I am new to Python and I would like to combine two arrays. I have two arrays: A and B and would like to get array C as follows:
A = [1, 2, 3, 4, 5]
B = [6, 7, 8, 9, 10]
Result array:
C = [(1, 6), (2, 7), (3, 8), (4, 9), (5, 10)]
2 Answers
Use zip, the built-in for pairwise iteration:
C = list(zip(A, B))
Note that you are not concatenating the two lists.
3 Comments
Pratheek Hb
Thanks for your solution. But, the following solution resulted in C= [('1', '6'), ('2', '7'), ('3', '8'), ('4', '9'), ('5', '10')] which is not correct. I need C = [(1, 6), (2, 7), (3, 8), (4, 9), (5, 10)]
Tomerikoo
@PratheekHb this means you have a list of strings and not ints as you presented in the question...
user2390182
Then your input was not what you showed in the question, but
['1', '2', '3', '4', '5'], etc. You can easily fix this: list(zip(map(int, A), map(int, B)))Without any dependency:
>>> A = [1, 2, 3, 4, 5]
>>> B = [6, 7, 8, 9, 10]
>>> C = [(A[i], B[i]) for i in range(len(A))]
>>> C
[(1, 6), (2, 7), (3, 8), (4, 9), (5, 10)]
4 Comments
ApproachingDarknessFish
Incorrect. 1.
len(A) is not an iterable. 2. If you change len(A) to range(len(A)), this approach will raise an IndexError if len(B) > len(A).
Blusky
Corrected with range (misread history). The behavior if
len(A) != len(B) has not been specified, so nothing wrong in that case.user2390182
Instructive solution.
zip is, however, not more of a dependency than range or len. The are all just built-ins.Pratheek Hb
The following solution resulted in C= [('1', '6'), ('2', '7'), ('3', '8'), ('4', '9'), ('5', '10')] which is not correct. I need C = [(1, 6), (2, 7), (3, 8), (4, 9), (5, 10)]
A and Bwould beA + B = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10].zipfunction is not as widely used as it should be, so the question already had its merit. Additionally, because of the unconventional description of 'concatenation', googling for it would have never yielded the desired results. Furthermore it was well asked, provided a nicely formatted example and described it very well. That is well asked question, especially for a newcommer! Welcome to stackoverflow!