I do a lot of work in number theory, especially with factors.
I often need just the two middle factors from a very long factor list/result, so I created a function that takes a factor list and returns just the middle two, or one, factors.
Nothing earth shattering, but I use it a lot, and thought that I would share it.
And those with better ways to offer, offer away. I am always interested.
# Imports #
import sys, os, importlib, inspect, locale, time, math
from datetime import datetime
locale.setlocale(locale.LC_ALL, '') # Set the locale for your system 'en_US.UTF-8'
# Main #
def main():
# Main Code
print("Middle Factors: {}".format(f_GetMiddleFactors([1, 7, 13, 91])))
print("Middle Factors: {}".format(f_GetMiddleFactors([1, 7, 49])))
#-----------Basement------------
# Functions #
# Normal Functions
# Get Middle Factor(s).
def f_GetMiddleFactors(v_FactorList):
"""Syntax: Takes LIST; Returns: LIST;
Desc.: Takes a factor LIST and returns a LIST of the middle factor(s).
Test: print("Middle Factors: {}".format(f_GetMiddleFactors([1, 7, 13, 91])))"""
v_MiddleFactors = [] ## Init v_MiddleFactors list
v_LengthList = len(v_FactorList) ## Get init factor list length
if v_LengthList % 2 == 0: ## If list length is even
v_MiddleFactors.append((str(v_FactorList[((v_LengthList/2)-1)]))) ## Extracts and appends to v_MiddleFactors the middle-left factor.
v_MiddleFactors.append((str(v_FactorList[(v_LengthList/2)]))) ## Extracts and appends to v_MiddleFactors the middle-right factor.
else: ## If list length is odd
v_MiddleFactors.append((str(v_FactorList[((v_LengthList/2))]))) ## Extracts and appends to v_MiddleFactors the middle factor.
return v_MiddleFactors
# Main Loop #
if __name__ == '__main__':
main()
Here is how you can code the same thing by using a PEP8 compliant coding style
# -*-coding: utf8-*-
# python 2 or 3
def get_middle_factors(factor_list):
"""Return a list of the term(s) in the middle of a list
A list of length 1 or 2 is returned, depending on the
length of the argument list being odd or even.
Example:
>>> get_middle_factors([1, 7, 13, 91])
[7, 13]
>>> get_middle_factors([1, 7, 91])
[7]
"""
n = len(factor_list)
q, r = divmod(n-1, 2)
return factor_list[q:q+1+r]
def main():
print("Middle Factors: {}".format(get_middle_factors([1, 7, 13, 91])))
print("Middle Factors: {}".format(get_middle_factors([1, 7, 49])))
if __name__ == '__main__':
main()
""" my output -->
Middle Factors: [7, 13]
Middle Factors: [7]
"""
PEP 8 is good for your code as it makes it easier to read, especially by trained python programmers, who may not take the time to decrypt code written with a significantly different style.
@BustACode we just want to help you with your coding style.
Python was written for readability, so yes PEP 8 is good to follow. See ...
https://www.python.org/dev/peps/pep-0008/
Also, don't use unnecessary imports. They just clutter your code and are somewhat misleading.
Remember ...
import this
The Ninja IDE helps you with PEP 8 style:
http://www.ninja-ide.org/
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