思考题
一行版：
[dict(zip(attributes, value)) for value in values]

循环版：
l = []
for value in values:
    d = {}
    for i in range(3):
        d[attributes[i]] = value[i]
    l.append(d)

# Ask
attributes = ['name', 'dob', 'gender']
values = [['jason', '2000-01-01', 'male'],
          ['mike', '1999-01-01', 'male'],
          ['nancy', '2001-02-01', 'female']]
# Answers1
lis = []
for i in values:
    print(i)
    print(dict(zip(attributes, i)))
    print(lis.append(dict(zip(attributes, i))))
# Answers2
[dict(zip(attributes, i)) for i in values]
# Answers3
list(map(lambda x: dict(zip(attributes, x)), values))
# Answers4
lis = []
for i in values:
    dic = {}
    for n, j in enumerate(i):
        dic[attributes[n]] = j
    lis.append(dic)
print(lis)

我没有想到用多行条件循环语句实现的方法。先留言，然后看别人怎么做的。

attributes = ['name', 'dob', 'gender']
values = [
['jason', '2000-01-01', 'male'],
['mike', '1999-01-01', 'male'],
['nancy', '2001-02-01', 'female']
]

# 多行代码版
list1 = [] # 建立空列表
for value in values: # 对值列表进行循环
    dict1 = {} # 建立空字典，方便后续字典键值对存储
    for index,key in enumerate(attributes): # 遍历键列表，返回元素与索引
        dict1[key] = value[index] # 键与值配对，组装成字典元素
    list1.append(dict1) # 将新字典添加到里列表中
print(list1) # 打印显示出完整列表

# 一行代码版
[{key:value[index] for index,key in enumerate(attributes)}for value in values]
# 最外层[]与上面“list1=[]”和“list1.append(dict1)”等价
# "{key:value[index] for index,key in enumerate(attributes)}"与上面"for value in values:"内代码等价
# 体会：先梳理逻辑，写出多行代码版，再回溯写出一行代码版，体现出Python的简洁优美

# 多行版本
combined = []
for value in values:
    temp_dict = {}
    for index, attr in enumerate(attributes):
        temp_dict[attr] = value[index]
    combined.append(temp_dict)

combined

# 单行版本
[{attr: value[index] for index, attr in enumerate(attributes)} for value in values]

attributes = ['name','dob','gender']
values = [['jason','2000-1-01','male'],['mike','1999-01-1','male'],['nancy','201-02-01','female']]
output = []
for v in values:
    dst = {}
    for k in range(len(v)):
        dst[attributes[k]] = v[k]
    output.append(dst)
print(output)

Python中没有switch，但是可以用字典来代替。
#不调用函数
level = {0:'red', 1:'yellow', 2:'green'}
try:
    level(id)
except:
    pass

#需要调用函数
def delete (id):
    '''delete the movie that id is 0'''
………
    ………
………
    ………
opt = {0:delete, 1:warning, 2:allow}

opt[ id ]( id )

通过反复查看文章内容跟参考留言中的评论，理解后输出。
虽然不是自己写的，也算是对自己的一种鼓励，至少清楚了逻辑。
#一行
print ([dict(zip(attributes,value))for value in values])

#多行输出
list1 = [] #建议空列表
for value in values: #对值列表进行循环
    dict1 = {} #简历空字典，后续对字典键值对存储
    for index ,key in enumerate(attributes):#遍历建列表，返回索引与元素
        dict1[key] = value[index] #键与值匹配对，组装成字典元素
    list1.append(dict1) #将字典添加到列表中
print(list1)

# general.
expected_output = []
for v in values:
    expected_output.append(dict(zip(attributes, v)))
print(expected_output)

# one line
print([dict(zip(attributes, v)) for v in values])

# 多行
output_dist = []
for v in values:
    dict_at = {}
    for i in range(len(attributes)):
        dict_at[attributes[i]] = v[i]
    output_dist.append(dict_at)
print(output_dist)

# 一行
print( [{ attributes[i]: value[i] for i in range(len(attributes)) } for value in values])

一行: [dict(zip(attributes, value)) for value in values]
多行:
l = []
for value in values:
    d = {}
    for i,j in zip(attributes, value):
        d[i] = j
    l.append(d)

写个长的吧
def get_list_dicts(attribs,vals):
    result=[]
    for l in values:
        temp_dict={}
        if len(l)==len(attributes):
            for i in range(len(attributes)):
                temp_dict[attributes[i]]=l[i]
        else:
            print('values列表中第{}组内值的个数'
                  '需与属性列表内的个数保持一致！'.format(values.index(l)+1))
            return None
        result.append(temp_dict)
    return result