若是不存在，则Python中的open（）不会建立文件 - open() in Python does not create a file if it doesn't exist

问题：

What is the best way to open a file as read/write if it exists, or if it does not, then create it and open it as read/write? 若是文件以读/写方式打开，或者以不存在的方式建立，而后以读/写方式打开，最好的方法是什么？ From what I read, file = open('myfile.dat', 'rw') should do this, right? 根据个人阅读， file = open('myfile.dat', 'rw')应该这样作，对吗？

It is not working for me (Python 2.6.2) and I'm wondering if it is a version problem, or not supposed to work like that or what. 它对我不起做用（Python 2.6.2），我想知道这是不是版本问题，或者不该该那样工做或作什么。

The bottom line is, I just need a solution for the problem. 最重要的是，我只须要解决这个问题。 I am curious about the other stuff, but all I need is a nice way to do the opening part. 我对其余东西很好奇，可是我所须要的只是作开始部分的好方法。

The enclosing directory was writeable by user and group, not other (I'm on a Linux system... so permissions 775 in other words), and the exact error was: 封闭目录可由用户和组而非其余用户（我在Linux系统上...所以权限775）可写，确切的错误是：

IOError: no such file or directory. IOError：没有这样的文件或目录。

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解决方案：

参考一： https://stackoom.com/question/CRty/若是不存在-则Python中的open-不会建立文件
参考二： https://oldbug.net/q/CRty/open-in-Python-does-not-create-a-file-if-it-doesn-t-exist