I have read the answers to the Slicing a list into n nearly-equal-length partitions [duplicate] question.
This is the accepted answer:
def partition(lst, n): division = len(lst) / float(n) return [ lst[int(round(division * i)): int(round(division * (i + 1)))] for i in xrange(n) ] I am wondering, how does one modify these solutions in order to randomly assign items to a partition as opposed to incremental assignment.
Call random.shuffle() on the list before partitioning it.
Complete 2018 solution (python 3.6):
import random def partition (list_in, n): random.shuffle(list_in) return [list_in[i::n] for i in range(n)] Beware! this may mutate your original list
shuffle input list.
First you randomize the list and then you split it in n nearly equal parts.
Shuffling the list doesn't preserve order. You could do something like this instead (pretty easy to adapt to more than two parts). Completely untested.
from __future__ import annotations from typing import TypeVar import random T = TypeVar("T") def partition_list(s: list[T]) -> tuple[list[T], list[T]]: """ Randomly partition a list into two lists, preserving order. The number to take is drawn from a uniform distribution. """ len_a = random.randint(0, len(s)) len_b = len(s) - len_a put_in_a = [True] * len_a + [False] * len_b random.shuffle(put_in_a) a: list[T] = [] b: list[T] = [] for val, in_a in zip(s, put_in_a): if in_a: a.append(val) else: b.append(val) return a, b The random partition that also preserves the order:
def partition_preserve_order(list_in, n): indices = list(range(len(list_in))) shuffle(indices) index_partitions = [sorted(indices[i::n]) for i in range(n)] return [[list_in[i] for i in index_partition] for index_partition in index_partitions] (that is we shuffle the indices then sort them within the partitions)
example:
random_partition_preserve_order(list('abcdefghijklmnopqrstuvxyz'), 3) # [ # ['c', 'd', 'g', 'm', 'p', 'r', 'v', 'x', 'y'], # ['b', 'e', 'h', 'k', 'o', 'q', 't', 'u'], # ['a', 'f', 'i', 'j', 'l', 'n', 's', 'z'] # ] 