测试工程师的一些面试题目（python）和总结

 1 def conversion(n,the_dict):
 2     for key,value in n.items():
 3         #如果value的值是字典，递归
 4         if isinstance(value, dict):
 5             conversion(value,the_dict)
 6         else:
 7            the_dict[key] = value
 8     return the_dict
 9 #实际上JSON对象应该要先用 json.loads(r)解码，转换为python的字典
10 #但是这里如果这样直接输入这种格式的数据，会默认为字典对象，就不解码了
11 r = {"a":"aa","b":"bb","c":{"d":"dd","e":"ee"}}
12 
13 the_dict = dict()
14 print(conversion(r,the_dict))

1 def factorial(n):
2     if n == 1:
3         return 1
4     return n*factorial(n-1)
5 
6 print(factorial(5))

1 the_str = ‘12,33,44‘
2 
3 #分割字符串，返回的是一个列表
4 the_list = the_str.split(‘,‘)
5 print(the_list)
6 
7 #将列表合并为字符串，‘,‘可以指定不同元素之间的分隔符
8 print(‘,‘.join(the_list))

 1 the_dict = {‘a‘: ‘aa‘, ‘b‘: ‘bb‘, ‘d‘: ‘dd‘, ‘e‘: ‘ee‘}
 2 
 3 #迭代字典的key值和value值
 4 for key,value in the_dict.items():
 5     print(key)
 6     print(value)
 7 #迭代字典的key值
 8 for key1 in the_dict.keys():
 9     print(key1)
10 
11 #迭代字典的value值
12 for value1 in the_dict.values():
13     print(value1)

 1 def student_achie(A):
 2     for i in range(len(A)):
 3         key = i
 4         for j in range(i+1,len(A)):
 5             if A[key].split(‘:‘)[1] < A[j].split(‘:‘)[1]:
 6                 key = j
 7         A[key],A[i] = A[i],A[key]
 8     return A
 9 
10 #学生姓名及学生成绩
11 stu = [‘张三:20‘,‘李四:70‘,‘王五:88‘,‘李六:40‘,‘王吉:55.5‘,]
12 
13 print(student_achie(stu))

1 #计算学生总成绩
2 def total_achie(A):
3     if len(A) == 1:
4         return float(A[0].split(‘:‘)[1])
5     return float(A.pop().split(‘:‘)[1]) + total_achie(A)
6 #学生姓名及学生成绩
7 stu = [‘张三:20‘,‘李四:70‘,‘王五:88‘,‘李六:40‘,‘王吉:55.5‘,]
8 
9 print(total_achie(stu))

 1 def fibonacci(n):
 2     the_list = []
 3     for i in range(n):
 4         if i == 0:
 5             the_list.append(1)
 6         elif i == 1:
 7             the_list.append(1)
 8         else:
 9             the_list.append(the_list[i-1] + the_list[i-2])
10     return the_list
11 #输出[1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
12 print(fibonacci(10))