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I know I can use io.StringIO or io.BytesIO to return an open file handle that can be written to and read from.

However, I'm looking for a way to make an in-memory region look like a named disk file that has not yet been opened.

The reason I want this is because there are routines which take the name of a disk file as an argument, and these routines then open the file and manipulate it. In some cases, I want the input or output for these routines to be a memory buffer, not a disk file, and for those routines, I can't pass an already-open file handle.

For example, one such routine is Image.save() from PIL, which expects a path name as its argument, not an already open file handle. When using that routine, I'd like the image data to be saved directly to a memory buffer, without any intermediate file IO being performed. There are also many other routines that take path names as arguments for which I would like this same behavior.

Is there any way to accomplish this in python?

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  • Image.save() will happily take a file pointer in place of a filename, so you can use BytesIO with it. Commented Jun 24, 2017 at 3:37
  • Hmm ... that failed when I tried it, but I'll try again. Perhaps I was making some other error. However, I still am looking for a general solution to this problem. I only used Image.save() as an example. Commented Jun 24, 2017 at 3:41
  • Yes, that's why I posted a comment rather than an answer. You could conceivably do what you want by overriding open and returning a BytesIO object for special paths (however you define "special") rather than a file. Commented Jun 24, 2017 at 3:46
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    Oh, OK. I see. Yes, I am thinking of something like that. I was just hoping that some package might already exist which does what i want. But if not, I'll write one. With regard specifically to Image.save(), it turns out that I have to previously set the name of the BytesIO buffer via something like byteiobuffer.name = "foo.jpg", because Image.save() looks at that name attribute in order to get the image type. Commented Jun 24, 2017 at 4:05
  • Check this out, it might be what you need: stackoverflow.com/questions/32321216/… Commented Jun 24, 2017 at 6:34

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Per the suggestion by Shmulik Asafi, I can use the memory file system within pyfilesystem: making a memory only fileobject in python with pyfilesystem

Also see http://pyfilesystem.org/

I have tested this, and it indeed seems to give me what I want.

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It would be useful if you added an example of how to create a file in memory and access it by name using pyfilesystem.

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