您的位置:首页 > 编程语言 > Python开发

python数据类型-字典的练习

2018-03-08 19:02 645 查看


自己做的:
dic = {'k1':'v1','k2':'v2','k3':'v3','k4':'v4'}
#1.循环所有k,value,key和value
# for k in dic:
# print(k)
# for v in dic:
# print(dic[v])
# for i in dic:
# print(i,dic[i])
#2.添加:
# dic['k5']='v5'
# print(dic)
#3.删除k1,v1
# dic.pop('k1')
# print(dic)
#4.删除k2
# print(dic)
# for k in dic :
# if k == 'k2':
# dic['k2']=''
# print(dic)
# else:
# print('None')
#5.'k2'对应的值
# print(dic['k2'])
# print(dic.get('k2'))
#6.获取k3值
# for k in dic :
# if k == 'k3':
# print(dic['k3'])
# else:
# print('None')
#7.
# dic2 = {'k1':'v111','a':'b'}
# dic2.update({'k2':'v2','k3':'v3'})
#
# print(dic2)
#8.
lis = [['k',['qwe',20,{'k1':[
4000
'tt'
,3,'1']},89]]]
#1,
# lis[0][1][2]['k1'][0]='TT'
# print(lis)
# 2.
# lis[0][1][2].update({'k1':['TT',3,'1']})
# print(lis)
# lis[0][1][2]['k1'][0]=(lis[0][1][2]['k1'][0]).upper()
# print(lis)
#2.
# lis[0][1][2]['k1'][1] = '100'
# print(lis)
#3.
# lis[0][1][2]['k1'][2] = 101
# print(lis)
# #11.
# li = [1,2,3,'a','b',4,'c']
# dic3 = {}
# dic3.update({'k1':[]})
# for index,i in enumerate(li) :
# if index%2 != 0:
# dic3['k1'].append(i)
# print(dic3)

内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: