How to do multiple arguments to map function where one remains the same in python?

Question

One option is a list comprehension:

[add(x, 2) for x in [1, 2, 3]]

More options:

a = [1, 2, 3]

import functools
map(functools.partial(add, y=2), a)

import itertools
map(add, a, itertools.repeat(2, len(a)))

The docs explicitly suggest this is the main use for itertools.repeat:

Make an iterator that returns object over and over again. Runs indefinitely unless the times argument is specified. Used as argument to map() for invariant parameters to the called function. Also used with zip() to create an invariant part of a tuple record.

And there's no reason for pass len([1,2,3]) as the times argument; map stops as soon as the first iterable is consumed, so an infinite iterable is perfectly fine:

>>> from operator import add
>>> from itertools import repeat
>>> list(map(add, [1,2,3], repeat(4)))
[5, 6, 7]

In fact, this is equivalent to the example for repeat in the docs:

>>> list(map(pow, range(10), repeat(2)))
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

This makes for a nice lazy-functional-language-y solution that's also perfectly readable in Python-iterator terms.

Use a list comprehension.

[x + 2 for x in [1, 2, 3]]

If you really, really, really want to use map, give it an anonymous function as the first argument:

map(lambda x: x + 2, [1,2,3])

Map can contain multiple arguments, the standard way is

map(add, a, b)

In your question, it should be

map(add, a, [2]*len(a))