Counting elements of multidimensional list according to patternsSpecifying 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
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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
$begingroup$
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
edited May 7 at 12:01
user64494
3,68511222
asked May 7 at 11:44
user57467user57467
663
$endgroup$
add a comment |
$begingroup$
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
edited May 7 at 12:01
user64494
3,68511222
asked May 7 at 11:44
user57467user57467
663
$endgroup$
$begingroup$
l2 = Tuples[a, 2]andl3 = Tuples[a, 3]are a bit simpler and more easily extended to higher dimensions.
$endgroup$
– Roman
May 7 at 11:58
add a comment |
$begingroup$
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
edited May 7 at 12:01
user64494
3,68511222
asked May 7 at 11:44
user57467user57467
663
$endgroup$
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
edited May 7 at 12:01
user64494
3,68511222
asked May 7 at 11:44
user57467user57467
663
edited May 7 at 12:01
user64494
3,68511222
asked May 7 at 11:44
user57467user57467
663
edited May 7 at 12:01
user64494
3,68511222
edited May 7 at 12:01
user64494
3,68511222
edited May 7 at 12:01
user64494
3,68511222
3,68511222
asked May 7 at 11:44
user57467user57467
663
asked May 7 at 11:44
user57467user57467
663
asked May 7 at 11:44
user57467user57467
663
663
$begingroup$
l2 = Tuples[a, 2]andl3 = Tuples[a, 3]are a bit simpler and more easily extended to higher dimensions.
$endgroup$
– Roman
May 7 at 11:58
add a comment |
$begingroup$
l2 = Tuples[a, 2]andl3 = 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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
answered May 7 at 11:49
kglrkglr
192k10213432
$endgroup$
$begingroup$
Thanks! So easy!
$endgroup$
– user57467
May 7 at 11:50
$begingroup$
@user57467, you are welcome.
$endgroup$
– kglr
May 7 at 11:53
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
answered May 7 at 11:49
kglrkglr
192k10213432
$endgroup$
$begingroup$
Thanks! So easy!
$endgroup$
– user57467
May 7 at 11:50
$begingroup$
@user57467, you are welcome.
$endgroup$
– kglr
May 7 at 11:53
add a comment |
$begingroup$
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
answered May 7 at 11:49
kglrkglr
192k10213432
$endgroup$
$begingroup$
Thanks! So easy!
$endgroup$
– user57467
May 7 at 11:50
$begingroup$
@user57467, you are welcome.
$endgroup$
– kglr
May 7 at 11:53
add a comment |
$begingroup$
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
answered May 7 at 11:49
kglrkglr
192k10213432
$endgroup$
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
answered May 7 at 11:49
kglrkglr
192k10213432
edited May 7 at 12:04
answered May 7 at 11:49
kglrkglr
192k10213432
answered May 7 at 11:49
kglrkglr
192k10213432
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
l2 = Tuples[a, 2]andl3 = Tuples[a, 3]are a bit simpler and more easily extended to higher dimensions.$endgroup$
– Roman
May 7 at 11:58