关于python：Pickle – 加载变量（如果存在）或创建并保存它

Pickle - Load variable if exists or create and save it

如果变量已经存在，是否有更好的方法用pickle加载它，如果不存在，是否创建并转储它？

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if os.path.isfile("var.pickle"):
foo = pickle.load( open("var.pickle","rb" ) )
else:
foo = 3
pickle.dump( foo, open("var.pickle","wb" ) )

你可以遵循EAFP原则，请求宽恕：

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import pickle

try:
foo = pickle.load(open("var.pickle","rb"))
except (OSError, IOError) as e:
foo = 3
pickle.dump(foo, open("var.pickle","wb"))

为了可重用性，我将它放在一个函数中，避免对文件上的控制流进行错误捕获，因为它效率较低，并且我将使用上下文管理器打开文件。

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import os
import pickle

def read_or_new_pickle(path, default):
if os.path.isfile(path):
with open(path,"rb") as f:
try:
return pickle.load(f)
except Exception: # so many things could go wrong, can't be more specific.
pass
with open(path,"wb") as f:
pickle.dump(default, f)
return default

号

用途：

1	foo = read_or_new_pickle(path="var.pickle", default=3)

foo返回3。

1	foo = read_or_new_pickle(path="var.pickle", default=4)

。

而foo仍然返回3。

诚然，下面的内容相当简短和优雅，但是太多事情可能会出错，你必须抓住一切(不相信我吗？试试这个：import io, pickle; pickle.load(io.BytesIO(b"\x00"))，玩二进制)：

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import pickle

def read_or_new_pickle(path, default):
try:
foo = pickle.load(open(path,"rb"))
except Exception:
foo = default
pickle.dump(foo, open(path,"wb"))
return foo

同样的用法。但我担心的是，如果文件是空的或格式不正确的，可能无法以足够快的速度关闭该文件，以避免在第二次打开时出错。因此，使用上下文管理器：

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import pickle

def read_or_new_pickle(path, default):
try:
with open(path,"rb") as f:
foo = pickle.load(f)
except Exception:
foo = default
with open(path,"wb") as f:
pickle.dump(foo, f)
return foo

。