JavaScript array of objects contains the same array data
7
I try to get all same data values into an array of objects. This is my input:
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
]
I need a result like:
["65d4ze"]
I try to loop on my object to get this output, but I'm completely lost... I don't know how to know if the result is into all data arrays.
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b);javascript arrays
edited yesterday
Peter Mortensen
13.8k1987113
asked yesterday
KamouloxKamoulox
413
New contributor
Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
show 3 more comments
7
I try to get all same data values into an array of objects. This is my input:
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
]
I need a result like:
["65d4ze"]
I try to loop on my object to get this output, but I'm completely lost... I don't know how to know if the result is into all data arrays.
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b);javascript arrays
edited yesterday
Peter Mortensen
13.8k1987113
asked yesterday
KamouloxKamoulox
413
New contributor
Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
So get all arrays elements that are present on alla?
– Eddie
yesterday
3
Possible duplicate of Simplest code for array intersection in javascript
– James Long
yesterday
@JamesLong I don't understand the duplication, why this answer can help me ?
– Kamoulox
yesterday
@Eddie I need to get all array into the key data. This key is present into all object of my principal array.
– Kamoulox
yesterday
@Kamoulox you have 2 arrays (a[0].data and a[1].data). You seem to be wanting an array of all the values that appear in both arrays. Unless I'm mis-reading your question
– James Long
yesterday
|
show 3 more comments
7
7
7
1
I try to get all same data values into an array of objects. This is my input:
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
]
I need a result like:
["65d4ze"]
I try to loop on my object to get this output, but I'm completely lost... I don't know how to know if the result is into all data arrays.
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b);javascript arrays
edited yesterday
Peter Mortensen
13.8k1987113
asked yesterday
KamouloxKamoulox
413
New contributor
Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I try to get all same data values into an array of objects. This is my input:
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
]
I need a result like:
["65d4ze"]
I try to loop on my object to get this output, but I'm completely lost... I don't know how to know if the result is into all data arrays.
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b);var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b);var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b);javascript arrays
javascript arrays
edited yesterday
Peter Mortensen
13.8k1987113
asked yesterday
KamouloxKamoulox
413
New contributor
Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Peter Mortensen
13.8k1987113
asked yesterday
KamouloxKamoulox
413
New contributor
Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Peter Mortensen
13.8k1987113
edited yesterday
Peter Mortensen
13.8k1987113
edited yesterday
Peter Mortensen
13.8k1987113
13.8k1987113
asked yesterday
KamouloxKamoulox
413
New contributor
Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
KamouloxKamoulox
413
asked yesterday
KamouloxKamoulox
413
413
New contributor
Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
So get all arrays elements that are present on alla?
– Eddie
yesterday
3
Possible duplicate of Simplest code for array intersection in javascript
– James Long
yesterday
@JamesLong I don't understand the duplication, why this answer can help me ?
– Kamoulox
yesterday
@Eddie I need to get all array into the key data. This key is present into all object of my principal array.
– Kamoulox
yesterday
@Kamoulox you have 2 arrays (a[0].data and a[1].data). You seem to be wanting an array of all the values that appear in both arrays. Unless I'm mis-reading your question
– James Long
yesterday
|
show 3 more comments
So get all arrays elements that are present on alla?
– Eddie
yesterday
3
Possible duplicate of Simplest code for array intersection in javascript
– James Long
yesterday
@JamesLong I don't understand the duplication, why this answer can help me ?
– Kamoulox
yesterday
@Eddie I need to get all array into the key data. This key is present into all object of my principal array.
– Kamoulox
yesterday
@Kamoulox you have 2 arrays (a[0].data and a[1].data). You seem to be wanting an array of all the values that appear in both arrays. Unless I'm mis-reading your question
– James Long
yesterday
So get all arrays elements that are present on all
a?– Eddie
yesterday
So get all arrays elements that are present on all
a?– Eddie
yesterday
3
3
Possible duplicate of Simplest code for array intersection in javascript
– James Long
yesterday
Possible duplicate of Simplest code for array intersection in javascript
– James Long
yesterday
@JamesLong I don't understand the duplication, why this answer can help me ?
– Kamoulox
yesterday
@JamesLong I don't understand the duplication, why this answer can help me ?
– Kamoulox
yesterday
@Eddie I need to get all array into the key data. This key is present into all object of my principal array.
