Pythonic way of manipulating nested list into nested chunked list

Question

I have a list whose nested list's size may vary with the multiple of 2. Currently, in this example, the nested list's length is 4.

a_list = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]

According to length, I am trying to break the list to get the following result in the best possible pythonic way:

a = [[1,2], [5,6], [9,10]]
b = [[3,4], [7,8], [11,12]]

and if nested list's length is 6, then

c = [[..], [..], [..]]

Its kind of a transpose of a nested list but with sets of 2 values in a single row not to be transposed.

Which values would go in the three lists of your second example? — alexis, Sep 29 '14 at 14:37
@alexis like if a_list = [[1,2,3,4,100,102],[5,6,7,8,103,104],[9,10,11,12,105,106]], then c = [[100,102],[103,104],[105,106]] — Ali Raza Bhayani, Sep 29 '14 at 14:45

score 6 · Accepted Answer · answered Sep 29 '14 at 14:34

6

Using list comprehension:

>>> a_list = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
>>> a = [x[:2] for x in a_list]
>>> b = [x[2:] for x in a_list]
>>> a
[[1, 2], [5, 6], [9, 10]]
>>> b
[[3, 4], [7, 8], [11, 12]]

More general solution:

>>> [[x[i:i+2] for x in a_list] for i in range(0, len(a_list[0]), 2)]
[[[1, 2], [5, 6], [9, 10]],
 [[3, 4], [7, 8], [11, 12]]]

answered Sep 29 '14 at 14:34

falsetru

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@thefourtheye, I added another code that is more general. – falsetru Sep 29 '14 at 14:35
1

I think it would be best to not use tuple unpacking in the second example, as the idea is that it can handle any length of sublist. Just store the results in a new nested list. – Roger Fan Sep 29 '14 at 14:37
@RogerFan, I updated the answer accordingly. Thank you for suggestion. – falsetru Sep 29 '14 at 14:39
@falsetru Nice one, my approach was also similar. But yours looks simple. – thefourtheye Sep 29 '14 at 14:51

score 2 · Answer 2 · answered Sep 29 '14 at 14:52

2

I'd hesitate to call this "pythonic", since it's pretty much illegible, but:

>>> a, b = zip(*(zip(*[iter(s)]*2) for s in a_list))
>>> a
((1, 2), (5, 6), (9, 10))
>>> b
((3, 4), (7, 8), (11, 12))

Also works for 6-item lists:

>>> a_list = [[1,2,3,4,100,102],[5,6,7,8,103,104],[9,10,11,12,105,106]]
>>> a, b, c = zip(*(zip(*[iter(s)]*2) for s in a_list))
>>> a
((1, 2), (5, 6), (9, 10))
>>> b
((3, 4), (7, 8), (11, 12))
>>> c
((100, 102), (103, 104), (105, 106))

answered Sep 29 '14 at 14:52

Zero Piraeus

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+1 for `zip(*[iter(s)]*2)`, I've never seen this idiom :-) I _would_ call your answer pythonic: Iterator solutions are pythonic, and this has fewer components than the slice solution-- as long as you speak iterator. – alexis Sep 29 '14 at 15:46
For the general-purpose solution, one can simply turn the result of `zip` into a list (in python 2 it already is a list) and save it in a variable. – alexis Sep 29 '14 at 15:51

thefourtheye · Answer 3 · 2014-09-29T14:54:03.047

Almost the same as falsetru's answer, but first the nested lists are split into chunks of size 2 and then all of them are zipped together.

>>> a_list = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]
>>> zip(*([j[i*2: i*2 + 2] for i in range(len(j) / 2)] for j in a_list))
[([1, 2], [5, 6], [9, 10]), ([3, 4], [7, 8], [11, 12])]

>>> a_list = [[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12]]
>>> zip(*([j[i*2: i*2 + 2] for i in range(len(j) / 2)] for j in a_list))
[([1, 2], [7, 8]), ([3, 4], [9, 10]), ([5, 6], [11, 12])]

>>> a_list = [[1,2,3,4,100,102],[5,6,7,8,103,104],[9,10,11,12,105,106]]
>>> zip(*([j[i*2: i*2 + 2] for i in range(len(j) / 2)] for j in a_list))
[([1, 2], [5, 6], [9, 10]), ([3, 4], [7, 8], [11, 12]), ([100, 102], [103, 104], [105, 106])]

Mazdak · Answer 4 · 2014-09-29T14:59:12.100

1

A Fast way is using numpy.hsplit :

>>> import numpy
>>> numpy.hsplit(numpy.array(a_list),2)
[array([[ 1,  2],[ 5,  6],[ 9, 10]]),array([[ 3,  4],[ 7,  8],[11, 12]])]

edited Sep 29 '14 at 14:59

answered Sep 29 '14 at 14:53

Mazdak

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score 1 · Answer 5 · edited May 23 '17 at 11:59

Since readability is pythonic, here's a simpler iterator-based solution (without the neat tricks that @Zero used to turn it into a one-liner):

First, an iterator that turns a list [1,2,3,4,5,6] into [(1, 2), (3, 4), (5, 6)].

def pairs(lst):
    it=iter(lst)
    return list(zip(it, it))   # Return a list of pairs drawn from the same iterator

The list a_list can be transformed into a list of such pair lists as follows:

a_list = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]    
pair_list = [ pairs(row) for row in a_list ]

Finally, we need to effectively transpose this list, making a list of the first pair/element from each sublist, another list of the second one, etc. A nice idiom for transposing a list is zip(*some_list). Let's use it to make the transformation requested by the OP:

a, b = zip(*pair_list)

or to collect any number of generated lists in one list:

results = list( zip(*pair_list) )

Feel free to pack these into a one-liner (though I wouldn't):

results = list(zip( *(pairs(row) for row in a_list) ))

Pythonic way of manipulating nested list into nested chunked list

5 Answers5