Find if two arrays are repeated in array and then select them
I have multiple arrays in a main/parent array like this:
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
here are the array's for simpler reading:
[1, 17]
[1, 17]
[1, 17]
[2, 12]
[5, 9]
[2, 12]
[6, 2]
[2, 12]
[2, 12]
I want to select the arrays that are repeated 3 or more times (> 3) and assign it to a variable. So in this example, var repeatedArrays would be [1, 17] and [2, 12].
So this should be the final result:
console.log(repeatedArrays);
>>> [[1, 17], [2, 12]]
I found something similar here but it uses underscore.js and lodash.
How could I it with javascript or even jquery (if need be)?
javascript jquery arrays object multidimensional-array
asked Nov 23 '18 at 20:59
Timmy BalkTimmy Balk
366
add a comment |
I have multiple arrays in a main/parent array like this:
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
here are the array's for simpler reading:
[1, 17]
[1, 17]
[1, 17]
[2, 12]
[5, 9]
[2, 12]
[6, 2]
[2, 12]
[2, 12]
I want to select the arrays that are repeated 3 or more times (> 3) and assign it to a variable. So in this example, var repeatedArrays would be [1, 17] and [2, 12].
So this should be the final result:
console.log(repeatedArrays);
>>> [[1, 17], [2, 12]]
I found something similar here but it uses underscore.js and lodash.
How could I it with javascript or even jquery (if need be)?
javascript jquery arrays object multidimensional-array
asked Nov 23 '18 at 20:59
Timmy BalkTimmy Balk
366
You could compare them against their JSON values, however,[1, 17]wouldn't match[17, 1]
– Get Off My Lawn
Nov 23 '18 at 21:03
Choose best answer for you and accept it by click on gray "check" button on its left side
– Kamil Kiełczewski
Nov 24 '18 at 12:37
add a comment |
I have multiple arrays in a main/parent array like this:
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
here are the array's for simpler reading:
[1, 17]
[1, 17]
[1, 17]
[2, 12]
[5, 9]
[2, 12]
[6, 2]
[2, 12]
[2, 12]
I want to select the arrays that are repeated 3 or more times (> 3) and assign it to a variable. So in this example, var repeatedArrays would be [1, 17] and [2, 12].
So this should be the final result:
console.log(repeatedArrays);
>>> [[1, 17], [2, 12]]
I found something similar here but it uses underscore.js and lodash.
How could I it with javascript or even jquery (if need be)?
javascript jquery arrays object multidimensional-array
asked Nov 23 '18 at 20:59
Timmy BalkTimmy Balk
366
I have multiple arrays in a main/parent array like this:
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
here are the array's for simpler reading:
[1, 17]
[1, 17]
[1, 17]
[2, 12]
[5, 9]
[2, 12]
[6, 2]
[2, 12]
[2, 12]
I want to select the arrays that are repeated 3 or more times (> 3) and assign it to a variable. So in this example, var repeatedArrays would be [1, 17] and [2, 12].
So this should be the final result:
console.log(repeatedArrays);
>>> [[1, 17], [2, 12]]
I found something similar here but it uses underscore.js and lodash.
How could I it with javascript or even jquery (if need be)?
javascript jquery arrays object multidimensional-array
javascript jquery arrays object multidimensional-array
asked Nov 23 '18 at 20:59
Timmy BalkTimmy Balk
366
asked Nov 23 '18 at 20:59
Timmy BalkTimmy Balk
366
asked Nov 23 '18 at 20:59
Timmy BalkTimmy Balk
366
asked Nov 23 '18 at 20:59
Timmy BalkTimmy Balk
366
asked Nov 23 '18 at 20:59
Timmy BalkTimmy Balk
366
366
You could compare them against their JSON values, however,[1, 17]wouldn't match[17, 1]
– Get Off My Lawn
Nov 23 '18 at 21:03
Choose best answer for you and accept it by click on gray "check" button on its left side
– Kamil Kiełczewski
Nov 24 '18 at 12:37
add a comment |
You could compare them against their JSON values, however,[1, 17]wouldn't match[17, 1]
– Get Off My Lawn
Nov 23 '18 at 21:03
Choose best answer for you and accept it by click on gray "check" button on its left side
– Kamil Kiełczewski
Nov 24 '18 at 12:37
You could compare them against their JSON values, however,
[1, 17] wouldn't match [17, 1]– Get Off My Lawn
Nov 23 '18 at 21:03
You could compare them against their JSON values, however,
[1, 17] wouldn't match [17, 1]– Get Off My Lawn
Nov 23 '18 at 21:03
Choose best answer for you and accept it by click on gray "check" button on its left side
– Kamil Kiełczewski
Nov 24 '18 at 12:37
Choose best answer for you and accept it by click on gray "check" button on its left side
– Kamil Kiełczewski
Nov 24 '18 at 12:37
add a comment |
6 Answers
6
active
oldest
votes
You could take a Map with stringified arrays and count, then filter by count and restore the arrays.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = Array
.from(array.reduce(
(map, array) =>
(json => map.set(json, (map.get(json) || 0) + 1))
(JSON.stringify(array)),
new Map
))
.filter(([, count]) => count > 2)
.map(([json]) => JSON.parse(json));
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }Filter with a map at wanted count.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(map => a =>
(json =>
(count => map.set(json, count) && !(2 - count))
(1 + map.get(json) || 1)
)
(JSON.stringify(a))
)
(new Map)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }Unique!
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(s => a => (j => !s.has(j) && s.add(j))(JSON.stringify(a)))
(new Set)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }answered Nov 23 '18 at 21:06
Nina ScholzNina Scholz
186k1596170
Awesome! This works! Is there any way to remove the duplicated arrays from the original array? So the final output ofvar arraywould be[[1, 17], [2, 12], [5, 9], [6, 2]]. I triedfilter()andindexOf()but it did not work
– Timmy Balk
Nov 23 '18 at 21:46
for that task, you may use aSetinstead of aMap.
– Nina Scholz
Nov 23 '18 at 21:55
Could you please edit your answer and show me? Do I usehas()?
– Timmy Balk
Nov 23 '18 at 21:59
@TimmyBalk just remove.filter(([, count]) => count > 2)line - and don't change anything else
– Kamil Kiełczewski
Nov 23 '18 at 22:00
@KamilKiełczewski Excellent! Thank you so much. I'm going to research into thefilter()function now, thanks again Nina and Kamil. But i'm assuming removing the.filterline is not the same as what Nina was saying?
– Timmy Balk
Nov 23 '18 at 22:02
|
show 6 more comments
You can use Object.reduce, Object.entries for this like below
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let res = Object.entries(
array.reduce((o, d) => {
let key = d.join('-')
o[key] = (o[key] || 0) + 1
return o
}, {}))
.flatMap(([k, v]) => v > 2 ? [k.split('-').map(Number)] : )
console.log(res)OR may be just with Array.filters
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let temp = {}
let res = array.filter(d => {
let key = d.join('-')
temp[key] = (temp[key] || 0) + 1
return temp[key] == 3
})
console.log(res)answered Nov 23 '18 at 21:11
Nitish NarangNitish Narang
2,9501815
1
First snipped don't meet requirements (return elements are strings not nubmers) - but the second snipped is brilliant, +1
– Kamil Kiełczewski
Nov 23 '18 at 21:58
Thank you @KamilKiełczewski. Updated first snippet to return Numbers.
– Nitish Narang
Nov 24 '18 at 4:19
add a comment |
Try this
array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
Time complexity O(n) (one array pass by filter function). Inspired by Nitish answer.
Explanation
The (r={}, a=>...) will return last expression after comma (which is a=>...) (e.g. (5,6)==6). In r={} we set once temporary object where we will store unique keys. In filter function a=>... in a we have current array element . In r[a] JS implicity cast a to string (e.g 1,17). Then in !(2-(r[a]=++r[a]|0)) we increase counter of occurrence element a and return true (as filter function value) if element a occurred 3 times. If r[a] is undefined the ++r[a] returns NaN, and further NaN|0=0 (also number|0=number). The r[a]= initialise first counter value, if we omit it the ++ will only set NaN to r[a] which is non-incrementable (so we need to put zero at init). If we remove 2- as result we get input array without duplicates - or alternatively we can also get this by a=>!(r[a]=a in r). If we change 2- to 1- we get array with duplicates only.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
var r= array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
console.log(JSON.stringify(r));answered Nov 23 '18 at 21:19
Kamil KiełczewskiKamil Kiełczewski
11.7k86894
add a comment |
For a different take, you can first sort your list, then loop through once and pull out the elements that meet your requirement. This will probably be faster than stringifying keys from the array even with the sort:
var arr = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]]
arr.sort((a, b) => a[0] - b[0] || a[1] - b[1])
// define equal for array
const equal = (arr1, arr2) => arr1.every((n, j) => n === arr2[j])
let GROUP_SIZE = 3
first = 0, last = 1, res =
while(last < arr.length){
if (equal(arr[first], arr[last])) last++
else {
if (last - first >= GROUP_SIZE) res.push(arr[first])
first = last
}
}
if (last - first >= GROUP_SIZE) res.push(arr[first])
console.log(res)answered Nov 23 '18 at 21:34
Mark MeyerMark Meyer
38.6k33159
Interesting approach, but is it really faster?
– Timmy Balk
Nov 23 '18 at 21:52
1
@TimmyBalk this isn't a criticism of the JSON approach -- I think it's good, but if my test is correct, it is slower (at least on my browser). Here's a jsperf test: jsperf.com/2json-v-sort-loop
– Mark Meyer
Nov 23 '18 at 21:55
add a comment |
ES6:
const repeatMap = {}
array.forEach(arr => {
const key = JSON.stringify(arr)
if (repeatMap[key]) {
repeatMap[key]++
} else {
repeatMap[key] = 1
}
})
const repeatedArrays = Object.keys(repeatMap)
.filter(key => repeatMap[key] >= 3)
.map(key => JSON.parse(key))
answered Nov 23 '18 at 21:11
Ben StewardBen Steward
1,135315
add a comment |
You could also do this with a single Array.reduce where you would only push to a result property if the length is equal to 3:
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
console.log(array.reduce((r,c) => {
let key = c.join('-')
r[key] = (r[key] || 0) + 1
r[key] == 3 ? r.result.push(c) : 0 // if we have a hit push to result
return r
}, { result: }).result) // print the result propertyanswered Nov 24 '18 at 7:27
AkrionAkrion
9,48511224
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could take a Map with stringified arrays and count, then filter by count and restore the arrays.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = Array
.from(array.reduce(
(map, array) =>
(json => map.set(json, (map.get(json) || 0) + 1))
(JSON.stringify(array)),
new Map
))
.filter(([, count]) => count > 2)
.map(([json]) => JSON.parse(json));
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }Filter with a map at wanted count.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(map => a =>
(json =>
(count => map.set(json, count) && !(2 - count))
(1 + map.get(json) || 1)
)
(JSON.stringify(a))
)
(new Map)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }Unique!
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(s => a => (j => !s.has(j) && s.add(j))(JSON.stringify(a)))
(new Set)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }answered Nov 23 '18 at 21:06
Nina ScholzNina Scholz
186k1596170
Awesome! This works! Is there any way to remove the duplicated arrays from the original array? So the final output ofvar arraywould be[[1, 17], [2, 12], [5, 9], [6, 2]]. I triedfilter()andindexOf()but it did not work
– Timmy Balk
Nov 23 '18 at 21:46
for that task, you may use aSetinstead of aMap.
– Nina Scholz
Nov 23 '18 at 21:55
Could you please edit your answer and show me? Do I usehas()?
– Timmy Balk
Nov 23 '18 at 21:59
@TimmyBalk just remove.filter(([, count]) => count > 2)line - and don't change anything else
– Kamil Kiełczewski
Nov 23 '18 at 22:00
@KamilKiełczewski Excellent! Thank you so much. I'm going to research into thefilter()function now, thanks again Nina and Kamil. But i'm assuming removing the.filterline is not the same as what Nina was saying?
– Timmy Balk
Nov 23 '18 at 22:02
|
show 6 more comments
You could take a Map with stringified arrays and count, then filter by count and restore the arrays.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = Array
.from(array.reduce(
(map, array) =>
(json => map.set(json, (map.get(json) || 0) + 1))
(JSON.stringify(array)),
new Map
))
.filter(([, count]) => count > 2)
.map(([json]) => JSON.parse(json));
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }Filter with a map at wanted count.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(map => a =>
(json =>
(count => map.set(json, count) && !(2 - count))
(1 + map.get(json) || 1)
)
(JSON.stringify(a))
)
(new Map)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }Unique!
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(s => a => (j => !s.has(j) && s.add(j))(JSON.stringify(a)))
(new Set)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }answered Nov 23 '18 at 21:06
Nina ScholzNina Scholz
186k1596170
Awesome! This works! Is there any way to remove the duplicated arrays from the original array? So the final output ofvar arraywould be[[1, 17], [2, 12], [5, 9], [6, 2]]. I triedfilter()andindexOf()but it did not work
– Timmy Balk
Nov 23 '18 at 21:46
for that task, you may use aSetinstead of aMap.
– Nina Scholz
Nov 23 '18 at 21:55
Could you please edit your answer and show me? Do I usehas()?
– Timmy Balk
Nov 23 '18 at 21:59
@TimmyBalk just remove.filter(([, count]) => count > 2)line - and don't change anything else
– Kamil Kiełczewski
Nov 23 '18 at 22:00
@KamilKiełczewski Excellent! Thank you so much. I'm going to research into thefilter()function now, thanks again Nina and Kamil. But i'm assuming removing the.filterline is not the same as what Nina was saying?
– Timmy Balk
Nov 23 '18 at 22:02
|
show 6 more comments
You could take a Map with stringified arrays and count, then filter by count and restore the arrays.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = Array
.from(array.reduce(
(map, array) =>
(json => map.set(json, (map.get(json) || 0) + 1))
(JSON.stringify(array)),
new Map
))
.filter(([, count]) => count > 2)
.map(([json]) => JSON.parse(json));
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }Filter with a map at wanted count.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(map => a =>
(json =>
(count => map.set(json, count) && !(2 - count))
(1 + map.get(json) || 1)
)
(JSON.stringify(a))
)
(new Map)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }Unique!
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(s => a => (j => !s.has(j) && s.add(j))(JSON.stringify(a)))
(new Set)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }answered Nov 23 '18 at 21:06
Nina ScholzNina Scholz
186k1596170
You could take a Map with stringified arrays and count, then filter by count and restore the arrays.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = Array
.from(array.reduce(
(map, array) =>
(json => map.set(json, (map.get(json) || 0) + 1))
(JSON.stringify(array)),
new Map
))
.filter(([, count]) => count > 2)
.map(([json]) => JSON.parse(json));
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }Filter with a map at wanted count.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(map => a =>
(json =>
(count => map.set(json, count) && !(2 - count))
(1 + map.get(json) || 1)
)
(JSON.stringify(a))
)
(new Map)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }Unique!
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(s => a => (j => !s.has(j) && s.add(j))(JSON.stringify(a)))
(new Set)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = Array
.from(array.reduce(
(map, array) =>
(json => map.set(json, (map.get(json) || 0) + 1))
(JSON.stringify(array)),
new Map
))
.filter(([, count]) => count > 2)
.map(([json]) => JSON.parse(json));
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = Array
.from(array.reduce(
(map, array) =>
(json => map.set(json, (map.get(json) || 0) + 1))
(JSON.stringify(array)),
new Map
))
.filter(([, count]) => count > 2)
.map(([json]) => JSON.parse(json));
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(map => a =>
(json =>
(count => map.set(json, count) && !(2 - count))
(1 + map.get(json) || 1)
)
(JSON.stringify(a))
)
(new Map)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(map => a =>
(json =>
(count => map.set(json, count) && !(2 - count))
(1 + map.get(json) || 1)
)
(JSON.stringify(a))
)
(new Map)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(s => a => (j => !s.has(j) && s.add(j))(JSON.stringify(a)))
(new Set)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(s => a => (j => !s.has(j) && s.add(j))(JSON.stringify(a)))
(new Set)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }answered Nov 23 '18 at 21:06
Nina ScholzNina Scholz
186k1596170
edited Nov 27 '18 at 8:39
answered Nov 23 '18 at 21:06
Nina ScholzNina Scholz
186k1596170
answered Nov 23 '18 at 21:06
Nina ScholzNina Scholz
186k1596170
answered Nov 23 '18 at 21:06
Nina ScholzNina Scholz
186k1596170
186k1596170
Awesome! This works! Is there any way to remove the duplicated arrays from the original array? So the final output ofvar arraywould be[[1, 17], [2, 12], [5, 9], [6, 2]]. I triedfilter()andindexOf()but it did not work
– Timmy Balk
Nov 23 '18 at 21:46
for that task, you may use aSetinstead of aMap.
– Nina Scholz
Nov 23 '18 at 21:55
Could you please edit your answer and show me? Do I usehas()?
– Timmy Balk
Nov 23 '18 at 21:59
@TimmyBalk just remove.filter(([, count]) => count > 2)line - and don't change anything else
– Kamil Kiełczewski
Nov 23 '18 at 22:00
@KamilKiełczewski Excellent! Thank you so much. I'm going to research into thefilter()function now, thanks again Nina and Kamil. But i'm assuming removing the.filterline is not the same as what Nina was saying?
– Timmy Balk
Nov 23 '18 at 22:02
|
show 6 more comments
Awesome! This works! Is there any way to remove the duplicated arrays from the original array? So the final output ofvar arraywould be[[1, 17], [2, 12], [5, 9], [6, 2]]. I triedfilter()andindexOf()but it did not work
– Timmy Balk
Nov 23 '18 at 21:46
for that task, you may use aSetinstead of aMap.
– Nina Scholz
Nov 23 '18 at 21:55
Could you please edit your answer and show me? Do I usehas()?
– Timmy Balk
Nov 23 '18 at 21:59
@TimmyBalk just remove.filter(([, count]) => count > 2)line - and don't change anything else
– Kamil Kiełczewski
Nov 23 '18 at 22:00
@KamilKiełczewski Excellent! Thank you so much. I'm going to research into thefilter()function now, thanks again Nina and Kamil. But i'm assuming removing the.filterline is not the same as what Nina was saying?
– Timmy Balk
Nov 23 '18 at 22:02
Awesome! This works! Is there any way to remove the duplicated arrays from the original array? So the final output of
var array would be [[1, 17], [2, 12], [5, 9], [6, 2]]. I tried filter() and indexOf() but it did not work– Timmy Balk
Nov 23 '18 at 21:46
Awesome! This works! Is there any way to remove the duplicated arrays from the original array? So the final output of
var array would be [[1, 17], [2, 12], [5, 9], [6, 2]]. I tried filter() and indexOf() but it did not work– Timmy Balk
Nov 23 '18 at 21:46
for that task, you may use a
Set instead of a Map.– Nina Scholz
Nov 23 '18 at 21:55
for that task, you may use a
Set instead of a Map.– Nina Scholz
Nov 23 '18 at 21:55
Could you please edit your answer and show me? Do I use
has()?– Timmy Balk
Nov 23 '18 at 21:59
Could you please edit your answer and show me? Do I use
has()?– Timmy Balk
Nov 23 '18 at 21:59
@TimmyBalk just remove
.filter(([, count]) => count > 2) line - and don't change anything else– Kamil Kiełczewski
Nov 23 '18 at 22:00
@TimmyBalk just remove
.filter(([, count]) => count > 2) line - and don't change anything else– Kamil Kiełczewski
Nov 23 '18 at 22:00
@KamilKiełczewski Excellent! Thank you so much. I'm going to research into the
filter() function now, thanks again Nina and Kamil. But i'm assuming removing the .filter line is not the same as what Nina was saying?– Timmy Balk
Nov 23 '18 at 22:02
@KamilKiełczewski Excellent! Thank you so much. I'm going to research into the
filter() function now, thanks again Nina and Kamil. But i'm assuming removing the .filter line is not the same as what Nina was saying?– Timmy Balk
Nov 23 '18 at 22:02
|
show 6 more comments
You can use Object.reduce, Object.entries for this like below
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let res = Object.entries(
array.reduce((o, d) => {
let key = d.join('-')
o[key] = (o[key] || 0) + 1
return o
}, {}))
.flatMap(([k, v]) => v > 2 ? [k.split('-').map(Number)] : )
console.log(res)OR may be just with Array.filters
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let temp = {}
let res = array.filter(d => {
let key = d.join('-')
temp[key] = (temp[key] || 0) + 1
return temp[key] == 3
})
console.log(res)answered Nov 23 '18 at 21:11
Nitish NarangNitish Narang
2,9501815
1
First snipped don't meet requirements (return elements are strings not nubmers) - but the second snipped is brilliant, +1
– Kamil Kiełczewski
Nov 23 '18 at 21:58
Thank you @KamilKiełczewski. Updated first snippet to return Numbers.
– Nitish Narang
Nov 24 '18 at 4:19
add a comment |
You can use Object.reduce, Object.entries for this like below
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let res = Object.entries(
array.reduce((o, d) => {
let key = d.join('-')
o[key] = (o[key] || 0) + 1
return o
}, {}))
.flatMap(([k, v]) => v > 2 ? [k.split('-').map(Number)] : )
console.log(res)OR may be just with Array.filters
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let temp = {}
let res = array.filter(d => {
let key = d.join('-')
temp[key] = (temp[key] || 0) + 1
return temp[key] == 3
})
console.log(res)answered Nov 23 '18 at 21:11
Nitish NarangNitish Narang
2,9501815
1
First snipped don't meet requirements (return elements are strings not nubmers) - but the second snipped is brilliant, +1
– Kamil Kiełczewski
Nov 23 '18 at 21:58
Thank you @KamilKiełczewski. Updated first snippet to return Numbers.
– Nitish Narang
Nov 24 '18 at 4:19
add a comment |
You can use Object.reduce, Object.entries for this like below
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let res = Object.entries(
array.reduce((o, d) => {
let key = d.join('-')
o[key] = (o[key] || 0) + 1
return o
}, {}))
.flatMap(([k, v]) => v > 2 ? [k.split('-').map(Number)] : )
console.log(res)OR may be just with Array.filters
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let temp = {}
let res = array.filter(d => {
let key = d.join('-')
temp[key] = (temp[key] || 0) + 1
return temp[key] == 3
})
console.log(res)answered Nov 23 '18 at 21:11
Nitish NarangNitish Narang
2,9501815
You can use Object.reduce, Object.entries for this like below
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let res = Object.entries(
array.reduce((o, d) => {
let key = d.join('-')
o[key] = (o[key] || 0) + 1
return o
}, {}))
.flatMap(([k, v]) => v > 2 ? [k.split('-').map(Number)] : )
console.log(res)OR may be just with Array.filters
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let temp = {}
let res = array.filter(d => {
let key = d.join('-')
temp[key] = (temp[key] || 0) + 1
return temp[key] == 3
})
console.log(res)var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let res = Object.entries(
array.reduce((o, d) => {
let key = d.join('-')
o[key] = (o[key] || 0) + 1
return o
}, {}))
.flatMap(([k, v]) => v > 2 ? [k.split('-').map(Number)] : )
console.log(res)var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let res = Object.entries(
array.reduce((o, d) => {
let key = d.join('-')
o[key] = (o[key] || 0) + 1
return o
}, {}))
.flatMap(([k, v]) => v > 2 ? [k.split('-').map(Number)] : )
console.log(res)var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let temp = {}
let res = array.filter(d => {
let key = d.join('-')
temp[key] = (temp[key] || 0) + 1
return temp[key] == 3
})
console.log(res)var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let temp = {}
let res = array.filter(d => {
let key = d.join('-')
temp[key] = (temp[key] || 0) + 1
return temp[key] == 3
})
console.log(res)answered Nov 23 '18 at 21:11
Nitish NarangNitish Narang
2,9501815
edited Nov 24 '18 at 4:08
answered Nov 23 '18 at 21:11
Nitish NarangNitish Narang
2,9501815
answered Nov 23 '18 at 21:11
Nitish NarangNitish Narang
2,9501815
answered Nov 23 '18 at 21:11
Nitish NarangNitish Narang
2,9501815
2,9501815
1
First snipped don't meet requirements (return elements are strings not nubmers) - but the second snipped is brilliant, +1
– Kamil Kiełczewski
Nov 23 '18 at 21:58
Thank you @KamilKiełczewski. Updated first snippet to return Numbers.
– Nitish Narang
Nov 24 '18 at 4:19
add a comment |
1
First snipped don't meet requirements (return elements are strings not nubmers) - but the second snipped is brilliant, +1
– Kamil Kiełczewski
Nov 23 '18 at 21:58
Thank you @KamilKiełczewski. Updated first snippet to return Numbers.
– Nitish Narang
Nov 24 '18 at 4:19
1
1
First snipped don't meet requirements (return elements are strings not nubmers) - but the second snipped is brilliant, +1
– Kamil Kiełczewski
Nov 23 '18 at 21:58
First snipped don't meet requirements (return elements are strings not nubmers) - but the second snipped is brilliant, +1
– Kamil Kiełczewski
Nov 23 '18 at 21:58
Thank you @KamilKiełczewski. Updated first snippet to return Numbers.
– Nitish Narang
Nov 24 '18 at 4:19
Thank you @KamilKiełczewski. Updated first snippet to return Numbers.
– Nitish Narang
Nov 24 '18 at 4:19
add a comment |
Try this
array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
Time complexity O(n) (one array pass by filter function). Inspired by Nitish answer.
Explanation
The (r={}, a=>...) will return last expression after comma (which is a=>...) (e.g. (5,6)==6). In r={} we set once temporary object where we will store unique keys. In filter function a=>... in a we have current array element . In r[a] JS implicity cast a to string (e.g 1,17). Then in !(2-(r[a]=++r[a]|0)) we increase counter of occurrence element a and return true (as filter function value) if element a occurred 3 times. If r[a] is undefined the ++r[a] returns NaN, and further NaN|0=0 (also number|0=number). The r[a]= initialise first counter value, if we omit it the ++ will only set NaN to r[a] which is non-incrementable (so we need to put zero at init). If we remove 2- as result we get input array without duplicates - or alternatively we can also get this by a=>!(r[a]=a in r). If we change 2- to 1- we get array with duplicates only.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
var r= array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
console.log(JSON.stringify(r));answered Nov 23 '18 at 21:19
Kamil KiełczewskiKamil Kiełczewski
11.7k86894
add a comment |
Try this
array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
Time complexity O(n) (one array pass by filter function). Inspired by Nitish answer.
Explanation
The (r={}, a=>...) will return last expression after comma (which is a=>...) (e.g. (5,6)==6). In r={} we set once temporary object where we will store unique keys. In filter function a=>... in a we have current array element . In r[a] JS implicity cast a to string (e.g 1,17). Then in !(2-(r[a]=++r[a]|0)) we increase counter of occurrence element a and return true (as filter function value) if element a occurred 3 times. If r[a] is undefined the ++r[a] returns NaN, and further NaN|0=0 (also number|0=number). The r[a]= initialise first counter value, if we omit it the ++ will only set NaN to r[a] which is non-incrementable (so we need to put zero at init). If we remove 2- as result we get input array without duplicates - or alternatively we can also get this by a=>!(r[a]=a in r). If we change 2- to 1- we get array with duplicates only.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
var r= array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
console.log(JSON.stringify(r));answered Nov 23 '18 at 21:19
Kamil KiełczewskiKamil Kiełczewski
11.7k86894
add a comment |
Try this
array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
Time complexity O(n) (one array pass by filter function). Inspired by Nitish answer.
Explanation
The (r={}, a=>...) will return last expression after comma (which is a=>...) (e.g. (5,6)==6). In r={} we set once temporary object where we will store unique keys. In filter function a=>... in a we have current array element . In r[a] JS implicity cast a to string (e.g 1,17). Then in !(2-(r[a]=++r[a]|0)) we increase counter of occurrence element a and return true (as filter function value) if element a occurred 3 times. If r[a] is undefined the ++r[a] returns NaN, and further NaN|0=0 (also number|0=number). The r[a]= initialise first counter value, if we omit it the ++ will only set NaN to r[a] which is non-incrementable (so we need to put zero at init). If we remove 2- as result we get input array without duplicates - or alternatively we can also get this by a=>!(r[a]=a in r). If we change 2- to 1- we get array with duplicates only.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
var r= array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
console.log(JSON.stringify(r));answered Nov 23 '18 at 21:19
Kamil KiełczewskiKamil Kiełczewski
11.7k86894
Try this
array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
Time complexity O(n) (one array pass by filter function). Inspired by Nitish answer.
Explanation
The (r={}, a=>...) will return last expression after comma (which is a=>...) (e.g. (5,6)==6). In r={} we set once temporary object where we will store unique keys. In filter function a=>... in a we have current array element . In r[a] JS implicity cast a to string (e.g 1,17). Then in !(2-(r[a]=++r[a]|0)) we increase counter of occurrence element a and return true (as filter function value) if element a occurred 3 times. If r[a] is undefined the ++r[a] returns NaN, and further NaN|0=0 (also number|0=number). The r[a]= initialise first counter value, if we omit it the ++ will only set NaN to r[a] which is non-incrementable (so we need to put zero at init). If we remove 2- as result we get input array without duplicates - or alternatively we can also get this by a=>!(r[a]=a in r). If we change 2- to 1- we get array with duplicates only.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
var r= array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
console.log(JSON.stringify(r));var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
var r= array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
console.log(JSON.stringify(r));var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
var r= array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
console.log(JSON.stringify(r));answered Nov 23 '18 at 21:19
Kamil KiełczewskiKamil Kiełczewski
11.7k86894
edited Feb 1 at 10:11
answered Nov 23 '18 at 21:19
Kamil KiełczewskiKamil Kiełczewski
11.7k86894
answered Nov 23 '18 at 21:19
Kamil KiełczewskiKamil Kiełczewski
11.7k86894
answered Nov 23 '18 at 21:19
Kamil KiełczewskiKamil Kiełczewski
11.7k86894
11.7k86894
add a comment |
add a comment |
For a different take, you can first sort your list, then loop through once and pull out the elements that meet your requirement. This will probably be faster than stringifying keys from the array even with the sort:
var arr = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]]
arr.sort((a, b) => a[0] - b[0] || a[1] - b[1])
// define equal for array
const equal = (arr1, arr2) => arr1.every((n, j) => n === arr2[j])
let GROUP_SIZE = 3
first = 0, last = 1, res =
while(last < arr.length){
if (equal(arr[first], arr[last])) last++
else {
if (last - first >= GROUP_SIZE) res.push(arr[first])
first = last
}
}
if (last - first >= GROUP_SIZE) res.push(arr[first])
console.log(res)answered Nov 23 '18 at 21:34
Mark MeyerMark Meyer
38.6k33159
Interesting approach, but is it really faster?
– Timmy Balk
Nov 23 '18 at 21:52
1
@TimmyBalk this isn't a criticism of the JSON approach -- I think it's good, but if my test is correct, it is slower (at least on my browser). Here's a jsperf test: jsperf.com/2json-v-sort-loop
– Mark Meyer
Nov 23 '18 at 21:55
add a comment |
For a different take, you can first sort your list, then loop through once and pull out the elements that meet your requirement. This will probably be faster than stringifying keys from the array even with the sort:
var arr = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]]
arr.sort((a, b) => a[0] - b[0] || a[1] - b[1])
// define equal for array
const equal = (arr1, arr2) => arr1.every((n, j) => n === arr2[j])
let GROUP_SIZE = 3
first = 0, last = 1, res =
while(last < arr.length){
if (equal(arr[first], arr[last])) last++
else {
if (last - first >= GROUP_SIZE) res.push(arr[first])
first = last
}
}
if (last - first >= GROUP_SIZE) res.push(arr[first])
console.log(res)answered Nov 23 '18 at 21:34
Mark MeyerMark Meyer
38.6k33159
Interesting approach, but is it really faster?
– Timmy Balk
Nov 23 '18 at 21:52
1
@TimmyBalk this isn't a criticism of the JSON approach -- I think it's good, but if my test is correct, it is slower (at least on my browser). Here's a jsperf test: jsperf.com/2json-v-sort-loop
– Mark Meyer
Nov 23 '18 at 21:55
add a comment |
For a different take, you can first sort your list, then loop through once and pull out the elements that meet your requirement. This will probably be faster than stringifying keys from the array even with the sort:
var arr = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]]
arr.sort((a, b) => a[0] - b[0] || a[1] - b[1])
// define equal for array
const equal = (arr1, arr2) => arr1.every((n, j) => n === arr2[j])
let GROUP_SIZE = 3
first = 0, last = 1, res =
while(last < arr.length){
if (equal(arr[first], arr[last])) last++
else {
if (last - first >= GROUP_SIZE) res.push(arr[first])
first = last
}
}
if (last - first >= GROUP_SIZE) res.push(arr[first])
console.log(res)answered Nov 23 '18 at 21:34
Mark MeyerMark Meyer
38.6k33159
For a different take, you can first sort your list, then loop through once and pull out the elements that meet your requirement. This will probably be faster than stringifying keys from the array even with the sort:
var arr = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]]
arr.sort((a, b) => a[0] - b[0] || a[1] - b[1])
// define equal for array
const equal = (arr1, arr2) => arr1.every((n, j) => n === arr2[j])
let GROUP_SIZE = 3
first = 0, last = 1, res =
while(last < arr.length){
if (equal(arr[first], arr[last])) last++
else {
if (last - first >= GROUP_SIZE) res.push(arr[first])
first = last
}
}
if (last - first >= GROUP_SIZE) res.push(arr[first])
console.log(res)var arr = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]]
arr.sort((a, b) => a[0] - b[0] || a[1] - b[1])
// define equal for array
const equal = (arr1, arr2) => arr1.every((n, j) => n === arr2[j])
let GROUP_SIZE = 3
first = 0, last = 1, res =
while(last < arr.length){
if (equal(arr[first], arr[last])) last++
else {
if (last - first >= GROUP_SIZE) res.push(arr[first])
first = last
}
}
if (last - first >= GROUP_SIZE) res.push(arr[first])
console.log(res)var arr = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]]
arr.sort((a, b) => a[0] - b[0] || a[1] - b[1])
// define equal for array
const equal = (arr1, arr2) => arr1.every((n, j) => n === arr2[j])
let GROUP_SIZE = 3
first = 0, last = 1, res =
while(last < arr.length){
if (equal(arr[first], arr[last])) last++
else {
if (last - first >= GROUP_SIZE) res.push(arr[first])
first = last
}
}
if (last - first >= GROUP_SIZE) res.push(arr[first])
console.log(res)answered Nov 23 '18 at 21:34
Mark MeyerMark Meyer
38.6k33159
answered Nov 23 '18 at 21:34
Mark MeyerMark Meyer
38.6k33159
answered Nov 23 '18 at 21:34
Mark MeyerMark Meyer
38.6k33159
answered Nov 23 '18 at 21:34
Mark MeyerMark Meyer
38.6k33159
38.6k33159
Interesting approach, but is it really faster?
– Timmy Balk
Nov 23 '18 at 21:52
1
@TimmyBalk this isn't a criticism of the JSON approach -- I think it's good, but if my test is correct, it is slower (at least on my browser). Here's a jsperf test: jsperf.com/2json-v-sort-loop
– Mark Meyer
Nov 23 '18 at 21:55
add a comment |
Interesting approach, but is it really faster?
– Timmy Balk
Nov 23 '18 at 21:52
1
@TimmyBalk this isn't a criticism of the JSON approach -- I think it's good, but if my test is correct, it is slower (at least on my browser). Here's a jsperf test: jsperf.com/2json-v-sort-loop
– Mark Meyer
Nov 23 '18 at 21:55
Interesting approach, but is it really faster?
– Timmy Balk
Nov 23 '18 at 21:52
Interesting approach, but is it really faster?
– Timmy Balk
Nov 23 '18 at 21:52
1
1
@TimmyBalk this isn't a criticism of the JSON approach -- I think it's good, but if my test is correct, it is slower (at least on my browser). Here's a jsperf test: jsperf.com/2json-v-sort-loop
– Mark Meyer
Nov 23 '18 at 21:55
@TimmyBalk this isn't a criticism of the JSON approach -- I think it's good, but if my test is correct, it is slower (at least on my browser). Here's a jsperf test: jsperf.com/2json-v-sort-loop
– Mark Meyer
Nov 23 '18 at 21:55
add a comment |
ES6:
const repeatMap = {}
array.forEach(arr => {
const key = JSON.stringify(arr)
if (repeatMap[key]) {
repeatMap[key]++
} else {
repeatMap[key] = 1
}
})
const repeatedArrays = Object.keys(repeatMap)
.filter(key => repeatMap[key] >= 3)
.map(key => JSON.parse(key))
answered Nov 23 '18 at 21:11
Ben StewardBen Steward
1,135315
add a comment |
ES6:
const repeatMap = {}
array.forEach(arr => {
const key = JSON.stringify(arr)
if (repeatMap[key]) {
repeatMap[key]++
} else {
repeatMap[key] = 1
}
})
const repeatedArrays = Object.keys(repeatMap)
.filter(key => repeatMap[key] >= 3)
.map(key => JSON.parse(key))
answered Nov 23 '18 at 21:11
Ben StewardBen Steward
1,135315
add a comment |
ES6:
const repeatMap = {}
array.forEach(arr => {
const key = JSON.stringify(arr)
if (repeatMap[key]) {
repeatMap[key]++
} else {
repeatMap[key] = 1
}
})
const repeatedArrays = Object.keys(repeatMap)
.filter(key => repeatMap[key] >= 3)
.map(key => JSON.parse(key))
answered Nov 23 '18 at 21:11
Ben StewardBen Steward
1,135315
ES6:
const repeatMap = {}
array.forEach(arr => {
const key = JSON.stringify(arr)
if (repeatMap[key]) {
repeatMap[key]++
} else {
repeatMap[key] = 1
}
})
const repeatedArrays = Object.keys(repeatMap)
.filter(key => repeatMap[key] >= 3)
.map(key => JSON.parse(key))
answered Nov 23 '18 at 21:11
Ben StewardBen Steward
1,135315
answered Nov 23 '18 at 21:11
Ben StewardBen Steward
1,135315
answered Nov 23 '18 at 21:11
Ben StewardBen Steward
1,135315
answered Nov 23 '18 at 21:11
Ben StewardBen Steward
1,135315
1,135315
add a comment |
add a comment |
You could also do this with a single Array.reduce where you would only push to a result property if the length is equal to 3:
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
console.log(array.reduce((r,c) => {
let key = c.join('-')
r[key] = (r[key] || 0) + 1
r[key] == 3 ? r.result.push(c) : 0 // if we have a hit push to result
return r
}, { result: }).result) // print the result propertyanswered Nov 24 '18 at 7:27
AkrionAkrion
9,48511224
add a comment |
You could also do this with a single Array.reduce where you would only push to a result property if the length is equal to 3:
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
console.log(array.reduce((r,c) => {
let key = c.join('-')
r[key] = (r[key] || 0) + 1
r[key] == 3 ? r.result.push(c) : 0 // if we have a hit push to result
return r
}, { result: }).result) // print the result propertyanswered Nov 24 '18 at 7:27
AkrionAkrion
9,48511224
add a comment |
You could also do this with a single Array.reduce where you would only push to a result property if the length is equal to 3:
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
console.log(array.reduce((r,c) => {
let key = c.join('-')
r[key] = (r[key] || 0) + 1
r[key] == 3 ? r.result.push(c) : 0 // if we have a hit push to result
return r
}, { result: }).result) // print the result propertyanswered Nov 24 '18 at 7:27
AkrionAkrion
9,48511224
You could also do this with a single Array.reduce where you would only push to a result property if the length is equal to 3:
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
console.log(array.reduce((r,c) => {
let key = c.join('-')
r[key] = (r[key] || 0) + 1
r[key] == 3 ? r.result.push(c) : 0 // if we have a hit push to result
return r
}, { result: }).result) // print the result propertyvar array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
console.log(array.reduce((r,c) => {
let key = c.join('-')
r[key] = (r[key] || 0) + 1
r[key] == 3 ? r.result.push(c) : 0 // if we have a hit push to result
return r
}, { result: }).result) // print the result propertyvar array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
console.log(array.reduce((r,c) => {
let key = c.join('-')
r[key] = (r[key] || 0) + 1
r[key] == 3 ? r.result.push(c) : 0 // if we have a hit push to result
return r
}, { result: }).result) // print the result propertyanswered Nov 24 '18 at 7:27
AkrionAkrion
9,48511224
answered Nov 24 '18 at 7:27
AkrionAkrion
9,48511224
answered Nov 24 '18 at 7:27
AkrionAkrion
9,48511224
answered Nov 24 '18 at 7:27
AkrionAkrion
9,48511224
9,48511224
add a comment |
add a comment |
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You could compare them against their JSON values, however,
[1, 17]wouldn't match[17, 1]– Get Off My Lawn
Nov 23 '18 at 21:03
Choose best answer for you and accept it by click on gray "check" button on its left side
– Kamil Kiełczewski
Nov 24 '18 at 12:37