How to covert a list into a list with sublists? python

Question

I'm making a little data base on python.

my_list = ['jack', 'liam', 'gary', 'poly']

I want to convert the list into this

result = [('jack', 'liam'),('liam', 'gary'),('gary', 'poly')]

Is there anyway i can do this?

Answer 1

Use list comprehension

• Iterate till second last value of the list
• get current and next value as a tuple.

my_list = ['jack', 'liam', 'gary', 'poly']
new_list = [(my_list[i],my_list[i+1]) for i in range(len(my_list)-1)]

print(new_list)
>> [('jack', 'liam'), ('liam', 'gary'), ('gary', 'poly')]

Or Use zip()

which will return consecutive combinations as tuple.

my_list = ['jack', 'liam', 'gary', 'poly']
new_list = list(zip(my_list, my_list[1:]))

print(new_list)
>> [('jack', 'liam'), ('liam', 'gary'), ('gary', 'poly')]

Answer 2

zip will return a tuple from two iterables; if you take the same list, but shifted as you wish (in this case, one element forward) you can have your expected result.

Also, the generator will exhaust on the shortest iterable (the second one).

>>> [(a,b) for (a,b) in zip(my_list, my_list[1:])]
[('jack', 'liam'), ('liam', 'gary'), ('gary', 'poly')]

Answer 3

my_list = ['jack', 'liam', 'gary', 'poly']

result_list = [(my_list[i],my_list[i+1]) for i in range(len(my_list) - 1)]

print(result_list)

Output

[('jack', 'liam'), ('liam', 'gary'), ('gary', 'poly')]

We can fetch 2 elements into a tuple till we iterate and reach the second last element.

Answer 4

result = []
my_list = ['jack', 'liam', 'gary', 'poly']
for i, item in enumerate(my_list):
   if i+1 == len(my_list): break
   result.append((item, my_list[i+1]))

Or more Pythonic and elegant way:

result = [(item, my_list[i+1]) for i, item in enumerate(my_list) if i+1 < len(my_list)] 

Answer 5

my_list = ['jack', 'liam', 'gary', 'poly']
result = []
for i in range(0, len(my_list)):
    if not i > len(my_list)-2:
        result.append((my_list[i], my_list[i+1]))

output

[('jack', 'liam'), ('liam', 'gary'), ('gary', 'poly')]