Print consecutive pair of subsets with increasing size from a given list of numbers

Question

Example 1,

Input: [a,b,c,d]

Output: [[a],[a,b],[b,c],[c,d],[a,b,c],[b,c,d],[a,b,c,d]]

Example 2,

Input: [1,2,3,4,5]

Output: [[1],[1,2],[2,3],[3,4],[4,5],[1,2,3],[2,3,4],[3,4,5],[1,2,3,4],[2,3,4,5],[1,2,3,4,5]]

In the same way, the number of elements in the pair starting from 1 increases till 'n'(size of the given list)

Is there a possible way of handling, a given list of any size (If possible in Python)

Extra Info:

I have tried this code, which returns pair of 3 elements in the output but I want to return first 1 element, next with 2 elements, till n-1 elements as I mentioned above in the example

Input:

listA = [51,23,11,45]
res = [[listA[i], listA[i + 1],listA[i + 2]] for i in range(len(listA) - 2)]
print("List with paired elements: \n",res)

Output:

List with paired elements: [[51, 23, 11], [23, 11, 45]]

Try: https://stackoverflow.com/questions/41626379/python-power-set-of-a-list — Andrej Kesely, Feb 17 '22 at 18:40
@AndrejKesely I don't think it is powerset based on the output. Why isn't the list containing [2],[3],[4],[5] based on your description Shobhan? — Richard K Yu, Feb 17 '22 at 18:41
@AndrejKesely Because it is a list meant to find the consecutive subpattern from the given list of numbers (a trend)... so single element does not make up it... So we can even ignore the first element or include the single elements as you said... Thank you for your comment Sir! — Shobhan Kumar.R, Feb 18 '22 at 04:53

score 0 · Answer 1 · answered Feb 17 '22 at 19:06

I am a bit confused about the output requirement, but here is something that produces the result I think you are looking for:

def get_list_of_size(l, n):
    return [l[i:i+n] for i in range(len(l)-n+1)]

def get_consecutive_ss(l):
    output=[]
    for i in range(len(l)):
        output+=get_list_of_size(l, i+1)
    return [output[0]]+output[len(l):]


l = [1,2,3,4,5]
l2=['a','b','c','d']

print(get_consecutive_ss(l))
print(get_consecutive_ss(l2))

Output:

[[1], [1, 2], [2, 3], [3, 4], [4, 5], [1, 2, 3], [2, 3, 4], [3, 4, 5], [1, 2, 3, 4], [2, 3, 4, 5], [1, 2, 3, 4, 5]]
[['a'], ['a', 'b'], ['b', 'c'], ['c', 'd'], ['a', 'b', 'c'], ['b', 'c', 'd'], ['a', 'b', 'c', 'd']]

The requirement is a list meant to find the consecutive subpattern from the given list of numbers (a trend)...And I was totally baffled by the requirement too... Thank You So Much for Your Answer!! — Shobhan Kumar.R, Feb 18 '22 at 05:01

mathfux · Accepted Answer · 2022-02-17T19:39:12.770

Fisrt way

You can do it automatically using a package more_itertools.substrings:

import more_itertools
list(more_itertools.substrings([2,3,5,7,11]))

Output:

[(2,), (3,), (5,), (7,), (11,), (2, 3), (3, 5), (5, 7), (7, 11), (2, 3, 5), (3, 5, 7), (5, 7, 11), (2, 3, 5, 7), (3, 5, 7, 11), (2, 3, 5, 7, 11)]

You could also check a source code of implementation of this method or make it yourself as an exercise.

Second way

It allows you to achieve exactly what you expect using more_itertools.sliding_window:

from itertools import chain
from more_itertools import sliding_window

x = [2,3,5,7,11]
first_part = [(x[0],)]
second_part = [sliding_window(x, n) for n in range(2, len(x)+1)]
list(chain(first_part, *second_part))

Output:

[(2,), (2, 3), (3, 5), (5, 7), (7, 11), (2, 3, 5), (3, 5, 7), (5, 7, 11), (2, 3, 5, 7), (3, 5, 7, 11), (2, 3, 5, 7, 11)]

Print consecutive pair of subsets with increasing size from a given list of numbers

2 Answers2