Count of 1D array if they are equal to the arithmetic mean of their neighbours

Question

I will like to find out the numbers in an array which is equal to the arithmetic mean of their immediate neighbors.The twist here is to use 0 if a[i-1] or a[i+1] does not exist.

For example:

a =  [2,4,6,8,2,5]
a[0] = (0 + 4)/2 = 2 ----count 1
a[1] = (2 + 6)/2 = 4---- count 2
a[5] = (2 + 0)/2 != 5 ----no count

If it is not, it should be a count of 0. I have written the below code but it won't work for all the arrays, like this one. How can I do it?

a = [100, 110, 120, 130, 140, 150, 160, 170, 180, 190]

My code:

def countMeans(a):
    count = 0
    for i in range(1, len(a)-1):
        if a[i-1]+ a[i+1] is 2 * a[i]:
            count += 1
    return count

See [is operator behaves unexpectedly with integers](https://stackoverflow.com/questions/306313/is-operator-behaves-unexpectedly-with-integers) — Tomerikoo, Nov 23 '20 at 08:19

score 1 · Accepted Answer · answered Nov 23 '20 at 09:16

As per your new update, use the following code:

def countMeans(a):
    count = 0
    if len(a) > 1:
        for i in range(0, len(a)):
            val1 = 0 if i == 0 else a[i-1]
            try:
                val2 = a[i+1]
            except:
                val2 = 0
            if val1 + val2 == 2 * a[i]:
                count += 1
    else:
        count = 1 if a[0] == 0 else 0
    return count

score 1 · Answer 2 · answered Nov 23 '20 at 09:33

Let me know if I got this right.

def countMeans(a):
    count = 0
    if (a[1]/2) == a[0]:
        count +=1
    for i in range(1, len(a)-1):
        if (a[i-1] + a[i+1])/2 == a[i]:
            count += 1
    if (a[-2]/2) == a[-1]:
        count +=1
    return count

a =  [2,4,6,8,2,5]
print (countMeans(a))

a = [100, 110, 120, 130, 140, 150, 160, 170, 180, 190]
print (countMeans(a))

The output for this is:

3
8

score 0 · Answer 3 · answered Nov 23 '20 at 08:13

0

Use == instead of is while comparing the values. Use the following snippet:

def countMeans(a):
    count = 0
    for i in range(1, len(a)-1):
        if a[i-1]+ a[i+1] == 2 * a[i]:
            count += 1
    return count

Remember is operator defines if both the variables point to the same object whereas the == sign checks if the values for the two variables are the same.

answered Nov 23 '20 at 08:13

theashwanisingla

417
3
12

Thanks. It improved but it didn't work if my array a = [0] or something like a = [2,4,6,3,2] – Abimbola Ojikutu Nov 23 '20 at 08:17
I didn't get your question. It prints correct in both cases. – theashwanisingla Nov 23 '20 at 08:42
Yes so, a = [0] should return a count of 1, but it returns a count of 0. Its neighbors will be assumed to be 0, like [0,0,0] so it should return 1. While a = [2,4,6,3,2], should return a total count of 2 – Abimbola Ojikutu Nov 23 '20 at 08:46
Absolutely not, code is not about to assume by its own. You need to provide some condition for it. And if you want the first letter to be included in your loop, you need to update your question, so that I can answer accordingly. – theashwanisingla Nov 23 '20 at 08:51
I've posted a new answer. Please check it. – theashwanisingla Nov 23 '20 at 09:19

Count of 1D array if they are equal to the arithmetic mean of their neighbours

3 Answers3