我有兩個字典 Dict1 和 Dict2。Dict1中每個鍵的值如下所示：Dict1 = { D1:[ C1, C2], D2: [C1], D3: [C2, C3], D4: [C3]}但是 Dict2 由來自 Dict1 的鍵組成。Dict 2 中的值是由元組組成的列表，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)])我有興趣創(chuàng)建 Dict3，它由 Dict1 中的鍵和列表中元組的第一個元素的值組成，這些值是 Dict2 中的值，如下所示。您可以忽略元組的第二個元素。例如，C1 將 [W1, W2, W3] 作為第一元素，而 C2 將 [W3, W4, W5] 作為第一元素。由于Dict1中的D1是D1：[C1，C2]，我想將C1和C2的列表組合為[W1，W2，W3，W4，W5]。所以最終期望的輸出是：Dict3= { D1: [W1, W2, W3, W4, W5], D2:[W1, W2, W3], D3:[W3, W4, W5, W6, W1, W2 ] }如何從Dict1和Dict2獲取Dict 3？這是我的代碼： 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)我得到的當前輸出是D1 ['W1', 'W2', 'W3']D1 ['W3', 'W4', 'W5']D2 ['W1', 'W2', 'W3']D3 ['W3', 'W4', 'W5']D3 ['W6', 'W1', 'W2']D4 ['W6', 'W1', 'W2']我想要的輸出應(yīng)該是像字典一樣： Dict3 = { 'D1':['W1', 'W2', 'W3', 'W4', 'W5'], 'D2': ['W1', 'W2', 'W3'], 'D3': ['W3', 'W4', 'W5', 'W6', 'W1', 'W2' ], 'D4': ['W6', 'W1', 'W2']}
3 回答
飲歌長嘯
TA貢獻1951條經(jīng)驗 獲得超3個贊
您不需要循環(huán)即可從中獲取項目Dict2。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)
慕森王
TA貢獻1777條經(jīng)驗 獲得超3個贊
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 = {k: [ii for i in v for ii, _ in Dict2[i]] for k, v in Dict1.items()}
print(Dict3)
印刷：
{'D1': ['W1', 'W2', 'W3', 'W3', 'W4', 'W5'], 'D2': ['W1', 'W2', 'W3'], 'D3': ['W3', 'W4', 'W5', 'W6', 'W1', 'W2'], 'D4': ['W6', 'W1', 'W2']}
犯罪嫌疑人X
TA貢獻2080條經(jīng)驗 獲得超4個贊
我不確定是否有更快的解決方案，但它可以滿足您的需求：
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 = {k: [t[0] for e in v for t in dict2[e]] for k, v in dict1.items()}
print(dict3)
輸出：
{'D1': ['W1', 'W2', 'W3', 'W3', 'W4', 'W5'], 'D2': ['W1', 'W2', 'W3'], 'D3': ['W3', 'W4', 'W5', 'W6', 'W1', 'W2'], 'D4': ['W6', 'W1', 'W2']}
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