implementation of the revenue sharing model #2

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dachary merged 2 commits from dachary/accountant:wip-share into master 2022-08-16 01:12:08 -04:00
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# py.test-3 --log-cli-level=DEBUG -v -k test_share_income share.py
EXPENSE = 1
#
# For a given income, return a list of the members who will not be paid in full
# and how much will remain in their expense balance.
#
# See https://forum.hostea.org/t/decision-revenue-sharing-model/92
#
# The members list given in argument is a list of pairs of
#
# ['member', EXPENSE]
#
# Where 'member' is the unique id of the member and EXPENSE is a
# positive integer.
#
# The income argument must be an integer lower or equal to the sum of
# all members EXPENSE.
#
# The returned member list is a subset of the members list given in
# argument and only contains the members that cannot be paid in full
# with the income given in argument. For instance:
#
# share_income(3, [['a', 2], ['b', 2]]) == [['b', 1]]
#
# Means that with an income of 3, only member 'a' can be paid in full and
# the remaining expense for member 'b' is 1.
#
def share_income(income, members):
if income <= 0:
#
# Recursion termination: no more income to share, none of the
# remaining members get anything
#
return members
if income <= len(members):
#
# The income is an integer number: when it cannot be divided
# among members (2/3 == 0), some of them get 1 and the others
# get nothing. If the income is consistently very low some
# members to never get paid but the amount of money involved
# would then be so low that no member would care.
#
share = 1
count = income
else:
#
# The share that each member will get at this stage of the
# recursion is the lowest expense. If there is not enough
# income to give each member the lowest expense, it is divided
# evenly.
#
share = min(int(income / len(members)), min(members, key=lambda m: m[EXPENSE])[EXPENSE])
count = len(members)
#
# remaining is the list of members that cannot be paid in full
# with the share.
#
remaining = []
for i in range(count):
m = members[i]
m[EXPENSE] -= share
if m[EXPENSE] > 0:
#
# If the member needs more to be paid in full, they are
# elligible to participate in the next recursion round.
#
remaining.append(m)
#
# This is the border case when income <= len(members): some
# members cannot get their share because there is not enough
# income to provide an equal share of 1 to everyone and were
# excluded from the loop above. They are added because they could
# not be paid in full.
#
remaining.extend(members[count:])
#
# Now that each member got an equal share (except for the border
# case above) recurse, but only with the members that expect to be
# paid more.
#
return share_income(income - share * count, remaining)
def test_share_income():
assert share_income(1, [['a', 1], ['b', 1]]) == [['b', 1]]
assert share_income(5, [['a', 10], ['b', 2]]) == [['a', 7]]
assert share_income(5, [['a', 2], ['b', 10], ['c', 1]]) == [['b', 8]]
assert share_income(5, [['a', 2], ['b', 10], ['c', 1], ['d', 40]]) == [['b', 9], ['d', 39]]
assert share_income(5, [['a', 2], ['b', 10], ['c', 1], ['d', 40], ['e', 3]]) == [['a', 1], ['b', 9], ['d', 39], ['e', 2]]
assert share_income(5, [['a', 2], ['b', 10], ['c', 1], ['d', 40], ['e', 3], ['f', 1]]) == [['a', 1], ['b', 9], ['d', 39], ['e', 2], ['f', 1]]