Pickle-加载变量（如果存在）或创建并保存

Question

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

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

Answer 1

您可以按照^{} principle和ask for forgiveness：

import pickle

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

Answer 2

我将它放在一个可重用的函数中，避免错误捕捉文件上的控制流，因为它效率较低，我将使用上下文管理器打开文件。

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

用法：

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

foo返回3

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

而foo仍然返回3。

不可否认，下面的文章很简短，很优雅，但是太多的事情可能会出错，你必须抓住所有的东西（不相信我吗？试试这个：import io, pickle; pickle.load(io.BytesIO(b"\x00"))并播放二进制文件）：

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。但我担心文件关闭的速度可能不够快，以避免在出现空文件或格式错误的情况下第二次打开时出错。所以使用上下文管理器：

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