I want to split a list (float or integer) according to the following conditions:
I have what splits a list into equal sizes by giving the number of subsamples.
The code I have laid my hand on which worked but does not give me what I want
import numpy as np
x = [1,2,3,4,5,6,7,8,9,10]
l = np.array_split(x,3)
output
[[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
I desire to have a list of subsample of all possibilities without duplication. that is, a combination of all unique possibilities when the list is split into 2, 3, 4, etc (no two or more subsamples will have the same element)).
I do not what to specify chunk number so that it will not be limited to such number.
Here is what I did manually
From the following series [1,2,3,4,5,6,7,8,9,10] I sliced it into all possible blocks as follows:
when splitting into 2
[1][2,3,4,5,6,7,8,9,10]
[1,2][3,4,5,6,7,8,9,10]
[1,2,3][4,5,6,7,8,9,10]
[1,2,3,4][5,6,7,8,9,10]
[1,2,3,4,5][6,7,8,9,10]
[1,2,3,4,5,6][7,8,9,10]
[1,2,3,4,5,6,7,8][9,10]
[1,2,3,4,5,6,7,8,9][10]
when splitting into 3
[1][2][3,4,5,6,7,8,9,10]
[1][2,3][4,5,6,7,8,9,10]
[1][2,3,4][5,6,7,8,9,10]
[1][2,3,4,5][6,7,8,9,10]
[1][2,3,4,5,6][7,8,9,10]
[1][2,3,4,5,6,7][8,9,10]
[1][2,3,4,5,6,7,8][9,10]
[1][2,3,4,5,6,7,8,9][10]
[1,2][3][4,5,6,7,8,9,10]
[1,2][3,4][5,6,7,8,9,10]
[1,2][3,4,5][6,7,8,9,10]
[1,2][3,4,5,6][7,8,9,10]
[1,2][3,4,5,6,7][8,9,10]
[1,2][3,4,5,6,7,8][9,10]
[1,2][3,4,5,6,7,8,9][10]
[1,2,3][4][5,6,7,8,9,10]
[1,2,3][4,5][6,7,8,9,10]
[1,2,3][4,5,6][7,8,9,10]
[1,2,3][4,5,6,7][8,9,10]
[1,2,3][4,5,6,7,8][9,10]
[1,2,3][4,5,6,7,8,9][10]
[1,2,3,4][5][6,7,8,9,10]
[1,2,3,4][5,6][7,8,9,10]
[1,2,3,4][5,6,7][8,9,10]
[1,2,3,4][5,6,7,8][9,10]
[1,2,3,4][5,6,7,8,9][10]
[1,2,3,4,5][6][7,8,9,10]
[1,2,3,4,5][6,7][8,9,10]
[1,2,3,4,5][6,7,8][9,10]
[1,2,3,4,5][6,7,8,9][10]
[1,2,3,4,5,6][7][8,9,10]
[1,2,3,4,5,6][7,8][9,10]
[1,2,3,4,5,6][7,8,9][10]
when splitting into 4
[1,2,3,4,5,6,7][8][9,10]
[1,2,3,4,5,6,7][8,9][10]
[1,2,3,4,5,6,7,8][9][10]
After all possible splitting into blocks, I removed all the single digit blocks and also removed all duplicated blocks.
[2,3,4,5,6,7,8,9,10]
[1,2,3,4,5,6,7,8,9]
[3,4,5,6,7,8,9,10]
[2,3]
[2,3,4]
[2,3,4,5]
[2,3,4,5,6]
[2,3,4,5,6,7]
[2,3,4,5,6,7,8]
[2,3,4,5,6,7,8,9]
[4,5,6,7,8,9,10]
[3,4]
[3,4,5]
[3,4,5,6]
[3,4,5,6,7]
[3,4,5,6,7,8]
[1,2][3,4,5,6,7,8,9]
[5,6,7,8,9,10]
[4,5]
[4,5,6]
[4,5,6,7]
[4,5,6,7,8]
[1,2,3][4,5,6,7,8,9]
[6,7,8,9,10]
[5,6]
[5,6,7]
[5,6,7,8]
[1,2,3,4][5,6,7,8,9]
[1,2,3,4,5][7,8,9,10]
[6,7]
[6,7,8]
[6,7,8,9]
[8,9,10]
[7,8]
[1,2,3,4,5,6][7,8,9]
[9,10]
[1,2,3,4,5,6,7][8,9]
[1,2,3,4,5,6,7,8]
Here, I gather all the possible chunks together.
This is what I desire as an output.
[[2,3,4,5,6,7,8,9,10], [1,2,3,4,5,6,7,8,9], [3,4,5,6,7,8,9,10], [2,3], [2,3,4], [2,3,4,5], [2,3,4,5,6], [2,3,4,5,6,7], [2,3,4,5,6,7,8], [2,3,4,5,6,7,8,9], [4,5,6,7,8,9,10],[3,4], [3,4,5], [3,4,5,6], [3,4,5,6,7], [3,4,5,6,7,8], [1,2], [3,4,5,6,7,8,9], [5,6,7,8,9,10], [4,5], [4,5,6], [4,5,6,7], [4,5,6,7,8], [1,2,3], [4,5,6,7,8,9], [6,7,8,9,10], [5,6], [5,6,7], [5,6,7,8], [1,2,3,4], [5,6,7,8,9], [1,2,3,4,5], [7,8,9,10], [6,7], [6,7,8], [6,7,8,9], [8,9,10], [7,8], [1,2,3,4,5,6], [7,8,9], [9,10], [1,2,3,4,5,6,7], [8,9], [1,2,3,4,5,6,7,8]]
It looks to me like the problem reduces to finding all substrings of length 2 or greater that leave at least one fragment of length 1. In other words, you won't have to enumerate every partition to find them.
def parts(thing):
result = []
for i in range(len(thing)):
for j in range(i + 1, len(thing) + 1):
if 1 < len(thing[i:j]) < len(thing):
result.append(thing[i:j])
return result
res = parts([*range(1,11)])
# res
# [[1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 6],
# [1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 8],
# [1, 2, 3, 4, 5, 6, 7, 8, 9], [2, 3], [2, 3, 4], [2, 3, 4, 5],
# [2, 3, 4, 5, 6], [2, 3, 4, 5, 6, 7], [2, 3, 4, 5, 6, 7, 8],
# [2, 3, 4, 5, 6, 7, 8, 9], [2, 3, 4, 5, 6, 7, 8, 9, 10], [3, 4], [3, 4, 5],
# [3, 4, 5, 6], [3, 4, 5, 6, 7], [3, 4, 5, 6, 7, 8], [3, 4, 5, 6, 7, 8, 9],
# [3, 4, 5, 6, 7, 8, 9, 10], [4, 5], [4, 5, 6], [4, 5, 6, 7], [4, 5, 6, 7, 8],
# [4, 5, 6, 7, 8, 9], [4, 5, 6, 7, 8, 9, 10], [5, 6], [5, 6, 7], [5, 6, 7, 8],
# [5, 6, 7, 8, 9], [5, 6, 7, 8, 9, 10], [6, 7], [6, 7, 8], [6, 7, 8, 9],
# [6, 7, 8, 9, 10], [7, 8], [7, 8, 9], [7, 8, 9, 10], [8, 9], [8, 9, 10],
# [9, 10]]