CODEWITHSUNDEEP
python
user2210605
user2210605

Reputation: 11

using map in python 3

Hi this code works in python 2.7 but not in python 3

import itertools
def product(a,b):
    return map(list, itertools.product(a, repeat=b)
print(sorted(product({0,1}, 3)))

it outputs

  [[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 1, 1], [1, 0, 0], [1, 0, 1], [1, 1, 0], [1, 1, 1]]

in python 2.7 but in python 3 it gives map object at 0x028DB2F0 does anyone know how to change this to work for python 3 so the output remains the same as in python 2.7

Upvotes: 1

Views: 1607

Answers (2)

James Henstridge
James Henstridge

Reputation: 43899

In Python 3, the map() builtin returns an iterator rather than a list, behaving somewhat like the Python 2 itertools.imap() function.

If you need a list, you can simply pass that iterator to list(). For example:

>>> x = map(lambda x: x + 1, [1, 2, 3])
>>> x
<map object at 0x7f8571319b90>
>>> list(x)
[2, 3, 4]

Upvotes: 1

Artsiom Rudzenka
Artsiom Rudzenka

Reputation: 29103

Simply wrap it by a list this way:

import itertools

def product(a,b):
    return list(map(list, itertools.product(a, repeat=b))

print(sorted(product({0,1}, 3)))

Find more Getting a map() to return a list in Python 3.x

In two words in Python 3 map

Return an iterator that applies function to every item of iterable, yielding the results.

While in python 2.7 it

Apply function to every item of iterable and return a list of the results.

Upvotes: 2

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