How would you do the equivalent of preprocessor directives in Python?

There's __debug__, which is a special value that the compiler does preprocess.

if __debug__:
  print "If this prints, you're not running python -O."
else:
  print "If this prints, you are running python -O!"

__debug__ will be replaced with a constant 0 or 1 by the compiler, and the optimizer will remove any if 0: lines before your source is interpreted.

I suspect you're gonna hate this answer. The way you do that in Python is

# code here
if DEBUG:
   #debugging code goes here
else:
   # other code here.

Since python is an interpreter, there's no preprocessing step to be applied, and no particular advantage to having a special syntax.

You can use the preprocessor in Python. Just run your scripts through the cpp (C-Preprocessor) in your bin directory. However I've done this with Lua and the benefits of easy interpretation have outweighed the more complex compilation IMHO.

You can just use the normal language constructs:

DEBUG = True
if DEBUG:
  # Define a function, a class or do some crazy stuff
  def f():
    return 23
else:
  def f():
    return 42

• Python if cannot eliminate elements from arrays.
• C precompilers do not handle #! or other lines beginning with # as needed.
• pypreprocessor seems to be python specific

Use a common m4 instead, like this:

ifelse(DEBUG,True,dnl`
  < do some code >
dnl,dnl`
  < do some other code >dnl
')

ifelse(
  M4_CPU,x86_64,`
    < do some code specific for M4_CPU >
',M4_CPU,arm,`
    < do some code specific for M4_CPU >
',M4_CPU,ppc64le,`
    < do some code specific for M4_CPU >
')

ifelse(
  M4_OS,windows,`
    < do some code specific for M4_OS >
  ',M4_OS,linux,`
    < do some code specific for M4_OS >
  ',M4_OS,android,`
    < do some code specific for M4_OS >
')

m4 -D DEBUG=True -D M4_OS=android -D M4_CPU=arm test.py.m4 > test.py

Use gpp - a general-purpose preprocessor.