How to iterate over a range of permutations? [duplicate]

Question

Say you have n items each ranging from 1-100. How can I get go over all possible variations within the range?

Example:

3 stocks A, B and C

Working to find possible portfolio allocation.

A - 0     0   0            1    2          1    1
B - 0     1   2      ...   0    0    ...   1    2
C - 100   99  98           99   98         98   97

Looking for an efficient way to get a matrix of all possible outcomes.

Sum should add up to 100 and cover all possible variations for n elements.

This is a pretty unclear question... Are you saying that you want all permutations of three numbers between 1 and 100 such that the sum of the three equals 100? — mgilson
– mgilson, Commented Sep 8, 2016 at 21:33
@mgilson Yes, that is correct. I'll update question. Looking for an efficient way to handle it though. And for n numbers, obviously — James Raitsev
– James Raitsev, Commented Sep 8, 2016 at 21:34
This sounds like you're trying to brute-force a problem that would be better solved by getting a more sophisticated library for it. — user2357112
– user2357112, Commented Sep 8, 2016 at 21:43
I think your problem has an answer here stackoverflow.com/questions/13988197/…. — Rockybilly
– Rockybilly, Commented Sep 8, 2016 at 21:47

Community · Accepted Answer · 2017-05-23 12:01:02Z

2

How I'd do it:

>>> import itertools
>>> cp = itertools.product(range(101),repeat=3)
>>> portfolios = list(p for p in cp if sum(p)==100)

But that creates unnecessary combinations. See discussions of integer partitioning to avoid that. E.g., Elegant Python code for Integer Partitioning

edited May 23, 2017 at 12:01

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answered Sep 8, 2016 at 21:46

Alan

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3 Comments

mgilson Over a year ago

This is O(N^3) -- It'll take a million iterations when you should be able to come up with a solution in about 5000 iterations (for N = 100).

Jaakko Over a year ago

The amount-of-used-code-lines efficiency is great.

Alan Over a year ago

@mgilson Agreed. That's why I provided the link for integer partitioning.

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