Python: [[1,2], [3,4], [5,6], [7,8]] are converted to [[1], [2,3], [4,5], [6,7 ], [8]] and vice versa

Question

my current pointer solutions will be

• ether through an iterator class that gives new compiled internal lists
• or through the iter function, which gives new compiled internal lists

is there any other better way to solve this problem?

Edit

@Glenn: good objection. I did not think about it because I experienced lists that were not ordered as I thought.

@ THC4k: Thank you for your decision. I recognized chain.from_iterable

@Mike DeSimone: Hm tested your solution, but something went wrong, maybe I missed something, ...

@Jamie and Odomontois: Thanks for pointing in more detail

my goal

, - /: (start, stop) - , .

: , (s1 s2 e2 e1)

:

• "taskList" ( - :).
• datetime: start .
• : "taskList" , ,
• 'taskList' , datetime

,

taskList = [(9:00,10:00),(9:30,11:00),(11:00,12:30),(13:30,14:00),(14:00,18:00)]

end:

result = [(9:00,12:30), (13:30,18:00)]

, "taskList" ,

taskListT1 = [(9:00,),(10:00,9:30),(11:00,11:00),(12:30,13:30),(14:00,14:00),(18:00,)]

(a, b), a >= b:

taskListT2 = [(9:00,),(12:30,13:30),(18:00,)]

:

result = [(9:00,12:30), (13:30,18:00)]

Answer 1

, :

# transform forwards
def transform_pairs( lst ):
    it = iter(lst)
    a,last = next(it)
    yield [a]
    for a,b in it:
        yield last, a
        last = b
    yield [last]

, .

, . , .

from itertools import chain

def transform( iterable, offset):
    it = chain.from_iterable(iterable) # turn it back to one long list.
    if offset:
        yield next(it), # the trailing `,` makes this a tuple.
    for item in it:
        try:
            x = next(it)
        except StopIteration: # there is no 2nd item left
            yield item,
        else:
             yield item, x # yield the pair

print list(transform(transform([[1,2],[3,4],[5,6],[7,8]], True), False))

Answer 2

", " ( OP ):

def compress_task_list(tasks):
    tasks = list(tasks)
    tasks.sort(key=lambda item: item[0]) # make sure list is in order by start time
    result = []
    first_start = tasks[0][0]
    final_stop = tasks[0][1]
    for start, stop in tasks[1:]:
        if start > final_stop:
            result.append((first_start, final_stop))
            first_start = start
            final_stop = stop
        elif stop > final_stop:
            final_stop = stop
    result.append((first_start, final_stop))
    return tuple(result)

if __name__ == '__main__':
    import unittest

    class Test_Compress_Task_List(unittest.TestCase):
        def test_01(self):
            "completely separate"
            initial = ((8.0, 9.5), (10.0, 12.0), (13.0, 15.5), (16.0, 17.0))
            expected = ((8.0, 9.5), (10.0, 12.0), (13.0, 15.5), (16.0, 17.0))
            self.assertEqual(compress_task_list(initial), expected)
        def test_02(self):
            "end equals start"
            initial = ((8.0, 9.5), (9.5, 12.0), (13.0, 15.5), (15.5, 17.0))
            expected = ((8.0, 12.0), (13.0, 17.0))
            self.assertEqual(compress_task_list(initial), expected)
        def test_03(self):
            "end equals start (with more empty times)"
            initial = ((8.0, 8.5), (8.5, 10.0), (10.25, 12.0), (12.5, 13.75), (13.75, 15.0), (15.25, 16.0), (16.0, 17.0))
            expected = ((8.0, 10.0), (10.25, 12.0), (12.5, 15.0), (15.25, 17.0))
            self.assertEqual(compress_task_list(initial), expected)
        def test_04(self):
            "out of order, cross-overs, and tasks completely inside other tasks"
            initial = ((8.0, 8.5), (8.0, 10.0), (10.25, 12.0), (10.0, 11.5), (13.0, 15.5), (14.0, 15.0), (16.0, 17.0))
            expected = ((8.0, 12.0), (13.0, 15.5), (16.0, 17.0))
            self.assertEqual(compress_task_list(initial), expected)

    unittest.main()

, Python .;)

Answer 3

, , Pythonic:

l = [[1,2], [3,4], [5,6], [7,8]]
o = []
last = []
for a, b in l:
    o.append(last+[a])
    last = [b]
o.append(last)

print o

[[1], [2, 3], [4, 5], [6, 7], [8]]

:

o = [[l[0][0]]]
for i in range(len(l)-1):
    o.append([l[i][1], l[i+1][0]])
o.append([l[-1][1]])

Answer 4

l = [[1,2], [3,4], [5,6], [7,8]]
m = [([] if i==0 else [l[i-1][1]] )+([] if i==len(l) else [l[i][0]]) for i in xrange(len(l)+1)]

Answer 5

:

pairs = [[1,2], [3,4], [5,6], [7,8]]
print pairs, '->',
transformed = ([[pairs[0][0]]]+
               [[a,b] for a,b in zip(
                   (second for first, second in pairs[:-1]),
                   (first for first, second in pairs[1:]))]+
               [[pairs[-1][-1]]]
               )
print transformed
""" Output:
[[1, 2], [3, 4], [5, 6], [7, 8]] -> [[1], [2, 3], [4, 5], [6, 7], [8]]
"""

Answer 6

This is a generator that can work with the input of a generator or list, so you do not need to store everything in memory:

def shift_pairs(inPairs):
    lastPair = None
    for pair in inPairs:
        if lastPair:
            yield [lastPair[1], pair[0]]
        else:
            yield [pair[0]]
        lastPair = pair
    yield [lastPair[1]]

I must point out that in Python, short fixed-length lists usually execute as tuples:

def shift_pairs(inPairs):
    lastPair = None
    for pair in inPairs:
        if lastPair:
            yield (lastPair[1], pair[0])
        else:
            yield (pair[0],)
        lastPair = pair
    yield (lastPair[1],)