Printing out combinations of two lists in python

Question

I am trying to print out all the possible combinations of two lists, so for example if list1 has values {1,2,3} and list two has values {A,B,C} I should get the following:

A=1 B=1 C=1
A=1 B=1 C=2
A=1 B=1 C=3
A=1 B=2 C=1
A=1 B=2 C=2
A=1 B=2 C=3
A=1 B=3 C=1
A=1 B=3 C=2
A=1 B=3 C=3
etc

For all of the possible values in the list. So the last line should be

A=3 B=3 C=3

Now this should work for abititary sized arrays. So if I have list1 = {1,2,3,4,5,6} list2 = {A,B,C,D,E,F}

I should get

A=1 B=1 C=1 D=1 E=1 F=1
A=1 B=1 C=1 D=1 E=1 F=2
A=1 B=1 C=1 D=1 E=1 F=3
A=1 B=1 C=1 D=1 E=1 F=4
A=1 B=1 C=1 D=1 E=1 F=5
A=1 B=1 C=1 D=1 E=1 F=6
A=1 B=1 C=1 D=1 E=2 F=1
A=1 B=1 C=1 D=1 E=2 F=2
A=1 B=1 C=1 D=1 E=2 F=3
A=1 B=1 C=1 D=1 E=2 F=4
A=1 B=1 C=1 D=1 E=2 F=5
A=1 B=1 C=1 D=1 E=2 F=6
etc

So in the first example there would be 3^3 = 27 combinations and I attempted to solve the problem like:

def printArrays():
    letters = {'A', 'B', 'C', 'D', 'E', 'F'}
    domain = {1, 2, 3, 4, 5, 6}
    myStr = ""
    for value in domain:
        for letter in letters:
            myStr += letter + "=" + str(value) + " "
        myStr += "\n"
    print(myStr)

Obviously it gives the wrong output since it gives me:

F=1 D=1 C=1 B=1 E=1 A=1 
F=2 D=2 C=2 B=2 E=2 A=2 
F=3 D=3 C=3 B=3 E=3 A=3 
F=4 D=4 C=4 B=4 E=4 A=4 
F=5 D=5 C=5 B=5 E=5 A=5 
F=6 D=6 C=6 B=6 E=6 A=6

What would be the correct way to solve this problem?

Note: That I would also want a solution that does not involve any libraries as I am trying to learn recursion so a recursive solution would be appreciated.

Oh... I did not notice. You used set { }. Just use list [] instead — Xin Cheng
– Xin Cheng, Commented Oct 13, 2022 at 4:15
the data stored in the set is unordered. While the data stored in the list is ordered. — Xin Cheng
– Xin Cheng, Commented Oct 13, 2022 at 4:16

Michael Delgado · Accepted Answer · 2022-10-13 04:59:10Z

4

what you're doing is iterating through letters * domain. What you want is to iterate though letters ^ len(domain).

I'll use itertools.product for this since it's built explicitly for combinatorial products:

import itertools

def printArrays():
    letters = {'A', 'B', 'C', 'D', 'E', 'F'}
    domain = {1, 2, 3, 4, 5, 6}
    myStr = ""
    for current in itertools.product(domain, repeat=len(letters)):
        for k, v in zip(letters, current):
            myStr += k + "=" + str(v) + " "
        myStr += "\n"
    print(myStr)

This prints what you're looking for:

E=1 D=1 F=1 B=1 A=1 C=1
E=1 D=1 F=1 B=1 A=1 C=2
E=1 D=1 F=1 B=1 A=1 C=3
E=1 D=1 F=1 B=1 A=1 C=4
E=1 D=1 F=1 B=1 A=1 C=5
E=1 D=1 F=1 B=1 A=1 C=6
E=1 D=1 F=1 B=1 A=2 C=1
E=1 D=1 F=1 B=1 A=2 C=2
E=1 D=1 F=1 B=1 A=2 C=3
E=1 D=1 F=1 B=1 A=2 C=4
E=1 D=1 F=1 B=1 A=2 C=5
E=1 D=1 F=1 B=1 A=2 C=6
E=1 D=1 F=1 B=1 A=3 C=1
...
E=6 D=6 F=6 B=6 A=6 C=2
E=6 D=6 F=6 B=6 A=6 C=3
E=6 D=6 F=6 B=6 A=6 C=4
E=6 D=6 F=6 B=6 A=6 C=5
E=6 D=6 F=6 B=6 A=6 C=6

edited Oct 13, 2022 at 4:59

answered Oct 13, 2022 at 4:07

Michael Delgado

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1 Comment

chen Over a year ago

Is there a way to do this without libraries as I am trying to learn recursion.

ILS · Accepted Answer · 2022-10-13 04:08:25Z

2

The value assignment is like Cartesian product of list1, so using itertools.product.

import itertools

list1 = [1, 2, 3]
list2 = ['A', 'B', 'C']
assignment_list = itertools.product(list1, repeat=len(list2))
buffer = ""
for assignment in assignment_list:
    for var, val in zip(list2, assignment):
        buffer += f"{var}={val} "
    buffer += "\n"
print(buffer)

answered Oct 13, 2022 at 4:08

ILS

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1 Comment

Michael Delgado Over a year ago

good call with using the repeat argument of itertools.product!

Michael Delgado · Accepted Answer · 2022-10-13 04:34:39Z

This addresses the problem with a recursive approach:

def add_domain_for_remaining_letters(letters, domain, prefix=""):
    # termination condition:
    # if no remaining letters, just return the prefix
    if len(letters) == 0:
        return prefix + "\n"

    # otherwise, loop over each i in the domain
    # and recurse for each remaining letter
    result = ""
    for i in domain:
        result += add_domain_for_remaining_letters(
            letters=letters[1:], domain=domain, prefix=f"{prefix}{letters[0]}={i} "
        )
    return result

This gives the desired output as a string:

In [20]: print(add_domain_for_remaining_letters(letters=["A", "B"], domain=[1, 2]))
A=1 B=1
A=1 B=2
A=2 B=1
A=2 B=2


In [21]: print(add_domain_for_remaining_letters(letters=["A", "B", "C", "D"], domain=[1, 2, 3]))
A=1 B=1 C=1 D=1
A=1 B=1 C=1 D=2
A=1 B=1 C=1 D=3
A=1 B=1 C=2 D=1
...
A=3 B=3 C=2 D=1
A=3 B=3 C=2 D=2
A=3 B=3 C=2 D=3
A=3 B=3 C=3 D=1
A=3 B=3 C=3 D=2
A=3 B=3 C=3 D=3

Mark · Accepted Answer · 2022-10-13 04:54:53Z

Notice that A=, B=, C= is the same for every row. As others have pointed out the numbers are just the cartesian product of the set of numbers. With that in mind, you could make a simple recursive function to handle the product, and then zip in the letters. Python generators allow you to do this in a memory efficient way — you don't need to store all the combinations in memory:

n = [1,2,3]
keys = ['A','B','C']

# Generate product
def rec_product(l, r):
    if r > 1:
        yield from ((*t, n) for t in rec_product(l, r-1) for n in l)
    else:
        yield from ((n,) for n in l)

# add the letters and print
for p in rec_product(n, len(n)):
    print(*(f'{k}={v}' for k, v in zip(keys, p)))

This will print:

A=1 B=1 C=1
A=1 B=1 C=2
A=1 B=1 C=3
…
A=3 B=3 C=2
A=3 B=3 C=3

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