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Counting elements of multidimensional list according to patterns


Specifying string patterns in DeleteCasesGenerating a list of cubefree numbersGenerating the set of all three-dimensional binary arrays distinct under rotation and reflection?Distinguishing between elements of a given listFlatten sublists within a bigger listList manipulation with random elements under limiting conditionsEfficiently Evaluate a Multivariable Function on a List of TuplesTuples of elements from list excluding anything with repeated valuesDeleting duplicates only if they are a certain kind of duplicateReformatting the pattern inside a list













2












$begingroup$


I have the tuples :



a = Range[0, 2];
l2 = Tuples[a, a]


which gives



0, 0, 0, 1, 0, 2, 1, 0, 1, 1, 1, 2, 2, 0, 2, 1, 2, 
2


I want to find all tuples which contain number 1 exactly once.
A possible solution would be :



l21 = Cases[l2, 1, x_ /; x != 1 | x_ /; x != 1, 1]


For the three - dimensional case it would be something like this :



l3 = Tuples[a, a, a];
Cases[l3, 1, x_ /; x != 1, y_ /; y != 1 | x_ /; x != 1, 1, 
 y_ /; y != 1 | x_ /; x != 1, y_ /; y != 1, 1]


What would be a shorter way to use patterns in Cases for higher
dimensions, e.g. 



Tuples[a,a,a,a,a]





list-manipulation







share|improve this question











$endgroup$











  • $begingroup$
    l2 = Tuples[a, 2] and l3 = Tuples[a, 3] are a bit simpler and more easily extended to higher dimensions.
    $endgroup$
    – Roman
    May 7 at 11:58
















2












$begingroup$


I have the tuples :



a = Range[0, 2];
l2 = Tuples[a, a]


which gives



0, 0, 0, 1, 0, 2, 1, 0, 1, 1, 1, 2, 2, 0, 2, 1, 2, 
2


I want to find all tuples which contain number 1 exactly once.
A possible solution would be :



l21 = Cases[l2, 1, x_ /; x != 1 | x_ /; x != 1, 1]


For the three - dimensional case it would be something like this :



l3 = Tuples[a, a, a];
Cases[l3, 1, x_ /; x != 1, y_ /; y != 1 | x_ /; x != 1, 1, 
 y_ /; y != 1 | x_ /; x != 1, y_ /; y != 1, 1]


What would be a shorter way to use patterns in Cases for higher
dimensions, e.g. 



Tuples[a,a,a,a,a]





list-manipulation







share|improve this question











$endgroup$











  • $begingroup$
    l2 = Tuples[a, 2] and l3 = Tuples[a, 3] are a bit simpler and more easily extended to higher dimensions.
    $endgroup$
    – Roman
    May 7 at 11:58














2












2








2





$begingroup$


I have the tuples :



a = Range[0, 2];
l2 = Tuples[a, a]


which gives



0, 0, 0, 1, 0, 2, 1, 0, 1, 1, 1, 2, 2, 0, 2, 1, 2, 
2


I want to find all tuples which contain number 1 exactly once.
A possible solution would be :



l21 = Cases[l2, 1, x_ /; x != 1 | x_ /; x != 1, 1]


For the three - dimensional case it would be something like this :



l3 = Tuples[a, a, a];
Cases[l3, 1, x_ /; x != 1, y_ /; y != 1 | x_ /; x != 1, 1, 
 y_ /; y != 1 | x_ /; x != 1, y_ /; y != 1, 1]


What would be a shorter way to use patterns in Cases for higher
dimensions, e.g. 



