我有一本這樣的字典:d = {'a':[{'a1':1},{'a2':5},{'a3':4}], 'b':[{'b1':0},{'b2':1},{'b3':2}], 'c':[{'c1':1},{'c2':2}]}我想按每個list值的總和(字典中每個項目的值)對它進行排序,以便得出:r = [('a', (10, [{'a1':1},{'a2':5},{'a3':4}])),('b', (3, [{'b1':0},{'b2':1},{'b3':2}])),# 'b' and 'c' have sum of '3', so they tied here('c', (3, [{'c1':1},{'c2':2}]))]我可以天真地做到這一點。我想知道如何以更Python化的方式完成此任務。我已經嘗試過了,但是由于明顯的原因而無法正常工作:sorted(sum(d.values()), key=d.get, reverse=True)
1 回答

POPMUISE
TA貢獻1765條經驗 獲得超5個贊
您可以嘗試以下方法:
d = {'a':[{'a1':1},{'a2':5},{'a3':4}], 'b':[{'b1':0},{'b2':1},{'b3':2}], 'c':[{'c1':1},{'c2':2}]}
new_d = {a:(sum(list(i.values())[0] for i in b), b) for a, b in d.items()}
final_result = sorted(new_d.items(), key=lambda x:x[-1][0], reverse=True)
輸出:
('a', (10, [{'a1': 1}, {'a2': 5}, {'a3': 4}])), ('c', (3, [{'c1': 1}, {'c2': 2}])), ('b', (3, [{'b1': 0}, {'b2': 1}, {'b3': 2}]))]
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