Splitting a list into uneven groups?

Question

I know how to split a list into even groups, but I\'m having trouble splitting it into uneven groups.

Essentially here is what I have: some list, let\'s call it

Answer 1

You can create an iterator and itertools.islice:

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

from itertools import islice
it = iter(mylist)

sliced =[list(islice(it, 0, i)) for i in seclist]

Which would give you:

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

Once i elements are consumed they are gone so we keep getting the next i elements.

Not sure what should happen with any remaining elements, if you want them added, you could add something like:

mylist = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 ,14]
seclist = [2, 4, 6]

from itertools import islice

it = iter(mylist)

slices = [sli for sli in (list(islice(it, 0, i)) for i in seclist)]
remaining = list(it)
if remaining:
    slices.append(remaining)
print(slices)

Which would give you:

 [[1, 2], [3, 4, 5, 6], [7, 8, 9, 10, 11, 12], [13, 14]]

Or in contrast if there were not enough, you could use a couple of approaches to remove empty lists, one an inner generator expression:

from itertools import islice

it = iter(mylist)
slices = [sli for sli in (list(islice(it, 0, i)) for i in seclist) if sli]

Or combine with itertools.takewhile:

 from itertools import islice, takewhile

it = iter(mylist)
slices = list(takewhile(bool, (list(islice(it, 0, i)) for i in seclist)))

Which for:

mylist = [1, 2, 3, 4, 5, 6]
seclist = [2, 4, 6,8]

would give you:

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

As opposed to:

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

What you use completely depends on your possible inouts and how you would like to handle the various possibilities.

Answer 2

This solution keeps track of how many items you've written. It will crash if the sum of the numbers in the second_list is longer than mylist

total = 0
listChunks = []
for j in range(len(second_list)):
    chunk_mylist = mylist[total:total+second_list[j]]
    listChunks.append(chunk_mylist)
    total += second_list[j]

After running this, listChunks is a list containing sublists with the lengths found in second_list.

Answer 3

A numpythonic approach:

>>> lst = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
>>> sec = [2, 4, 5]
>>> np.split(lst, np.cumsum(sec))
[array([0, 1]), array([2, 3, 4, 5]), array([ 6,  7,  8,  9, 10]), array([11])]

And here is a Python3.X approach using itertool.accumulate():

>>> lst = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
>>> sec = [2,4,6]
>>> from itertools import accumulate
>>> sec = list(accumulate(sec_lst))
>>> sec = [0] + sec + [None] if sec[0] != 0 else sec + [None]
>>> 
>>> [lst[i:j] for i, j in zip(sec, sec[1:])]
[[0, 1], [2, 3, 4, 5], [6, 7, 8, 9, 10], [11]]

Answer 4

Using list-comprehensions together with slicing and sum() function (all basic and built-in tools of python):

mylist = [1,2,3,4,5,6,7,8,9,10]
seclist = [2,4,6]

[mylist[sum(seclist[:i]):sum(seclist[:i+1])] for i in range(len(seclist))]

#output:
[[1, 2], [3, 4, 5, 6], [7, 8, 9, 10]]

---

If seclist is very long and you wish to be more efficient use numpy.cumsum() first:

import numpy as np
cumlist = np.hstack((0, np.cumsum(seclist)))
[mylist[cumlist[i]:cumlist[i+1]] for i in range(len(cumlist)-1)]

and get the same results

Answer 5

Padriac has the best solution IMO, but I'll chime in with this hacky one liner that requires no imports:

>>> L = [1,2,3,4,5,6,7,8,9,10] # source list - could be any iterable
>>> S = [2,3,4] # group sizes - could be any iterable

>>> [ [ [ next(i) for k in range(j) ] for j in iter(S) ] for i in [iter(L)] ][0]
[[1, 2], [3, 4, 5], [6, 7, 8, 9]]

Answer 6

subgroups = []
start=0
for i in second_list:
  subgroups.append(mylist[start:start + i])
  start = i + start

At the end subgroups will contain the desired lists

Example run:

>>> mylist = [1,2,3,4,5,6,7,8,9,10,11,12]
>>> second_list = [2,4,5,9]
>>> subgroups = []
>>> start=0
>>> for i in second_list:
...    subgroups.append(mylist[start:start + i])
...    start = i + start
...
>>> subgroups
[[1, 2], [3, 4, 5, 6], [7, 8, 9, 10, 11], [12]]