'Python split list into n chunks
I know this question has been covered many times but my requirement is different.
I have a list like: range(1, 26). I want to divide this list into a fixed number n. Assuming n = 6.
>>> x
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]
>>> l = [ x [i:i + 6] for i in range(0, len(x), 6) ]
>>> l
[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18], [19, 20, 21, 22, 23, 24], [25]]
As you can see I didn't get 6 chunks (six sublists with elements of original list). How do I divide a list in such a way that I get exactly n chunks which may be even or uneven
Solution 1:[1]
Use numpy
>>> import numpy
>>> x = range(25)
>>> l = numpy.array_split(numpy.array(x),6)
or
>>> import numpy
>>> x = numpy.arange(25)
>>> l = numpy.array_split(x,6);
You can also use numpy.split but that one throws in error if the length is not exactly divisible.
Solution 2:[2]
The solution(s) below have many advantages:
- Uses generator to yield the result.
- No imports.
- Lists are balanced (you never end up with 4 lists of size 4 and one list of size 1 if you split a list of length 17 into 5).
def chunks(l, n):
"""Yield n number of striped chunks from l."""
for i in range(0, n):
yield l[i::n]
The code above produces the below output for l = range(16) and n = 6:
[0, 6, 12]
[1, 7, 13]
[2, 8, 14]
[3, 9, 15]
[4, 10]
[5, 11]
If you need the chunks to be sequential instead of striped use this:
def chunks(l, n):
"""Yield n number of sequential chunks from l."""
d, r = divmod(len(l), n)
for i in range(n):
si = (d+1)*(i if i < r else r) + d*(0 if i < r else i - r)
yield l[si:si+(d+1 if i < r else d)]
Which for l = range(16) and n = 6 produces:
[0, 1, 2]
[3, 4, 5]
[6, 7, 8]
[9, 10, 11]
[12, 13]
[14, 15]
See this stackoverflow link for more information on the advantages of generators.
Solution 3:[3]
If order doesn't matter:
def chunker_list(seq, size):
return (seq[i::size] for i in range(size))
print(list(chunker_list([1, 2, 3, 4, 5], 2)))
>>> [[1, 3, 5], [2, 4]]
print(list(chunker_list([1, 2, 3, 4, 5], 3)))
>>> [[1, 4], [2, 5], [3]]
print(list(chunker_list([1, 2, 3, 4, 5], 4)))
>>> [[1, 5], [2], [3], [4]]
print(list(chunker_list([1, 2, 3, 4, 5], 5)))
>>> [[1], [2], [3], [4], [5]]
print(list(chunker_list([1, 2, 3, 4, 5], 6)))
>>> [[1], [2], [3], [4], [5], []]
Solution 4:[4]
more_itertools.divide is one approach to solve this problem:
import more_itertools as mit
iterable = range(1, 26)
[list(c) for c in mit.divide(6, iterable)]
Output
[[ 1, 2, 3, 4, 5], # remaining item
[ 6, 7, 8, 9],
[10, 11, 12, 13],
[14, 15, 16, 17],
[18, 19, 20, 21],
[22, 23, 24, 25]]
As shown, if the iterable is not evenly divisible, the remaining items are distributed from the first to the last chunk.
See more about the more_itertools library here.
Solution 5:[5]
My answer is to simply use python built-in Slice:
# Assume x is our list which we wish to slice
x = range(1, 26)
# Assume we want to slice it to 6 equal chunks
result = []
for i in range(0, len(x), 6):
slice_item = slice(i, i + 6, 1)
result.append(x[slice_item])
# Result would be equal to
[[0,1,2,3,4,5], [6,7,8,9,10,11], [12,13,14,15,16,17],[18,19,20,21,22,23], [24, 25]]
Solution 6:[6]
Try this:
from __future__ import division
import math
def chunked(iterable, n):
""" Split iterable into ``n`` iterables of similar size
Examples::
>>> l = [1, 2, 3, 4]
>>> list(chunked(l, 4))
[[1], [2], [3], [4]]
>>> l = [1, 2, 3]
>>> list(chunked(l, 4))
[[1], [2], [3], []]
>>> l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> list(chunked(l, 4))
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]
"""
chunksize = int(math.ceil(len(iterable) / n))
return (iterable[i * chunksize:i * chunksize + chunksize]
for i in range(n))
It returns an iterator instead of a list for efficiency (I'm assuming you want to loop over the chunks), but you can replace that with a list comprehension if you want. When the number of items is not divisible by number of chunks, the last chunk is smaller than the others.
EDIT: Fixed second example to show that it doesn't handle one edge case
Solution 7:[7]
Hint:
- x is the string to be split.
k is number of chunks
n = len(x)/k [x[i:i+n] for i in range(0, len(x), n)]
Solution 8:[8]
Here take my 2 cents..
from math import ceil
size = 3
seq = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
chunks = [
seq[i * size:(i * size) + size]
for i in range(ceil(len(seq) / size))
]
# [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11]]
Solution 9:[9]
One way would be to make the last list uneven and the rest even. This can be done as follows:
>>> x
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]
>>> m = len(x) // 6
>>> test = [x[i:i+m] for i in range(0, len(x), m)]
>>> test[-2:] = [test[-2] + test[-1]]
>>> test
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16], [17, 18, 19, 20], [21, 22, 23, 24, 25]]
Solution 10:[10]
Assuming you want to divide into n chunks:
n = 6
num = float(len(x))/n
l = [ x [i:i + int(num)] for i in range(0, (n-1)*int(num), int(num))]
l.append(x[(n-1)*int(num):])
This method simply divides the length of the list by the number of chunks and, in case the length is not a multiple of the number, adds the extra elements in the last list.
Solution 11:[11]
If you want to have the chunks as evenly sized as possible:
def chunk_ranges(items: int, chunks: int) -> List[Tuple[int, int]]:
"""
Split the items by best effort into equally-sized chunks.
If there are fewer items than chunks, each chunk contains an item and
there are fewer returned chunk indices than the argument `chunks`.
:param items: number of items in the batch.
:param chunks: number of chunks
:return: list of (chunk begin inclusive, chunk end exclusive)
"""
assert chunks > 0, \
"Unexpected non-positive chunk count: {}".format(chunks)
result = [] # type: List[Tuple[int, int]]
if items <= chunks:
for i in range(0, items):
result.append((i, i + 1))
return result
chunk_size, extras = divmod(items, chunks)
start = 0
for i in range(0, chunks):
if i < extras:
end = start + chunk_size + 1
else:
end = start + chunk_size
result.append((start, end))
start = end
return result
Test case:
def test_chunk_ranges(self):
self.assertListEqual(chunk_ranges(items=8, chunks=1),
[(0, 8)])
self.assertListEqual(chunk_ranges(items=8, chunks=2),
[(0, 4), (4, 8)])
self.assertListEqual(chunk_ranges(items=8, chunks=3),
[(0, 3), (3, 6), (6, 8)])
self.assertListEqual(chunk_ranges(items=8, chunks=5),
[(0, 2), (2, 4), (4, 6), (6, 7), (7, 8)])
self.assertListEqual(chunk_ranges(items=8, chunks=6),
[(0, 2), (2, 4), (4, 5), (5, 6), (6, 7), (7, 8)])
self.assertListEqual(
chunk_ranges(items=8, chunks=7),
[(0, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8)])
self.assertListEqual(
chunk_ranges(items=8, chunks=9),
[(0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8)])
Solution 12:[12]
In cases, where your list contains elements of different types or iterable objects that store values of different types (f.e. some elements are integers, and some are strings), if you use array_split function from numpy package to split it, you will get chunks with elements of same type:
import numpy as np
data1 = [(1, 2), ('a', 'b'), (3, 4), (5, 6), ('c', 'd'), ('e', 'f')]
chunks = np.array_split(data1, 3)
print(chunks)
# [array([['1', '2'],
# ['a', 'b']], dtype='<U11'), array([['3', '4'],
# ['5', '6']], dtype='<U11'), array([['c', 'd'],
# ['e', 'f']], dtype='<U11')]
data2 = [1, 2, 'a', 'b', 3, 4, 5, 6, 'c', 'd', 'e', 'f']
chunks = np.array_split(data2, 3)
print(chunks)
# [array(['1', '2', 'a', 'b'], dtype='<U11'), array(['3', '4', '5', '6'], dtype='<U11'),
# array(['c', 'd', 'e', 'f'], dtype='<U11')]
If you would like to have initial types of elements in chunks after splitting of list, you can modify source code of array_split function from numpy package or use this implementation:
from itertools import accumulate
def list_split(input_list, num_of_chunks):
n_total = len(input_list)
n_each_chunk, extras = divmod(n_total, num_of_chunks)
chunk_sizes = ([0] + extras * [n_each_chunk + 1] + (num_of_chunks - extras) * [n_each_chunk])
div_points = list(accumulate(chunk_sizes))
sub_lists = []
for i in range(num_of_chunks):
start = div_points[i]
end = div_points[i + 1]
sub_lists.append(input_list[start:end])
return (sub_list for sub_list in sub_lists)
result = list(list_split(data1, 3))
print(result)
# [[(1, 2), ('a', 'b')], [(3, 4), (5, 6)], [('c', 'd'), ('e', 'f')]]
result = list(list_split(data2, 3))
print(result)
# [[1, 2, 'a', 'b'], [3, 4, 5, 6], ['c', 'd', 'e', 'f']]
Solution 13:[13]
This solution is based on the zip "grouper" pattern from the Python 3 docs. The small addition is that if N does not divide the list length evenly, all the extra items are placed into the first chunk.
import itertools
def segment_list(l, N):
chunk_size, remainder = divmod(len(l), N)
first, rest = l[:chunk_size + remainder], l[chunk_size + remainder:]
return itertools.chain([first], zip(*[iter(rest)] * chunk_size))
Example usage:
>>> my_list = list(range(10))
>>> segment_list(my_list, 2)
[[0, 1, 2, 3, 4], (5, 6, 7, 8, 9)]
>>> segment_list(my_list, 3)
[[0, 1, 2, 3], (4, 5, 6), (7, 8, 9)]
>>>
The advantages of this solution are that it preserves the order of the original list, and is written in a functional style that lazily evaluates the list only once when called.
Note that because it returns an iterator, the result can only be consumed once. If you want the convenience of a non-lazy list, you can wrap the result in list:
>>> x = list(segment_list(my_list, 2))
>>> x
[[0, 1, 2, 3, 4], (5, 6, 7, 8, 9)]
>>> x
[[0, 1, 2, 3, 4], (5, 6, 7, 8, 9)]
>>>
Solution 14:[14]
I would simply do (let's say you want n chunks)
import numpy as np
# convert x to numpy.ndarray
x = np.array(x)
l = np.array_split(x, n)
It works and it's only 2 lines.
Example:
# your list
x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]
# amount of chunks you want
n = 6
x = np.array(x)
l = np.array_split(x, n)
print(l)
>> [array([1, 2, 3, 4, 5]), array([6, 7, 8, 9]), array([10, 11, 12, 13]), array([14, 15, 16, 17]), array([18, 19, 20, 21]), array([22, 23, 24, 25])]
And if you want a list of list:
l = [list(elem) for elem in l]
print(l)
>> [[1, 2, 3, 4, 5], [6, 7, 8, 9], [10, 11, 12, 13], [14, 15, 16, 17], [18, 19, 20, 21], [22, 23, 24, 25]]
Solution 15:[15]
x=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]
chunk = len(x)/6
l=[]
i=0
while i<len(x):
if len(l)<=4:
l.append(x [i:i + chunk])
else:
l.append(x [i:])
break
i+=chunk
print l
#output=[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16], [17, 18, 19, 20], [21, 22, 23, 24, 25]]
Solution 16:[16]
arr1=[-20, 20, -10, 0, 4, 8, 10, 6, 15, 9, 18, 35, 40, -30, -90, 99]
n=4
final = [arr1[i * n:(i + 1) * n] for i in range((len(arr1) + n - 1) // n )]
print(final)
Output:
[[-20, 20, -10, 0], [4, 8, 10, 6], [15, 9, 18, 35], [40, -30, -90, 99]]
Solution 17:[17]
This function will return the list of lists with the set maximum amount of values in one list (chunk).
def chuncker(list_to_split, chunk_size):
list_of_chunks =[]
start_chunk = 0
end_chunk = start_chunk+chunk_size
while end_chunk <= len(list_to_split)+chunk_size:
chunk_ls = list_to_split[start_chunk: end_chunk]
list_of_chunks.append(chunk_ls)
start_chunk = start_chunk +chunk_size
end_chunk = end_chunk+chunk_size
return list_of_chunks
Example:
ls = list(range(20))
chuncker(list_to_split = ls, chunk_size = 6)
output:
[[0, 1, 2, 3, 4, 5], [6, 7, 8, 9, 10, 11], [12, 13, 14, 15, 16, 17], [18, 19]]
Solution 18:[18]
I came up with the following solution:
l = [x[i::n] for i in range(n)]
For example:
n = 6
x = list(range(26))
l = [x[i::n] for i in range(n)]
print(l)
Output:
[[0, 6, 12, 18, 24], [1, 7, 13, 19, 25], [2, 8, 14, 20], [3, 9, 15, 21], [4, 10, 16, 22], [5, 11, 17, 23]]
As you can see, the output consists from n chunks, which have roughly the same number of elements.
How it works?
The trick is to use list slice step (the number after two semicolons) and to increment the offset of stepped slicing. First, it takes every n element starting from the first, then every n element starting from the second and so on. This completes the task.
Solution 19:[19]
This accepts generators, without consuming it at once. If we know the size of the generator, the binsize can be calculated by max(1, size // n_chunks).
from time import sleep
def chunks(items, binsize):
lst = []
for item in items:
lst.append(item)
if len(lst) == binsize:
yield lst
lst = []
if len(lst) > 0:
yield lst
def g():
for item in [1, 2, 3, 4, 5, 6, 7]:
print("accessed:", item)
sleep(1)
yield item
for a in chunks(g(), 3):
print("chunk:", list(a), "\n")
Solution 20:[20]
For people looking for an answer in python 3(.6) without imports.
x is the list to be split.
n is the length of chunks.
L is the new list.
n = 6
L = [x[i:i + int(n)] for i in range(0, (n - 1) * int(n), int(n))]
#[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18], [19, 20, 21, 22, 23, 24], [25]]
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow