在python中从一个字典到另一个字典的迭代

Question

我有两本字典Dict1和DICT2。Dict1中每个键的值如下所示：

Dict1 = { D1:[ C1, C2], D2: [C1], D3: [C2, C3], D4: [C3]}

但是Dict2由键组成，这些键是DICT1中的值。DICT2中的值是由元组组成的列表，如下所示。

Dict2 = { C1: [ (W1, 1 ), (W2, 5),  (W3, 4) ], C2: [ (W3, 2), (W4, 6) ,  (W5, 7) ],  C3:[ (W6,0),( W1,4), ( W2, 8)], C4: [(W7,3 ),(W8, 8)])

Dict3= { D1: [W1, W2, W3, W4, W5], D2:[W1, W2, W3], D3:[W3, W4, W5, W6, W1, W2 ] }

        Dict1 = { 'D1':[ 'C1', 'C2'], 'D2': ['C1'], 'D3': ['C2', 'C3'], 'D4': ['C3']}

        Dict2 = { 'C1': [ ('W1', 1 ), ('W2', 5),  ('W3', 4) ], 'C2': [ ('W3', 2), ('W4', 6) ,  ('W5', 7) ],  'C3':[ ('W6',0),( 'W1',4), ( 'W2', 8)], 'C4': [('W7',3 ),('W8', 8)]}


        for d, c_list in Dict1.items():
            for c in c_list:
                for cw, w_tup_list in Dict2.items():
                    if (c == cw):
                        w_list = []
                        for w_tup in w_tup_list:
                           w_list.append(w_tup[0])
                        print (d, w_list)

我得到的当前输出是

D1 ['W1', 'W2', 'W3']
D1 ['W3', 'W4', 'W5']
D2 ['W1', 'W2', 'W3']
D3 ['W3', 'W4', 'W5']
D3 ['W6', 'W1', 'W2']
D4 ['W6', 'W1', 'W2']

我想要的输出应该是字典式的：

    Dict3 = { 'D1':['W1', 'W2', 'W3', 'W4', 'W5'], 'D2': ['W1', 'W2', 'W3'], 'D3': ['W3', 'W4', 'W5', 'W6', 'W1', 'W2' ], 'D4': ['W6', 'W1', 'W2']}

Answer 1

从dict2获取项不需要循环。只需使用dict2[c]。

Dict1 = { 'D1':[ 'C1', 'C2'], 'D2': ['C1'], 'D3': ['C2', 'C3'], 'D4': ['C3']}
Dict2 = { 'C1': [ ('W1', 1 ), ('W2', 5),  ('W3', 4) ], 'C2': [ ('W3', 2), ('W4', 6) ,  ('W5', 7) ],  'C3':[ ('W6',0),( 'W1',4), ( 'W2', 8)], 'C4': [('W7',3 ),('W8', 8)]}
Dict3 = {}

for d, c_list in Dict1.items():
    for c in c_list:
        w = [item[0] for item in Dict2[c]]
        Dict3[d] = Dict3.get(d, []) + w

print(Dict3)