CODEWITHSUNDEEP
python
assda
assda

Reputation: 3

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

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?

Upvotes: 0

Views: 100

Answers (5)

Hamza usman ghani
Hamza usman ghani

Reputation: 2243

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')]

Upvotes: 1

williamr21
williamr21

Reputation: 147

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')]

Upvotes: 0

AmineBTG
AmineBTG

Reputation: 697

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)]

Upvotes: 0

Stan11
Stan11

Reputation: 284

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.

Upvotes: 0

crissal
crissal

Reputation: 2647

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')]

Upvotes: 0

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