I'm looking for an efficient way to achieve this:
you have a list of numbers 1.....n (typically: 1..5 or 1..7 or so - reasonably small, but can vary from case to case)
you need all combinations of all lengths for those numbers, e.g. all combinations of just one number ({1}, {2}, .... {n}), then all combinations of two distinct numbers ({1,2}, {1,3}, {1,4} ..... {n-1, n} ), then all combinations fo three of those numbers ({1,2,3}, {1,2,4}) and so forth
Basically, within the group, the order is irrelevant, so {1,2,3} is equivalent to {1,3,2} - it's just a matter of getting all groups of x numbers from that list
Seems like there ought to be a simple algorithm for this - but I have searched in vain so far. Most combinatorics and permutation algorithms seems to a) take order into account (e.g. 123 is not equal to 132), and they always seems to operate on a single string of characters or numbers....
Anyone have a great, nice'n'quick algorithm up their sleeve??
Thanks!
Add a Custom Column to and name it List1. Enter the formula =List1. Expand out the new List1 column and then Close & Load the query to a table. The table will have all the combinations of items from both lists and we saved on making a custom column in List1 and avoided using a merge query altogether!
Joining queries to find all combinations of two lists Once the queries from the tables are ready, go to Data > Get Data > Combine Queries > Merge to open the Merge dialog of Power Query. Select each table in the drop downs. Click on the column for each table to select them.
Not my code, but you're looking for the powerset. Google gave me this solution, which seems t work:
public IEnumerable<IEnumerable<T>> GetPowerSet<T>(List<T> list)
{
return from m in Enumerable.Range(0, 1 << list.Count)
select
from i in Enumerable.Range(0, list.Count)
where (m & (1 << i)) != 0
select list[i];
}
Source: http://rosettacode.org/wiki/Power_set#C.23
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