positive" respectively.

If they are going to use arrays then this is hopelessly slow — roganjosh, Dec 06 '20 at 19:33
Whether or not it is "hopeless" depends on the exact aims of the script/program. This is a simple solution that works well and answers the question. The OP has not specified any high performance requirements. — Alexander Kalian, Dec 06 '20 at 19:40

score 0 · Accepted Answer · answered Dec 06 '20 at 19:30

0

You are not modifying count variable, because you use comparison operator instead of assignment. Replace count==count+1 by count=count+1 to fix it.

answered Dec 06 '20 at 19:30

fdermishin

3,519
3
24
45

score 0 · Answer 3 · answered Dec 06 '20 at 20:00

0

Since you always count the first number as 1 time, you may try this numpy code:

arr = np.array([0.7,0.5,0.6,-0.1,0.9,0.6,0.4,-0.2,-0.3,0.2,0.3,-0.3,-0.9,-0.7,0.1,0.2])
change_count = np.count_nonzero(np.diff(np.sign(arr))) + 1

Out[122]: 7

answered Dec 06 '20 at 20:00

Andy L.

24,909
4
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Counting elements of an array in Python

3 Answers3