Question

当我试图修改我的列表然后加载它时，我收到错误说：

Traceback (most recent call last):
  File "C:\Users\T\Desktop\pickle_process\pickle_process.py", line 16, in <module>
    print (library[1])
IndexError: string index out of range

请建议解决方案我的代码：

import pickle

library = []

with open ("LibFile.pickle", "ab") as lib:
    user = input("give the number")
    print ("Pickling")
    library.append(user)
    pickle.dump(user, lib)
    lib.close()

lib = open("LibFile.pickle", "rb")
library = pickle.load(lib)
for key in library:
    print (library[0])
    print (library[1])

Answer 1

这与酸洗无关。我将编写新的示例代码，说明它无法正常工作的原因。

library = []
library.append("user_input_goes_here")
print(library[0])
# OUTPUT: "user_input_goes_here")
print(library[1])
# IndexError occurs here.

你只是在空列表中附加一件事。为什么你认为有两个要素？：）

如果您多次执行此操作，则会失败，因为您在模式'ab'而不是'wb'中打开了pickle文件。每次写信时都应该覆盖泡菜。

import pickle

library = ["index zero"]
def append_and_pickle(what_to_append,what_to_pickle):
    what_to_pickle.append(what_to_append)
    with open("testname.pkl", "wb") as picklejar:
        pickle.dump(what_to_pickle, picklejar)
        # no need to close with a context manager

append_and_pickle("index one", library)
with open("testname.pkl","rb") as picklejar:
    library = pickle.load(picklejar)

print(library[1])
# OUTPUT: "index one"

这可能看似违反直觉，因为您已经＆＃34;追加＆＃34;列表，但请记住，一旦你腌制一个对象它不再是一个列表，它就是一个pickle文件。当您向列表中添加元素时，您实际上并未附加到FILE，而是您正在更改对象本身！这意味着你需要完全改变文件中写的内容，以便它附加额外的元素来描述这个新对象。

Answer 2

您正在迭代load函数返回的对象，并且由于某种原因您尝试通过索引访问该对象。变化：

for key in library:
    print (library[0])
    print (library[1])

为：

for key in library:
    print key

Library[1]不存在，因此string index out of range错误。

腌制清单 - 错误

2 个答案: