## How can I make a combined power function faster?

I made this function pretty quickly to find the combined powers of everything in a list, but I'm pretty sure there is a way to make it faster. It returns a 2 item list, with the combined powers and the list formatted for export.

#The input for this function is a list of numbers (but string data type) def find(x): total = int(x[0]) for i in range (1,len(x)): total = total ** int(x[i]) value = [total,'^'.join(x)] return value

This will calculate the values faster than what you currently have:

import functools import operator l = [2, 3, 4] functools.reduce(operator.pow, l) ## 4096

If you want to show the chain of values in the list too, as in the original post, you can define a function e.g. like this one:

def func(vals): vals_string = '^'.join(str(val) for val in vals) total = functools.reduce(operator.pow, vals) return [total, vals_string]

Usage example:

l = [2, 3, 4, 5] result = func(l) print(result) ## [1152921504606846976, '2^3^4^5']

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You want to avoid doing a stack of exponentiations, as those are expensive. Multiplication is cheaper, so you can save computing power by multiplying all the values to the right of the first one, and then raising the first to that power.

from functools import reduce from operator import mul from typing import List def find(x: List[int]) -> int: return x[0]**reduce(mul, x[1:], 1)

You could also do

def find(i: int, *updog: int) -> int: return i ** reduce(mul, updog, 1)

and call it like `find(*[2,1,2,3,4])`

or `find(2, 1, 2, 3)`

.

Either way, this uses the function signature to provide a stricter guarantee that the initial value is set, rather than having undefined behavior if the list is empty.

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You can use NumPy's `power.accumulate()`

to have cumulative exponents. **However**, you will soon run into large numbers for a large input list. The `[-1]`

here gives you the cumulative of all the elements

import numpy as np def find(x): total = np.power.accumulate(x)[-1] value = [total,'^'.join(map(str, inp))] return value x = np.array([2,3,4,5]) print (find(x)) # [1152921504606846976, '2^3^4^5']

**Modular Exponentiation in Python,** Pow function calculates in O(log n) time in python but it takes a lot of time when the value of xy and then mod it with p to get (xy) % p evaluated. the (%) operator takes a lot of time, so a Fast Modular Exponentiation is used. of Non- Zero elements in Python list � Python | Merge Python key values to list� Power query will then show all the items available to the workbook. This will include all the power query functions! If you click on any of the functions listed, you’ll be taken to a mini help guide for that function. The function name is shown. There is a brief description of what the function does. You can test out the function on your data.

You can use `functools.reduce()`

:

from functools import reduce import operator reduce(operator.pow, list)

What `reduce()`

does is it applies the first parameter to every element in the second.

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##### Comments

- If it is already a list of integers, why convert to integer every time?
- line 5 raises:
`TypeError: sequence item 0: expected str instance, int found`

when I use input`[1,2,3,4]`

- If you are looking for help improving code that already works, consider codereview.stackexchange.com
- This appears to be written to take a list of string representations of integers, not a list of integers. Also, it's backwards from the usual convention for powers -
`2^3^4`

usually means`2^(3^4)`

, but you're computing`(2^3)^4`

. - Why are you computing this, anyway? If you're trying to solve some programming challenge, computing giant power towers at all is probably not the way to go, even if the problem description has a power tower in it.
- I think you forgot to handle the case where
`len(x) == 1`

. Other than that, this is a valid simplification. - Adding the extra argument for
`reduce`

is definitely good. I'm not so sure about the empty-`x`

handling - I would be more inclined to raise an exception. I don't see a strong argument for`find([]) == 1`

making sense. - @user2357112 I have supplied an alternative approach that separates out the first value. I think I prefer it because it makes it really stinking obvious and strict that that first value has to exist for the thing to make sense. Sure, we could raise an
`IndexError`

, but a`TypeError: find() missing 1 required positional argument`

is much clearer IMO. - First, NumPy ufuncs have a
`reduce`

method that's more appropriate than`accumulate`

. Second, the results quickly overflow the largest data type NumPy offers. - @user2357112: Hmm, I see. Any workaround to specify the upper limit of datatype to avoid overflow and still stick to accumulate?
- the description of
`reduce()`

is how i'd describe what`map()`

does... note that`reduce()`

works on*binary*functions, whereas map doesn't have to. from the official docs for functools.reduce(): "Apply function of two arguments cumulatively to the items of sequence, from left to right, so as to reduce the sequence to a single value."