Python 3 Map function is not Calling up function

Question

Why doesn\'t following code print anything:

#!/usr/bin/python3
class test:
    def do_someting(self,value):
        print(value)
        return value

    de         


        

Answer 1

map() returns an iterator, and will not process elements until you ask it to.

Turn it into a list to force all elements to be processed:

list(map(self.do_someting,range(10)))

or use collections.deque() with the length set to 0 to not produce a list if you don't need the map output:

from collections import deque

deque(map(self.do_someting, range(10)))

but note that simply using a for loop is far more readable for any future maintainers of your code:

for i in range(10):
    self.do_someting(i)

Answer 2

I just want to add the following:

With multiple iterables, the iterator stops when the shortest iterable is exhausted [ https://docs.python.org/3.4/library/functions.html#map ]

Python 2.7.6 (default, Mar 22 2014, 22:59:56)

>>> list(map(lambda a, b: [a, b], [1, 2, 3], ['a', 'b']))
[[1, 'a'], [2, 'b'], [3, None]]

Python 3.4.0 (default, Apr 11 2014, 13:05:11)

>>> list(map(lambda a, b: [a, b], [1, 2, 3], ['a', 'b']))
[[1, 'a'], [2, 'b']]

That difference makes the answer about simple wrapping with list(...) not completely correct

The same could be achieved with:

>>> import itertools
>>> [[a, b] for a, b in itertools.zip_longest([1, 2, 3], ['a', 'b'])]
[[1, 'a'], [2, 'b'], [3, None]]

Answer 3

Before Python 3, map() returned a list, not an iterator. So your example would work in Python 2.7.

list() creates a new list by iterating over its argument. ( list() is NOT JUST a type conversion from say tuple to list. So list(list((1,2))) returns [1,2]. ) So list(map(...)) is backwards compatible with Python 2.7.