let say i have 1 multidimensional array and i want to exclude values that not equal in javascript.
here is the example array.
var filter = ["big_number", "odds_number"];
var arrays = {
"first" : {
"big_number" : [50,51,52],
"odds_number" : [39,41,51,53]
},
"second" : {
"big_number" : [61,62,63,64,65,70,72,73],
"odds_number" : [13,15,17,19,61,63,65,73]
}
};
i want to convert that array to be like this.
var new_arrays = {
"first" : [51],
"second" : [61,63,65,73]
};
here is my code
var newArray = {
"first" : [],
"second" : []
};
for (var k in arrays){
if (arrays.hasOwnProperty(k)) {
for(var f=0; f<filter.length; f++) {
newArray[k].push(arrays[k][filter[f]].filter(value => -1 !== arrays[k][filter[f]].indexOf(value))));
}
}
}
console.log(newArray);
actually i could do this code
var newArray = {
"first" : [],
"second" : []
};
for (var k in arrays){
if (arrays.hasOwnProperty(k)) {
newArray[k].push(arrays[k]["big_number"].filter(value => -1 !== arrays[k]["odds_number"].indexOf(value))));
}
}
console.log(newArray);
but i need to convert it through filter variable.
i could not use filter[0] and filter[1], because that values could change dynamically and could be more than 2 values in array.
asked Feb 20, 2019 at 10:43
Willyanto Halim
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why down vote my question?Willyanto Halim– Willyanto Halim2019年02月20日 10:58:04 +00:00Commented Feb 20, 2019 at 10:58
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I guess that's because you haven't shown any attempt yourself. And I personally don't see the pattern here.Jonas Wilms– Jonas Wilms2019年02月20日 11:12:01 +00:00Commented Feb 20, 2019 at 11:12
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@JonasWilms i have mentioned that i could done it if there were 2 arrays separated.. so should i need to put code with 2 arrays separated?Willyanto Halim– Willyanto Halim2019年02月20日 11:14:59 +00:00Commented Feb 20, 2019 at 11:14
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@FZs could you read the question first before you said it duplicate question?Willyanto Halim– Willyanto Halim2019年02月20日 11:45:57 +00:00Commented Feb 20, 2019 at 11:45
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@FZs is it clear for you for that updated question?Willyanto Halim– Willyanto Halim2019年02月20日 12:05:06 +00:00Commented Feb 20, 2019 at 12:05
4 Answers 4
You could loop through the keys and update the values using filter and includes:
var arrays={"first":{"big_number":[50,51,52],"odds_number":[39,41,51,53]},"second":{"big_number":[61,62,63,64,65,70,72,73],"odds_number":[13,15,17,19,61,63,65,73]}};
for (let key in arrays) {
arrays[key] = arrays[key]["big_number"]
.filter(n => arrays[key]["odds_number"].includes(n));
}
console.log(arrays)If you don't want to mutate the original object then use Object.entries and reduce:
var arrays={"first":{"big_number":[50,51,52],"odds_number":[39,41,51,53]},"second":{"big_number":[61,62,63,64,65,70,72,73],"odds_number":[13,15,17,19,61,63,65,73]}};
const newObject = Object.entries(arrays).reduce((r, [key, {big_number, odds_number}]) => {
r[key] = big_number.filter(n => odds_number.includes(n));
return r
}, {})
console.log(newObject)If you have more than 2 array properties, you can do something like this: Get all the arrays using Object.values and then use reduce to run the previous code recursively
var arrays = {
"first": {
"big_number": [50, 51, 52],
"odds_number": [39, 41, 51, 53],
"another_key": [41, 51, 53]
},
"second": {
"big_number": [61, 62, 63, 64, 65, 70, 72, 73],
"odds_number": [13, 15, 17, 19, 61, 63, 65, 73],
"another_key": [63, 65]
}
};
for (let key in arrays) {
arrays[key] = Object.values(arrays[key])
.reduce((a, b) => a.filter(c => b.includes(c)))
}
console.log(arrays)
answered Feb 20, 2019 at 10:48
adiga
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12 Comments
Willyanto Halim
sorry, i forget to put filter variable.. it should loop through that variable.. i have update the question..
adiga
What do you mean by
filter variable? Instead of arrays[key]["big_number"], do arrays[key][filter[0]] and filter[1] for odds_numberWillyanto Halim
yes.. filter variable value is ["big_number", "odds_number"] and the variable could change dynamically through checkbox.. so i want to loop it through that filter variable..
adiga
Do as mentioned in the previous comment.
adiga
@WillyantoHalim Intersection is getting items common to all arrays. If one of them is empty, then result will be empty array. This is valid
|
Here is a little intersection snippet:
function intersect(a,b){
b.slice()
return a.filter(item=>{
if(b.includes(item)){
b.splice(b.indexOf(item),1)
return true
}
})
}
Using that, you can do this easily:
function intersect(a,b){
b.slice()
return a.filter(item=>{
if(b.includes(item)){
b.splice(b.indexOf(item),1)
return true
}
})
}
var filter = ["big_number", "odds_number"];
var output={}
var arrays = {
"first" : {
"big_number" : [50,51,52],
"odds_number" : [39,41,51,53]
},
"second" : {
"big_number" : [61,62,63,64,65,70,72,73],
"odds_number" : [13,15,17,19,61,63,65,73]
}
};
for(x in arrays){
output[x]=arrays[x][filter[0]]
for(let i=1;i<filter.length;i++){
output[x]=intersect(output[x],arrays[x][filter[i]])
}
}
console.log (output)
answered Feb 20, 2019 at 10:53
FZs
18.8k13 gold badges47 silver badges56 bronze badges
2 Comments
Willyanto Halim
you can have a look for the updated question and remove that duplicate question tag.. thanks
FZs
@WillyantoHalim I updated my answer to be dynamic. And sorry for the duplicate mark, I can't remove that, but only you can see; simply ignore it. Happy coding!
use Object.entries to get keys and values and then use reduce
var arrays = {
"first" : {
"big_number" : [50,51,52],
"odds_number" : [39,41,51,53]
},
"second" : {
"big_number" : [61,62,63,64,65,70,72,73],
"odds_number" : [13,15,17,19,61,63,65,73]
}
};
const output =Object.entries(arrays).reduce((accu, [key, {big_number}]) => {
if(!accu[key]) accu[key] = [];
big_number.forEach(num => {
if(num%2 !==0)
accu[key].push(num);
})
return accu;
}, {});
console.log(output);
answered Feb 20, 2019 at 10:50
random
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1 Comment
Willyanto Halim
sorry, i forget to put filter variable.. it should loop through that variable.. i have update the question.
You can get the unique values from both the arrays using Set and then using filter get only the common values.
var arrays = {"first": {"big_number": [50, 51, 52],"odds_number": [39, 41, 51, 53]},"second": {"big_number": [61, 62, 63, 64, 65, 70, 72, 73],"odds_number": [13, 15, 17, 19, 61, 63, 65, 73]}},
result = Object.keys(arrays).reduce((r,k) => {
let setB = new Set(arrays[k]["big_number"]);
r[k] = [...new Set(arrays[k]["odds_number"])].filter(x => setB.has(x));
return r;
},{});
console.log(result).as-console-wrapper { max-height: 100% !important; top: 0; }
answered Feb 20, 2019 at 10:52
Hassan Imam
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1 Comment
Willyanto Halim
sorry, i forget to put filter variable.. it should loop through that variable.. i have update the question.
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