How do I call a function twice or more times consecutively?

Question

Is there a short way to call a function twice or more consecutively in Python? For example:

do()
do()
do()

maybe like:

3*do         


        

Answer 1

You can use itertools.repeat with operator.methodcaller to call the __call__ method of the function N times. Here is an example of a generator function doing it:

from itertools import repeat
from operator import methodcaller


def call_n_times(function, n):
    yield from map(methodcaller('__call__'), repeat(function, n))

Example of usage:

import random
from functools import partial

throw_dice = partial(random.randint, 1, 6)
result = call_n_times(throw_dice, 10)
print(list(result))
# [6, 3, 1, 2, 4, 6, 4, 1, 4, 6]

Answer 2

Here is an approach that doesn't require the use of a for loop or defining an intermediate function or lambda function (and is also a one-liner). The method combines the following two ideas:

• calling the iter() built-in function with the optional sentinel argument, and

• using the itertools recipe for advancing an iterator n steps (see the recipe for consume()).

Putting these together, we get:

next(islice(iter(do, object()), 3, 3), None)

(The idea to pass object() as the sentinel comes from this accepted Stack Overflow answer.)

And here is what this looks like from the interactive prompt:

>>> def do():
...   print("called")
... 
>>> next(itertools.islice(iter(do, object()), 3, 3), None)
called
called
called

Answer 3

You could define a function that repeats the passed function N times.

def repeat_fun(times, f):
    for i in range(times): f()

If you want to make it even more flexible, you can even pass arguments to the function being repeated:

def repeat_fun(times, f, *args):
    for i in range(times): f(*args)

Usage:

>>> def do():
...   print 'Doing'
... 
>>> def say(s):
...   print s
... 
>>> repeat_fun(3, do)
Doing
Doing
Doing
>>> repeat_fun(4, say, 'Hello!')
Hello!
Hello!
Hello!
Hello!