All combinations of grouped Python dicts

Question

I have a list of lists containing dict objects. Each of the nested lists are made up by an array of objects(dict) that define a part. I need to figure out how I can generate a list containing all the part combinations possible.

For example here is my list of lists grouped by parts.

options = [
    [
        # seat part
        {'index': 0, 'name': 'Seat A', ...},
        {'index': 1, 'name': 'Seat B', ...},
        {'index': 2, 'name': 'Seat C', ...},
        {'index': 3, 'name': 'Seat D', ...},
        {'index': 4, 'name': 'Seat E', ...}
    ],
    [
        # legs part
        {'index': 0, 'name': 'Legs A', ...},
        {'index': 1, 'name': 'Legs B', ...},
        {'index': 2, 'name': 'Legs C', ...}
    ],
    [
        # pillows part
        {'index': 0, 'name': 'Pillows A', ...},
        {'index': 1, 'name': 'Pillows B', ...},
        {'index': 2, 'name': 'Pillows C', ...}
    ]
]

In this particular case I would expect it to generate 45 variations. The solution can’t be hard coded to work for only 3 options of parts, sometimes there may be 2 different types of parts, other times there could be more. The number of variations per part can vary as you see above as some have 3 others have more or less.

I would expect the results to just be a list containing a list of all variations.

variations = [
    [{'index': 0, 'name': 'Seat A', ...}, {'index': 0, 'name': 'Legs A', ...}, {'index': 0, 'name': 'Pillows A', ...}]
    [{'index': 0, 'name': 'Seat A', ...}, {'index': 0, 'name': 'Legs A', ...}, {'index': 0, 'name': 'Pillows B', ...}]
    [{'index': 0, 'name': 'Seat A', ...}, {'index': 0, 'name': 'Legs A', ...}, {'index': 0, 'name': 'Pillows C', ...}]
    [{'index': 0, 'name': 'Seat A', ...}, {'index': 0, 'name': 'Legs B', ...}, {'index': 0, 'name': 'Pillows A', ...}]
    [{'index': 0, 'name': 'Seat A', ...}, {'index': 0, 'name': 'Legs B', ...}, {'index': 0, 'name': 'Pillows B', ...}]
    [{'index': 0, 'name': 'Seat A', ...}, {'index': 0, 'name': 'Legs B', ...}, {'index': 0, 'name': 'Pillows C', ...}]
    [{'index': 0, 'name': 'Seat A', ...}, {'index': 0, 'name': 'Legs C', ...}, {'index': 0, 'name': 'Pillows A', ...}]
    [{'index': 0, 'name': 'Seat A', ...}, {'index': 0, 'name': 'Legs C', ...}, {'index': 0, 'name': 'Pillows B', ...}]
    [{'index': 0, 'name': 'Seat A', ...}, {'index': 0, 'name': 'Legs C', ...}, {'index': 0, 'name': 'Pillows C', ...}]
    ...
    [{'index': 0, 'name': 'Seat E', ...}, {'index': 0, 'name': 'Legs A', ...}, {'index': 0, 'name': 'Pillows A', ...}]
    [{'index': 0, 'name': 'Seat E', ...}, {'index': 0, 'name': 'Legs A', ...}, {'index': 0, 'name': 'Pillows B', ...}]
    [{'index': 0, 'name': 'Seat E', ...}, {'index': 0, 'name': 'Legs A', ...}, {'index': 0, 'name': 'Pillows C', ...}]
    [{'index': 0, 'name': 'Seat E', ...}, {'index': 0, 'name': 'Legs B', ...}, {'index': 0, 'name': 'Pillows A', ...}]
    [{'index': 0, 'name': 'Seat E', ...}, {'index': 0, 'name': 'Legs B', ...}, {'index': 0, 'name': 'Pillows B', ...}]
    [{'index': 0, 'name': 'Seat E', ...}, {'index': 0, 'name': 'Legs B', ...}, {'index': 0, 'name': 'Pillows C', ...}]
    [{'index': 0, 'name': 'Seat E', ...}, {'index': 0, 'name': 'Legs C', ...}, {'index': 0, 'name': 'Pillows A', ...}]
    [{'index': 0, 'name': 'Seat E', ...}, {'index': 0, 'name': 'Legs C', ...}, {'index': 0, 'name': 'Pillows B', ...}]
    [{'index': 0, 'name': 'Seat E', ...}, {'index': 0, 'name': 'Legs C', ...}, {'index': 0, 'name': 'Pillows C', ...}]
]

