New answers tagged filtering
5
Values @ GroupBy[list, Most, Mean]
{{1, 1, (a + b)/2}, {1, 2, (c + d)/2}, {1, 3, e}}
Also
Mean /@ GatherBy[list, Most]
{{1, 1, (a + b)/2}, {1, 2, (c + d)/2}, {1, 3, e}}
3
Is this faster and works correctly for all your cases?
list={{1,1,a},{1,1,b},{1,2,c},{1,2,d},{1,3,e}};
Mean/@Split[Sort[list],Most[#1]==Most[#2]&]
(*{{1, 1, (a + b)/2}, {1, 2, (c + d)/2}, {1, 3, e}}*)
4
Converting them to associations and using Merge might be a good solution:
In[1]:= KeyValueMap[Append,
Merge[<|#[[1 ;; 2]] -> #[[3]]|> & /@ {{1, 1, a}, {1, 1, b}, {1, 2, c}, {1, 2, d}, {1, 3, e}}, Mean]]
Out[1]= {{1, 1, (a + b)/2}, {1, 2, (c + d)/2}, {1, 3, e}}
2
Testing on your sample data dataset:
r1 = DeleteMissing[dataset, 1, 1]
Takes 0.000530041 second
r2 = dataset[Select[Not@AnyTrue[#, MissingQ] &]]
Takes 0.000412052 second
r3 = dataset[Select[Not@MatchQ[#, KeyValuePattern[{_ -> _?MissingQ}]] &]]
Takes 0.000428904 second
All have the same result:
r1 == r2 == r3
(*Out: True *)
1
one way to do this is:
Dataset@DeleteCases[
Normal[dataset],
<|___, _ -> Missing["Unmatched"], ___|>
];
which takes $2.32$ seconds on bigDataset (in my question)
Benchmarking @BenIzd 's methods and my own
Top 50 recent answers are included
