Get the first element of each tuple in a list in Python

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00:00 Question
00:33 Accepted answer (Score 238)
01:11 Answer 2 (Score 41)
01:24 Answer 3 (Score 24)
01:42 Answer 4 (Score 13)
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Full question
https://stackoverflow.com/questions/2241...

Accepted answer links:
[list comprehension]: http://docs.python.org/3/tutorial/datast...

Answer 3 links:
[map]: https://docs.python.org/2/library/functi...
[operator.itemgetter]: https://docs.python.org/2/library/operat...

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Tags
#python #python27 #syntax

#avk47

ACCEPTED ANSWER

res_list = [x[0] for x in rows]

Below is a demonstration:

>>> rows = [(1, 2), (3, 4), (5, 6)]
>>> [x[0] for x in rows]
[1, 3, 5]
>>>

Alternately, you could use unpacking instead of x[0]:

res_list = [x for x,_ in rows]

Below is a demonstration:

>>> lst = [(1, 2), (3, 4), (5, 6)]
>>> [x for x,_ in lst]
[1, 3, 5]
>>>

Both methods practically do the same thing, so you can choose whichever you like.

If you don't want to use list comprehension by some reasons, you can use map and operator.itemgetter:

>>> from operator import itemgetter
>>> rows = [(1, 2), (3, 4), (5, 6)]
>>> map(itemgetter(1), rows)
[2, 4, 6]
>>>

You can use list comprehension:

res_list = [i[0] for i in rows]

This should make the trick

res_list = [x[0] for x in rows]

For a discussion on why to prefer comprehensions over higher-order functions such as map, go to http://www.artima.com/weblogs/viewpost.jsp?thread=98196.