当两个 python 脚本访问单个 pickle 文件时，如何避免读取错误？

Question

我需要在从两个不同来源启动的两个 Python 脚本之间传递数据。我看到我的记录器出现读取错误。有没有更好的方法？

程序 A： 大约每分钟定期写入 pickle 文件。

def cacheData(filepath, data):
    #create a cPickle file to cache the current data
    try:  
        outFile = open(filepath,'wb')  
        #cPickle the current data to a file  
        cPickle.dump(data, outFile)  
        outFile.close()   
    except Exception, e:  
        logger.warn("Error creating cache file")

程序B:是由用户启动的已编译可执行文件。它读取 pickle 文件以启动一些代码。

def readCachedObj(filepath):
    #read cPickle file and return data as object
    try:
        inFile = open(filepath,'rb')
        cache = cPickle.load(inFile)
        inFile.close()
        return cache
    except Exception, e:
        logger.warn("Error reading cached data cPickle")

更新 1

def replace(src, dst):
    win32api.MoveFileEx(src, dst, win32con.MOVEFILE_REPLACE_EXISTING)    

def cacheData(filepath, data):
    #create a cPickle file to cache the current data
    try:  
        tmpfile = str(uuid.uuid4()) + '.tmp'
        outFile = open(tmpfile,'wb')  
        #cPickle the current data to a temp file  
        cPickle.dump(data, outFile)  
        outFile.close()  
        #replace the pickle file with the new temp file
        replace(tmpfile, filepath)    
        #remove any extraneous temp files
        for f in glob.glob("*.tmp"):
            os.remove(f)        
    except Exception, e:  
        logger.warn("Error creating cache file")  

def readCachedObj(filepath):
    #read cPickle file and return data as object
    try:     
        inFile = open(filepath,'rb')
        cache = cPickle.load(inFile)
        inFile.close()
        return cache
    except Exception, e:
        logger.warn("Error reading cached data cPickle")

Answer 1

永远不要覆盖现有文件。相反，写入 new 文件，然后在成功关闭它后执行（原子）重命名。

Answer 2

这似乎适合我。通过创建新的临时文件然后使用文件覆盖，cacheData 函数在处理最终的 pickle 文件时会更加快捷。

def replace(src, dst):
    win32api.MoveFileEx(src, dst, win32con.MOVEFILE_REPLACE_EXISTING)    

def cacheData(filepath, data):
    #create a cPickle file to cache the current data
    try:  
        tmpfile = str(uuid.uuid4()) + '.tmp'
        outFile = open(tmpfile,'wb')  
        #cPickle the current data to a temp file  
        cPickle.dump(data, outFile)  
        outFile.close()  
        #replace the pickle file with the new temp file
        replace(tmpfile, filepath)    
        #remove any extraneous temp files
        for f in glob.glob("*.tmp"):
            os.remove(f)        
    except Exception, e:  
        logger.warn("Error creating cache file")  

def readCachedObj(filepath):
    #read cPickle file and return data as object
    try:     
        inFile = open(filepath,'rb')
        cache = cPickle.load(inFile)
        inFile.close()
        return cache
    except Exception, e:
        logger.warn("Error reading cached data cPickle")

当两个 python 脚本访问单个 pickle 文件时，如何避免读取错误？

How can I avoid read errors when two python scripts access a single pickle file?

python

pickle

更新 1

这似乎适合我。通过创建新的临时文件然后使用文件覆盖，cacheData 函数在处理最终的 pickle 文件时会更加快捷。