Python Dictionaries

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　　dict函数

 

　　基本的字典操作
　　len(d)返回d中项(key-value)的数量
　　d[k]返回关联到键k上的值
　　d[k]=v将值v关联到键k上
　　del d[k]删除键为k的项
　　k in d检查d中是否有含有键为k的项
　　来看个简单的例子

databases = {
"Oracle" : {
"username" : "oracleapp",
"password" : "oracleapp123"
},
"Db2" : {
"username" : "db2app",
"password" : "db2app123"
},
"MongoDB" : {
"username" : "mongodbapp",
"password" : "mongodbapp123"
}
}
labels = {
"username" : "DB Username",
"password" : "DB Password"
}
name = raw_input("DB Name : ")
request = raw_input("DB Username(u) or DB Password(p)? ")
if request == "u" : key = "username"
if request == "p" : key = "password"
if name in databases :
print "%s database's %s is %s." % (name, labels[key], databases[name][key])
　　运行结果如下所示：

 
　　字典的格式化字符串

 
　　字典方法：clear、copy、fromkeys、get

　　接下来改写上面那个代码，使用get方法来访问实体

databases = {
"Oracle" : {
"username" : "oracleapp",
"password" : "oracleapp123"
},
"DB2" : {
"username" : "db2app",
"password" : "db2app123"
},
"MongoDB" : {
"username" : "mongodbapp",
"password" : "mongodbapp123"
}
}
labels = {
"username" : "DB Username",
"password" : "DB Password"
}
name = raw_input("DB Name : ")
request = raw_input("DB Username(u) or DB Password(p)? ")

if request == "u" :
key = "username"
elif request == "p" :
key = "password"
else :
key = request

db = databases.get(name, {})
label = labels.get(key, key)
result = db.get(key, "not available")

print "%s database's %s is %s." % (name, label, result)
　　运行结果如下所示：

 

　　has_key、items、iteritems、keys、iterkeys、values、itervalues

 

　　pop、popitem、setdefault、update

Summary