How to get the name of function in if statement?

Question

How to get the name of function in if statement?
Is there any easy solution?
My case:

if is_wedge(df):
    fname = is_wedge.__name__
    img_path = Path(main_dir / 'images' / fname / img1.jpeg')

I think this is a duplicate question. Try [Determine function name from within that function (without using traceback)](https://stackoverflow.com/questions/5067604/determine-function-name-from-within-that-function-without-using-traceback) — anosha_rehan, Mar 30 '22 at 11:39
@anosha_rehan: Not a dup — there's nothing indicating this code is inside a function. — martineau, Mar 30 '22 at 11:41
@jfaccioni: If one knows enough to write `my_function.__name__` then obviously one already knows the name of the function. — martineau, Mar 30 '22 at 11:42
Batjl: Your question makes little sense — how could the `print` know it was being called because some function returned `True` when it was called earlier? — martineau, Mar 30 '22 at 11:45
@martineau well yes, but the function is already known anyway if you're calling it in the `if` statement... — jfaccioni, Mar 30 '22 at 11:46
Very strange requirement, can you explain an example use-case? — matszwecja, Mar 30 '22 at 11:50
`is_wedge` ≠ `iswedge`. Question still not making much sense. Also looks like OP is using pandas but hasn't mention it. — martineau, Mar 30 '22 at 12:06
Does this answer your question? [Determine function name from within that function (without using traceback)](https://stackoverflow.com/questions/5067604/determine-function-name-from-within-that-function-without-using-traceback) — Paco, Mar 30 '22 at 16:57

score 0 · Answer 1 · answered Mar 30 '22 at 14:04

Could you use inspect.stack?

import inspect

def iswedge():
    stack = inspect.stack()    # Get stack info
    caller_info = stack[0]    # Get caller info
    fname = caller_info[3]    # Get function name
    img_path = Path(main_dir / 'images' / fname / img1.jpeg')

How to get the name of function in if statement?

1 Answers1