Question

我有2个字典，一个将代码构建为bldg名称

Dct_1 = {'1': ['Bldg 1'], '2': ['Bldg 2'], '3': ['Bldg 3'], '4': ['Bldg 4'], '5': ['Bldg 5']...})

第二个将建筑代码与楼层和租户联系起来

Dct_2 = {'1': [('Floor 0', 'Ten 1'), ('Floor 1', 'Ten 2'), ('Floor 3', 'Ten 3')], '2': [('Floor 1', 'Ten A'), ('Floor 2', 'Ten B')...]}

我正在尝试编写一个函数来创建第3个dct，以将楼层和租户与bldg名称相关联。

bldg_floor_tenant_dct = {'Bldg 1': [('Floor 0', 'Ten 1'), ('Floor 1', 'Ten 2'), ('Floor 3', 'Ten 3')], 'Bldg 2': [('Floor 1', 'Ten A'), ('Floor 2', 'Ten B')] ...}

是我想要的。

我尝试过以下两个函数，第一个返回错误而第二个没有任何操作

bldg_floor_tenant_dct = {}
for facilities_code, building_name in Dct_1.iteritems():
    floors = Dct_2[facilities_code]
    for k(floor_num, floor_dept) in floors:
        b_data = bldg_floor_tenant_dct.get(building_name, [])
        b_data.append({floor_num: floor_num,
            dept: floor_dept})

print bldg_floor_tenant_dct

错误：

  File "<stdin>", line 3
SyntaxError: can't assign to function call

和第二个：

bldg_floor_tenant_dct = defaultdict(list)
for keys in Dct_1:
    if Dct_1.keys() == Dct_2.keys():
        bldg_floor_tenant_dct[Dct_1.values()].append(Dct_2.values())
        print bldg_floor_tenant_dct
        break

我缺少什么或者有更好的方式来写这个？

Answer 1

你可以使用字典理解。缩短Dct_1和Dct_2的演示：

>>> Dct_1 = {'1': ['Bldg 1'], '2': ['Bldg 2']}
>>> Dct_2 = {'1': [('Floor 0', 'Ten 1'), ('Floor 1', 'Ten 2'), ('Floor 3', 'Ten 3')], '2': [('Floor 1', 'Ten A'), ('Floor 2', 'Ten B')]}
>>> {v[0]:Dct_2[k] for k,v in Dct_1.items()}
{'Bldg 2': [('Floor 1', 'Ten A'), ('Floor 2', 'Ten B')], 'Bldg 1': [('Floor 0', 'Ten 1'), ('Floor 1', 'Ten 2'), ('Floor 3', 'Ten 3')]}

Answer 2

问题在于这一行：

for k(floor_num, floor_dept) in floors:

k(floor_num, floor_dept)看起来像是python的一个函数，你不能以这种方式为函数赋值。而是尝试

for floor_num, floor_dept in floors:

Answer 3

我不确定为什么，但你的解决方案似乎很复杂。我会这样写的：

Dct_1 = {'1': ['Bldg 1'], '2': ['Bldg 2'], '3': ['Bldg 3'], '4': ['Bldg 4'], '5': ['Bldg 5']}
Dct_2 = {'1': [('Floor 0', 'Ten 1'), ('Floor 1', 'Ten 2'), ('Floor 3', 'Ten 3')], '2': [('Floor 1', 'Ten A'), ('Floor 2', 'Ten B')]}

result = {}

for key, value in Dct_2.iteritems():
    for item in value:
        if key in Dct_1 and Dct_1[key]:
            result[item] = Dct_1[key]

print result

将2个词典合并为一个Python

3 个答案: