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 "", line 1, in
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
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