同時對資料做轉換和換算
阿新 • • 發佈：2018-11-16
1.如果需要對資料進行換算，例如sum(),min(),max(),但是首先得對資料做轉換和篩選，，有一個很優雅的方式解決
score= [5, 7, 18, 19, 29, 39, 40, 41, 51, 57] s=sum(x*x for x in score) print(s)
2.實際應用於一些題目測評答案分值的計算
answer={ "1":0, "2":1, "3":0, "4":1, "5":0, "6":1, "7":0, "8":1, "9":0, "10":1, "11":0,"12":1, "13":0, "14":1, "15":0, "16":1, "17":0, "18":1, "19":0, "20":1, "21":0, "22":1, "23":0, "24":1, "25":0, "26":1, "27":0, "28":1, "29":0, "30":1, "31":0, "32":1, "33":0, "34":1, "35":0, "36":1, "37":0, "38":1, "39":0, "40":1, "41":0, "42":1, "43":0, "44":1, "45":0, "46":1, "47":0, "48":1, "49":0, "50":1, "51":0, "52":1, "53":0, "54":1, "55":0, "56":1, "57":0, "58":1, "59":0, "60":1 } testQuestions= [{'_id': '常規型(C)', 'score': [5, 7, 18, 19, 29, 39, 40, 41, 51, 57], 'max': 10}, {'_id': '社會型(S)', 'score': [1, 12, 15, 26, 27, 37, 45, 52, 53, 59], 'max': 10}, {'_id': '企業型(E)', 'score': [3, 11, 16, 24, 25, 28, 35, 38, 46, 60], 'max': 10}, {'_id': '研究型(I)', 'score': [6, 8, 20, 21, 30, 31, 42, 55, 56, 58], 'max': 10}, {'_id': '現實型(R)', 'score': [2, 13, 14, 22, 23, 36, 43, 44, 47, 48], 'max': 10}, {'_id': '藝術型(A)', 'score': [4, 9, 10, 17, 32, 33, 34, 49, 50, 54], 'max': 10}] #需求是對testQuestions列表中score列表中題號找到answer中對應的值，計算出_id 的總分數，說明白點就是算_id 的分值，現在我的score欄位只是存著題號，要根據題號，去answer中找到分數累加， #我的方法阿是： for item in testQuestions: s=sum(answer[str(x)] for x in item["score"]) print(s) item["score"]=s #輸出結果testQuestions：{'_id': '常規型(C)', 'score': 2, 'max': 10}, {'_id': '社會型(S)', 'score': 3, 'max': 10}, {'_id': '企業型(E)', 'score': 6, 'max': 10}, {'_id': '研究型(I)', 'score': 7, 'max': 10}, {'_id': '現實型(R)', 'score': 6, 'max': 10}, {'_id': '藝術型(A)', 'score': 6, 'max': 10}]