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deleted 22 characters in body

edited Dec 9, 2016 at 16:06

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This is my first time posting! But sinceSince you asked for a Pythonic solution and I see none so far, I propose:

new_L = [sum(L[:i+1]) for i in range(len(L))]

It's certainly less efficient than an accumulator -- it's O((n^2)/2)\$O(\frac{n^2}{2})\$ vs O(n)\$O(n)\$ -- but it uses a list comprehension as you suggested.

This is my first time posting! But since you asked for a Pythonic solution and I see none so far, I propose:

new_L = [sum(L[:i+1]) for i in range(len(L))]

It's certainly less efficient than an accumulator -- it's O((n^2)/2) vs O(n) -- but it uses a list comprehension as you suggested.

Since you asked for a Pythonic solution and I see none so far, I propose:

new_L = [sum(L[:i+1]) for i in range(len(L))]

It's certainly less efficient than an accumulator -- it's \$O(\frac{n^2}{2})\$ vs \$O(n)\$ -- but it uses a list comprehension as you suggested.

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answered Dec 9, 2016 at 16:04

Matthew Cotton

This is my first time posting! But since you asked for a Pythonic solution and I see none so far, I propose:

new_L = [sum(L[:i+1]) for i in range(len(L))]

It's certainly less efficient than an accumulator -- it's O((n^2)/2) vs O(n) -- but it uses a list comprehension as you suggested.