Combination without repetition javascript [duplicate]










3
















This question already has an answer here:



  • JavaScript - Generating combinations from n arrays with m elements

    8 answers



Given



[
["blue", "red"],
[1, 2],
[true, false],
]


How can I get the possible combinations in javascript?:



blue, 1, true
blue, 1, false
blue, 2, true
blue, 2, false
red, 1, true
red, 1, false
red, 2, true
red, 2, false


Order doesn't matter.










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marked as duplicate by Always Sunny, Graham, Foo, Hassan Imam, Nina Scholz javascript
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Nov 15 '18 at 7:59


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






















    3
















    This question already has an answer here:



    • JavaScript - Generating combinations from n arrays with m elements

      8 answers



    Given



    [
    ["blue", "red"],
    [1, 2],
    [true, false],
    ]


    How can I get the possible combinations in javascript?:



    blue, 1, true
    blue, 1, false
    blue, 2, true
    blue, 2, false
    red, 1, true
    red, 1, false
    red, 2, true
    red, 2, false


    Order doesn't matter.










    share|improve this question













    marked as duplicate by Always Sunny, Graham, Foo, Hassan Imam, Nina Scholz javascript
    Users with the  javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.

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    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.




















      3












      3








      3









      This question already has an answer here:



      • JavaScript - Generating combinations from n arrays with m elements

        8 answers



      Given



      [
      ["blue", "red"],
      [1, 2],
      [true, false],
      ]


      How can I get the possible combinations in javascript?:



      blue, 1, true
      blue, 1, false
      blue, 2, true
      blue, 2, false
      red, 1, true
      red, 1, false
      red, 2, true
      red, 2, false


      Order doesn't matter.










      share|improve this question















      This question already has an answer here:



      • JavaScript - Generating combinations from n arrays with m elements

        8 answers



      Given



      [
      ["blue", "red"],
      [1, 2],
      [true, false],
      ]


      How can I get the possible combinations in javascript?:



      blue, 1, true
      blue, 1, false
      blue, 2, true
      blue, 2, false
      red, 1, true
      red, 1, false
      red, 2, true
      red, 2, false


      Order doesn't matter.





      This question already has an answer here:



      • JavaScript - Generating combinations from n arrays with m elements

        8 answers







      javascript combinations permutation






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 15 '18 at 6:03









      Mauro AguilarMauro Aguilar

      216213




      216213




      marked as duplicate by Always Sunny, Graham, Foo, Hassan Imam, Nina Scholz javascript
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      Nov 15 '18 at 7:59


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      marked as duplicate by Always Sunny, Graham, Foo, Hassan Imam, Nina Scholz javascript
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      Nov 15 '18 at 7:59


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























          1 Answer
          1






          active

          oldest

          votes


















          1














          this was quite complex, and lots of fun, so thanks for asking!



          as here we can have m element arrays of arbitary size,






          var a = [
          ["blue", "red"],
          [1, 2],
          [true, false],
          ]

          function allPossibleCombinations(items, isCombination=false)
          // finding all possible combinations of the last 2 items
          // remove those 2, add these combinations
          // isCombination shows if the last element is itself part of the combination series
          if(items.length == 1)
          return items[0]

          else if(items.length == 2)
          var combinations =
          for (var i=0; i<items[1].length; i++)
          for(var j=0; j<items[0].length; j++)
          if(isCombination)
          // clone array to not modify original array
          var combination = items[1][i].slice();
          combination.push(items[0][j]);

          else
          var combination = [items[1][i], items[0][j]];

          combinations.push(combination);


          return combinations

          else if(items.length > 2)
          var last2 = items.slice(-2);
          var butLast2 = items.slice(0, items.length - 2);
          last2 = allPossibleCombinations(last2, isCombination);
          butLast2.push(last2)
          var combinations = butLast2;
          return allPossibleCombinations(combinations, isCombination=true)



          console.log(allPossibleCombinations(a));
          console.log(allPossibleCombinations(a).length);








          share|improve this answer

























          • pretty impressive, works just fine, thank you.

            – Mauro Aguilar
            Nov 15 '18 at 9:57

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          this was quite complex, and lots of fun, so thanks for asking!



          as here we can have m element arrays of arbitary size,






          var a = [
          ["blue", "red"],
          [1, 2],
          [true, false],
          ]

          function allPossibleCombinations(items, isCombination=false)
          // finding all possible combinations of the last 2 items
          // remove those 2, add these combinations
          // isCombination shows if the last element is itself part of the combination series
          if(items.length == 1)
          return items[0]

          else if(items.length == 2)
          var combinations =
          for (var i=0; i<items[1].length; i++)
          for(var j=0; j<items[0].length; j++)
          if(isCombination)
          // clone array to not modify original array
          var combination = items[1][i].slice();
          combination.push(items[0][j]);

          else
          var combination = [items[1][i], items[0][j]];

          combinations.push(combination);


          return combinations

          else if(items.length > 2)
          var last2 = items.slice(-2);
          var butLast2 = items.slice(0, items.length - 2);
          last2 = allPossibleCombinations(last2, isCombination);
          butLast2.push(last2)
          var combinations = butLast2;
          return allPossibleCombinations(combinations, isCombination=true)



          console.log(allPossibleCombinations(a));
          console.log(allPossibleCombinations(a).length);








          share|improve this answer

























          • pretty impressive, works just fine, thank you.

            – Mauro Aguilar
            Nov 15 '18 at 9:57















          1














          this was quite complex, and lots of fun, so thanks for asking!



          as here we can have m element arrays of arbitary size,






          var a = [
          ["blue", "red"],
          [1, 2],
          [true, false],
          ]

          function allPossibleCombinations(items, isCombination=false)
          // finding all possible combinations of the last 2 items
          // remove those 2, add these combinations
          // isCombination shows if the last element is itself part of the combination series
          if(items.length == 1)
          return items[0]

          else if(items.length == 2)
          var combinations =
          for (var i=0; i<items[1].length; i++)
          for(var j=0; j<items[0].length; j++)
          if(isCombination)
          // clone array to not modify original array
          var combination = items[1][i].slice();
          combination.push(items[0][j]);

          else
          var combination = [items[1][i], items[0][j]];

          combinations.push(combination);


          return combinations

          else if(items.length > 2)
          var last2 = items.slice(-2);
          var butLast2 = items.slice(0, items.length - 2);
          last2 = allPossibleCombinations(last2, isCombination);
          butLast2.push(last2)
          var combinations = butLast2;
          return allPossibleCombinations(combinations, isCombination=true)



          console.log(allPossibleCombinations(a));
          console.log(allPossibleCombinations(a).length);








          share|improve this answer

























          • pretty impressive, works just fine, thank you.

            – Mauro Aguilar
            Nov 15 '18 at 9:57













          1












          1








          1







          this was quite complex, and lots of fun, so thanks for asking!



          as here we can have m element arrays of arbitary size,






          var a = [
          ["blue", "red"],
          [1, 2],
          [true, false],
          ]

          function allPossibleCombinations(items, isCombination=false)
          // finding all possible combinations of the last 2 items
          // remove those 2, add these combinations
          // isCombination shows if the last element is itself part of the combination series
          if(items.length == 1)
          return items[0]

          else if(items.length == 2)
          var combinations =
          for (var i=0; i<items[1].length; i++)
          for(var j=0; j<items[0].length; j++)
          if(isCombination)
          // clone array to not modify original array
          var combination = items[1][i].slice();
          combination.push(items[0][j]);

          else
          var combination = [items[1][i], items[0][j]];

          combinations.push(combination);


          return combinations

          else if(items.length > 2)
          var last2 = items.slice(-2);
          var butLast2 = items.slice(0, items.length - 2);
          last2 = allPossibleCombinations(last2, isCombination);
          butLast2.push(last2)
          var combinations = butLast2;
          return allPossibleCombinations(combinations, isCombination=true)



          console.log(allPossibleCombinations(a));
          console.log(allPossibleCombinations(a).length);








          share|improve this answer















          this was quite complex, and lots of fun, so thanks for asking!



          as here we can have m element arrays of arbitary size,






          var a = [
          ["blue", "red"],
          [1, 2],
          [true, false],
          ]

          function allPossibleCombinations(items, isCombination=false)
          // finding all possible combinations of the last 2 items
          // remove those 2, add these combinations
          // isCombination shows if the last element is itself part of the combination series
          if(items.length == 1)
          return items[0]

          else if(items.length == 2)
          var combinations =
          for (var i=0; i<items[1].length; i++)
          for(var j=0; j<items[0].length; j++)
          if(isCombination)
          // clone array to not modify original array
          var combination = items[1][i].slice();
          combination.push(items[0][j]);

          else
          var combination = [items[1][i], items[0][j]];

          combinations.push(combination);


          return combinations

          else if(items.length > 2)
          var last2 = items.slice(-2);
          var butLast2 = items.slice(0, items.length - 2);
          last2 = allPossibleCombinations(last2, isCombination);
          butLast2.push(last2)
          var combinations = butLast2;
          return allPossibleCombinations(combinations, isCombination=true)



          console.log(allPossibleCombinations(a));
          console.log(allPossibleCombinations(a).length);








          var a = [
          ["blue", "red"],
          [1, 2],
          [true, false],
          ]

          function allPossibleCombinations(items, isCombination=false)
          // finding all possible combinations of the last 2 items
          // remove those 2, add these combinations
          // isCombination shows if the last element is itself part of the combination series
          if(items.length == 1)
          return items[0]

          else if(items.length == 2)
          var combinations =
          for (var i=0; i<items[1].length; i++)
          for(var j=0; j<items[0].length; j++)
          if(isCombination)
          // clone array to not modify original array
          var combination = items[1][i].slice();
          combination.push(items[0][j]);

          else
          var combination = [items[1][i], items[0][j]];

          combinations.push(combination);


          return combinations

          else if(items.length > 2)
          var last2 = items.slice(-2);
          var butLast2 = items.slice(0, items.length - 2);
          last2 = allPossibleCombinations(last2, isCombination);
          butLast2.push(last2)
          var combinations = butLast2;
          return allPossibleCombinations(combinations, isCombination=true)



          console.log(allPossibleCombinations(a));
          console.log(allPossibleCombinations(a).length);





          var a = [
          ["blue", "red"],
          [1, 2],
          [true, false],
          ]

          function allPossibleCombinations(items, isCombination=false)
          // finding all possible combinations of the last 2 items
          // remove those 2, add these combinations
          // isCombination shows if the last element is itself part of the combination series
          if(items.length == 1)
          return items[0]

          else if(items.length == 2)
          var combinations =
          for (var i=0; i<items[1].length; i++)
          for(var j=0; j<items[0].length; j++)
          if(isCombination)
          // clone array to not modify original array
          var combination = items[1][i].slice();
          combination.push(items[0][j]);

          else
          var combination = [items[1][i], items[0][j]];

          combinations.push(combination);


          return combinations

          else if(items.length > 2)
          var last2 = items.slice(-2);
          var butLast2 = items.slice(0, items.length - 2);
          last2 = allPossibleCombinations(last2, isCombination);
          butLast2.push(last2)
          var combinations = butLast2;
          return allPossibleCombinations(combinations, isCombination=true)



          console.log(allPossibleCombinations(a));
          console.log(allPossibleCombinations(a).length);






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 15 '18 at 7:57

























          answered Nov 15 '18 at 7:50









          Aditya ShankarAditya Shankar

          37418




          37418












          • pretty impressive, works just fine, thank you.

            – Mauro Aguilar
            Nov 15 '18 at 9:57

















          • pretty impressive, works just fine, thank you.

            – Mauro Aguilar
            Nov 15 '18 at 9:57
















          pretty impressive, works just fine, thank you.

          – Mauro Aguilar
          Nov 15 '18 at 9:57





          pretty impressive, works just fine, thank you.

          – Mauro Aguilar
          Nov 15 '18 at 9:57





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