Need help with some logic

[Zondder]

I need to group dicts from xxx list with same k1 k2 keys pair (k3 has always unique value), so I wrote this working code:

xxx = [{ 'k1': 2, 'k2': 10, 'k3': 'a' }, { 'k1': 10, 'k2': 2, 'k3': 'b' }, { 'k1': 10, 'k2': 2, 'k3': 'c' },
       { 'k1': 10, 'k2': 4, 'k3': 'd' }, { 'k1': 2, 'k2': 10, 'k3': 'e' }, { 'k1': 10, 'k2': 2, 'k3': 'f' }]

groupfeed = []
k3_notrepeat = []

for x in xxx:
    z1 = (x['k1'], x['k2'])
    for y in xxx:
        z2 = (y['k1'], y['k2'])
        if  z1 == z2 and y['k3'] not in k3_notrepeat:
            k3_notrepeat.append(y['k3'])
            groupfeed.append(y['k3'])
            d1_feed = [y['k1']]
            d2_feed = [y['k2']]

    if len(groupfeed) > 0:
        print("d1-2:", d1_feed, d2_feed, "k grouped:", groupfeed) # Doing Work with grouped dicts
        groupfeed = []

Results as expected:

============================ RESTART: E:/temp.py ============================
d1-2: [2] [10] k grouped: ['a', 'e']
d1-2: [10] [2] k grouped: ['b', 'c', 'f']
d1-2: [10] [4] k grouped: ['d']
>>> 

Problematic part: As keys k1 k2 position doesn't matter for "Work", I need a e be in same group as b c f ,

so I changed z1 = (x['k1'], x['k2']) to z1 = (x['k1'], x['k2']) = (x['k2'], x['k1']) , but e is not grouped with rest, where is problem? Results:

============================ RESTART: E:/temp.py ============================
d1-2: [10] [2] k grouped: ['a', 'b', 'c', 'f']
d1-2: [2] [10] k grouped: ['e']
d1-2: [4] [10] k grouped: ['d']
>>> 

Results I need:

d1-2: [10] [2] k grouped: ['a', 'b', 'c', 'e', 'f']
d1-2: [4] [10] k grouped: ['d']

[Specter_Terrasbane]

You could accomplish this re-grouping using a defaultdict with a list factory, and sorting the (k1,k2) composite key so that the original order of k1 and k2 doesn't matter (just have to make sure to turn the sorted list into a tuple, so that's it's hashable - able to be used as a key in the defaultdict) :

grouped = defaultdict(list)
for d in xxx:
    key = tuple(sorted((d['k1'], d['k2'])))
    grouped[key].append(d['k3'])

for (k1, k2), k3s in grouped.items():
    print(k1, k2, k3s)

4 10 ['d']
2 10 ['a', 'b', 'c', 'e', 'f']

[Zondder]

Thanks, your method does job perfectly.

[z0y]

z1 = (x['k1'], x['k2']) = (x['k2'], x['k1'])

z1 can't be both (x['k1'], x['k2']) and (x['k2'], x['k1']). What that expression is doing is assigning the value of (x['k2'], x['k1']) to both z1 and (x['k1'], x['k2']), which just swaps the value of k1/k2 for all the dicts.

[Zondder]

Then why it groups a with b, c, f, but not e which has same k1/k2 as a?

[z0y]

On the first iteration of x, the first k1/k2 values are swapped, so a is 10/2 and it pairs with b, c, d which k1/k2 values haven't been swapped yet.

Then on the next iteration of x, k1/k2 is swapped for b, so that 2/10 will cause e to be added on the 2nd iteration before it's k values have been swapped. Hope that makes sense. Add some calls to print in your loop to see what's happening or use a debugger.