Asked By: JokerMartini

||

Source

Answer 1

variations = [
[{‘index’: 0, ‘name’: ‘Seat A’, …}, {‘index’: 0, ‘name’: ‘Legs A’, …}, {‘index’: 0, ‘name’: ‘Pillows A’, …}]
[{‘index’: 0, ‘name’: ‘Seat A’, …}, {‘index’: 0, ‘name’: ‘Legs A’, …}, {‘index’: 0, ‘name’: ‘Pillows B’, …}]
[{‘index’: 0, ‘name’: ‘Seat A’, …}, {‘index’: 0, ‘name’: ‘Legs A’, …}, {‘index’: 0, ‘name’: ‘Pillows C’, …}]
[{‘index’: 0, ‘name’: ‘Seat A’, …}, {‘index’: 0, ‘name’: ‘Legs B’, …}, {‘index’: 0, ‘name’: ‘Pillows A’, …}]
[{‘index’: 0, ‘name’: ‘Seat A’, …}, {‘index’: 0, ‘name’: ‘Legs B’, …}, {‘index’: 0, ‘name’: ‘Pillows B’, …}]
[{‘index’: 0, ‘name’: ‘Seat A’, …}, {‘index’: 0, ‘name’: ‘Legs B’, …}, {‘index’: 0, ‘name’: ‘Pillows C’, …}]
[{‘index’: 0, ‘name’: ‘Seat A’, …}, {‘index’: 0, ‘name’: ‘Legs C’, …}, {‘index’: 0, ‘name’: ‘Pillows A’, …}]
[{‘index’: 0, ‘name’: ‘Seat A’, …}, {‘index’: 0, ‘name’: ‘Legs C’, …}, {‘index’: 0, ‘name’: ‘Pillows B’, …}]
[{‘index’: 0, ‘name’: ‘Seat A’, …}, {‘index’: 0, ‘name’: ‘Legs C’, …}, {‘index’: 0, ‘name’: ‘Pillows C’, …}]
…
[{‘index’: 0, ‘name’: ‘Seat E’, …}, {‘index’: 0, ‘name’: ‘Legs A’, …}, {‘index’: 0, ‘name’: ‘Pillows A’, …}]
[{‘index’: 0, ‘name’: ‘Seat E’, …}, {‘index’: 0, ‘name’: ‘Legs A’, …}, {‘index’: 0, ‘name’: ‘Pillows B’, …}]
[{‘index’: 0, ‘name’: ‘Seat E’, …}, {‘index’: 0, ‘name’: ‘Legs A’, …}, {‘index’: 0, ‘name’: ‘Pillows C’, …}]
[{‘index’: 0, ‘name’: ‘Seat E’, …}, {‘index’: 0, ‘name’: ‘Legs B’, …}, {‘index’: 0, ‘name’: ‘Pillows A’, …}]
[{‘index’: 0, ‘name’: ‘Seat E’, …}, {‘index’: 0, ‘name’: ‘Legs B’, …}, {‘index’: 0, ‘name’: ‘Pillows B’, …}]
[{‘index’: 0, ‘name’: ‘Seat E’, …}, {‘index’: 0, ‘name’: ‘Legs B’, …}, {‘index’: 0, ‘name’: ‘Pillows C’, …}]
[{‘index’: 0, ‘name’: ‘Seat E’, …}, {‘index’: 0, ‘name’: ‘Legs C’, …}, {‘index’: 0, ‘name’: ‘Pillows A’, …}]
[{‘index’: 0, ‘name’: ‘Seat E’, …}, {‘index’: 0, ‘name’: ‘Legs C’, …}, {‘index’: 0, ‘name’: ‘Pillows B’, …}]
[{‘index’: 0, ‘name’: ‘Seat E’, …}, {‘index’: 0, ‘name’: ‘Legs C’, …}, {‘index’: 0, ‘name’: ‘Pillows C’, …}]
]

Answered By: Shashi Bhushan Kumar

Answer 2

Here’s a two-liner that provides the outcome you seek by performing a Cartesian Product of the elements:

from itertools import product

# We use the unpack operator (*) to convert the sub-lists as
# individual sources that are paired against each other
# using the `product` function to perform the cartesian product.
variations = [list(p) for p in product(*options)]

References:

https://docs.python.org/3/library/itertools.html#itertools.product

Answered By: jrd1

All combinations of grouped Python dicts

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