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List of square roots python - py7 - Apr-05-2017
So the question is:
I have to do a little program where the function receives a argument x, who is an integer and returns a list of the sum of the first n square roots.
Like [sqrt(1), sqrt(1) + sqrt(2), sqrt(1) + sqrt(2) + sqrt(3), ....]
I've made this line of code but it is just returning the sum
return sum([x**0.5 for x in range(n+1)])
RE: List of square roots python - ichabod801 - Apr-06-2017
To get the cumulative sums of a list of length n:
[sum(the_list[:index]) for index in range(1, n+1)]One problem is that your list is n+1 long. That's because you're including the square root of zero.
RE: List of square roots python - py7 - Apr-08-2017
(Apr-06-2017, 01:39 AM)ichabod801 Wrote: To get the cumulative sums of a list of length n:
[sum(the_list[:index]) for index in range(1, n+1)]One problem is that your list is n+1 long. That's because you're including the square root of zero.
And what is 'the_list' ?
The only thing that the function receives is an integer number (n).
RE: List of square roots python - wavic - Apr-08-2017
It returns what you want.
Here is the sum of the list without any additional math:
>>> sum([x for x in range(11)]) 55Here is the sum of list's elements with x**0.5 applied on each of them
>>> sum([x**0.5 for x in range(11)]) 22.4682781862041Notice the result?
RE: List of square roots python - ichabod801 - Apr-08-2017
(Apr-08-2017, 08:33 PM)py7 Wrote: And what is 'the_list'
The list of square roots that you've already generated.
RE: List of square roots python - py7 - Apr-08-2017
(Apr-08-2017, 09:39 PM)ichabod801 Wrote: 1
[sum(the_list[:index]) for index in range(1, n+1)]
it only returns a list --> [0,0,0]
RE: List of square roots python - ichabod801 - Apr-08-2017
Output:>>> the_list = [1, 2, 3]
>>> n = 3
>>> [sum(the_list[:index]) for index in range(1, n+1)]
[1, 3, 6]