CODEWITHSUNDEEP
python
shane
shane

Reputation: 1863

Combine list of lists to another list of lists by a certain length in python

I'm looking for a one liner to convert

[[1], [1, 1], [1, 1, 1], [1], [1], [1], [1, 1, 1, 1]]

to

[[1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1, 1]]

where the algorithm combines the lists up to a certain threshold length.

I currently have this

batched = []
batch = []
for l in lists:
    batch.extend(l)
    if len(batch) > threshold:
        batched.append(batch)
        batch = []

Upvotes: 2

Views: 231

Answers (5)

shane
shane

Reputation: 1863

I appreciate all the help, and I learned that a one liner is not be best for this problem as it is ugly and unreadable and probably inefficient. That being said I did come up with this which seems to work from a different slightly cleaner approach.

>>> from functools import reduce
>>> l = [[1], [2, 3], [4, 5, 6], [7], [8], [9], [10, 11, 12, 13]]
>>> t = 3
>>> b = reduce(lambda p, n: p[:-1] + [p[-1] + n] if len(p[-1]) + len(n) <= t or not p[-1] else p + [n], l, [[]])
>>> b
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12, 13]]

Upvotes: 0

niemmi
niemmi

Reputation: 17263

There's nothing wrong with your original implementation but if you insist oneliner here's one ugly option:

from itertools import accumulate, chain, groupby

THRESHOLD = 3
l = [[1], [1, 1], [1, 1, 1], [1], [1], [1], [1, 1, 1, 1]]
res = [[y for x in g for y in x[1]]
       for k, g in groupby(zip(chain([0], accumulate(len(x) for x in l)), l),
                           lambda x: x[0] // THRESHOLD)]
print(res)

Output:

[[1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1, 1]]

The idea is to generate list of (item count so far, sublist) tuples and group them by dividing count by THRESHOLD.

>>> temp = list(zip(chain([0], accumulate(len(x) for x in l)), l))
>>> temp
[(0, [1]), (1, [1, 1]), (3, [1, 1, 1]), (6, [1]), (7, [1]), (8, [1]), (9, [1, 1, 1, 1])]
>>> groups = [list(g) for k, g in groupby(temp, lambda x: x[0] // THRESHOLD)]
>>> groups
[[(0, [1]), (1, [1, 1])], [(3, [1, 1, 1])], [(6, [1]), (7, [1]), (8, [1])], [(9, [1, 1, 1, 1])]]
>>> [[y for x in g for y in x[1]] for g in groups]
[[1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1, 1]]

Upvotes: 5

miradulo
miradulo

Reputation: 29690

This may not be ideal, but it would be my attempt. The idea is to chain your lists together with itertools.chain , then pull from the chain with itertools.islice and append a new list until you cannot any longer. It admittedly isn't a one liner. 

from itertools import chain, islice
def grouper(n, li):
    it = chain(*li)
    out_l = []
    while True:
        chunk = list(islice(it, n))
        if len(chunk) < n:
            if chunk:
                out_l[-1] += chunk
            return out_l
        out_l.append(chunk)

Demo

In[238]: orig = [[1], [1, 1], [1, 1, 1], [1], [1], [1], [1, 1, 1, 1]]
In[239]: grouper(3, orig)
Out[239]: [[1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1, 1]]
In[240]: grouper(4, orig)
Out[240]: [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 1]]
In[241]: grouper(5, orig)
Out[241]: [[1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1]]
Out[242]: grouper(1, orig)
Out[242]: [[1], [1], [1], [1], [1], [1], [1], [1], [1], [1], [1], [1], [1]]

Upvotes: 3

roganjosh
roganjosh

Reputation: 13175

Cracked it but only because I'm stubborn. It's really ugly and inefficient though and maybe there's a cleaner way but, even if there is, it's not worth it.

a = [[1], [1, 1], [1, 1, 1], [2], [1], [1], [1, 1, 1, 1]]

b = [[[item for sublist in a for item in sublist][x+(y*3)] for x in range(1,4)] for y in range(0, (len([i for j in a for i in j])/3))]

EDIT: Because I was testing this in Python 2.7 I missed the fact that division works differently in Python 3. Thanks @nexus66 for pointing out a modification (which just makes it even longer!).

c = [[[item for sublist in a for item in sublist][x+(y*3)] for x in range(1,4)] for y in range(0, int((len([i for j in a for i in j])/3)))]

Upvotes: 4

boot-scootin
boot-scootin

Reputation: 12515

This is an ugly one-liner... It's not exactly what you asked for (pretty close), but maybe it'll give you some idea of how to approach it...

arr = [[1], [1, 1], [1, 1, 1], [1], [1], [1], [1, 1, 1, 1]]

threshold = 4

[[x for y in arr for x in y][i:i+threshold] 
     for i, _ in enumerate([x for y in arr for x in y])
       if i % threshold == 0]

Out[31]:
[[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1], [1]]

Upvotes: 0

Related Questions