Tuples[a,a,a,a,a]





list-manipulation







share|improve this question











$endgroup$




I have the tuples :



a = Range[0, 2];
l2 = Tuples[a, a]


which gives



0, 0, 0, 1, 0, 2, 1, 0, 1, 1, 1, 2, 2, 0, 2, 1, 2, 
2


I want to find all tuples which contain number 1 exactly once.
A possible solution would be :



l21 = Cases[l2, 1, x_ /; x != 1 | x_ /; x != 1, 1]


For the three - dimensional case it would be something like this :



l3 = Tuples[a, a, a];
Cases[l3, 1, x_ /; x != 1, y_ /; y != 1 | x_ /; x != 1, 1, 
 y_ /; y != 1 | x_ /; x != 1, y_ /; y != 1, 1]


What would be a shorter way to use patterns in Cases for higher
dimensions, e.g. 



Tuples[a,a,a,a,a]





list-manipulation




list-manipulation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited May 7 at 12:01









user64494

3,68511222




3,68511222










asked May 7 at 11:44









user57467user57467

663




663











  • $begingroup$
    l2 = Tuples[a, 2] and l3 = Tuples[a, 3] are a bit simpler and more easily extended to higher dimensions.
    $endgroup$
    – Roman
    May 7 at 11:58

















  • $begingroup$
    l2 = Tuples[a, 2] and l3 = Tuples[a, 3] are a bit simpler and more easily extended to higher dimensions.
    $endgroup$
    – Roman
    May 7 at 11:58
















$begingroup$
l2 = Tuples[a, 2] and l3 = Tuples[a, 3] are a bit simpler and more easily extended to higher dimensions.
$endgroup$
– Roman
May 7 at 11:58





$begingroup$
l2 = Tuples[a, 2] and l3 = Tuples[a, 3] are a bit simpler and more easily extended to higher dimensions.
$endgroup$
– Roman
May 7 at 11:58











1 Answer
1






active

oldest

votes


















4












$begingroup$

Select[l2, Count[#, 1] == 1 &]
Pick[l2, Counts[#][1] & /@ l2, 1]
Pick[l2, Total@Transpose[1-Unitize[l2-1]], 1]
Cases[Except[1] ..., 1, Except[1] ...]@l2



0, 1, 1, 0, 1, 2, 2, 1 




Select[l3, Count[#, 1] == 1 &]
Pick[l3, Counts[#][1] & /@ l3, 1]
Pick[l3, Total@Transpose[1-Unitize[l3-1]], 1]
Cases[Except[1] ..., 1, Except[1] ...]@l3



0, 0, 1, 0, 1, 0, 0, 1, 2, 0, 2, 1, 1, 0, 0, 1, 0,
2, 1,
   2, 0, 1, 2, 2, 2, 0, 1, 2, 1, 0, 2, 1, 2, 2, 2, 1







share|improve this answer











$endgroup$












  • $begingroup$
    Thanks! So easy!
    $endgroup$
    – user57467
    May 7 at 11:50










  • $begingroup$
    @user57467, you are welcome.
    $endgroup$
    – kglr
    May 7 at 11:53











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Select[l2, Count[#, 1] == 1 &]
Pick[l2, Counts[#][1] & /@ l2, 1]
Pick[l2, Total@Transpose[1-Unitize[l2-1]], 1]
Cases[Except[1] ..., 1, Except[1] ...]@l2



0, 1, 1, 0, 1, 2, 2, 1 




Select[l3, Count[#, 1] == 1 &]
Pick[l3, Counts[#][1] & /@ l3, 1]
Pick[l3, Total@Transpose[1-Unitize[l3-1]], 1]
Cases[Except[1] ..., 1, Except[1] ...]@l3



0, 0, 1, 0, 1, 0, 0, 1, 2, 0, 2, 1, 1, 0, 0, 1, 0,
2, 1,
   2, 0, 1, 2, 2, 2, 0, 1, 2, 1, 0, 2, 1, 2, 2, 2, 1







share|improve this answer











$endgroup$












  • $begingroup$
    Thanks! So easy!
    $endgroup$
    – user57467
    May 7 at 11:50










  • $begingroup$
    @user57467, you are welcome.
    $endgroup$
    – kglr
    May 7 at 11:53