– Kamoulox
yesterday
@Eddie I need to get all array into the key data. This key is present into all object of my principal array.
– Kamoulox
yesterday
@Kamoulox you have 2 arrays (a[0].data and a[1].data). You seem to be wanting an array of all the values that appear in both arrays. Unless I'm mis-reading your question
– James Long
yesterday
@Kamoulox you have 2 arrays (a[0].data and a[1].data). You seem to be wanting an array of all the values that appear in both arrays. Unless I'm mis-reading your question
– James Long
yesterday
|
show 3 more comments
5 Answers
5
active
oldest
votes
11
You could map data and get the common values with Array#map, Array#reduce, Array#filter, Set and Set#has.
var array = [{ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] }, { name: "Bar", id: "321", data: ["65d4ze", "894ver81"] }],
key = 'data',
common = array
.map(o => o[key])
.reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));
console.log(common);answered yesterday
Nina ScholzNina Scholz
193k15107178
I have some troubles to understand this piece of codeb.filter(Set.prototype.has, new Set(a)). But it's working perfectly, thanks
– Kamoulox
yesterday
5
filterhas two parameters, one foir the callback, which is here a prototype function ofSet,haswhich checks a value against a set. at this point, the method does not works because it need an instance ofSet. the second parameter isthisArg, where you can hand over an object, which is used asthisin the callback. mabe a different use makes ir a bet more clear. you could gate the same with a boundthisby using this as callback:Set.prototype.has.bind(new Set(a)), which is now a function withthis.
– Nina Scholz
yesterday
With arraycommon = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));
– Pranav C Balan
yesterday
@PranavCBalan,Array#includesuses a second parameterfromIndexand gets this value with the index of the callback ...
– Nina Scholz
yesterday
@NinaScholz : oops that's right, that would make issues :)
– Pranav C Balan
yesterday
add a comment |
3
You can use the Array#filter method. Filter the first array by checking if a value is present in all other object properties (arrays), using the Array#every method to check if a value is present in all remaining arrays.
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
];
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
console.log(res)edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
Pranav C BalanPranav C Balan
87.6k1391117
add a comment |
1
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = {};
a.forEach(function(i) {
i.data.forEach(function(j) {
if (!b.hasOwnProperty(j)) {
b[j] = 0;
}
b[j] = b[j] + 1;
});
});
c =
for (var i in b) {
if (b.hasOwnProperty(i)) {
if (b[i] > 1) {
c.push(i)
}
}
}
console.log(c);answered yesterday
Shree TiwariShree Tiwari
287112
add a comment |
1
Use the flat function in the array:
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b.flat());edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
thelastwormthelastworm
1428
Thanks for your answer, that a great code. Butflatis not working on my browser
– Kamoulox
yesterday
add a comment |
1
You could use reduce and concat on each data array, and check the count of each item.
In the end, you check whether all objects across the array contain that item and return it if yes.
Note that this function works if you want to extract the item that has the same occurrence across all objects in the array.
If an item has duplicates, but does not fulfill the above condition, it would not be extracted.
let a = [{name: "Foo",id: "123",data: ["65d4ze", "65h8914d"]},{name: "Bar",id: "321",data: ["65d4ze", "894ver81"]}]
let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = {};
let result = ;
for (var i = 0; i < arr.length; i++) {
var num = arr[i];
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
for (let i in counts) {
if (counts[i] === a.length) {
result.push(i)
}
}
console.log(result)edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
tnkhtnkh
17210
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
11
You could map data and get the common values with Array#map, Array#reduce, Array#filter, Set and Set#has.
var array = [{ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] }, { name: "Bar", id: "321", data: ["65d4ze", "894ver81"] }],
key = 'data',
common = array
.map(o => o[key])
.reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));
console.log(common);answered yesterday
Nina ScholzNina Scholz
193k15107178
I have some troubles to understand this piece of codeb.filter(Set.prototype.has, new Set(a)). But it's working perfectly, thanks
– Kamoulox
yesterday
5
filterhas two parameters, one foir the callback, which is here a prototype function ofSet,haswhich checks a value against a set. at this point, the method does not works because it need an instance ofSet. the second parameter isthisArg, where you can hand over an object, which is used asthisin the callback. mabe a different use makes ir a bet more clear. you could gate the same with a boundthisby using this as callback:Set.prototype.has.bind(new Set(a)), which is now a function withthis.
