How to merge lists into a list of tuples?
Question:
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.
Answers:
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)]
Youre looking for the builtin function zip.
In python 3.0 zip returns a zip object. You can get a list out of it by calling list(zip(a, b)).
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)<len(list_2):
list_1.append('')
card = (list_1[i], list_2[i])
continue
deck.append(card)
break
>>>
>>> #and the result should be:
>>> print deck
>>> [('Ace', 'Spades'), ('King', 'Clubs'), ('', 'Diamonds')]
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
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))
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]
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))]
Or map with unpacking:
>>> list(map(lambda *x: x, list_a, list_b))
[(1, 5), (2, 6), (3, 7), (4, 8)]
>>>
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]]
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.
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)]
Youre looking for the builtin function zip.
In python 3.0 zip returns a zip object. You can get a list out of it by calling list(zip(a, b)).
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)<len(list_2):
list_1.append('')
card = (list_1[i], list_2[i])
continue
deck.append(card)
break
>>>
>>> #and the result should be:
>>> print deck
>>> [('Ace', 'Spades'), ('King', 'Clubs'), ('', 'Diamonds')]
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
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))
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]
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))]
Or map with unpacking:
>>> list(map(lambda *x: x, list_a, list_b))
[(1, 5), (2, 6), (3, 7), (4, 8)]
>>>
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]]