How to divide a list in equal size of a certain number and get the average of values in Python?

Question

I have a list of 248 values that looks like this: lst = [10.0, 7.4, 8.2, 8.8, 7.4,.... ] I want to get the average of every 8 value. Such as it will be divided to 31 sub lists and will have the average of every list.

This is what I have for now but I have to write this 31 times if I use this method. Can someone point me to an easier and more efficient way?

num1 = 0
for i in range (8):
    num += float(lst[i])
print(num1/8)


num2 = 0

for i in range (8, 16):
    num += float(lst[i])
print(num2/8)

[sum(x)/8 for x in [numbers[x:x+8] for x in range(0, len(numbers), 8)]] — Yanirmr, Nov 08 '20 at 15:08

Aplet123 · Accepted Answer · 2020-11-08T15:05:11.137

1

Just use a loop:

averages = []
for start in range(0, len(lst), 8):
    num = 0
    for i in range(min(8, len(lst) - start)):
        num += lst[start + i]
    averages.append(num / 8)

edited Nov 08 '20 at 15:05

answered Nov 08 '20 at 15:00

Aplet123

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Thanks, but it gives the same number for each value in 'lst' such as this: [7.6125, 7.6125, 7.6125, 7.6125, 7.6125, 7.6125, 7.6125, 7.6125, ...] – S. Acarsoy Nov 08 '20 at 15:04
Sorry, I had a minor mistake that I've fixed. – Aplet123 Nov 08 '20 at 15:05

How to divide a list in equal size of a certain number and get the average of values in Python?

1 Answers1