[已解决]使用pickle无法输出问题

季夏。。。

1. import pickle
   
2. my_list = [123,3.14,'wangyuq',['another,list']]
   
3. pickle_file = open('my_list.pkl','wb')
   
4. pickle.dump(my_list,pickle_file)
   
5. pickle_file.close()
   
6. 
   
7. pickle_file = open('my_list.pkl','rb')
   
8. print(pickle_file)
   

复制代码

我想请教一下各位老手，这样运行之后一直显示：<_io.BufferedReader name='my_list.pkl'>
改成：然后利用my_list2 = pickle.load(pickle_file) print(my_list2)就可以成功输出了。
这个为什么不能用pickle_file = open('my_list.pkl','rb')    print(pickle_file)直接输出啊？

最佳答案

月排行榜 / 总排行榜

jackz007

2022-9-29 11:03:02

1. import pickle
   
2. my_list = [123,3.14,'wangyuq',['another,list']]
   
3. pickle_file = open('my_list.pkl','wb')
   
4. pickle.dump(my_list,pickle_file)
   
5. pickle_file.close()
   
6. 
   
7. pickle_file = open('my_list.pkl','rb')
   
8. s = pickle . load(pickle_file)
   
9. pickle_file.close()
   
10. print(s)

复制代码

跳转到最佳答案楼层

specail

pickle_file = open('my_list.pkl','rb')
这里的pickle_file 是文件对象啊，你读出内容，当然是要读对象里面的数据

jackz007

1. import pickle
   
2. my_list = [123,3.14,'wangyuq',['another,list']]
   
3. pickle_file = open('my_list.pkl','wb')
   
4. pickle.dump(my_list,pickle_file)
   
5. pickle_file.close()
   
6. 
   
7. pickle_file = open('my_list.pkl','rb')
   
8. s = pickle . load(pickle_file)
   
9. pickle_file.close()
   
10. print(s)

复制代码