How to generate all possible arrangements of numbers from a list in Python?

Problem: Given a list of 3 numbers [4, 8, 15] generate a list of all possible arrangements of the numbers.

[4, 8, 15]

That's 3*3*3 = 27 unique entries from what I can gather. Something like:

3*3*3 = 27

4,4,4
4,4,8
4,4,15
4,8,4
4,8,8
4,8,15
4,15,4
4,15,8
4,15,15
...


I tried using itertools.permutations and itertools.combinations but I can't get all 27 values.

itertools.permutations

itertools.combinations

list(itertools.permutations([4,8,15],3)) for example only prints 6 values:

list(itertools.permutations([4,8,15],3))

[(4, 8, 15), (4, 15, 8), (8, 4, 15), (8, 15, 4), (15, 4, 8), (15, 8, 4)]

[(4, 8, 15), (4, 15, 8), (8, 4, 15), (8, 15, 4), (15, 4, 8), (15, 8, 4)]

Is there something that's available out of the box or is this more of a "roll your own" problem?

I tried ...

list(itertools.product([4, 8, 15], repeat=3))

3 Answers
3

you are confusing permutations with product:

permutations

product

len(list(itertools.permutations([4,8,15],3)))
# return 6
len(list(itertools.product([4,8,15], repeat=3)))
# return 27


The answer is still in itertools. The function called product does the trick; it takes two arguments: first is the iterable which has the usable elements, second is the amount of times the iterable can repeat itself.

product

itertools.product([4,8,15],repeat=3) would return the permutations you want in your example.

itertools.product([4,8,15],repeat=3)

The reason permutations or combinations don't work for you is because they don't allow items to repeat themselves; product calculates the cartesian product, which allows for repetition of items.

permutations

combinations

product

product

itertools.product(*iterables, repeat=1)

3

iterables

This code prints what you requested :)

list = [4,8,15]

for i in range(len(list)):
for j in range (len(list)):
for k in range (len(list)):
print ("("+str(list[i]) +","+str(list[j])+","+str(list[k])+")n")


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