python 3 map/lambda method with 2 inputs

Question

I have a dictionary like the following in python 3:

ss = {\'a\':\'2\', \'b\',\'3\'}

I want to convert all he values to int using map

Answer 1

ss.items() will give an iterable, which gives tuples on every iteration. In your lambda function, you have defined it to accept two parameters, but the tuple will be treated as a single argument. So there is no value to be passed to the second parameter.

1. You can fix it like this

   print(list(map(lambda args: int(args[1]), ss.items())))
   # [3, 2]
   

2. If you are ignoring the keys anyway, simply use ss.values() like this

   print(list(map(int, ss.values())))
   # [3, 2]
   

3. Otherwise, as suggested by Ashwini Chaudhary, using itertools.starmap,

   from itertools import starmap
   print(list(starmap(lambda key, value: int(value), ss.items())))
   # [3, 2]
   

4. I would prefer the List comprehension way

   print([int(value) for value in ss.values()])
   # [3, 2]
   

In Python 2.x, you could have done that like this

print map(lambda (key, value): int(value), ss.items())

This feature is called Tuple parameter unpacking. But this is removed in Python 3.x. Read more about it in PEP-3113