Fill a list with a for loop python

Question

Mine is not really a problem but the search for a solution, if any. And knowing this language necessarily exists. I have an integer 'n' and an integer 'p'. What I'm trying to do is divide n into digits, raise them to p and put them in a list. Later I will sum () the list. (Below is the code)

tot = sum([int(x)**p for x in str(n)])

The code works perfectly without any problems, the only thing I miss to add is that at each iteration, 'p' must be increase by 1. I have tried many things, and looked at a few other methods, but cannot find the correct syntax for adding this step. I really hope that some of you can help me find this solution. Many thanks in advance

Answer 1

Use enumerate to get an index which you can use to increase p by:

tot = sum([int(x) ** (p + i) for i, x in enumerate(str(n))])

If you want to start at p + 1 instead of p + 0 , use

tot = sum([int(x) ** (p + i) for i, x in enumerate(str(n), start=1)])

BTW, you don't even need to create the list. Remove the brackets to save some memory:

tot = sum(int(x) ** (p + i) for i, x in enumerate(str(n)))

Answer 2

You can use enumerate() inside your list comprehension to achieve this as:

>>> n = 12345
>>> p = 2

>>> sum([int(x)**(p+i) for i, x in enumerate(str(n))])
16739

enumerate() returns a tuple containing a count (from start which defaults to 0 ) and the values obtained from iterating over iterable.

You can pass start as 1 , if you want to start with p+1 as:

>>> sum([int(x)**(p+i) for i, x in enumerate(str(n), start=1)])
82481

Answer 3

A potential solution based on a for loop should be

tot = []

p = <whatever>

for x in str(n):
   temp = int(x) ** p 
   tot.append(temp)
   p = p + 1

sum(tot)