我正在寻找一种算法来分割一定数量的块之间的数字。 真实的例子是:根据任务总数为所有工人分配初始容量。
例如,如果有3个人和6个任务。每个人都有初始容量
= 6 / 3 = 2
我想出了一个实现。但是,我不确定它是实现它的最佳方式。
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))
输出:
[3, 3]
[2, 2, 3]
[3, 4, 4]
[3, 3, 4, 4, 4]
这个还有其他有效方法吗?
答案 0 :(得分:0)
也许:
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
答案 1 :(得分:0)
这个怎么样?
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
答案 2 :(得分:0)
我的解决方案是:
def cap_distr(ts, ps):
l = [ts // ps for _ in range(ps)]
for i in range(ts % ps):
l[i] += 1
return l
试验:
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:如果你不介意导入numpy,它甚至会缩短一行:
from numpy import ones
def cap_distr(ts, ps):
arr = ones(ps) * (ts // ps)
arr[:ts % ps] += 1
return arr.astype(int)