同时对数据做转换和换算

1.如果需要对数据进行换算，例如sum(),min(),max(),但是首先得对数据做转换和筛选，，有一个很优雅的方式解决

score= [5, 7, 18, 19, 29, 39, 40, 41, 51, 57]
s=sum(x*x for x in score)
print(s)

2.实际应用于一些题目测评答案分值的计算

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': '常规型(C)', 'score': [5, 7, 18, 19, 29, 39, 40, 41, 51, 57], 'max': 10}, {'_id': '社会型(S)', 'score': [1, 12, 15, 26, 27, 37, 45, 52, 53, 59], 'max': 10}, {'_id': '企业型(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': '现实型(R)', 'score': [2, 13, 14, 22, 23, 36, 43, 44, 47, 48], 'max': 10}, {'_id': '艺术型(A)', 'score': [4, 9, 10, 17, 32, 33, 34, 49, 50, 54], 'max': 10}]
#需求是对testQuestions列表中score列表中题号找到answer中对应的值，计算出_id 的总分数，说明白点就是算_id 的分值，现在我的score字段只是存着题号，要根据题号，去answer中找到分数累加，
#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
#输出结果testQuestions：{'_id': '常规型(C)', 'score': 2, 'max': 10}, {'_id': '社会型(S)', 'score': 3, 'max': 10}, {'_id': '企业型(E)', 'score': 6, 'max': 10}, {'_id': '研究型(I)', 'score': 7, 'max': 10}, {'_id': '现实型(R)', 'score': 6, 'max': 10}, {'_id': '艺术型(A)', 'score': 6, 'max': 10}]