Removing duplicates from an output base on the sum of two numbers

Question

I'm very new to Python so please be kind!:) I'm just messing about at the moment but I can't find anything specific on this particular issue.

I'm looking to remove duplicates from an output based upon the the sum of two numbers. I have the following lists of numbers and the short code where I can print the numbers that multiply to give 24.

A = 1,2,3,4,5,6,8
B = 1,2,3,4,5,6,8

for a in A:
    for b in B:
        if (a * b) == 24:
            print (a, b, a+b)

The outcomes here are:

3 8 11
4 6 10
6 4 10
8 3 11

In the output there are four combinations that give 24, two of which are permutations of each other. Is there any way that I can deduplicate the permutations to get "unique" outputs? One of the ways that I thought to do this was by adding a+b and trying to remove the duplicates in output based on this number but I can't think of a way to do this. Anyone got any ideas how I could do this?

Answer 1

Create a set and store combinations of (a,b) , (b,a) in it if printing. Don't print if (a,b) already in set:

A = 1,2,3,4,5,6,8
B = 1,2,3,4,5,6,8

got = set() # collect (a,b) and (b,a) that you printed already to avoid dupes
for a in A:
    for b in B:
        if (a * b) == 24:
            if (a,b) in got:
                continue
            got.add((a,b))  # need to add (a,b) and (b,a) to avoid printing 
            got.add((b,a))  # (b,a) after you already printed (a,b)
            print (a, b, a+b)

Output:

3 8 11
4 6 10

---

Lets compare this and olivn s approach - I removed both print statements:

A = 1,2,3,4,5,6,8 
B = 1,2,3,4,5,6,8 

def ol():
    res = set()
    for a in A:
        for b in B:
            if a * b == 24:
                if {a, b} not in res:
                    # print(a, b, a + b)
                    res.add(frozenset((a, b)))

def pa():
    got = set() # collect (a,b) and (b,a) that you printed already to avoid dupes
    for a in A:
        for b in B:
            if (a * b) == 24:
                if (a,b) in got:
                    continue
                got.add((a,b))  # need to add (a,b) and (b,a) to avoid printing 
                got.add((b,a))  # (b,a) after you already printed (a,b)
                # print (a, b, a+b)
 

from timeit import timeit

print("Olivn:", timeit(ol, number=10000))
print("this: ", timeit(pa, number=10000))

Output:

Olivn: 0.60943335
this:  0.6066992629999999

Answer 2

You may create keep track of the numbers with results as 24 and can avoid the duplicate numbers as follow

A = 1,2,3,4,5,6,8
B = 1,2,3,4,5,6,8
pass_list: list = []

for a in A:
    for b in B:
        if (a * b) == 24 and b not in pass_list:
            pass_list.append(a)
            print (a, b, a+b)

Output

3 8 11
4 6 10

---

Just extending @patrick's comparison with 1000000 counts, this answer seems bit more faster.

Olivn: 9.081725899999999
Patrick:  7.241247899999999
this:  5.5779484

Olivn: 9.3896246
Patrick:  7.960243800000001
this:  5.6804922

Olivn: 10.6548058
Patrick:  8.282727000000001
this:  7.1877061

Olivn: 11.1385039
Patrick:  9.226674899999999
this:  8.4106494

---

A = 1,2,3,4,5,6,8 
B = 1,2,3,4,5,6,8 

def ol():
    res = set()
    for a in A:
        for b in B:
            if a * b == 24:
                if {a, b} not in res:
                    # print(a, b, a + b)
                    res.add(frozenset((a, b)))

def pa():
    got = set() # collect (a,b) and (b,a) that you printed already to avoid dupes
    for a in A:
        for b in B:
            if (a * b) == 24:
                if (a,b) in got:
                    continue
                got.add((a,b))  # need to add (a,b) and (b,a) to avoid printing 
                got.add((b,a))  # (b,a) after you already printed (a,b)
                # print (a, b, a+b)
 

def pr():
    pass_list: list = []

    for a in A:
        for b in B:
            if (a * b) == 24 and b not in pass_list:
                pass_list.append(a)
                #print (a, b, a+b)


from timeit import timeit

print("Olivn:", timeit(ol, number=1000000))
print("Patrick: ", timeit(pa, number=1000000))
print("this: ", timeit(pr, number=1000000))

Answer 3

@PatrickArtner decided to not follow my recommendations in comments under his answer so here is correct (imho) way how you can apply set (and frozenset as well) :

A = 1,2,3,4,5,6,8
B = 1,2,3,4,5,6,8

res = set()
for a in A:
    for b in B:
        if a * b == 24:
            if {a, b} not in res:
                print(a, b, a + b)
                res.add(frozenset((a, b)))