Question

比方说，我想将多个项目（n）分配到固定大小（x）的数组中。困难的部分是我必须使用灵活性数组来分发项目。

假设x = 4，n = 11和flexibility = [20, 20, 30, 30]与len(flexibility) == x一起。

我的问题是：如何使用f中定义的百分比将n个元素分布在长度等于x的数组中？

最后我想要的是这样的

n = 11
x = 4
flexibility = [20, 20, 30, 30]
distributed = distribute_elements_in_slots(n, x, flexibility)
print(distributed)
# distributed = [2, 2, 3, 4]

在弹性值相等的情况下，最终结果将取决于我们决定应用所有项目的规则。在前一种情况下，使用[2，2，3，4]和[2，2，4，3]最终结果会很好。

编辑：我想要的方法示例如下：

def distribute_elements_in_slots(n, x, flexibility=[25,25,25,25]):
    element_in_slots = []

    element_per_percentage = x / 100

    for i in range(x):
        element_in_slots.append(round(slots_per_point_percentage * flexibility[i])

编辑2 ：我找到的解决方案之一是：

def distribute_elements_in_slots(n, x, flexibility=[25,25,25,25]):
    element_in_slots = [f * n / 100 for f in flexibility]

    carry = 0
    for i in range(len(element_in_slots)):
        element = element_in_slots[i] + carry
        element_in_slot[i] = floor(element)
        carry = element- floor(element)

    if np.sum(element_in_slots) < n:
        # Here the carry is almost 1
        max_index = element_in_slots.index(max(flexibiliyt))
        appointments_per_slot[max_index] = appointments_per_slot[max_index] + 1

这将根据灵活性阵列几乎均匀地分配插槽。

Answer 1

您需要做的是根据数组中给定的百分比对数字11进行分割，因此最初它变为percentage * number(11)。然后我们得到余数，然后将其分配到您认为是最后一个元素的某个位置。

In [10]: [i*n/100 for i in f]
Out[10]: [2.2, 2.2, 3.3, 3.3]

In [11]: b=[i*n/100 for i in f]

In [12]: rem = sum(b) - sum(map(int,b))


In [13]: rem
Out[13]: 1.0

In [24]: b= list(map(int,b))

In [26]: b[-1] +=rem

In [27]: b
Out[27]: [2, 2, 3, 4.0]

希望有帮助。：）

Answer 2

就像阿尔宾·保罗（Albin Paul）一样，我们需要为每个老虎机百分比分配整数数量。剩余的食物需要分配，首先要分配最大的食物。

def distribute_elements_in_slots(total, slots, pct):
    # Compute proportional distribution by given percentages.
    distr = [total * pct[i] / 100 for i in range(slots)]
    # Truncate each position and store the difference in a new list.
    solid = [int(elem) for elem in distr]
    short = [distr[i] - solid[i] for i in range(slots)]
    print(distr)
    print(solid)
    print(short)

    # allocate leftovers
    leftover = int(round(sum(short)))
    print(leftover)
    # For each unallocated item,
    #   find the neediest slot, and put an extra there.
    for i in range(leftover):
        shortest = short.index(max(short))
        solid[shortest] += 1
        short[shortest] = 0
        print("Added 1 to slot", shortest)

    return solid


n = 11
x = 4
flexibility = [20, 20, 30, 30]
distributed = distribute_elements_in_slots(n, x, flexibility)
print(distributed)
# distributed = [2, 2, 3, 4]

输出：

[2.2, 2.2, 3.3, 3.3]
[2, 2, 3, 3]
[0.2, 0.2, 0.3, 0.3]
1
Added 1 to slot 2
[2, 2, 4, 3]

根据百分比分配元素

2 个答案: