Formatting an exception in python to include all but the last 'n' traceback frames

Question

Assume you had a setup like this:

def a():
    b()

def b():
    c()

def c():
    d()

def d():
    e()

Attempting to call a() would result in the following traceback:

Traceback (most recent call last):
  File "<pyshell#181>", line 1, in <module>
    a()
  File "<pyshell#87>", line 2, in a
    b()
  File "<pyshell#90>", line 2, in b
    c()
  File "<pyshell#93>", line 2, in c
    d()
  File "<pyshell#96>", line 2, in d
    e()
NameError: name 'e' is not defined

Is there any way to format the exception so that it only includes the last n frames in the traceback? For example, if n = 2, the traceback would look like this:

Traceback (most recent call last):
  File "<pyshell#93>", line 2, in c
    d()
  File "<pyshell#96>", line 2, in d
    e()
NameError: name 'e' is not defined

I've tinkered with it for a little bit and can't figure out a way to do it.

Answer 1

Starting from Python 3.5, functions from the traceback module support negative limits (the original proposal was inspired by your question and approved by Guido):

import traceback

n = 2

try:
    a()
except:
    traceback.print_exc(limit=-n) # `n` last traceback entries

outputs

Traceback (most recent call last):
  File "/home/vaultah/test.py", line 8, in c
    d()
  File "/home/vaultah/test.py", line 11, in d
    e()
NameError: name 'e' is not defined

You can replicate this behavior even if you use older Python

import sys, traceback

n = 2

try:
    a()
except:
    ex = traceback.format_exception(*sys.exc_info())
    sys.stderr.write(''.join([ex[0]] + ex[-n-1:]))
    # or print(*[ex[0]] + ex[-n-1:], sep='', end='', file=sys.stderr)

The output will be exactly the same.