2 -9 字典练习题解答

解答

dic = {'k1':'v1','k2':'v2','k3':'v3'}
#1
for k in dic:
    print(k)
"""
k1
k2
k3
"""
#2
for v in dic.values():
    print(v)
"""
v1
v2
v3
"""
#3
for k , v in dic.items():
    print(k,v)
"""
k1 v1
k2 v2
k3 v3
"""
#4
dic['k4']='v4'
print(dic)
#{'k1': 'v1', 'k2': 'v2', 'k3': 'v3', 'k4': 'v4'}

#5
dic.pop("k1")
print(dic)
#{'k2': 'v2', 'k3': 'v3', 'k4': 'v4'}

#6
print(dic.pop('k5',None))
#None

#7
print(dic['k2'])
#v2

#8
print(dic.get('k6'))
#None

#9
dic = {'k1':'v1','k2':'v2','k3':'v3'}
dic2 = {'k1':'v111','a':'b'}
dic2.update(dic)
print(dic2)
#{'k1': 'v1', 'a': 'b', 'k2': 'v2', 'k3': 'v3'}

#10 -1
lis = [['k',['qwe',20,{'k1':['tt',3,1]},89],'ab']]
lis[0][1][2]['k1'][0] = lis[0][1][2].get('k1')[0].upper()#swapcase()
print(lis)
#[['k', ['qwe', 20, {'k1': ['TT', 3, 1]}, 89], 'ab']]

lis[0][1][2]['k1'][0] = 'TT'
print(lis)
#[['k', ['qwe', 20, {'k1': ['TT', 3, 1]}, 89], 'ab']]

#10 -2
lis[0][1][2]['k1'][1] = "100"
print(lis)
#[['k', ['qwe', 20, {'k1': ['tt', '100', 1]}, 89], 'ab']]

lis[0][1][2].get('k1')[1] = str(lis[0][1][2].get('k1')[1]+97)
print(lis)
#[['k', ['qwe', 20, {'k1': ['tt', '100', 1]}, 89], 'ab']]

#10 -3
lis[0][1][2].get('k1')[2] = lis[0][1][2].get("k1")[2]+100
print(lis)
#[['k', ['qwe', 20, {'k1': ['tt', 3, 101]}, 89], 'ab']]

lis[0][1][2].get("k1")[2] = 101
print(lis)
#[['k', ['qwe', 20, {'k1': ['tt', 3, 101]}, 89], 'ab']]

#11
li = [1,2,3,'a','b',4,'c']
#dic = {}#没有K1
#dic = {'k1':[]}#有k1
dic = {'k1':['e','f']}#有k1
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"]) == list:
        for i in li:
            if li.index(i) % 2 == 0:
                dic['k1'].append(i)
print(dic)
#{'k1': ['e', 'f', 1, 3, 'b', 'c']}

 

转载于:https://www.cnblogs.com/Mobai-c/p/10144904.html