how to loop through each element in list and increment by one-CodePudding

Hello i want to ask how to loop through each element and increment every time by one this is what i want first i want to sum 0 , 1, 3 ,6 , 10 and after that sum can somebody help me about that i don't know how to tell if it is loop through each element or iterate.It should look like these examples.I am sorry!

ls = [0, 1, 3, 6, 10]

ls = [1, 3, 6, 10]

ls = [3, 6, 10]

ls = [6, 10]

ls = [10]

ls = []

Here's the problem who i want to solve it : https://www.codewars.com/kata/5ce399e0047a45001c853c2b/train/python

I tried this but it doesn't work

 def parts_sums(ls):
    length_list = len(ls)
    for i in range(0,length_list 1):
        return length_list

CodePudding user response：

Note that there is a built-in function sum() in Python that does that job probably better than any code you can write in Python.

sum([0, 1, 3, 6, 10])

However, if you want to practice writing your sum function by iterating through a list and summing all the elements, this is how you do it.

def my_sum(ls):
    result = 0
    for i in range(len(ls)):
        result  = ls[i]
    return result

First of all, you need to initialize a variable to hold your result. The range() function generates all values from 0 to x. The for-loop assigns all values generated by the range function to i in order and executes the indented block below. The = assignment increments the left-hand side variable by the right-hand side expression value. At last, we return the result.

And if you prefer using a while-loop,

def my_sum(ls):
    result = 0
    i = 0
    while i < len(ls):
        result  = ls[i]
        i  = 1
    return result

It's always good to consult Python documentation when you are not sure how to use its built-in function.

If you want the accumulated sum of all items reversed, you can take a look at the accumulate function in itertools.

from itertools import accumulate
def parts_sums(ls):
    return list(accumulate(ls[::-1]))[::-1]   [0]

Or if you want to implement with a loop,

def parts_sums(ls):
    result = []
    part_sum = 0
    for item in ls[::-1]:
        result.append(part_sum)
        part_sum  = item
    result.append(part_sum)
    
    return result[::-1]

Or if you want to do it without reversing the list (say if you want to yield the results)

def parts_sums(ls):
    result = []
    part_sum = sum(ls)
    for item in ls:
        result.append(part_sum)
        part_sum -= item
    result.append(part_sum)
    
    return result

Note the algorithm is still O(n), not that time complexity matters in this case.

CodePudding user response：

For the question you have mentioned in codewars, you need to loop it thro 2 loops and keep reducing the first element in the inner loop for sum.

def parts_sums(ls):
    # your code
    sum = []
    for i in range(len(ls)):
        sum_temp =0
        for j in range(i,len(ls)):
            sum_temp  = ls[j]
        sum.append(sum_temp)
    sum.append(0)   # for the final empty list
    return sum
print(parts_sums([0, 1, 3, 6, 10]))

CodePudding user response：

This test will check the execution time too. So you need to be fast.

Naive approach

You can use sum or create your own sum.

def parts_sums(ls):
    return [
        sum(ls[i:])
        for i in range(len(ls)   1)
    ]

But i means you'll need to loop in a list twice. So it will be slow.

Sneaky approach

In a list like [a, b, c, d, e] you are calculating:

[a b c d e, a b c d, a b c, a b, a, 0]. So let's start from last to first element. [0, a, a b, a b c, a b c d, a b c d e]. Now we see a cumulative iteration: So get loop in the list, get the element, sum it with last element of the result ([0]) list and add it as the last element to the result list. Lastly reverse the result.

def parts_sums(ls):
    res = [0]
    for i in range(len(ls)-1, -1, -1):
        res.append(res[-1]   ls[i])
        
    return res[::-1]