What is the Pythonic approach to achieve the following?
# Original lists:
list_a = [1, 2, 3, 4]
list_b = [5, 6, 7, 8]
# List of tuples from 'list_a' and 'list_b':
list_c = [[1,5], [2,6], [3,7], [4,8]]
Each member of list_c is a tuple, whose first member is from list_a and the second is from list_b.
asked Mar 9, 2010 at 7:51
0
In Python 2:
>>> list_a = [1, 2, 3, 4]
>>> list_b = [5, 6, 7, 8]
>>> zip[list_a, list_b]
[[1, 5], [2, 6], [3, 7], [4, 8]]
In Python 3:
>>> list_a = [1, 2, 3, 4]
>>> list_b = [5, 6, 7, 8]
>>> list[zip[list_a, list_b]]
[[1, 5], [2, 6], [3, 7], [4, 8]]
mrgloom
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answered Mar 9, 2010 at 7:52
YOUYOU
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5
In python
3.0 zip returns a zip object. You can get a list out of it by calling list[zip[a, b]].
jamylak
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answered Feb 28, 2011 at 19:26
LodewijkLodewijk
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You can use map lambda
a = [2,3,4]
b = [5,6,7]
c = map[lambda x,y:[x,y],a,b]
This will also work if there lengths of original lists do not match
answered Jul 4, 2015 at 5:49
Dark KnightDark Knight
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6
Youre looking for the builtin function zip.
answered Mar 9, 2010 at 7:55
MizipzorMizipzor
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I am not sure if this a pythonic way or not but this seems simple if both lists have the same number of elements :
list_a = [1, 2, 3, 4]
list_b = [5, 6, 7, 8]
list_c=[[list_a[i],list_b[i]] for i in range[0,len[list_a]]]
Jee Mok
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answered Sep 11, 2018 at 7:51
VipinVipin
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The output which you showed in problem statement is not the tuple but list
list_c = [[1,5], [2,6], [3,7], [4,8]]
check for
type[list_c]
considering you want the result as tuple out of list_a and list_b, do
tuple[zip[list_a,list_b]]
answered May 12, 2016 at 14:52
cyborgcyborg
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I know this is an old question and was already answered, but for some reason, I still wanna post this alternative solution. I know it's easy to just find out which built-in function does the "magic" you need, but it doesn't hurt to know you can do it by yourself.
>>> list_1 = ['Ace', 'King']
>>> list_2 = ['Spades', 'Clubs', 'Diamonds']
>>> deck = []
>>> for i in range[max[[len[list_1],len[list_2]]]]:
while True:
try:
card = [list_1[i],list_2[i]]
except IndexError:
if len[list_1]>len[list_2]:
list_2.append['']
card = [list_1[i],list_2[i]]
elif len[list_1]>>
>>> #and the result should be:
>>> print deck
>>> [['Ace', 'Spades'], ['King', 'Clubs'], ['', 'Diamonds']]
answered Feb 17, 2014 at 10:01
KrugerKruger
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Or map with unpacking:
>>> list[map[lambda *x: x, list_a, list_b]]
[[1, 5], [2, 6], [3, 7], [4, 8]]
>>>
answered Nov 9, 2021 at 5:39
U12-ForwardU12-Forward
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One alternative without using zip:
list_c = [[p1, p2] for idx1, p1 in enumerate[list_a] for idx2, p2 in enumerate[list_b] if idx1==idx2]
In case one wants to get not only tuples 1st with 1st, 2nd with 2nd... but all possible combinations of the 2 lists, that would be done with
list_d = [[p1, p2] for p1 in list_a for p2 in list_b]
answered May 15, 2018 at 13:57
J0ANMMJ0ANMM
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Like me, if anyone needs to convert it to list of lists [2D lists] instead of list of tuples, then you could do the following:
list[map[list, list[zip[list_a, list_b]]]]
It should return a 2D List as follows:
[[1, 5],
[2, 6],
[3, 7],
[4, 8]]
answered Jan 15 at 6:44