– Nina Scholz
yesterday
With arraycommon = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));
– Pranav C Balan
yesterday
@PranavCBalan,Array#includesuses a second parameterfromIndexand gets this value with the index of the callback ...
– Nina Scholz
yesterday
@NinaScholz : oops that's right, that would make issues :)
– Pranav C Balan
yesterday
add a comment |
11
You could map data and get the common values with Array#map, Array#reduce, Array#filter, Set and Set#has.
var array = [{ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] }, { name: "Bar", id: "321", data: ["65d4ze", "894ver81"] }],
key = 'data',
common = array
.map(o => o[key])
.reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));
console.log(common);answered yesterday
Nina ScholzNina Scholz
193k15107178
I have some troubles to understand this piece of codeb.filter(Set.prototype.has, new Set(a)). But it's working perfectly, thanks
– Kamoulox
yesterday
5
filterhas two parameters, one foir the callback, which is here a prototype function ofSet,haswhich checks a value against a set. at this point, the method does not works because it need an instance ofSet. the second parameter isthisArg, where you can hand over an object, which is used asthisin the callback. mabe a different use makes ir a bet more clear. you could gate the same with a boundthisby using this as callback:Set.prototype.has.bind(new Set(a)), which is now a function withthis.
– Nina Scholz
yesterday
With arraycommon = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));
– Pranav C Balan
yesterday
@PranavCBalan,Array#includesuses a second parameterfromIndexand gets this value with the index of the callback ...
– Nina Scholz
yesterday
@NinaScholz : oops that's right, that would make issues :)
– Pranav C Balan
yesterday
add a comment |
11
11
11
You could map data and get the common values with Array#map, Array#reduce, Array#filter, Set and Set#has.
var array = [{ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] }, { name: "Bar", id: "321", data: ["65d4ze", "894ver81"] }],
key = 'data',
common = array
.map(o => o[key])
.reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));
console.log(common);answered yesterday
Nina ScholzNina Scholz
193k15107178
You could map data and get the common values with Array#map, Array#reduce, Array#filter, Set and Set#has.
var array = [{ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] }, { name: "Bar", id: "321", data: ["65d4ze", "894ver81"] }],
key = 'data',
common = array
.map(o => o[key])
.reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));
console.log(common);var array = [{ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] }, { name: "Bar", id: "321", data: ["65d4ze", "894ver81"] }],
key = 'data',
common = array
.map(o => o[key])
.reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));
console.log(common);var array = [{ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] }, { name: "Bar", id: "321", data: ["65d4ze", "894ver81"] }],
key = 'data',
common = array
.map(o => o[key])
.reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));
console.log(common);answered yesterday
Nina ScholzNina Scholz
193k15107178
answered yesterday
Nina ScholzNina Scholz
193k15107178
answered yesterday
Nina ScholzNina Scholz
193k15107178
answered yesterday
Nina ScholzNina Scholz
193k15107178
193k15107178
I have some troubles to understand this piece of codeb.filter(Set.prototype.has, new Set(a)). But it's working perfectly, thanks
– Kamoulox
yesterday
5
filterhas two parameters, one foir the callback, which is here a prototype function ofSet,haswhich checks a value against a set. at this point, the method does not works because it need an instance ofSet. the second parameter isthisArg, where you can hand over an object, which is used asthisin the callback. mabe a different use makes ir a bet more clear. you could gate the same with a boundthisby using this as callback:Set.prototype.has.bind(new Set(a)), which is now a function withthis.
– Nina Scholz
yesterday
With arraycommon = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));
– Pranav C Balan
yesterday
@PranavCBalan,Array#includesuses a second parameterfromIndexand gets this value with the index of the callback ...
– Nina Scholz
yesterday
@NinaScholz : oops that's right, that would make issues :)
– Pranav C Balan
yesterday
add a comment |
I have some troubles to understand this piece of codeb.filter(Set.prototype.has, new Set(a)). But it's working perfectly, thanks
– Kamoulox
yesterday
5
filterhas two parameters, one foir the callback, which is here a prototype function ofSet,haswhich checks a value against a set. at this point, the method does not works because it need an instance ofSet. the second parameter isthisArg, where you can hand over an object, which is used asthisin the callback. mabe a different use makes ir a bet more clear. you could gate the same with a boundthisby using this as callback:Set.prototype.has.bind(new Set(a)), which is now a function withthis.
– Nina Scholz
yesterday
With arraycommon = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));
– Pranav C Balan
yesterday
@PranavCBalan,Array#includesuses a second parameterfromIndexand gets this value with the index of the callback ...
– Nina Scholz
yesterday
@NinaScholz : oops that's right, that would make issues :)
– Pranav C Balan
yesterday
I have some troubles to understand this piece of code
b.filter(Set.prototype.has, new Set(a)). But it's working perfectly, thanks– Kamoulox
yesterday
I have some troubles to understand this piece of code
b.filter(Set.prototype.has, new Set(a)). But it's working perfectly, thanks– Kamoulox
yesterday
5
5
filter has two parameters, one foir the callback, which is here a prototype function of Set, has which checks a value against a set. at this point, the method does not works because it need an instance of Set. the second parameter is thisArg, where you can hand over an object, which is used as this in the callback. mabe a different use makes ir a bet more clear. you could gate the same with a bound this by using this as callback: Set.prototype.has.bind(new Set(a)), which is now a function with this.– Nina Scholz
yesterday
filter has two parameters, one foir the callback, which is here a prototype function of Set, has which checks a value against a set. at this point, the method does not works because it need an instance of Set. the second parameter is thisArg, where you can hand over an object, which is used as this in the callback. mabe a different use makes ir a bet more clear. you could gate the same with a bound this by using this as callback: Set.prototype.has.bind(new Set(a)), which is now a function with this.– Nina Scholz
yesterday
With array
common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));– Pranav C Balan
yesterday
With array
common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));– Pranav C Balan
yesterday
@PranavCBalan,
Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...– Nina Scholz
yesterday
@PranavCBalan,
Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...– Nina Scholz
yesterday
@NinaScholz : oops that's right, that would make issues :)
– Pranav C Balan
yesterday
@NinaScholz : oops that's right, that would make issues :)
– Pranav C Balan
yesterday
add a comment |
3
You can use the Array#filter method. Filter the first array by checking if a value is present in all other object properties (arrays), using the Array#every method to check if a value is present in all remaining arrays.
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
];
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
console.log(res)edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
Pranav C BalanPranav C Balan
87.6k1391117
add a comment |
3
You can use the Array#filter method. Filter the first array by checking if a value is present in all other object properties (arrays), using the Array#every method to check if a value is present in all remaining arrays.
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
];
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
console.log(res)edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
Pranav C BalanPranav C Balan
87.6k1391117
add a comment |
3
3
3
You can use the Array#filter method. Filter the first array by checking if a value is present in all other object properties (arrays), using the Array#every method to check if a value is present in all remaining arrays.
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
];
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
console.log(res)edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
Pranav C BalanPranav C Balan
87.6k1391117
You can use the Array#filter method. Filter the first array by checking if a value is present in all other object properties (arrays), using the Array#every method to check if a value is present in all remaining arrays.
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
];
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
console.log(res)var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
];
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
console.log(res)var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
];
let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));
console.log(res)edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
Pranav C BalanPranav C Balan
87.6k1391117
edited yesterday
Peter Mortensen
13.8k1987113
edited yesterday
Peter Mortensen
13.8k1987113
edited yesterday
Peter Mortensen
13.8k1987113
13.8k1987113
answered yesterday
Pranav C BalanPranav C Balan
87.6k1391117
answered yesterday
Pranav C BalanPranav C Balan
87.6k1391117
answered yesterday
Pranav C BalanPranav C Balan
87.6k1391117
87.6k1391117
add a comment |
add a comment |
1
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = {};
a.forEach(function(i) {
i.data.forEach(function(j) {
if (!b.hasOwnProperty(j)) {
b[j] = 0;
}
b[j] = b[j] + 1;
});
});
c =
for (var i in b) {
if (b.hasOwnProperty(i)) {
if (b[i] > 1) {
c.push(i)
}
}
}
console.log(c);answered yesterday
Shree TiwariShree Tiwari
287112
add a comment |
1
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = {};
a.forEach(function(i) {
i.data.forEach(function(j) {
if (!b.hasOwnProperty(j)) {
b[j] = 0;
}
b[j] = b[j] + 1;
});
});
c =
for (var i in b) {
if (b.hasOwnProperty(i)) {
if (b[i] > 1) {
c.push(i)
}
}
}
console.log(c);answered yesterday
Shree TiwariShree Tiwari
287112
add a comment |
1
1
1
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = {};
a.forEach(function(i) {
i.data.forEach(function(j) {
if (!b.hasOwnProperty(j)) {
b[j] = 0;
}
b[j] = b[j] + 1;
});
});
c =
for (var i in b) {
if (b.hasOwnProperty(i)) {
if (b[i] > 1) {
c.push(i)
}
}
}
console.log(c);answered yesterday
Shree TiwariShree Tiwari
287112
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = {};
a.forEach(function(i) {
i.data.forEach(function(j) {
if (!b.hasOwnProperty(j)) {
b[j] = 0;
}
b[j] = b[j] + 1;
});
});
c =
for (var i in b) {
if (b.hasOwnProperty(i)) {
if (b[i] > 1) {
c.push(i)
}
}
}
console.log(c);var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = {};
a.forEach(function(i) {
i.data.forEach(function(j) {
if (!b.hasOwnProperty(j)) {
b[j] = 0;
}
b[j] = b[j] + 1;
});
});
c =
for (var i in b) {
if (b.hasOwnProperty(i)) {
if (b[i] > 1) {
c.push(i)
}
}
}
console.log(c);var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = {};
a.forEach(function(i) {
i.data.forEach(function(j) {
if (!b.hasOwnProperty(j)) {
b[j] = 0;
}
b[j] = b[j] + 1;
});
});
c =
for (var i in b) {
if (b.hasOwnProperty(i)) {
if (b[i] > 1) {
c.push(i)
}
}
}
console.log(c);answered yesterday
Shree TiwariShree Tiwari
287112
answered yesterday
Shree TiwariShree Tiwari
287112
answered yesterday
Shree TiwariShree Tiwari
287112
answered yesterday
Shree TiwariShree Tiwari
287112
287112
add a comment |
add a comment |
1
Use the flat function in the array:
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b.flat());edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
thelastwormthelastworm
1428
Thanks for your answer, that a great code. Butflatis not working on my browser
– Kamoulox
yesterday
add a comment |
1
Use the flat function in the array:
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b.flat());edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
thelastwormthelastworm
1428
Thanks for your answer, that a great code. Butflatis not working on my browser
– Kamoulox
yesterday
add a comment |
1
1
1
Use the flat function in the array:
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b.flat());edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
thelastwormthelastworm
1428
Use the flat function in the array:
var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b.flat());var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b.flat());var a = [{
name: "Foo",
id: "123",
data: ["65d4ze", "65h8914d"]
},
{
name: "Bar",
id: "321",
data: ["65d4ze", "894ver81"]
}
],
b = ,
c =
a.forEach(function(object) {
b.push(object.data.map(function(val) {
return val;
})
);
});
console.log(b.flat());edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
thelastwormthelastworm
1428
edited yesterday
Peter Mortensen
13.8k1987113
edited yesterday
Peter Mortensen
13.8k1987113
edited yesterday
Peter Mortensen
13.8k1987113
13.8k1987113
answered yesterday
thelastwormthelastworm
1428
answered yesterday
thelastwormthelastworm
1428
answered yesterday
thelastwormthelastworm
1428
1428
Thanks for your answer, that a great code. Butflatis not working on my browser
– Kamoulox
yesterday
add a comment |
Thanks for your answer, that a great code. Butflatis not working on my browser
– Kamoulox
yesterday
Thanks for your answer, that a great code. But
flat is not working on my browser– Kamoulox
yesterday
Thanks for your answer, that a great code. But
flat is not working on my browser– Kamoulox
yesterday
add a comment |
1
You could use reduce and concat on each data array, and check the count of each item.
In the end, you check whether all objects across the array contain that item and return it if yes.
Note that this function works if you want to extract the item that has the same occurrence across all objects in the array.
If an item has duplicates, but does not fulfill the above condition, it would not be extracted.
let a = [{name: "Foo",id: "123",data: ["65d4ze", "65h8914d"]},{name: "Bar",id: "321",data: ["65d4ze", "894ver81"]}]
let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = {};
let result = ;
for (var i = 0; i < arr.length; i++) {
var num = arr[i];
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
for (let i in counts) {
if (counts[i] === a.length) {
result.push(i)
}
}
console.log(result)edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
tnkhtnkh
17210
add a comment |
1
You could use reduce and concat on each data array, and check the count of each item.
In the end, you check whether all objects across the array contain that item and return it if yes.
Note that this function works if you want to extract the item that has the same occurrence across all objects in the array.
If an item has duplicates, but does not fulfill the above condition, it would not be extracted.
let a = [{name: "Foo",id: "123",data: ["65d4ze", "65h8914d"]},{name: "Bar",id: "321",data: ["65d4ze", "894ver81"]}]
let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = {};
let result = ;
for (var i = 0; i < arr.length; i++) {
var num = arr[i];
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
for (let i in counts) {
if (counts[i] === a.length) {
result.push(i)
}
}
console.log(result)edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
tnkhtnkh
17210
add a comment |
1
1
1
You could use reduce and concat on each data array, and check the count of each item.
In the end, you check whether all objects across the array contain that item and return it if yes.
Note that this function works if you want to extract the item that has the same occurrence across all objects in the array.
If an item has duplicates, but does not fulfill the above condition, it would not be extracted.
let a = [{name: "Foo",id: "123",data: ["65d4ze", "65h8914d"]},{name: "Bar",id: "321",data: ["65d4ze", "894ver81"]}]
let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = {};
let result = ;
for (var i = 0; i < arr.length; i++) {
var num = arr[i];
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
for (let i in counts) {
if (counts[i] === a.length) {
result.push(i)
}
}
console.log(result)edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
tnkhtnkh
17210
You could use reduce and concat on each data array, and check the count of each item.
In the end, you check whether all objects across the array contain that item and return it if yes.
Note that this function works if you want to extract the item that has the same occurrence across all objects in the array.
If an item has duplicates, but does not fulfill the above condition, it would not be extracted.
let a = [{name: "Foo",id: "123",data: ["65d4ze", "65h8914d"]},{name: "Bar",id: "321",data: ["65d4ze", "894ver81"]}]
let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = {};
let result = ;
for (var i = 0; i < arr.length; i++) {
var num = arr[i];
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
for (let i in counts) {
if (counts[i] === a.length) {
result.push(i)
}
}
console.log(result)let a = [{name: "Foo",id: "123",data: ["65d4ze", "65h8914d"]},{name: "Bar",id: "321",data: ["65d4ze", "894ver81"]}]
let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = {};
let result = ;
for (var i = 0; i < arr.length; i++) {
var num = arr[i];
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
for (let i in counts) {
if (counts[i] === a.length) {
result.push(i)
}
}
console.log(result)let a = [{name: "Foo",id: "123",data: ["65d4ze", "65h8914d"]},{name: "Bar",id: "321",data: ["65d4ze", "894ver81"]}]
let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = {};
let result = ;
for (var i = 0; i < arr.length; i++) {
var num = arr[i];
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
for (let i in counts) {
if (counts[i] === a.length) {
result.push(i)
}
}
console.log(result)edited yesterday
Peter Mortensen
13.8k1987113
answered yesterday
tnkhtnkh
17210
edited yesterday
Peter Mortensen
13.8k1987113
edited yesterday
Peter Mortensen
13.8k1987113
edited yesterday
Peter Mortensen
13.8k1987113
13.8k1987113
answered yesterday
tnkhtnkh
17210
answered yesterday
tnkhtnkh
17210
answered yesterday
tnkhtnkh
17210
17210
add a comment |
add a comment |
Kamoulox is a new contributor. Be nice, and check out our Code of Conduct.
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Kamoulox is a new contributor. Be nice, and check out our Code of Conduct.
Kamoulox is a new contributor. Be nice, and check out our Code of Conduct.
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So get all arrays elements that are present on all
a?– Eddie
yesterday
3
Possible duplicate of Simplest code for array intersection in javascript
– James Long
yesterday
@JamesLong I don't understand the duplication, why this answer can help me ?
– Kamoulox
yesterday
@Eddie I need to get all array into the key data. This key is present into all object of my principal array.
– Kamoulox
yesterday
@Kamoulox you have 2 arrays (a[0].data and a[1].data). You seem to be wanting an array of all the values that appear in both arrays. Unless I'm mis-reading your question
– James Long
yesterday