Better algorithm to riffle shuffle (or interleave) multiple lists of varying lengths
I like to watch my favorite TV shows on the go. I have all episodes of each show I'm following in my playlist. Not all shows consist of the same number of episodes. Unlike some who prefer marathons, I like to interleave episodes of one show with those of another.
For example, if I have a show called ABC with 2 episodes, and a show called XYZ with 4 episodes, I would like my playlist to look like:
XYZe1.mp4
ABCe1.mp4
XYZe2.mp4
XYZe3.mp4
ABCe2.mp4
XYZe4.mp4
One 开发者_高级运维way to generate this interleaved playlist is to represent each show as a list of episodes and perform a riffle shuffle on all shows. One could write a function that would compute, for each episode, its position on a unit-time interval (between 0.0 and 1.0 exclusive, 0.0 being beginning of season, 1.0 being end of season), then sort all episodes according to their position.
I wrote the following simple function in Python 2.7 to perform an in-shuffle:
def riffle_shuffle(piles_list):
scored_pile = ((((item_position + 0.5) / len(pile), len(piles_list) - pile_position), item) for pile_position, pile in enumerate(piles_list) for item_position, item in enumerate(pile))
shuffled_pile = [item for score, item in sorted(scored_pile)]
return shuffled_pile
To get the playlist for the above example, I simply need to call:
riffle_shuffle([['ABCe1.mp4', 'ABCe2.mp4'], ['XYZe1.mp4', 'XYZe2.mp4', 'XYZe3.mp4', 'XYZe4.mp4']])
This works fairly well most of the time. However, there are cases where results are non-optimal--two adjacent entries in the playlist are episodes from the same show. For example:
>>> riffle_shuffle([['ABCe1', 'ABCe2'], ['LMNe1', 'LMNe2', 'LMNe3'], ['XYZe1', 'XYZe2', 'XYZe3', 'XYZe4', 'XYZe5']])
['XYZe1', 'LMNe1', 'ABCe1', 'XYZe2', 'XYZe3', 'LMNe2', 'XYZe4', 'ABCe2', 'LMNe3', 'XYZe5']
Notice that there are two episodes of 'XYZ' that appear side-by-side. This situation can be fixed trivially (manually swap 'ABCe1' with 'XYZe2').
I am curious to know if there are better ways to interleave, or perform riffle shuffle, on multiple lists of varying lengths. I would like to know if you have solutions that are simpler, more efficient, or just plain elegant.
Solution proposed by belisarius (thanks!):
import itertools
def riffle_shuffle_belisarius(piles_list):
def grouper(n, iterable, fillvalue=None):
args = [iter(iterable)] * n
return itertools.izip_longest(fillvalue=fillvalue, *args)
if not piles_list:
return []
piles_list.sort(key=len, reverse=True)
width = len(piles_list[0])
pile_iters_list = [iter(pile) for pile in piles_list]
pile_sizes_list = [[pile_position] * len(pile) for pile_position, pile in enumerate(piles_list)]
grouped_rows = grouper(width, itertools.chain.from_iterable(pile_sizes_list))
grouped_columns = itertools.izip_longest(*grouped_rows)
shuffled_pile = [pile_iters_list[position].next() for position in itertools.chain.from_iterable(grouped_columns) if position is not None]
return shuffled_pile
Example run:
>>> riffle_shuffle_belisarius([['ABCe1', 'ABCe2'], ['LMNe1', 'LMNe2', 'LMNe3'], ['XYZe1', 'XYZe2', 'XYZe3', 'XYZe4', 'XYZe5']])
['XYZe1', 'LMNe1', 'XYZe2', 'LMNe2', 'XYZe3', 'LMNe3', 'XYZe4', 'ABCe1', 'XYZe5', 'ABCe2']
A deterministic solution (ie not random)
Sort your shows by decreasing number of episodes.
Select the biggest and arrange a matrix with the number of columns corresponding to the number of episodes of this one, filled in the following way:
A A A A A A <- First show consist of 6 episodes
B B B B C C <- Second and third show - 4 episodes each
C C D D <- Third show 2 episodes
Then collect by columns
{A,B,C}, {A,B,C}, {A,B,D}, {A,B,D}, {A,C}, {A,C}
Then Join
{A,B,C,A,B,C,A,B,D,A,B,D,A,C,A,C}
And now assign sequential numbers
{A1, B1, C1, A2, B2, C2, A3, B3, D1, A4, B4, D2, A5, C3, A6, C4}
Edit
Your case
[['A'] * 2, ['L'] * 3, ['X'] * 5])
X X X X X
L L L A A
-> {X1, L1, X2, L2, X3, L3, X4, A1, X5, A2}
Edit 2
As no Python here, perhaps a Mathematica code may be of some use:
l = {, , ,}; (* Prepare input *)
l[[1]] = {a, a, a, a, a, a};
l[[2]] = {b, b, b, b};
l[[3]] = {c, c, c, c};
l[[4]] = {d, d};
le = Length@First@l;
k = DeleteCases[ (*Make the matrix*)
Flatten@Transpose@Partition[Flatten[l], le, le, 1, {Null}], Null];
Table[r[i] = 1, {i, k}]; (*init counters*)
ReplaceAll[#, x_ :> x[r[x]++]] & /@ k (*assign numbers*)
->{a[1], b[1], c[1], a[2], b[2], c[2], a[3], b[3], d[1], a[4], b[4],
d[2], a[5], c[3], a[6], c[4]}
My try:
program, play = [['ABCe1.mp4', 'ABCe2.mp4'],
['XYZe1.mp4', 'XYZe2.mp4', 'XYZe3.mp4', 'XYZe4.mp4',
'XYZe5.mp4', 'XYZe6.mp4', 'XYZe7.mp4'],
['OTHERe1.mp4', 'OTHERe2.mp4']], []
start_part = 3
while any(program):
m = max(program, key = len)
if (len(play) >1 and
play[-1][:start_part] != m[0][:start_part] and
play[-2].startswith(play[-1][:start_part])):
play.insert(-1, m.pop(0))
else:
play.append(m.pop(0))
print play
This would ensure that there is at least 1 and no more than 2 other episodes between two successive episodes of a show:
- While there are more than 3 shows, chain two shortest (i.e. having least episodes) shows together end-to-end.
- Let A be the longest show and B and C the other two.
- If B is shorter than A, pad it with
None
's at the end - If C is shorter than A, pad it with
None
's at the beginning - Shuffled playlist is
[x for x in itertools.chain(zip(A,B,C)) if x is not None]
This will ensure true shuffle i.e. a different result each time, with no contiguous items as much as possible.
The one you ask probably could return a few (1, 2) results limited by your requests.
from random import choice, randint
from operator import add
def randpop(playlists):
pl = choice(playlists)
el = pl.pop(randint(0, len(pl) -1))
return pl, el
def shuffle(playlists):
curr_pl = None
while any(playlists):
try:
curr_pl, el = randpop([pl for pl in playlists if pl and pl != curr_pl])
except IndexError:
break
else:
yield el
for el in reduce(add, playlists):
yield el
raise StopIteration
if __name__ == "__main__":
sample = [
'A1 A2 A3 A4'.split(),
'B1 B2 B3 B4 B5'.split(),
'X1 X2 X3 X4 X5 X6'.split()
]
for el in shuffle(sample):
print(el)
Edit:
Given episodes order is mandatory just simplify randpop
function:
def randpop(playlists):
pl = choice(playlists)
el = pl.pop(0)
return pl, el
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