Listing all permutations of RGB in Python

Question

I am trying to make a list of all possible 27 3-character permutations of RGB. My function is as follows:

def make_big_RGB():
    rgb = ('r', 'g', 'b')
    comb_list = []
    current_combination = list(rgb) #start with combination (r, g, b)
    for x in range(3):
        current_combination[0] = rgb[x]
        for y in range(3):
            current_combination[1] = rgb[y]
            for z in range(3):
                current_combination[2] = rgb[z]
                comb_list.append(current_combination)
                print('Added' + str(current_combination))
    for _ in comb_list:
        print(_)
make_big_RGB()

The result is as follows:

Added['r', 'r', 'r']
Added['r', 'r', 'g']
Added['r', 'r', 'b']
Added['r', 'g', 'r']
Added['r', 'g', 'g']
Added['r', 'g', 'b']
Added['r', 'b', 'r']
Added['r', 'b', 'g']
Added['r', 'b', 'b']
Added['g', 'r', 'r']
Added['g', 'r', 'g']
Added['g', 'r', 'b']
Added['g', 'g', 'r']
Added['g', 'g', 'g']
Added['g', 'g', 'b']
Added['g', 'b', 'r']
Added['g', 'b', 'g']
Added['g', 'b', 'b']
Added['b', 'r', 'r']
Added['b', 'r', 'g']
Added['b', 'r', 'b']
Added['b', 'g', 'r']
Added['b', 'g', 'g']
Added['b', 'g', 'b']
Added['b', 'b', 'r']
Added['b', 'b', 'g']
Added['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']
['b', 'b', 'b']

The last for loop prints out the wanted result, but when I subsequently try to print the whole list, the result is somehow a list of 27 items of [b, b, b]. I don't understand why.

You've appended 27 references to the one `current_combination` list object to `comb_list`. — PM 2Ring, Feb 08 '18 at 13:44
jp's solution is good. But just in case you want to keep your code, changing the line where you append to this: `comb_list.append(current_combination.copy())` should work as well. — Arne, Feb 08 '18 at 13:44
There aren't 27 permutations of RGB. There are 27 3-character combinations. There are 6 permutations. — khelwood, Feb 08 '18 at 13:47
Possible duplicate of [generating permutations with repetitions in python](https://stackoverflow.com/questions/3099987/generating-permutations-with-repetitions-in-python) — Georgy, Feb 08 '18 at 13:55
You keep modifying the same `current_combination` list object and appending a reference to that _one_ object to `comb_list`. So when you print `comb_list` all you'll see is 27 copies of whatever the current contents of `current_combination` are. — PM 2Ring, Feb 08 '18 at 14:00
Try this: put `current_combination[:] = 'xyz'` just after the end of your main loops but before your `for _ in comb_list:` printing loop. The code will then print 27 lines of `['x', 'y', 'z']` — PM 2Ring, Feb 08 '18 at 14:04
@Georgy Sure, your linked question shows how to get the Cartesian product. But it doesn't address the Hallvard's main question: why does `comb_list` end up containing a bunch of duplicate sublists? — PM 2Ring, Feb 08 '18 at 14:07
Here's a very closely related question: https://stackoverflow.com/questions/5280799/list-append-changing-all-elements-to-the-appended-item — PM 2Ring, Feb 08 '18 at 14:34
@khelwood After doing a little recap, I think you’ve got the two mixed up — Hallvard, Feb 09 '18 at 00:38
@khelwood I agree that there are 6 permutations of the set RGB, but likewise there are 27 3-character permutations, and 10 3-character combinations:['r', 'r', 'r'] ['b', 'b', 'b'] ['g', 'g', 'g'] ['r', 'g', 'b'] ['r', 'r', 'g'] ['r', 'r', 'b'] ['r', 'g', 'g'] ['g', 'g', 'b'] ['r', 'b', 'b'] ['g', 'b', 'b']) — Hallvard, Feb 09 '18 at 15:02
@khelwood That's OK, but if you at some point want to understand the rest of it, I recommend trying out http://www.statisticshowto.com/calculators/permutation-calculator-and-combination-calculator/ — Hallvard, Feb 10 '18 at 13:20
The reason is that you said there were 27 3-character combinations (i.e. order not mattering) of RGB. That is not correct. There are 10, namely the ones I listed. There are 27 3-character permutations (i.e. order mattering). Here is a simple proof: In the list of 27, you will find [r, r, g], [r, g, r] and [g, r, r]. They are three different permutations of one single combination. — Hallvard, Feb 10 '18 at 23:42
@khelwood You have obviously failed to read my early comment where I agreed with you on that, but disagreed with your erroneous statement about “27 combinations”. — Hallvard, Feb 12 '18 at 16:36
@khelwood How do you go from "27 permutations, but 10 combinations" to "27 equals 10"? — Hallvard, Feb 12 '18 at 16:50

jpp · Answer 1 · 2018-02-08T13:58:20.483

4

This should work:

from itertools import product

list(product('rgb', repeat=3))

edited Feb 08 '18 at 13:58

answered Feb 08 '18 at 13:43

jpp

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2

You do not need to apply `list` to `'rgb'`, you can do without. – Elmex80s Feb 08 '18 at 13:51
1

Or in Python 3, you can just do `[*product('rgb', repeat=3)]`. In earlier versions, you can do `list(product('rgb', repeat=3))`. – PM 2Ring Feb 08 '18 at 13:53

score 1 · Accepted Answer · answered Feb 08 '18 at 13:51

The problem is that with

comb_list.append(current_combination)

you are appending the same list again and again to the result list. Thus changing a value in that list, e.g. with current_combination[0] = rgb[x] will also change the value in all the "other" lists.

You could fix this by creating a new list from that list prior to insertion:

comb_list.append(list(current_combination))

Or just create that list from the three elements directly when adding it to the result:

for x in rgb:
    for y in rgb:
        for z in rgb:
            comb_list.append([x, y, z])

Or just use a list-comprehension for the whole thing:

comb_list = [[x, y, z] for x in rgb for y in rgb for z in rgb]

Or use itertools.product, as described in the other answer.

Ah, now I understand. Thank you for the explanation! – Hallvard Feb 08 '18 at 14:01 — Hallvard, Feb 08 '18 at 14:01

score 0 · Answer 3 · answered Feb 08 '18 at 13:54

0

One Solution:

import itertools

result = itertools.product("rgb",repeat = 3)    
print(list(result))

answered Feb 08 '18 at 13:54

Gab

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Listing all permutations of RGB in Python

3 Answers3