Get caller function's absolute path to its file?

Question

If there is method t1 in file a.py and there is a file b.py, which calls method t1 from a.py file. How do I get full/absolute path to b.py file inside t1 method?

With inspect module (just like here: how to get the caller's filename, method name in python), I can get relative path to file, but it seems it does not contain absolute path (or there is some other attribute object, to access to get it?).

As an example:

a.py:

def t1():
    print('callers absolute path')

b.py:

from a import t1
t1()  # should print absolute path for `b.py`

So you're asking how to turn a relative path into an absolute path...? — Aran-Fey, Apr 02 '19 at 07:51
@DirtyBit no. To get `a.py` path, you simply use `__file__` or `sys.path[0]`. But I need caller's path, not the called method's path. — Andrius, Apr 02 '19 at 07:55
@Aran-Fey If it does not return absolute path, then yes. I wonder if using combining with `os.path.realpath` would return correct absolute path all the time? Cause `realpath` function combines current working directory with relative path. — Andrius, Apr 02 '19 at 07:57
`__file__` is always a valid path, i.e. yes, it's relative to the current working directory. (Unless you've changed the CWD in your code, but then it's your own fault...) — Aran-Fey, Apr 02 '19 at 07:58

score 5 · Answer 1 · answered Apr 02 '19 at 08:15

import os
import inspect


def get_cfp(real: bool = False) -> str:
    """Return caller's current file path.

    Args:
        real: if True, returns full path, otherwise relative path
            (default: {False})
    """
    frame = inspect.stack()[1]
    p = frame[0].f_code.co_filename
    if real:
        return os.path.realpath(p)
    return p

Running from another module:

from module import my_module
p1 = my_module.get_cfp()
p2 = my_module.get_cfp(real=True)
print(p1)
print(p2)

Prints:

test_path/my_module_2.py
/home/user/python-programs/test_path/my_module_2.py

`frame = inspect.stack()[1]` `p = frame[0].f_code.co_filename` - very useful lines. After some edits I included this in my local proprietary package in order to save some files in the same folder where calling script is (instead of changing relative path using `os` all the time or passing whole path string). Thanks! — Soren V. Raben, Mar 19 '22 at 15:11

DirtyBit · Answer 2 · 2019-04-02T08:20:28.027

Using sys._getframe():

a1.py:

import sys
def t1():
    print(sys._getframe().f_code)

a2.py:

from a1 import t1
t1()  # should print absolute path for `b.py`

Hence:

py -m a2.py

OUTPUT:

<code object t1 at 0x0000029BF394AB70, file "C:\Users\dirtybit\PycharmProjects\a1.py", line 2>

EDIT:

Using inspect:

a1.py:

import inspect
def t1():
    print("Caller: {}".format(inspect.getfile(inspect.currentframe())))

a2.py:

from a1 import t1
t1()  # should print absolute path for `b.py`

OUTPUT:

Caller: C:\Users\dirtybit\PycharmProjects\a1.py

score 0 · Answer 3 · answered Apr 02 '19 at 07:50

0

You can get it with the os module in python.

>>> import a
>>> os.path.abspath(a.__file__)

answered Apr 02 '19 at 07:50

Jim Todd

1,488
1
11
15

This will give you path for current file, not for callers file. – Andrius Apr 02 '19 at 07:58

Mikkel Madsen · Answer 4 · 2019-04-02T08:02:11.013

0

Using the os module you can do the following:

a.py

import os

def t1(__file__):
    print(os.path.abspath(__file__))

b.py

from a import t1
t1(__file__)  # shoult print absolute path for `b.py`

With this, you could call t1(__file__ and get the absolute path for any file.

edited Apr 02 '19 at 08:02

answered Apr 02 '19 at 07:57

Mikkel Madsen

1
2

I know, but thats not what I need. I need to get that path inside `a.py`. It should be the job for `a.py` to provide that, not for `b.py`. – Andrius Apr 02 '19 at 08:00
print(os.path.abspath("b.py")) – Mikkel Madsen Apr 02 '19 at 08:10
Well, you still need to provide callers filename, so no good:) – Andrius Apr 02 '19 at 08:13

ymentha14 · Answer 5 · 2020-05-14T14:03:50.603

The trick consists in recovering both the current working directory and the relative path w.r.t that directory to the caller file (here b.py). The join does the rest.

a.py:

    import os
    import sys
    def t1():
        namespace = sys._getframe(1).f_globals
        cwd = os.getcwd()
        rel_path = namespace['__file__']
        abs_path= os.path.join(cwd,rel_path)
        print('callers absolute path!',abs_path)

b.py:

    from a import t1
    t1()  # prints absolute path for `b.py`

The trick does not work for jupyter notebooks unfortunately..

Get caller function's absolute path to its file?

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