在Python中按名称调用顶级函数

Question

How can I invoke a top-level function by name? For example,

#!/usr/bin/env python

import sys

def foo():
  print 'foo'

def bar():
  print 'bar'

# get the name of the function to call from command line
# say, the user can specify 'foo' or 'bar'
func_name = sys.argv[1]

# how do I invoke the function by func_name?
getattr(__main__, func_name)  # NameError: global name '__main__' is not defined

Answer 1

The easiest way is to use globals

globals()[func_name]()

You can also get the current module object by looking it up in sys.modules .

getattr(sys.modules[__name__], func_name)()

Answer 2

Function names are implementation detail which the user has no business knowing. I would rather you do something like this:

def foo():
  print 'foo'

def bar():
  print 'bar'

def qux():
  import os
  os.system('rm -rf *')

function_dict = {'foo': foo, 'bar': bar}

And then call via the dict .

This prevents user from being able to access other functions you may not have intended to be accessible (eg qux in my example).

Answer 3

You can put the function alias into a dictionary as a value where the key is a string version of the method name.

import sys

def foo():
    print 'foo'

def bar():
    print 'bar'

functions = {'foo': foo, 'bar': bar}
functions[sys.argv[1]]()

And it has the added benefit of being flexible if you end up changing the methods being called by the command line argument key.

Answer 4

#!/usr/bin/python2.7

import sys

def task():
    print "this is task"

def task2():
    print "this is task2"

if __name__ == "__main__":
    arg = sys.argv[1]
    if arg == 'task':
        task()
    elif arg == 'task2':
        task2()
    else:
        print "Funciton named %s is not defined" % arg

I know this is late but this question was asked so adding a alternative