How to divide a list into n equal parts, python

Question

Given (any) list of words lst I should divide it into 10 equal parts.

x = len(lst)/10

how to give these parts variable names?

In the output I need 10 variables (part1, part2... part10) with x number of words in it.

Answer 1

One-liner returning a list of lists, given a list and the chunk size:

>>> lol = lambda lst, sz: [lst[i:i+sz] for i in range(0, len(lst), sz)]

Testing:

>>> x = range(20, 36)
>>> print x
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35]

>>> lol(x, 4)
[[20, 21, 22, 23], 
 [24, 25, 26, 27], 
 [28, 29, 30, 31], 
 [32, 33, 34, 35]]

>>> lol(x, 7)
[[20, 21, 22, 23, 24, 25, 26], 
 [27, 28, 29, 30, 31, 32, 33], 
 [34, 35]]

Update:

I think the question is really asking is a function which, given a list and a number, returns a list containing $(number) lists, with the items of the original list evenly distributed. So your example of lol(x, 7) should really return [[20,21,22], [23,24,25], [26,27], [28,29], [30,31], [32,33], [34,35]]. – markrian

Well, in this case, you can try:

def slice_list(input, size):
    input_size = len(input)
    slice_size = input_size / size
    remain = input_size % size
    result = []
    iterator = iter(input)
    for i in range(size):
        result.append([])
        for j in range(slice_size):
            result[i].append(iterator.next())
        if remain:
            result[i].append(iterator.next())
            remain -= 1
    return result

I'm sure this can be improved but I'm feeling lazy. :-)

>>> slice_list(x, 7)
[[20, 21, 22], [23, 24, 25], 
 [26, 27], [28, 29], 
 [30, 31], [32, 33], 
 [34, 35]]

Answer 2

To achieve the same result as Paulo's update (divide a list into n chunks with size only differing by 1), the following is an elegant solution using recursion.

def divide(lst, n):
    p = len(lst) // n
    if len(lst)-p > 0:
        return [lst[:p]] + divide(lst[p:], n-1)
    else:
        return [lst]

Example:

lst = list(range(13))
print divide(lst,5) # [[0, 1], [2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]]