I want to write code that returns a new list where the i^th element is the sum of the first i 1 elements of the original list. For example, for the list [1, 2, 3], the result is [1, 3, 6].

So I've written this:

def cumulative_sum(numbers: list[int]) -> list[int]:
    """Returns the cumulative sum of the given list of numbers."""
    cum_sum = []
    for t in numbers:
        cum_sum.append(sum(numbers[:t]))
    return cum_sum


print(cumulative_sum([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]))

However, when I run this I get [1, 18, 9, 13, 18, 24, 31, 39, 48, 58], where there is an inexplicable 18 and the sequence seems to only work after the first 3 values.

CodePudding user response：

You are mixing indexes and elements when iterating

You can do something like this:

def cumulative_sum(numbers: list[int]) -> list[int]:
    """Returns the cumulative sum of the given list of numbers."""
    #initialize with the first element
    cum_sum = [numbers[0]]

    # start iterating with the second element
    for i in range(1, len(numbers)):
        # add the current element to the previous cumsum
        cum_sum.append(cum_sum[-1]   numbers[i])
    return cum_sum


print(cumulative_sum([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]))

CodePudding user response：

It often makes sense to implement such functions as generator,

def cumulative_sum(numbers):
    """Returns the cumulative sum of the given list of numbers."""
    total = 0
    for t in numbers:
        total  = t
        yield total

so you can loop over the result lazily for i in cumulative_sum([1, 2, 3]): or turn it into a list by list(cumulative_sum([1, 2, 3])) if you really need to. This function, however, is exactly the same as itertools.accumulate so you might as well use that:

from itertools import accumulate 

# [1, 3, 6]
print(list(accumulate([1, 2, 3])))