4












$begingroup$

Select[l2, Count[#, 1] == 1 &]
Pick[l2, Counts[#][1] & /@ l2, 1]
Pick[l2, Total@Transpose[1-Unitize[l2-1]], 1]
Cases[Except[1] ..., 1, Except[1] ...]@l2



0, 1, 1, 0, 1, 2, 2, 1 




Select[l3, Count[#, 1] == 1 &]
Pick[l3, Counts[#][1] & /@ l3, 1]
Pick[l3, Total@Transpose[1-Unitize[l3-1]], 1]
Cases[Except[1] ..., 1, Except[1] ...]@l3



0, 0, 1, 0, 1, 0, 0, 1, 2, 0, 2, 1, 1, 0, 0, 1, 0,
2, 1,
   2, 0, 1, 2, 2, 2, 0, 1, 2, 1, 0, 2, 1, 2, 2, 2, 1







share|improve this answer











$endgroup$












  • $begingroup$
    Thanks! So easy!
    $endgroup$
    – user57467
    May 7 at 11:50










  • $begingroup$
    @user57467, you are welcome.
    $endgroup$
    – kglr
    May 7 at 11:53













4












4








4





$begingroup$

Select[l2, Count[#, 1] == 1 &]
Pick[l2, Counts[#][1] & /@ l2, 1]
Pick[l2, Total@Transpose[1-Unitize[l2-1]], 1]
Cases[Except[1] ..., 1, Except[1] ...]@l2



0, 1, 1, 0, 1, 2, 2, 1 




Select[l3, Count[#, 1] == 1 &]
Pick[l3, Counts[#][1] & /@ l3, 1]
Pick[l3, Total@Transpose[1-Unitize[l3-1]], 1]
Cases[Except[1] ..., 1, Except[1] ...]@l3



0, 0, 1, 0, 1, 0, 0, 1, 2, 0, 2, 1, 1, 0, 0, 1, 0,
2, 1,
   2, 0, 1, 2, 2, 2, 0, 1, 2, 1, 0, 2, 1, 2, 2, 2, 1







share|improve this answer











$endgroup$



Select[l2, Count[#, 1] == 1 &]
Pick[l2, Counts[#][1] & /@ l2, 1]
Pick[l2, Total@Transpose[1-Unitize[l2-1]], 1]
Cases[Except[1] ..., 1, Except[1] ...]@l2



0, 1, 1, 0, 1, 2, 2, 1 




Select[l3, Count[#, 1] == 1 &]
Pick[l3, Counts[#][1] & /@ l3, 1]
Pick[l3, Total@Transpose[1-Unitize[l3-1]], 1]
Cases[Except[1] ..., 1, Except[1] ...]@l3



0, 0, 1, 0, 1, 0, 0, 1, 2, 0, 2, 1, 1, 0, 0, 1, 0,
2, 1,
   2, 0, 1, 2, 2, 2, 0, 1, 2, 1, 0, 2, 1, 2, 2, 2, 1








share|improve this answer














share|improve this answer



share|improve this answer








edited May 7 at 12:04

























answered May 7 at 11:49









kglrkglr

192k10213432




192k10213432











  • $begingroup$
    Thanks! So easy!
    $endgroup$
    – user57467
    May 7 at 11:50










  • $begingroup$
    @user57467, you are welcome.
    $endgroup$
    – kglr
    May 7 at 11:53
















  • $begingroup$
    Thanks! So easy!
    $endgroup$
    – user57467
    May 7 at 11:50










  • $begingroup$
    @user57467, you are welcome.
    $endgroup$
    – kglr
    May 7 at 11:53















$begingroup$
Thanks! So easy!
$endgroup$
– user57467
May 7 at 11:50




$begingroup$
Thanks! So easy!
$endgroup$
– user57467
May 7 at 11:50












$begingroup$
@user57467, you are welcome.
$endgroup$
– kglr
May 7 at 11:53




$begingroup$
@user57467, you are welcome.
$endgroup$
– kglr
May 7 at 11:53

















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