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Results tagged with tuples
Search options answers only
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user 125
4
votes
What is the identity for Tuples?
Use TagSetDelayed to define a function that behaves as desired:
ClearAll[iDentity]
iDentity /: {iDentity[___], a : {__}} := iDentity[a]
iDentity /: Tuples[iDentity[a_]] := a
Tuples[{iDentity[], {q}} … }] := a
altTuples[x_] := Tuples[x]
altTuples[{{x, y}, {1, 2}}]
{{x, 1}, {x, 2}, {y, 1}, {y, 2}}
altTuples[{tuplesIdentity, {1, 2}}]
{1, 2}
altTuples[{{}, {1, 2}}]
{1, 2} …
- 401k
5
votes
Accepted
How to list all possible 3-tuples with entries of the 3 tuples from 2 different sets?
{a, b} = {Range[4], Range[5, 8]};
triples = Tuples[{a, a, b}]
{{1, 1, 5}, {1, 1, 6}, {1, 1, 7}, {1, 1, 8}, {1, 2, 5}, {1, 2, 6}, {1, 2, 7},
{1, 2, 8}, {1, 3, 5}, {1, 3, 6}, {1, 3, 7}, {1, 3, 8}, { …
- 401k
2
votes
Making figure from list of tuples
arrow = Arrow[{{0, -1/2}, {0, 1/2}}];
(1) Use Tuples[{-1, 1}, n] (instead of Tuples[{0, 1}, n]) to obtain a list of directions, (2) Partition that list to get a matrix of desired dimensions, (3) Use … that matrix to Scale and Translate the graphics primitive arrow:
ClearAll[arrowTable]
arrowTable[n_, k_, len_, as_, opts : OptionsPattern[] ] :=
Module[{m = Partition[Join @@ Tuples[{-1, 1}, n], k …
- 401k
1
vote
Accepted
Referencing data in sets of sets
You can use ReplaceAll:
rules = {{0, 1, 1} -> setA, {1, 1, 0} -> setB};
data /. rules
{setA, setA, setB, setA, setB}
Alternatively, define a function that evaluates to setA for {0,1,1} and to set …
- 401k
5
votes
How do I get a list of all possible sums in a list nested list?
Few additional alternatives:
Distribute[foo @@ rn, List, foo, List, Plus]
Flatten @ Outer[Plus, ## & @@ rn]
Activate @ Tuples[Inactive[Plus] @@ rn] …
- 401k
4
votes
Accepted
How to find repeating elements in the following grid?
If we know that tuples is constructed from ngrids input grids and if each grid has at least one non-repeating element, we can get the duplicates and their counts in each of the grids using
extractDuplicates … =
Map[l |-> Select[GreaterThan[1]] @ Normalize[#, Min] & @ Counts[#[[All,l]]]] @
Range @ #2 &;
ngrids = 3;
extractDuplicates[tuples, ngrids]
{<|0.2 -> 2|>, <||>, <|0.35 -> 2|>}
We can get …
- 401k
4
votes
generating tuples of ones and zeroes with a fixed number of ones
ClearAll[L2]
L2[n_, m_] := Permute[PadRight[ConstantArray[1, m], n], SymmetricGroup @ n]
L2[3, 2]
{{1, 1, 0}, {1, 0, 1}, {0, 1, 1}}
L2[4, 2]
{{1, 1, 0, 0}, {1, 0, 1, 0}, {1, 0, 0, 1}, {0, 1, …
- 401k
2
votes
Accepted
Joining each element of second tuples to the end of each element of the first tuples
tup4a = Join @@@ Tuples[{tup1, tup3}];
tup4a // Short
{{0,-1,-1,0,-1,-1,d,0,0},{0,-1,-1,0,-1,-1,d,0,1},<<2912>>,
{1,1,1,1,1,1,d,-1,1},{1,1,1,1,1,1,d,-1,-1}}
tup4b = Distribute[{tup1, tup3}, List … , List, List, Join];
tup4a == tup4b
True
"to get that directly from tup1 and tup2":
tup4c = Join @@@ Tuples[{tup1, {{d}}, tup2}];
tup4a == tup4c
True …
- 401k
3
votes
generating tuples of ones and zeroes with a fixed number of ones
ClearAll[f0]
f0 = Module[{ss = MapIndexed[Thread[{#2[[1]], #}] &, Subsets[Range@#, {#2}]]},
SparseArray[Join @@ ss -> 1, {Length@ss, #}]] &;
Examples:
f0[3, 2]
f0[3, 2] // Normal
{{1, 1, 0} …
- 401k