Question

共有三个词典： trucks_dic，items_dic和loaded_trucks

如何从“卡车和物品字典”的键和值中替换卡车的键和值？

truck_dic = {6: [21.0, 7.0, 7.0],
             7: [23.0, 7.0, 7.0],
             8: [27.0, 7.0, 7.0],
             9: [20.0, 7.0, 7.0],
            10: [24.0, 7.0, 7.0],
            11: [28.0, 8.0, 8.0],
            12: [32.0, 8.0, 8.0],
            13: [32.0, 9.0, 10.0],
            14: [32.0, 9.0, 10.0],
            15: [20.0, 8.0, 8.0],
            16: [20.0, 8.0, 8.0]}

Items_dic = {0: [4.6, 4.3, 4.3],
 1: [4.6, 4.3, 4.3],
 2: [6.0, 5.6, 9.0],
 3: [8.75, 5.6, 6.6],
 4: [6.0, 5.16, 6.6],
 5: [6.0, 5.6, 9.0],
 6: [8.75, 5.6, 6.6],
 7: [4.6, 4.3, 4.3],
 8: [6.0, 5.6, 9.0],
 9: [8.75, 5.6, 6.6],
 10: [6.0, 5.16, 6.6]}

loaded_trucks = {'23.0x7.0x7.0': ['4.60x4.30x4.30, 4.60x4.30x4.30, 
                                  6.60x6.00x5.16, 6.60x6.00x5.16'],
                 '27.0x7.0x7.0': ['4.60x4.30x4.30, 9.00x6.00x5.60, 
                                   8.75x6.60x5.60'],
                 '20.0x8.0x8.0': ['9.00x6.00x5.60, 8.75x6.60x5.60'],
                 '21.0x7.0x7.0': ['9.00x6.00x5.60, 8.75x6.60x5.60']}

它给了

loaded_trucks = {'7': ['0', '7', '4', '10'],
                 '8': ['1', '5', '9'],
                 '16': ['2', '3'],
                 '6': ['6', '8']}

PS：loaded_trucks中的尺寸混乱

Answer 1

truck_dic = {6: [21.0, 7.0, 7.0],
             7: [23.0, 7.0, 7.0],
             8: [27.0, 7.0, 7.0],
             9: [20.0, 7.0, 7.0],
            10: [24.0, 7.0, 7.0],
            11: [28.0, 8.0, 8.0],
            12: [32.0, 8.0, 8.0],
            13: [32.0, 9.0, 10.0],
            14: [32.0, 9.0, 10.0],
            15: [20.0, 8.0, 8.0],
            16: [20.0, 8.0, 8.0]}

Items_dic = {0: [4.6, 4.3, 4.3],
 1: [4.6, 4.3, 4.3],
 2: [6.0, 5.6, 9.0],
 3: [8.75, 5.6, 6.6],
 4: [6.0, 5.16, 6.6],
 5: [6.0, 5.6, 9.0],
 6: [8.75, 5.6, 6.6],
 7: [4.6, 4.3, 4.3],
 8: [6.0, 5.6, 9.0],
 9: [8.75, 5.6, 6.6],
 10: [6.0, 5.16, 6.6]}

loaded_trucks = {'23.0x7.0x7.0': ['4.60x4.30x4.30, 4.60x4.30x4.30, 6.60x6.00x5.16, 6.60x6.00x5.16'],
                 '27.0x7.0x7.0': ['4.60x4.30x4.30, 9.00x6.00x5.60, 8.75x6.60x5.60'],
                 '20.0x8.0x8.0': ['9.00x6.00x5.60, 8.75x6.60x5.60'],
                 '21.0x7.0x7.0': ['9.00x6.00x5.60, 8.75x6.60x5.60']}

# Build dimensions to truck and item mappings
truck_dict = {tuple(sorted(v)): str(k) for k, v in truck_dic.items()}

items_dict = dict()
for k, v in Items_dic.items():
    items_dict.setdefault(tuple(sorted(v)), []).append(str(k))

# Generate result
result = dict()
def standardize(s):
    return tuple(sorted(float(v) for v in s.split('x')))

for k, v in loaded_trucks.items():
    # Note that the values in loaded_trucks are lists of a ONE str
    # and not a list of strs of dimensions.
    items = v[0].split(',')
    items = [i.strip() for i in items]

    print(items)
    print([standardize(i) for i in items])
    k = standardize(k)
    result[truck_dict[k]] = [items_dict[standardize(i)].pop() for i in items]

输出result是

{'7': ['7', '1', '10', '4'],
 '8': ['0', '8', '9'],
 '16': ['5', '6'],
 '6': ['2', '3']}

请注意，相同尺寸的项目可以互换。例如，第5项和第8项具有相同的尺寸，因此可以互换。

此外，我们不在乎单个尺寸的顺序是否重新排序，因为这仅表示该项目已旋转。例如，尺寸为“ 4.60x4.30x4.30”的商品被视为与尺寸为“ 4.30x4.30x4.60”的另一商品相同。

最后，每个项目仅分配了一次，但是我们假设原始loaded_trucks词典中的项目都在Items_dic中进行了说明，因此不会出现短缺或剩余的问题。列表将产生错误。

从多个字典中检索数据

1 个答案: