issue with extracting combination of values from multiple nested arrays
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trying to iterate over multiple nested arrays to get every single combination of possible values in a new array.
example:
[
['a1', 'a2'],
['b1', 'b2'],
['c1', 'c2']
]
output:
[
['a1'],
['a2'],
['b1'],
['b2'],
['c1'],
['c2'],
['a1', 'b1'],
['a1', 'b2'],
['a1', 'c1'],
['a1', 'c2'],
['a2', 'b1'],
['a2', 'b2'],
['a2', 'c1'],
['a2', 'c2'],
['b1', 'c1'],
['b1', 'c2'],
['b2', 'c1'],
['b2', 'c2']
]
any idea how I can achieve this, perhaps I need to split arrays in the first place ?
javascript arrays
asked Nov 15 '18 at 12:16
user1751287user1751287
193114
|
show 4 more comments
trying to iterate over multiple nested arrays to get every single combination of possible values in a new array.
example:
[
['a1', 'a2'],
['b1', 'b2'],
['c1', 'c2']
]
output:
[
['a1'],
['a2'],
['b1'],
['b2'],
['c1'],
['c2'],
['a1', 'b1'],
['a1', 'b2'],
['a1', 'c1'],
['a1', 'c2'],
['a2', 'b1'],
['a2', 'b2'],
['a2', 'c1'],
['a2', 'c2'],
['b1', 'c1'],
['b1', 'c2'],
['b2', 'c1'],
['b2', 'c2']
]
any idea how I can achieve this, perhaps I need to split arrays in the first place ?
javascript arrays
asked Nov 15 '18 at 12:16
user1751287user1751287
193114
Do you want every single combination or only combinations with 1 and 2 elements in them?
– VLAZ
Nov 15 '18 at 12:17
I think you should 1. Post any attempts of code that you've written and then ask about a specific problem that you're having, and 2. Explain in a little more detail as to how your input maps to that output
– Khauri McClain
Nov 15 '18 at 12:19
Just clarifying: so any combinations with 1..N members.
– VLAZ
Nov 15 '18 at 12:19
1
You have this similar question but i'm not flagging as duplicate because i'm not sure I understand the logic - why does the first 6 elements in the output are valid, but not['b1', 'a1']
– Alon Eitan
Nov 15 '18 at 12:20
I need exactly as above, not all combinations as ['a1', 'b1'] and ['b1', 'a1']
– user1751287
Nov 15 '18 at 12:21
|
show 4 more comments
trying to iterate over multiple nested arrays to get every single combination of possible values in a new array.
example:
[
['a1', 'a2'],
['b1', 'b2'],
['c1', 'c2']
]
output:
[
['a1'],
['a2'],
['b1'],
['b2'],
['c1'],
['c2'],
['a1', 'b1'],
['a1', 'b2'],
['a1', 'c1'],
['a1', 'c2'],
['a2', 'b1'],
['a2', 'b2'],
['a2', 'c1'],
['a2', 'c2'],
['b1', 'c1'],
['b1', 'c2'],
['b2', 'c1'],
['b2', 'c2']
]
any idea how I can achieve this, perhaps I need to split arrays in the first place ?
javascript arrays
asked Nov 15 '18 at 12:16
user1751287user1751287
193114
trying to iterate over multiple nested arrays to get every single combination of possible values in a new array.
example:
[
['a1', 'a2'],
['b1', 'b2'],
['c1', 'c2']
]
output:
[
['a1'],
['a2'],
['b1'],
['b2'],
['c1'],
['c2'],
['a1', 'b1'],
['a1', 'b2'],
['a1', 'c1'],
['a1', 'c2'],
['a2', 'b1'],
['a2', 'b2'],
['a2', 'c1'],
['a2', 'c2'],
['b1', 'c1'],
['b1', 'c2'],
['b2', 'c1'],
['b2', 'c2']
]
any idea how I can achieve this, perhaps I need to split arrays in the first place ?
javascript arrays
javascript arrays
asked Nov 15 '18 at 12:16
user1751287user1751287
193114
asked Nov 15 '18 at 12:16
user1751287user1751287
193114
asked Nov 15 '18 at 12:16
user1751287user1751287
193114
asked Nov 15 '18 at 12:16
user1751287user1751287
193114
asked Nov 15 '18 at 12:16
user1751287user1751287
193114
193114
Do you want every single combination or only combinations with 1 and 2 elements in them?
– VLAZ
Nov 15 '18 at 12:17
I think you should 1. Post any attempts of code that you've written and then ask about a specific problem that you're having, and 2. Explain in a little more detail as to how your input maps to that output
– Khauri McClain
Nov 15 '18 at 12:19
Just clarifying: so any combinations with 1..N members.
– VLAZ
Nov 15 '18 at 12:19
1
You have this similar question but i'm not flagging as duplicate because i'm not sure I understand the logic - why does the first 6 elements in the output are valid, but not['b1', 'a1']
– Alon Eitan
Nov 15 '18 at 12:20
I need exactly as above, not all combinations as ['a1', 'b1'] and ['b1', 'a1']
– user1751287
Nov 15 '18 at 12:21
|
show 4 more comments
Do you want every single combination or only combinations with 1 and 2 elements in them?
– VLAZ
Nov 15 '18 at 12:17
I think you should 1. Post any attempts of code that you've written and then ask about a specific problem that you're having, and 2. Explain in a little more detail as to how your input maps to that output
– Khauri McClain
Nov 15 '18 at 12:19
Just clarifying: so any combinations with 1..N members.
– VLAZ
Nov 15 '18 at 12:19
1
You have this similar question but i'm not flagging as duplicate because i'm not sure I understand the logic - why does the first 6 elements in the output are valid, but not['b1', 'a1']
– Alon Eitan
Nov 15 '18 at 12:20
I need exactly as above, not all combinations as ['a1', 'b1'] and ['b1', 'a1']
– user1751287
Nov 15 '18 at 12:21
Do you want every single combination or only combinations with 1 and 2 elements in them?
– VLAZ
Nov 15 '18 at 12:17
Do you want every single combination or only combinations with 1 and 2 elements in them?
– VLAZ
Nov 15 '18 at 12:17
I think you should 1. Post any attempts of code that you've written and then ask about a specific problem that you're having, and 2. Explain in a little more detail as to how your input maps to that output
– Khauri McClain
Nov 15 '18 at 12:19
I think you should 1. Post any attempts of code that you've written and then ask about a specific problem that you're having, and 2. Explain in a little more detail as to how your input maps to that output
– Khauri McClain
Nov 15 '18 at 12:19
Just clarifying: so any combinations with 1..N members.
– VLAZ
Nov 15 '18 at 12:19
Just clarifying: so any combinations with 1..N members.
– VLAZ
Nov 15 '18 at 12:19
1
1
You have this similar question but i'm not flagging as duplicate because i'm not sure I understand the logic - why does the first 6 elements in the output are valid, but not
['b1', 'a1']– Alon Eitan
Nov 15 '18 at 12:20
You have this similar question but i'm not flagging as duplicate because i'm not sure I understand the logic - why does the first 6 elements in the output are valid, but not
['b1', 'a1']– Alon Eitan
Nov 15 '18 at 12:20
I need exactly as above, not all combinations as ['a1', 'b1'] and ['b1', 'a1']
– user1751287
Nov 15 '18 at 12:21
I need exactly as above, not all combinations as ['a1', 'b1'] and ['b1', 'a1']
– user1751287
Nov 15 '18 at 12:21
|
show 4 more comments
3 Answers
3
active
oldest
votes
You could take an recursive approach by handing over the next indices and collect the temporary arrays.
function getCombinations(array, max)
function iter(i, j, temp)
if (array[i] && j >= array[i].length)
j = 0;
i++;
if (!array[i]
var result = ;
iter(0, 0, );
return result;
var array = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']],
result = getCombinations(array, 2);
console.log(result.map(a => a.join(' ')));.as-console-wrapper max-height: 100% !important; top: 0; answered Nov 15 '18 at 13:46
Nina ScholzNina Scholz
198k15110181
add a comment |
You could create recursive function using two nested for loops and one parameter to keep current row iteration count so you can start next for loop from there.
const data = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']]
const res =
function comb(data, n = 0, prev = )
for (var i = n; i < data.length; i++)
for (var j = 0; j < data[i].length; j++)
let el = data[i][j]
let arr = prev.concat(el);
if (arr.length <= data[i].length) res.push(arr)
comb(data, i + 1, arr)
comb(data)
console.log(JSON.stringify(res))answered Nov 15 '18 at 13:46
Nenad VracarNenad Vracar
73.6k126085
nice idea with nested loops.
– Nina Scholz
Nov 15 '18 at 14:00
@Nina Scholz Thank you, not sure if it works correctly in other scenarios with different array lengths.
– Nenad Vracar
Nov 15 '18 at 14:01
i took a limit for max and exit the recursion.
– Nina Scholz
Nov 15 '18 at 14:02
add a comment |
If you need the order too (so single elements first, pairs last), that can be done too, just it is going to be a bit longer than the other solutions:
var org=[['a1', 'a2'],['b1', 'b2'],['c1', 'c2']];
var res=;
// singles
org.forEach(arr =>
arr.forEach(elem =>
res.push([elem]);
);
);
// pairs
var start=0;
var end=res.length;
org.forEach(arr =>
start+=arr.length;
arr.forEach(elem =>
for(var i=start;i<end;i++)
res.push([elem,res[i][0]]);
);
);
console.log(JSON.stringify(res));//singles part is a nested pair of loops, pushing all the elements, and //pairs traverses the original array again, just it makes use of the already "flattened" result, end simply keeps the number of single elements (as the length of the array will grow) and start always jumps to the beginning of the next sub-array (so 'a1','a2'-like pairs are not generated, as it was required).
answered Nov 15 '18 at 14:19
tevemadartevemadar
4,7182827
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could take an recursive approach by handing over the next indices and collect the temporary arrays.
function getCombinations(array, max)
function iter(i, j, temp)
if (array[i] && j >= array[i].length)
j = 0;
i++;
if (!array[i]
var result = ;
iter(0, 0, );
return result;
var array = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']],
result = getCombinations(array, 2);
console.log(result.map(a => a.join(' ')));.as-console-wrapper max-height: 100% !important; top: 0; answered Nov 15 '18 at 13:46
Nina ScholzNina Scholz
198k15110181
add a comment |
You could take an recursive approach by handing over the next indices and collect the temporary arrays.
function getCombinations(array, max)
function iter(i, j, temp)
if (array[i] && j >= array[i].length)
j = 0;
i++;
if (!array[i]
var result = ;
iter(0, 0, );
return result;
var array = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']],
result = getCombinations(array, 2);
console.log(result.map(a => a.join(' ')));.as-console-wrapper max-height: 100% !important; top: 0; answered Nov 15 '18 at 13:46
Nina ScholzNina Scholz
198k15110181
add a comment |
You could take an recursive approach by handing over the next indices and collect the temporary arrays.
function getCombinations(array, max)
function iter(i, j, temp)
if (array[i] && j >= array[i].length)
j = 0;
i++;
if (!array[i]
var result = ;
iter(0, 0, );
return result;
var array = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']],
result = getCombinations(array, 2);
console.log(result.map(a => a.join(' ')));.as-console-wrapper max-height: 100% !important; top: 0; answered Nov 15 '18 at 13:46
Nina ScholzNina Scholz
198k15110181
You could take an recursive approach by handing over the next indices and collect the temporary arrays.
function getCombinations(array, max)
function iter(i, j, temp)
if (array[i] && j >= array[i].length)
j = 0;
i++;
if (!array[i]
var result = ;
iter(0, 0, );
return result;
var array = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']],
result = getCombinations(array, 2);
console.log(result.map(a => a.join(' ')));.as-console-wrapper max-height: 100% !important; top: 0; function getCombinations(array, max)
function iter(i, j, temp)
if (array[i] && j >= array[i].length)
j = 0;
i++;
if (!array[i]
var result = ;
iter(0, 0, );
return result;
var array = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']],
result = getCombinations(array, 2);
console.log(result.map(a => a.join(' ')));.as-console-wrapper max-height: 100% !important; top: 0; function getCombinations(array, max)
function iter(i, j, temp)
if (array[i] && j >= array[i].length)
j = 0;
i++;
if (!array[i]
var result = ;
iter(0, 0, );
return result;
var array = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']],
result = getCombinations(array, 2);
console.log(result.map(a => a.join(' ')));.as-console-wrapper max-height: 100% !important; top: 0; answered Nov 15 '18 at 13:46
Nina ScholzNina Scholz
198k15110181
edited Nov 15 '18 at 13:59
answered Nov 15 '18 at 13:46
Nina ScholzNina Scholz
198k15110181
answered Nov 15 '18 at 13:46
Nina ScholzNina Scholz
198k15110181
answered Nov 15 '18 at 13:46
Nina ScholzNina Scholz
198k15110181
198k15110181
add a comment |
add a comment |
You could create recursive function using two nested for loops and one parameter to keep current row iteration count so you can start next for loop from there.
const data = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']]
const res =
function comb(data, n = 0, prev = )
for (var i = n; i < data.length; i++)
for (var j = 0; j < data[i].length; j++)
let el = data[i][j]
let arr = prev.concat(el);
if (arr.length <= data[i].length) res.push(arr)
comb(data, i + 1, arr)
comb(data)
console.log(JSON.stringify(res))answered Nov 15 '18 at 13:46
Nenad VracarNenad Vracar
73.6k126085
nice idea with nested loops.
– Nina Scholz
Nov 15 '18 at 14:00
@Nina Scholz Thank you, not sure if it works correctly in other scenarios with different array lengths.
– Nenad Vracar
Nov 15 '18 at 14:01
i took a limit for max and exit the recursion.
– Nina Scholz
Nov 15 '18 at 14:02
add a comment |
You could create recursive function using two nested for loops and one parameter to keep current row iteration count so you can start next for loop from there.
const data = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']]
const res =
function comb(data, n = 0, prev = )
for (var i = n; i < data.length; i++)
for (var j = 0; j < data[i].length; j++)
let el = data[i][j]
let arr = prev.concat(el);
if (arr.length <= data[i].length) res.push(arr)
comb(data, i + 1, arr)
comb(data)
console.log(JSON.stringify(res))answered Nov 15 '18 at 13:46
Nenad VracarNenad Vracar
73.6k126085
nice idea with nested loops.
– Nina Scholz
Nov 15 '18 at 14:00
@Nina Scholz Thank you, not sure if it works correctly in other scenarios with different array lengths.
– Nenad Vracar
Nov 15 '18 at 14:01
i took a limit for max and exit the recursion.
– Nina Scholz
Nov 15 '18 at 14:02
add a comment |
You could create recursive function using two nested for loops and one parameter to keep current row iteration count so you can start next for loop from there.
const data = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']]
const res =
function comb(data, n = 0, prev = )
for (var i = n; i < data.length; i++)
for (var j = 0; j < data[i].length; j++)
let el = data[i][j]
let arr = prev.concat(el);
if (arr.length <= data[i].length) res.push(arr)
comb(data, i + 1, arr)
comb(data)
console.log(JSON.stringify(res))answered Nov 15 '18 at 13:46
Nenad VracarNenad Vracar
73.6k126085
You could create recursive function using two nested for loops and one parameter to keep current row iteration count so you can start next for loop from there.
const data = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']]
const res =
function comb(data, n = 0, prev = )
for (var i = n; i < data.length; i++)
for (var j = 0; j < data[i].length; j++)
let el = data[i][j]
let arr = prev.concat(el);
if (arr.length <= data[i].length) res.push(arr)
comb(data, i + 1, arr)
comb(data)
console.log(JSON.stringify(res))const data = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']]
const res =
function comb(data, n = 0, prev = )
for (var i = n; i < data.length; i++)
for (var j = 0; j < data[i].length; j++)
let el = data[i][j]
let arr = prev.concat(el);
if (arr.length <= data[i].length) res.push(arr)
comb(data, i + 1, arr)
comb(data)
console.log(JSON.stringify(res))const data = [['a1', 'a2'], ['b1', 'b2'], ['c1', 'c2']]
const res =
function comb(data, n = 0, prev = )
for (var i = n; i < data.length; i++)
for (var j = 0; j < data[i].length; j++)
let el = data[i][j]
let arr = prev.concat(el);
if (arr.length <= data[i].length) res.push(arr)
comb(data, i + 1, arr)
comb(data)
console.log(JSON.stringify(res))answered Nov 15 '18 at 13:46
Nenad VracarNenad Vracar
73.6k126085
answered Nov 15 '18 at 13:46
Nenad VracarNenad Vracar
73.6k126085
answered Nov 15 '18 at 13:46
Nenad VracarNenad Vracar
73.6k126085
answered Nov 15 '18 at 13:46
Nenad VracarNenad Vracar
73.6k126085
73.6k126085
nice idea with nested loops.
– Nina Scholz
Nov 15 '18 at 14:00
@Nina Scholz Thank you, not sure if it works correctly in other scenarios with different array lengths.
– Nenad Vracar
Nov 15 '18 at 14:01
i took a limit for max and exit the recursion.
– Nina Scholz
Nov 15 '18 at 14:02
add a comment |
nice idea with nested loops.
– Nina Scholz
Nov 15 '18 at 14:00
@Nina Scholz Thank you, not sure if it works correctly in other scenarios with different array lengths.
– Nenad Vracar
Nov 15 '18 at 14:01
i took a limit for max and exit the recursion.
– Nina Scholz
Nov 15 '18 at 14:02
nice idea with nested loops.
– Nina Scholz
Nov 15 '18 at 14:00
nice idea with nested loops.
– Nina Scholz
Nov 15 '18 at 14:00
@Nina Scholz Thank you, not sure if it works correctly in other scenarios with different array lengths.
– Nenad Vracar
Nov 15 '18 at 14:01
@Nina Scholz Thank you, not sure if it works correctly in other scenarios with different array lengths.
– Nenad Vracar
Nov 15 '18 at 14:01
i took a limit for max and exit the recursion.
– Nina Scholz
Nov 15 '18 at 14:02
i took a limit for max and exit the recursion.
– Nina Scholz
Nov 15 '18 at 14:02
add a comment |
If you need the order too (so single elements first, pairs last), that can be done too, just it is going to be a bit longer than the other solutions:
var org=[['a1', 'a2'],['b1', 'b2'],['c1', 'c2']];
var res=;
// singles
org.forEach(arr =>
arr.forEach(elem =>
res.push([elem]);
);
);
// pairs
var start=0;
var end=res.length;
org.forEach(arr =>
start+=arr.length;
arr.forEach(elem =>
for(var i=start;i<end;i++)
res.push([elem,res[i][0]]);
);
);
console.log(JSON.stringify(res));//singles part is a nested pair of loops, pushing all the elements, and //pairs traverses the original array again, just it makes use of the already "flattened" result, end simply keeps the number of single elements (as the length of the array will grow) and start always jumps to the beginning of the next sub-array (so 'a1','a2'-like pairs are not generated, as it was required).
answered Nov 15 '18 at 14:19
tevemadartevemadar
4,7182827
add a comment |
If you need the order too (so single elements first, pairs last), that can be done too, just it is going to be a bit longer than the other solutions:
var org=[['a1', 'a2'],['b1', 'b2'],['c1', 'c2']];
var res=;
// singles
org.forEach(arr =>
arr.forEach(elem =>
res.push([elem]);
);
);
// pairs
var start=0;
var end=res.length;
org.forEach(arr =>
start+=arr.length;
arr.forEach(elem =>
for(var i=start;i<end;i++)
res.push([elem,res[i][0]]);
);
);
console.log(JSON.stringify(res));//singles part is a nested pair of loops, pushing all the elements, and //pairs traverses the original array again, just it makes use of the already "flattened" result, end simply keeps the number of single elements (as the length of the array will grow) and start always jumps to the beginning of the next sub-array (so 'a1','a2'-like pairs are not generated, as it was required).
answered Nov 15 '18 at 14:19
tevemadartevemadar
4,7182827
add a comment |
If you need the order too (so single elements first, pairs last), that can be done too, just it is going to be a bit longer than the other solutions:
var org=[['a1', 'a2'],['b1', 'b2'],['c1', 'c2']];
var res=;
// singles
org.forEach(arr =>
arr.forEach(elem =>
res.push([elem]);
);
);
// pairs
var start=0;
var end=res.length;
org.forEach(arr =>
start+=arr.length;
arr.forEach(elem =>
for(var i=start;i<end;i++)
res.push([elem,res[i][0]]);
);
);
console.log(JSON.stringify(res));//singles part is a nested pair of loops, pushing all the elements, and //pairs traverses the original array again, just it makes use of the already "flattened" result, end simply keeps the number of single elements (as the length of the array will grow) and start always jumps to the beginning of the next sub-array (so 'a1','a2'-like pairs are not generated, as it was required).
answered Nov 15 '18 at 14:19
tevemadartevemadar
4,7182827
If you need the order too (so single elements first, pairs last), that can be done too, just it is going to be a bit longer than the other solutions:
var org=[['a1', 'a2'],['b1', 'b2'],['c1', 'c2']];
var res=;
// singles
org.forEach(arr =>
arr.forEach(elem =>
res.push([elem]);
);
);
// pairs
var start=0;
var end=res.length;
org.forEach(arr =>
start+=arr.length;
arr.forEach(elem =>
for(var i=start;i<end;i++)
res.push([elem,res[i][0]]);
);
);
console.log(JSON.stringify(res));//singles part is a nested pair of loops, pushing all the elements, and //pairs traverses the original array again, just it makes use of the already "flattened" result, end simply keeps the number of single elements (as the length of the array will grow) and start always jumps to the beginning of the next sub-array (so 'a1','a2'-like pairs are not generated, as it was required).
var org=[['a1', 'a2'],['b1', 'b2'],['c1', 'c2']];
var res=;
// singles
org.forEach(arr =>
arr.forEach(elem =>
res.push([elem]);
);
);
// pairs
var start=0;
var end=res.length;
org.forEach(arr =>
start+=arr.length;
arr.forEach(elem =>
for(var i=start;i<end;i++)
res.push([elem,res[i][0]]);
);
);
console.log(JSON.stringify(res));var org=[['a1', 'a2'],['b1', 'b2'],['c1', 'c2']];
var res=;
// singles
org.forEach(arr =>
arr.forEach(elem =>
res.push([elem]);
);
);
// pairs
var start=0;
var end=res.length;
org.forEach(arr =>
start+=arr.length;
arr.forEach(elem =>
for(var i=start;i<end;i++)
res.push([elem,res[i][0]]);
);
);
console.log(JSON.stringify(res));answered Nov 15 '18 at 14:19
tevemadartevemadar
4,7182827
edited Nov 15 '18 at 14:24
answered Nov 15 '18 at 14:19
tevemadartevemadar
4,7182827
answered Nov 15 '18 at 14:19
tevemadartevemadar
4,7182827
answered Nov 15 '18 at 14:19
tevemadartevemadar
4,7182827
4,7182827
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Do you want every single combination or only combinations with 1 and 2 elements in them?
– VLAZ
Nov 15 '18 at 12:17
I think you should 1. Post any attempts of code that you've written and then ask about a specific problem that you're having, and 2. Explain in a little more detail as to how your input maps to that output
– Khauri McClain
Nov 15 '18 at 12:19
Just clarifying: so any combinations with 1..N members.
– VLAZ
Nov 15 '18 at 12:19
1
You have this similar question but i'm not flagging as duplicate because i'm not sure I understand the logic - why does the first 6 elements in the output are valid, but not
['b1', 'a1']– Alon Eitan
Nov 15 '18 at 12:20
I need exactly as above, not all combinations as ['a1', 'b1'] and ['b1', 'a1']
– user1751287
Nov 15 '18 at 12:21