Question

我为我的大学CS课分配了这个问题。总的来说，它在字典和集合以及编码方面还是新的。所以我正在为将userInput分配给字典的部分寻找一种更pythonic的方式。这可以满足我的需求，但除此之外，它会变得乏味。是的，我从CodeAcademy看到了与此功能类似的其他问题，但这些问题没有我的问题的答案。预先感谢您的任何投入。保重。

gradeBook = {
  "student1" : {
  "name": "Andy", 
  "homework": [90.0, 85.0, 78.0, 92.0], 
  "quizzes": [85.0, 87.0, 90.0, 75.0, 95.0], 
  "tests": [85.0, 65.0]
  },
"student2" : {
  "name": "Brad", 
  "homework": [88.0, 78.0, 86.0, 99.0], 
  "quizzes": [88.0, 82.0, 100.0, 74.0, 82.0], 
  "tests": [90.0, 92.0]
  }, 
"student3" : {
  "name": "Billy",
  "homework": [81.0, 79.0, 80.0, 50.0], 
  "quizzes": [80.0, 78.0, 95.0, 80.0, 84.0], 
  "tests": [95.0, 93.0]
  }}
for k, v in gradeBook.items():
  print()
  for k in v:
    print(k + ':', v[k])
print()

userInput = input("Which students' average would you like to view: \nAndy, Brad, or Billy? \n")
print()

#assigning userInput to a students dictionary
if userInput == "Andy":
  userInput1 = gradeBook["student1"]["homework"]
  userInput2 = gradeBook["student1"]["quizzes"]
  userInput3 = gradeBook["student1"]["tests"]
elif userInput == "Brad":
  userInput1 = gradeBook["student2"]["homework"]
  userInput2 = gradeBook["student2"]["quizzes"]
  userInput3 = gradeBook["student2"]["tests"]
else:
  userInput1 = gradeBook["student3"]["homework"]
  userInput2 = gradeBook["student3"]["quizzes"]
  userInput3 = gradeBook["student3"]["tests"]

def get_average(userInput):
  averageHW = sum(userInput1) / len(userInput1)
  averageQ = sum(userInput2) / len(userInput2)
  averageT = sum(userInput3) / len(userInput3)
  print(averageHW, averageQ, averageT) #normal average
  wAverageHW = averageHW * 0.10
  wAverageQ = averageQ * 0.30
  wAverageT = averageT * 0.60
  print(wAverageHW, wAverageQ, wAverageT) #weighted average

get_average(userInput)

Answer 1

如果您要按学生姓名进行查找，则需要重新输入字典的密码。

gradebookByStudent = {gradebook[k][‘name’]: gradebook[k] for k in gradebook}

#get your student name via user input 

print(‘HW average is: ‘ + mean(gradebookByStudent[userInput][‘homework’]))

这使您可以查询O（log（n））最坏的情况，即遍历字典时的平均值O（1）与O（n），可节省大量时间。

Answer 2

您可以使用元组拆包法来遍历字典 for key, student in gradebook.items(): If student ['name'] == userinput: getaverage(student)

类似这样的东西更像pythonic，使您可以同时浏览字典的键和值部分

将userInput分配给字典的更多pythonic方法？

2 个答案: