Use map() function with string

Question

Is there a way to use map() function with a string instead of a list? Or the map() function is meant only to work with lists?

For instance, ignoring the content of the lambda function, this code returns a map object and not a string:

def rot_13(string):
    alph = 'abcdefghijklmnopqrstuwxyz'
    return str(map(lambda i: alph[(alph.find(i)+13) % len(alph)], string))

map returns a generator in Python3, try: return "".join(list(map(lambda i: alph[(alph.find(i)+13) % len(alph)], string))) — Maurice Meyer
– Maurice Meyer, Commented May 23, 2018 at 15:27
You can turn a string into a list with something like a_list = list(a_string). — martineau
– martineau, Commented May 23, 2018 at 15:40
Yes. map always returns a map object. No matter the iterable you are napping over. It is meant to work with any iterable. — juanpa.arrivillaga
– juanpa.arrivillaga, Commented May 23, 2018 at 16:06

AKX · Accepted Answer · 2018-05-23 15:29:55Z

10

Using list/generator comprehensions is preferable (more Pythonic) to map():

def rot_13(string):
    alph = 'abcdefghijklmnopqrstuwxyz'
    return ''.join(alph[(alph.find(i)+13) % len(alph)] for i in string)

answered May 23, 2018 at 15:29

AKX

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Comments

Netwave · Accepted Answer · 2018-05-23 15:44:18Z

8

python3 map returns a generator, so you have to consume it, using str.join is the easiest way:

"".join(map(lambda i: alph[(alph.find(i)+13) % len(alph)], string))

It is similar to this code:

my_string = ""
for l in map(lambda i: alph[(alph.find(i)+13) % len(alph)], string):
    my_string += l

answered May 23, 2018 at 15:44

Netwave

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map does not return a generator, it returns a map-object, which is an iterator but not a generator.