Grouping lists within lists in Python 3

Question

I have a list of lists of strings like so:

List1 = [
          [\'John\', \'Doe\'], 
          [\'1\',\'2\',\'3\'], 
          [\'Henry\', \'Doe\'], 
          [         


        

Answer 1

This should do what you want assuming you always want to take pairs of the inner lists together.

list1 = [['John', 'Doe'], ['1','2','3'], ['Henry', 'Doe'], ['4','5','6']] 
output = [list(pair) for pair in zip(list1[::2], list1[1::2])]

It uses zip, which gives you tuples, but if you need it exactly as you've shown, in lists, the outer list comprehension does that.

Answer 2

Here it is in 8 lines. I used tuples rather than lists because it's the "correct" thing to do:

def pairUp(iterable):
    """
        [1,2,3,4,5,6] -> [(1,2),(3,4),(5,6)]
    """
    sequence = iter(iterable)
    for a in sequence:
        try:
            b = next(sequence)
        except StopIteration:
            raise Exception('tried to pair-up %s, but has odd number of items' % str(iterable))
        yield (a,b)

---

Demo:

>>> list(pairUp(range(0)))    
[]

>>> list(pairUp(range(1)))
Exception: tried to pair-up [0], but has odd number of items

>>> list(pairUp(range(2)))
[(0, 1)]

>>> list(pairUp(range(3)))
Exception: tried to pair-up [0, 1, 2], but has odd number of items

>>> list(pairUp(range(4)))
[(0, 1), (2, 3)]

>>> list(pairUp(range(5)))
Exception: tried to pair-up [0, 1, 2, 3, 4], but has odd number of items

---

Concise method:

zip(sequence[::2], sequence[1::2])
# does not check for odd number of elements

Answer 3

List1 = [['John', 'Doe'], ['1','2','3'],
         ['Henry', 'Doe'], ['4','5','6'],
         ['Bob', 'Opoto'], ['10','11','12']]

def pairing(iterable):
    it = iter(iterable)
    itn = it.next
    for x in it :
        yield (x,itn())     

# The generator pairing(iterable) yields tuples:  

for tu in pairing(List1):
    print tu  

# produces:  

(['John', 'Doe'], ['1', '2', '3'])
(['Henry', 'Doe'], ['4', '5', '6'])
(['Bob', 'Opoto'], ['8', '9', '10'])    

# If you really want a yielding of lists:

from itertools import imap
# In Python 2. In Python 3, map is a generator
for li in imap(list,pairing(List1)):
    print li

# or defining pairing() precisely so:

def pairing(iterable):
    it = iter(iterable)
    itn = it.next
    for x in it :
        yield [x,itn()]

# produce   

[['John', 'Doe'], ['1', '2', '3']]
[['Henry', 'Doe'], ['4', '5', '6']]
[['Bob', 'Opoto'], ['8', '9', '10']]

Edit: Defining a generator function isn't required, you can do the pairing of a list on the fly:

List1 = [['John', 'Doe'], ['1','2','3'],
         ['Henry', 'Doe'], ['4','5','6'],
         ['Bob', 'Opoto'], ['8','9','10']]

it = iter(List1)
itn = it.next
List1 = [ [x,itn()] for x in it]