有两种方法
当知道字典的键时：
unit_rooms={ 3:{301:[1,80],302:[1,80],303:[2,90],304:[2,90]},
4:{401:[1,80],402:[1,80],403:[2,90],404:[2,90]},
5:{501:[1,80],502:[1,80],503:[2,90],504:[2,90]}
}
for i in range(3,6):
rooms=unit_rooms[i]
print(rooms)
运行结果：
当不知道字典的键时：
unit_rooms = {3:{301:[1,80],302:[1,80],303:[2,90],304:[2,90]},
4:{401:[1,80],402:[1,80],403:[2,90],404:[2,90]},
5:{501:[1,80],502:[1,80],503:[2,90],504:[2,90]}
}
for rooms in unit_rooms.values():
print(rooms)
运行结果：
获取字典的值：
for value in DictName.values(): # value的名字可以自行另取 # DictName是要遍历的字典的名称 # .values():是固定的用法
获取键值：
for k,v in DictName.items(): #遍历字典的键值对，k对应键，v对应值 #k,v 的名字可以自己取，DictName是字典名
unit_rooms = {3:{301:[1,80],302:[1,80],303:[2,90],304:[2,90]},
4:{401:[1,80],402:[1,80],403:[2,90],404:[2,90]},
5:{501:[1,80],502:[1,80],503:[2,90],504:[2,90]}
}
for sub_dict in unit_rooms.values():
# 遍历大字典的值，即小字典sub_dict
for room,info in sub_dict.items():
print('房间号：%d,方向：%d,面积：%d'%(room,info[0],info[1]))
运行结果：
方向1代表南北，方向2代表东西
把数字替换掉：
unit_rooms = {3:{301:[1,80],302:[1,80],303:[2,90],304:[2,90]},
4:{401:[1,80],402:[1,80],403:[2,90],404:[2,90]},
5:{501:[1,80],502:[1,80],503:[2,90],504:[2,90]}
}
for sub_dict in unit_rooms.values():
for room,info in sub_dict.items():
dire = ['', '南北', '东西']
#建立一个列表，第0个元素为空，第1个元素为'南北'，第2个元素为'东西'
print(dire[info[0]])
举例：
unit_rooms = {3:{301:[1,80],302:[1,80],303:[2,90],304:[2,90]},
4:{401:[1,80],402:[1,80],403:[2,90],404:[2,90]},
5:{501:[1,80],502:[1,80],503:[2,90],504:[2,90]}
}
for sub_dict in unit_rooms.values():
for room,info in sub_dict.items():
dire = ['', '南北', '东西']
print('户室号：%d 朝向：%s 面积：%d' % (room,dire[info[0]],info[1]))
运行结果：
附：字典dic最大值对应的键
步骤：（1）用max()函数找出最大的值maxValue；（2）套用一的自定义函数
def getKey(dic, value):
if value not in dic.values():
return None
result = set()
for key in dic:
if dic[key]==value:
result.add(key)
return result
dic = {'a':2, 'b':1, 'c':10, 'd':10}
maxValue=max(dic.values())
result = getKey(dic,maxValue)
>> {'c','d'}
总结
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原文链接：https://blog.csdn.net/Nurbiya_K/article/details/104650418
