Python 2 vs 3: Lambda Operator

Question

Has there been a change in the way lambda functions work between Python 2 and 3? I ran the code in python 2 and it works fine, but fails in Python 3 which I am trying to port my code into in order take advantage of a 3rd party module.

pos_io_tagged = map(lambda ((word, pos_tag), io_tag):
    (word, pos_tag, io_tag), zip(zip(words, pos_tags), io_tags))

I have researched multiple questions on Stackoverflow and read a couple of articles such as this but still can't find the answer. Is there any resources that I can view?

And the way you can use argument assignment has also changed; unpacking is no longer supported. — Martijn Pieters, Apr 01 '15 at 16:08
Check out http://www.diveintopython3.net/porting-code-to-python-3-with-2to3.html#tuple_params for info on passing tuples to `lambda` functions. — MattDMo, Apr 01 '15 at 16:10
I'm going to go the assumption you got a `SyntaxError: invalid syntax` on the `lambda` signature here. In future, **always** include the full error message! — Martijn Pieters, Apr 01 '15 at 16:10
`lambda wpio: wpio[0] + (wpio[1],)` would do the same thing without tuple argument unpacking. — Martijn Pieters, Apr 01 '15 at 16:12

score 9 · Accepted Answer · answered Apr 01 '15 at 16:10

Your problem is that you are using parentheses () with your lambda expression, which will confuse it. Try the following:

pos_io_tagged = map(lambda word, pos_tag, io_tag:
    (word, pos_tag, io_tag), zip(zip(words, pos_tags), io_tags))

Look here for more information.

Python 2 vs 3: Lambda Operator

1 Answers1