一个简单的python项目，总和和平均值

Question

I use python 3 and I have to make a program that creates all the multiples of 5 to 100,then print them as well as their average.I have tried this but...

for i in range(0, 101):
    if i % 5 == 0:
        x = 0
        x += 1
        y = sum(i)
        z = y/x 
        print(i)
        print(z)

When I try this i get :

 Traceback (most recent call last):
  File "C:/Users/Chriskaf/Desktop/(3)count.py", line 5, in <module>
    y = sum(i)
TypeError: 'int' object is not iterable

Thanks for you time :)

Answer 1

A possible solution would be to create a list where you save all values. Then apply the sum and average :

total = []
for i in range(0, 101):
    if i % 5 == 0:
        total.append(i)

total_sum = sum(total)
total_average = sum(total)/len(total)

Answer 2

try this:

counter = 0

sum_numbers = 0

for number in range(1,101):

    if number % 5 ==0:

        counter += 1

        sum_numbers += number

        avg = sum_numbers / float(counter)

        print (number, avg)

Output:

(5, 5.0)
(10, 7.5)
(15, 10.0)
(20, 12.5)
(25, 15.0)
(30, 17.5)
(35, 20.0)
(40, 22.5)
(45, 25.0)
(50, 27.5)
(55, 30.0)
(60, 32.5)
(65, 35.0)
(70, 37.5)
(75, 40.0)
(80, 42.5)
(85, 45.0)
(90, 47.5)
(95, 50.0)
(100, 52.5)

Your code had a couple of issues.

• the x counter was reset on every iteration (it was always set to 0, and after that to 1)
• missuse of sum (error code here) - you cant sum a single digit (which is what i is) - the sum function description is: sum(sequence[, start]) -> value Return the sum of a sequence of numbers (NOT strings) plus the value of parameter 'start' (which defaults to 0). When the sequence is empty, return start.
• you are dividing y by an int (would result in nearest round int, I believe you want float results)

Hope that helps!

Answer 3

Even better, use the "step" parameter of range to get exactly the numbers you want. Also, don't re-compute the sum.

total = [i for i in range(5, 101, 5)]
total_sum = sum(total)
total average = total_sum / len(total)