字典中的键使用时必须满足一下两个条件:
1、每个键只能对应一个项,也就是说,一键对应多个值时不允许的(列表、元组和其他字典的容器对象除外)。当有键发生冲突时(即字典键重复赋值),取最后的赋值。
复制代码 代码如下:>>> myuniversity_dict = {\’name\’:\’yuanyuan\’, \’age\’:18, \’age\’:19, \’age\’:20, \’schoolname\’:Chengdu, \’schoolname\’:Xinxiang}
Traceback (most recent call last):
File \”
NameError: name \’Chengdu\’ is not defined
>>> myuniversity_dict = {\’name\’:\’yuanyuan\’, \’age\’:18, \’age\’:19, \’age\’:20, \’schoolname\’:\’Chengdu\’, \’schoolname\’:\’Xinxiang\’}
>>> myuniversity_dict
{\’age\’: 20, \’name\’: \’yuanyuan\’, \’schoolname\’: \’Xinxiang\’}
>>>
2、键必须是可哈希的,像列表和字典这样的可变类型,由于他们是不可哈希的,所以不能作为字典的键。
为什么呢?—— 解释器调用哈希函数,根据字典中键的值来计算存储你的数据的位置。如果键是可变对象,可以对键本身进行修改,那么当键发生变化时,哈希函数会映射到不同的地址来存储数据,这样哈希函数就不可能可靠地存储或获取相关的数据; 选择可哈希键的原因就是他们的值不能被改变。摘抄python 核心编程(第二版)的一个实例如下:
#!/usr/bin/env python
db = {}
def newuser():
prompt = \'login desired: \'
while True:
name = raw_input(prompt)
if db.has_key(name):
prompt = \'name taken, try another\\n\'
continue
else:
break
pwd = raw_input(\'passwd: \')
db[name] = pwd
def olduser():
name = raw_input(\'login: \')
pwd = raw_input(\'passwd: \')
passwd = db.get(name)
if passwd == pwd:
print \'welcome back\', name
else:
print \'login incorrect\'
def showmenu():
prompt = \"\"\"
(N)ew User Login
(E)xisting User Login
(Q)uit
Enter choice:\"\"\"
done = False
while not done:
chosen = False
while not chosen:
try:
choice = raw_input(prompt).strip()[0].lower()
except:
choice = \'q\'
print \'\\nYou picked: [%s]\' % choice
if choice not in \'neq\':
print \'invalid option, try again\'
else:
chosen = True
if choice == \'q\':done = True
if choice == \'n\':newuser()
if choice == \'e\':olduser()
if __name__ == \'__main__\':
showmenu()
运行结果:
[root@localhost src]# python usrpw.py (N)ew User Login (E)xisting User Login (Q)uit Enter choice:n You picked: [n] login desired: root passwd: 1 (N)ew User Login (E)xisting User Login (Q)uit Enter choice:n You picked: [n] login desired: root name taken, try another