Python - 附加到腌制列表

Question

我正在努力在腌制文件中附加一个列表。这是代码：

 #saving high scores to a pickled file

import pickle

first_name = input("Please enter your name:")
score = input("Please enter your score:")

scores = []
high_scores = first_name, score
scores.append(high_scores)

file = open("high_scores.dat", "ab")
pickle.dump(scores, file)
file.close()

file = open("high_scores.dat", "rb")
scores = pickle.load(file)
print(scores)
file.close()

我第一次运行代码时，它会打印姓名和分数。

我第二次运行代码时，它打印了 2 个名字和 2 个分数。

我第三次运行代码时，它打印了名字和分数，但它用我输入的第三个名字和分数覆盖了第二个名字和分数。我只是想让它继续添加名字和分数。我不明白为什么要保存第一个名字并覆盖第二个名字。

原文由 Charlie 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

在附加到它之前，您需要先从数据库（即您的 pickle 文件）中提取列表。

 import pickle
import os

high_scores_filename = 'high_scores.dat'

scores = []

# first time you run this, "high_scores.dat" won't exist
#   so we need to check for its existence before we load
#   our "database"
if os.path.exists(high_scores_filename):
    # "with" statements are very handy for opening files.
    with open(high_scores_filename,'rb') as rfp:
        scores = pickle.load(rfp)
    # Notice that there's no "rfp.close()"
    #   ... the "with" clause calls close() automatically!

first_name = input("Please enter your name:")
score = input("Please enter your score:")

high_scores = first_name, score
scores.append(high_scores)

# Now we "sync" our database
with open(high_scores_filename,'wb') as wfp:
    pickle.dump(scores, wfp)

# Re-load our database
with open(high_scores_filename,'rb') as rfp:
    scores = pickle.load(rfp)

print(scores)

原文由 dogwynn 发布，翻译遵循 CC BY-SA 3.0 许可协议

Answer 2

如果您想写入和读取 pickled 文件，您可以为列表中的每个条目多次调用转储。每次转储时，都会将分数附加到腌制文件中，每次加载时都会读取下一个分数。

 >>> import pickle as dill
>>>
>>> scores = [('joe', 1), ('bill', 2), ('betty', 100)]
>>> nscores = len(scores)
>>>
>>> with open('high.pkl', 'ab') as f:
…   _ = [dill.dump(score, f) for score in scores]
...
>>>
>>> with open('high.pkl', 'ab') as f:
...   dill.dump(('mary', 1000), f)
...
>>> # we added a score on the fly, so load nscores+1
>>> with open('high.pkl', 'rb') as f:
...     _scores = [dill.load(f) for i in range(nscores + 1)]
...
>>> _scores
[('joe', 1), ('bill', 2), ('betty', 100), ('mary', 1000)]
>>>

您的代码失败的原因很可能是您将原始的 scores 替换为未经处理的分数列表。因此，如果添加了任何新乐谱，您会把它们记在记忆中。

 >>> scores
[('joe', 1), ('bill', 2), ('betty', 100)]
>>> f = open('high.pkl', 'wb')
>>> dill.dump(scores, f)
>>> f.close()
>>>
>>> scores.append(('mary',1000))
>>> scores
[('joe', 1), ('bill', 2), ('betty', 100), ('mary', 1000)]
>>>
>>> f = open('high.pkl', 'rb')
>>> _scores = dill.load(f)
>>> f.close()
>>> _scores
[('joe', 1), ('bill', 2), ('betty', 100)]
>>> blow away the old scores list, by pointing to _scores
>>> scores = _scores
>>> scores
[('joe', 1), ('bill', 2), ('betty', 100)]

所以它更像是 scores 的 python 名称参考问题，而不是 pickle 问题。 Pickle 只是实例化一个新列表并调用它 scores （在你的情况下），然后它垃圾收集任何东西 scores 之前指向。

 >>> scores = 1
>>> f = open('high.pkl', 'rb')
>>> scores = dill.load(f)
>>> f.close()
>>> scores
[('joe', 1), ('bill', 2), ('betty', 100)]

原文由 Mike McKerns 发布，翻译遵循 CC BY-SA 3.0 许可协议