[英]Algorithm for distributing a number between certain number of chunks
我正在尋找一種算法,用於在一定數量的塊之間拆分數字。 真正的例子是:根據任務總數為所有工作人員分配初始能力。
例如,如果有3個人和6個任務。 每個將具有初始容量= 6 / 3 = 2
我想出一個實現。 但是,我不確定這是否是實現它的最佳方法。
def capacity_distribution(task_size, people_size):
"""
It distributes initial capacity to each person.
"""
div = task_size // people_size
remainder = task_size % people_size
first_chunk_size = people_size - remainder
second_chunk_size = people_size - first_chunk_size
first_capacity_list = []
second_capacity_list = []
first_capacity_list = [div for _ in range(first_chunk_size)]
second_capacity_list = [div + 1 for _ in range(second_chunk_size)]
first_capacity_list.extend(second_capacity_list)
return first_capacity_list
print(capacity_distribution(6, 2))
print(capacity_distribution(7, 3))
print(capacity_distribution(11, 3))
print(capacity_distribution(18, 5))
輸出:
[3, 3]
[2, 2, 3]
[3, 4, 4]
[3, 3, 4, 4, 4]
還有其他有效的方法嗎?
也許:
def capacity_distribution(task_size, people_size):
each = math.floor(task_size/people_size) # everyone gets this many
extra = task_size % people_size # this many get 1 extra
distribution = [each for x in range(people_size)]
for x in range(people_size):
if x < extra:
distribution[x] += 1
return distribution
這個怎么樣?
def capacity_distribution(task_size, people_size):
modulus = task_size % people_size
div = task_size // people_size
dist = []
if modulus:
dist.append(div)
dist.extend(capacity_distribution(task_size-div, people_size-1))
else:
dist.extend([div]*people_size)
return dist
我的解決方案是:
def cap_distr(ts, ps):
l = [ts // ps for _ in range(ps)]
for i in range(ts % ps):
l[i] += 1
return l
測試:
for test in [(6, 2), (7, 3), (11, 3), (18, 5)]: print(cap_distr(*test))
[3, 3]
[3, 2, 2]
[4, 4, 3]
[4, 4, 4, 3, 3]
PS:如果您不介意導入numpy,它甚至會短一行:
from numpy import ones
def cap_distr(ts, ps):
arr = ones(ps) * (ts // ps)
arr[:ts % ps] += 1
return arr.astype(int)
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