Size of an open file object

Solution 1:

$ ls -la chardet-1.0.1.tgz
-rwxr-xr-x 1 vinko vinko 179218 2008-10-20 17:49 chardet-1.0.1.tgz
$ python
Python 2.5.1 (r251:54863, Jul 31 2008, 22:53:39)
[GCC 4.1.2 (Ubuntu 4.1.2-0ubuntu4)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> f = open('chardet-1.0.1.tgz','rb')
>>> f.seek(0,2)
>>> f.tell()
179218L

Adding ChrisJY's idea to the example

>>> import os
>>> os.fstat(f.fileno()).st_size
179218L
>>>        

Note: Based on the comments, f.seek(0, 2) is must before calling f.tell(), without which it would return a size of 0. The reason is that f.seek(0, 2) moves the file object's position to the end of the file.

Solution 2:

Well, if the file object support the tell method, you can do:

current_size = f.tell()

That will tell you were it is currently writing. If you write in a sequential way this will be the size of the file.

Otherwise, you can use the file system capabilities, i.e. os.fstat as suggested by others.

Solution 3:

If you have the file descriptor, you can use fstat to find out the size, if any. A more generic solution is to seek to the end of the file, and read its location there.

Solution 4:

Another solution is using StringIO "if you are doing in-memory operations".

with open(file_path, 'rb') as x:
    body = StringIO()
    body.write(x.read())
    body.seek(0, 0)

Now body behaves like a file object with various attributes like body.read().

body.len gives the file size.