How to remove subarray with a common element in another array in python?

Question

ListA = [['abc','15','2021-10-02 08:53:29'],
         ['def','10','2021-10-01 07:52:19'],
         ['abc','15','2021-10-02 09:53:29'],
         ['def','10','2021-10-01 06:52:19'],
         ['gfc','10','2021-10-01 07:52:19']]

ListB = ['abc','def']

Since ListB has 'abc' and 'def', the script should

compare subarrays in array A which have "abc":
1. find the latest date. eg.'2021-10-02 09:53:29'
2. remove subarray which has "abc" but the date is earlier than the latest date
compare subarrays in array A which have "def" a) find the latest date. eg. '2021-10-01 07:52:19' b) remove subarray which has "def" but the date is earlier than the latest date

The final output should be

A = [['def','10','2021-10-01 07:52:19'],
     ['abc','15','2021-10-02 09:53:29'],
     ['gfc','10','2021-10-01 07:52:19']]

How to do this in Python?

Hi! Welcome to Stack Overflow! You are very welcome to be here, we love having your here. It's just your question that we are having troubles welcoming, because we are not a code writing service. So you are welcome and we welcome you, please feel that you are welcomed, but your question is going to get closed and deleted. If you do not abide by these guidelines. — The Singularity
– The Singularity, Commented Oct 5, 2021 at 12:03
Please provide what you have already tried and which part you are stuck with — Robin Gertenbach
– Robin Gertenbach, Commented Oct 5, 2021 at 12:04
Also I assume the last sublist is not intended to be a nested sublist with a missing closed bracket so I changed it to be a list of lists — Robin Gertenbach
– Robin Gertenbach, Commented Oct 5, 2021 at 12:07

mozway · Accepted Answer · 2021-10-05 12:26:39Z

2

You could sort the array by key and date, then groupby and take the item with highest date if key is in ListB else, the whole array. Finally flatten using chain:

from itertools import groupby, chain

list(chain(*([list(g)[-1]] if k in ListB else list(g)
             for k,g in groupby(sorted(ListA, key=lambda x: (x[0], x[-1])), lambda x: x[0])
            )
          ))

output:

[['abc', '15', '2021-10-02 09:53:29'],
 ['def', '10', '2021-10-01 07:52:19'],
 ['gfc', '10', '2021-10-01 07:52:19']]

NB. this is not preserving the original order of the lists

keeping order

If order is important, it is possible to save it (here using enumerate) and to reorder after filtering:

from itertools import groupby, chain

[
i[1] for i in
sorted(chain(*([list(g)[-1]] if k in ListB else list(g)
             for k,g in groupby(sorted(enumerate(ListA), key=lambda x: (x[1][0], x[1][-1])), lambda x: x[1][0])
            )
          ))
]

output:

[['def', '10', '2021-10-01 07:52:19'],
 ['abc', '15', '2021-10-02 09:53:29'],
 ['gfc', '10', '2021-10-01 07:52:19']]

edited Oct 5, 2021 at 12:26

answered Oct 5, 2021 at 12:21

mozway

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3 Comments

mozway Over a year ago

@liuxu you can consider marking the answer as accepted if your problem is solved ;)

liuxu Over a year ago

Thanks @mozway, but can you explain more about the script? I want to understand more about it :)

mozway Over a year ago

@liuxu what have you not understood? Group by key (second item), if the key is in ListB take the last (most recent) element, else the whole list. Finally, flatten the lists to original format. Let me know if anything is unclear, but the best might be to try running the code from in to out to see the intermediates

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