Pickle

Question

我正在尝试将多个词典的列表保存在 pickle 文件中，并且能够随时在列表中存储新的附加词典并将其保存在 pickle 文件中，然后将 pickle 文件作为列表加载字典。到目前为止，代码的行为如下：

第一次尝试在列表中追加字典时，pickle 文件如下所示（数据保存成功）：

Running info()......

===================================================
Existing data: []
===================================================

Running save_score()......

Inside try: {}
============================================
Existing data: [{}]
New data: {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}
Saved Data: [{}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}]
===================================================

我第二次尝试在列表中附加字典，pickle 文件如下所示：

Running info()......

===================================================
Existing data: [[{}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}]]
===================================================

Running save_score()......

Inside try: [{}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}]
============================================
Existing data: [[{}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}], [{}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}]]
New data: {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}
Saved Data: [[{}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}], [{}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}], {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}]
===================================================

如您所见（字典中的值在上面的示例中不是问题），而不是附加到现有列表，每次都会在列表中创建一个新列表。到目前为止，我还没有设法解决这个问题并获得预期的输出。

预期的输出应该是：

Saved Data: [{}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}, {}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}]

相反，输出是：

Saved Data: [[{}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}], [{}, {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}], {'ComputerName': 'computer name', 'PlayerName': 'user 1', 'ComputerScore': 3, 'PlayerScore': 1}]

代码如下：

import pickle

data = []
computer_name = "computer name"
username = "user 1"
computer_score = 3
human_score = 1

def save_score():
    try:
        existing_data = pickle.load(open("w6_project.p", "rb"))
        print("Inside try: {}".format(existing_data))
        data.append(existing_data)
    except (OSError, IOError) as e:
        pickle.dump({}, open("w6_project.p", "wb"))
    new_data = {"ComputerName": computer_name, "PlayerName": username, "ComputerScore": computer_score, "PlayerScore": human_score}
    print("============================================")
    print("Existing data: {}".format(data))
    data.append(new_data)
    print("New data: {}".format(new_data))
    pickle.dump(data, open("w6_project.p", "wb"))  # Save the data into a pickle file
    print("Saved Data: {}".format(data))
    print("===================================================")
    pass

def info():
    try:
        existing_data = pickle.load(open("w6_project.p", "rb"))
        data.append(existing_data)
    except (OSError, IOError) as e:
        pickle.dump({}, open("w6_project.p", "wb"))
    print("===================================================")
    print("Existing data: {}".format(data))
    print("===================================================")
    pass

def main():
    print("Running info()......\n")
    info()
    print("\nRunning save_score()......\n")
    save_score()

if __name__ == "__main__":
    main()

Answer 1

问题是您有两个列表 - data 和 existing_data - 并且您以错误的方式加入它们。使用 append() 可以将一个列表放在另一个列表中。你应该使用 extend() 加入他们

 data.extend(existing_data)

或更短

 data += existing_data

但是如果你在加载 pickle 后添加新数据，那么你应该简单地直接赋值给变量

 data = pickle.load(...)

编辑： 最小工作示例

import pickle

computer_name = "computer name"
username = "user 1"
computer_score = 3
human_score = 1

def save_score():
    
    try:
        data = pickle.load(open("w6_project.p", "rb"))
        print("Inside try: {}".format(data))
    except (OSError, IOError) as e:
        data = []
        
    new_data = {"ComputerName": computer_name, "PlayerName": username, "ComputerScore": computer_score, "PlayerScore": human_score}
    data.append(new_data)

    pickle.dump(data, open("w6_project.p", "wb"))  # Save the data into a pickle file

    print("============================================")
    print("Existing Data: {}".format(data))
    print("New Data: {}".format(new_data))
    print("Saved Data: {}".format(data))
    print("Length Data: {}".format( len(data) ))
    print("===================================================")

def info():
    try:
        data = pickle.load(open("w6_project.p", "rb"))
    except (OSError, IOError) as e:
        data = []
        
    print("===================================================")
    print("Existing Data: {}".format(data))
    print("Length Data: {}".format( len(data) ))
    print("===================================================")

def main():
    print("Running info()......\n")
    info()

    print("\nRunning save_score()......\n")
    save_score()

if __name__ == "__main__":
    main()

Pickle - 无法正确存储附加字典列表

Pickle - Unable to correctly store appended dictionary to list

python

dictionary

list