Comparing list elements

Question

I make a simple code:

arr =  [4, 9, 5, 3, 2, 10]
kpmp = [] # Empty list to store the comparison

for i in arr:
    if i>0:
        x = 0
        j = i
        while j>=0:
            j-=1
            if arr[i] > arr[j]:
                x+=1
    kpmp.append(x)
print(kpmp)

But, I got this error:

Traceback (most recent call last):
  File "python", line 11, in <module>
IndexError: list index out of range

Which is this line if arr[i] > arr[j]:

The expected output I'd like to print is a list [0, 0, 1, 3, 4, 0]
Explanation:
arr = [4, 9, 5, 3, 2, 10]
1. 4 has nothing to compare so the result is 0
2. 9 > 4 the result is 0
3. 4 < 5 < 9 the result is 1
4. 3 < 4 < 5 < 9 the result is 3
5. 2 < 3 < 4 < 5 < 9 the result is 4
6. Well, 10 is the greatest in the list so the result is 0

I got stuck here, kinda basic knowledge with Python, though.
Thanks for your help!

Answer 1

You are currently using elements of list as indices, which will cause an out of range error if they are greater than the length of your list.

Since arr contains 9, this will throw an out of range error. Also, you need to define x before your if statement, or else you will get a variable not found error.

Here is a different solution to your problem using a list comprehension and slicing:

x = [4, 9, 5, 3, 2, 10]
final = [sum(k > j for k in x[:i]) for i, j in enumerate(x)]

# [0, 0, 1, 3, 4, 0]

Explanation:

sum(k > j for k in x[:i])

This counts all previous values that are greater than the current value. x[:i] gets all elements before the current element in your list.