How to dynamically define functions?

Question

I have functions like this:

def activate_field_1():
   print 1

def activate_field_2():
   print 2

def activate_field_3():
   print 3

How do I define activate_field_[x] for x=1:10, without typing out each one of them? I'd much rather pass a parameter, of course, but for my purposes this is not possible.

You say in a comment that you want to define them statically but then you mention that doing so will waste space in the source file. Which is it? — aaronasterling, Sep 11 '10 at 02:58
What is a "dynamic function"? Can you explain the intended meaning of this phrase? — Anderson Green, May 14 '13 at 21:37
This question appears to be a duplicate: http://stackoverflow.com/questions/11291242/python-dynamically-create-function-at-runtime — Anderson Green, May 14 '13 at 21:40

score 25 · Accepted Answer · edited Jul 23 '16 at 00:03

25

Do you want to define these individually in your source file, statically? Then your best option would be to write a script to generate them.

If on the other hand you want these functions at runtime you can use a higher order function. For e.g.

>>> def make_func(value_to_print):
...     def _function():
...         print value_to_print
...     return _function
...
>>> f1 = make_func(1)
>>> f1()
1
>>> f2 = make_func(2)
>>> f2()
2

You can generate a list of these and store, again at runtime.

>>> my_functions = [make_func(i) for i in range(1, 11)]
>>> for each in my_functions:
...     each()
...
1
2
3
...

edited Jul 23 '16 at 00:03

endolith

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answered Sep 10 '10 at 19:29

Manoj Govindan

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Yeah, I wanted to define them statically in the source file. I can script them, though so many functions wastes space. Thanks for the explanation on how to do it at runtime. – atp Sep 10 '10 at 19:36

martineau · Answer 2 · 2019-09-06T17:38:38.390

11

Here's something that produces function names exactly like you wanted (and is a little simpler than the Dynamic/runtime method creation's accepted answer mentioned in @Goutham's now deleted answer):

FUNC_TEMPLATE = """def activate_field_{0}(): print({0})"""
for x in range(1, 11): exec(FUNC_TEMPLATE.format(x))

>>> activate_field_1()
1
>>> activate_field_7()
7

In Python versions 3.6+, it can be written as shown below using so-called f-string literals:

for x in range(1, 11): exec(f"""def activate_field_{x}(): print({x})""")

edited Sep 06 '19 at 17:38

answered Sep 11 '10 at 23:26

martineau

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Exactly what I needed to dynamically generate test case functions for unit testing, thanks. – May 25 '17 at 03:33
2

This was very nice, I was able to define functions with decorators too. – Eduardo Santana Sep 06 '19 at 17:30
1

Good to remember that this approach is not suitable for production code. – Jourdan Rodrigues Mar 17 '20 at 20:07
@JourdanRodrigues why not? Anything that works is suitable. – t3chb0t Aug 16 '23 at 15:19

score 5 · Answer 3 · answered Sep 10 '10 at 19:35

5

You may put new symbols into the dictionary of current variable bindings returned by vars():

for i in range(1, 11):
    def f(x):
        def g():
            print x
        return g
    vars()['activate_field_%d' % i] = f(i)

>>> activate_field_3()
3

But this trick is generally not recommented unless you definitely sure you need it.

answered Sep 10 '10 at 19:35

Andrey Vlasovskikh

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1

You really _don't_ need to [re]define `f()` 10 times… – martineau Jul 27 '20 at 23:50

score 3 · Answer 4 · answered Sep 10 '10 at 19:51

Maybe you could adapt this recipe for your needs.

from functools import partial
class FunctionPool:
    def __init__(self,function):
        self.function = function
    def __getitem__(self,item):
        return partial(self.function,item)

>>> @FunctionPool
def func(item,param):
    print "function#{item} called with {param}".format(
        item = item,
        param = param )
>>> f = func[2]
>>> f(3)
function#2 called with 3

How to dynamically define functions?

4 Answers4

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