Question

是否有更好的方法可以加载pickle变量（如果已存在）或创建变量并将其转储（如果不存在）？

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

您可以关注EAFP 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

我把它放在一个可重用的函数中，避免错误捕获文件上的控制流，因为效率较低，我会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)

且foo仍会返回3。

不可否认，以下内容相当短暂而优雅，但有太多事情可能出错，你必须抓住一切（不相信我？试试这个：{{1}并使用二进制文件）：

import io, pickle; pickle.load(io.BytesIO(b"\x00"))

相同用法。但我担心文件可能没有足够快地关闭，以避免在文件空或格式错误时第二次打开它时出错。所以使用上下文管理器：

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

Pickle - 加载变量（如果存在）或创建并保存它

2 个答案: