# 作业 #写代码：有如下字典 # 按照要求实现每一个功能 # dict = {"k1":"v1","k2":"v2","k3":"v3"} # 1、请循环遍历出所有的key
dic = {"k1":"v1","k2":"v2","k3":"v3"} re = dic.keys() for i in re: print(i,end=' ') # # 2, 请循环遍历出所有的value
dic = {"k1":"v1","k2":"v2","k3":"v3"} re = dic.values() for i in re: print(i,end=' ') # 3、请循环遍历出所有的key和value
dic = {"k1":"v1","k2":"v2","k3":"v3"} re = dic.items() for k,v in re: print(k,v,end=' ') # 4、请在字典中增加一个键值对,"k4":"v4"，输出添加后的字典
dic = {"k1":"v1","k2":"v2","k3":"v3"} dic['k4'] = 'v4'
print(dic) # 5、请删除字典中键值对"k1":"v1",并输出删除后的结果
dic = {"k1":"v1","k2":"v2","k3":"v3"} dic.pop('k1') print(dic) # 6、请删除字典中键"k5"对应的键值对，如果字典中不存在键"k5",则不报错，返回None
dic = {"k1":"v1","k2":"v2","k3":"v3"} re = dic.pop('k5',None) print(re) # 7、请获取字典中"k2"对应的值
dic = {"k1":"v1","k2":"v2","k3":"v3"} print(dic['k2']) # 8、请获取字典中"k6"对应的值,如果不存在，则不报错，并且让其返回None。
dic = {"k1":"v1","k2":"v2","k3":"v3"} re = dic.pop('k6',None) print(re) # 9、现有dict2 = {"k1":"v11","a":"b"},通过一行操作使dict2 = {"k1":"v1","k2":"v2","k3":"v3","a":"b"}
dict2 = {"k1":"v11","a":"b"} dic = {"k1":"v1","k2":"v2","k3":"v3"} dict2.update(dic) print(dict2) # 10.现有一个列表li = [1,2,3,'a',4,'c'], # 有一个字典(此字典是动态生成的，你并不知道他里面有多少键值对，所以用dic={}模拟字典； # 现在需要完成这样的操作：如果该字典没有"k1"这个键，那就创建 # 这个"k1"键和对应的值(该键对应的值为空列表)，并将列表li中的索引位为奇数对应的元素， # 添加到"k1"这个键对应的空列表中。如果该字典中有"k1"这个键，且k1对应的value是列表类型。 # 那就将该列表li中的索引位为奇数对应的元素，添加到"k1"，这个键对应的值中。
dic = {} li = [1,2,3,'a',4,'c'] if 'k1' not in dic: dic.setdefault('k1',[]) for i in li: if li.index(i)%2 == 1: dic['k1'].append(i) else: if type(dic['k1']) == type([]): for i in li: if li.index(i)%2 == 1: dic['k1'].append(i) else: print('字典k1,的value值类型不是列表，无法添加') print(dic) # 12现在有如下字典，完成一下需求：
dic = { 'name':'汪峰', 'age':48, 'wife':[{'name':'国际章','age':38}], 'children':{'girl_first':'小苹果','girl_second':'小怡','girl_three':'顶顶'} } # 1. 获取汪峰的名字。 # 2.获取这个字典：{'name':'国际章','age':38}。 # 3. 获取汪峰妻子的名字。 # 4. 获取汪峰的第三个孩子名字。 # 1.
print(dic['name']) # 2
print(dic['wife']) # 3
print(dic['wife'][0]['name']) # 4
print(dic['children']['girl_three']) #