Python hoist: immediate locals only

The implementation of hoist in my earlier snippet returned the values of local variables of all functions called by the hoisted function. This new version only returns the locals of the hoisted function.

As before, you can use it like this:

def f(x):
  y = x + 1
  z = y * 2
  return z + 1

f(4) == 11
hoist(f)(4) == {'x': 4, 'y': 5, 'z': 10, 'return': 11}

Here’s another usage example:

def f(x):
  y = messy_g(x + 1)
  return y + 1

def messy_g(z):
  q = z * 9
  w = q / 3
  return w + 1

f(1) == 8
hoist(f)(1) == {'x': 1, 'y': 7, 'return': 8}

Notice in this example that the local variables used by messy_g are not returned by hoist(f). Only f’s local variables are returned.

Here’s the implementation.

import functools
import inspect
import sys

def make_trace(results, f, original_trace_fn):
  def trace_local(frame, event, arg):
    # event: 'call', 'line', 'return', 'exception' or 'opcode'
    if event == 'line':
      arg_info = inspect.getargvalues(frame)
      results.update(arg_info.locals.copy())
    if event == 'return':
      arg_info = inspect.getargvalues(frame)
      values = arg_info.locals.copy()
      results.update(values)
      results['return'] = arg
    if event == 'call':
      return original_trace_fn

  def trace_global(frame, event, arg):
    if event == 'call':
      info = inspect.getframeinfo(frame)
      info_parent = inspect.getframeinfo(frame.f_back)
      if info.function == f.__name__ and info_parent.function == 'hoisted_f__':
        return trace_local
  return trace_global


def hoist(f):

  @functools.wraps(f)
  def hoisted_f__(*args, **kwargs):
    original_trace_fn = sys.gettrace()

    results = {}
    trace_fn = make_trace(results, f, original_trace_fn)

    sys.settrace(trace_fn)
    f(*args, **kwargs)
    sys.settrace(original_trace_fn)

    return results

  return hoisted_f__

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