Producing a list of permutations from an array

Question

This is probably straightforward but I'm new to this sort of programming and struggling to wrap my head around what I need to do.

I'm trying to build a list of filters to apply to an image. I have seven filters I can apply and each filter can either be 0 or 1 (on or off). The filters are:

filters = [
    'Exposure',
    'Noise',
    'Pressure',
    'XQ Mix',
    'Invert',
    'Desaturate',
    'Equalise'
]

What I want to produce is every possible permutation of these filters in their states. There should be 128 possible permutations (2^7) but when I run the following code I get 5080 permutations:

perms = permutations(filters)
perm_count = 0
for p in perms:
    print(p)
    perm_count = perm_count + 1

print str(perm_count) + ' total permutations'

I'm likely using the wrong method – all this is doing is shuffling the sequence of filters, which I don't care about.

I tried updating the filter list to have two items for each filter, eg. ['Exposure0', 'Exposure1'] etc, but running combinations(filters, 7) against this gives me duplicate values (eg. both on and off states in the same list).

I'm struggling here – can anyone give me a nudge in the right direction for approaching something like this? Looking in the docs, something like product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy seems along the right lines but I still can't wrap my head around it. Help appreciated.

Note: you can use `len(list(...))`. Oh, and explicitly state `itertools` plz. — Karoly Horvath, Nov 17 '15 at 10:51
@KarolyHorvath I tried using `len(list(...))` initially (and again just now to check) and it returns 0 – that was why I added the counter. — Matt Andrews, Nov 17 '15 at 10:54
That's.. how should I say it... *interesting*... works here. — Karoly Horvath, Nov 17 '15 at 10:56
Your machine isn't special. It works there too. Feel free to post a separate question ;) — Karoly Horvath, Nov 17 '15 at 11:46
What would you do if someone says 2+2 is 5? Why is this unfriendly or passive-aggressive? — Karoly Horvath, Nov 17 '15 at 11:56
The downvote icon for the question says "this question does not show any research effort; it is unclear or not useful". Me replying here about what an (unrelated) method call does has no bearing on my question. You should read this article, maybe: https://medium.com/@johnslegers/the-decline-of-stack-overflow-7cb69faa575d — Matt Andrews, Nov 17 '15 at 12:09
I know that article. Let me ask you something: does my profile suggest I like what's going on here? :) — Karoly Horvath, Nov 17 '15 at 12:18

score 3 · Accepted Answer · edited May 23 '17 at 10:27

3

What you are looking for as far as I understand is called a powerset. Here are some implementations for this method: Getting the subsets of a set in Python

edited May 23 '17 at 10:27

Community

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answered Nov 17 '15 at 10:45

SurDin

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Thank you – this was exactly what I wanted – effectively the missing items in the set are the "off" values and I can work out my filters. Much appreciated, thanks! – Matt Andrews Nov 17 '15 at 11:13

score 2 · Answer 2 · answered Nov 17 '15 at 11:17

2

You are using permutations the wrong way. Look at the documentation:

>>> permutations('ABCD', 2)
AB AC AD BA BC BD CA CB CD DA DB DC

You are just mixing your filters values and creating all the possibilities, which is 7! = 5040.

What you want is the product of (0, 1) seven times.

>>> p = product((0, 1), repeat=7)
>>> print(len(p))
128
# [(0, 0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0, 1), ...]

answered Nov 17 '15 at 11:17

Delgan

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Thank you – have already accepted the other answer (because it gives me the actual list of "on" filters, whereas here I'd still need to associate the list of 0/1s with the filter sequence again. Thanks though, I've learned something about `itertools` today. – Matt Andrews Nov 17 '15 at 11:37
1

@MattAndrews Sure, the main thing is that you have resolved your issue, I hope you better understand how to use `product()` and `permutations()` now. – Delgan Nov 17 '15 at 11:40

Producing a list of permutations from an array

2 Answers2