Merge two arrays of objects by ID
-3
I have two arrays like this:
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
Now I want a result like this:
result = [ {name: "apple", prices: [{id: 1, price: 50}, {id: 1, price: 40}]}, {name: "orange", prices: [{id: 2, price: 30}]}, {name: "others", prices: [{id: null, price: 80}]}]
I want to map the elements of the array a to the name of the second array b on the basis of their ids.
javascript arrays json reactjs
asked Nov 24 '18 at 21:30
Nitesh RanjanNitesh Ranjan
10116
add a comment |
-3
I have two arrays like this:
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
Now I want a result like this:
result = [ {name: "apple", prices: [{id: 1, price: 50}, {id: 1, price: 40}]}, {name: "orange", prices: [{id: 2, price: 30}]}, {name: "others", prices: [{id: null, price: 80}]}]
I want to map the elements of the array a to the name of the second array b on the basis of their ids.
javascript arrays json reactjs
asked Nov 24 '18 at 21:30
Nitesh RanjanNitesh Ranjan
10116
is the result for apple's prices supposed to be[{id: 1, price: 50}, {id: 1, price: 40}]instead?
– Shawn Andrews
Nov 24 '18 at 21:36
this is just an example of the real problem i have. i did give a bad example but yes i want exactly what i wrote.
– Nitesh Ranjan
Nov 24 '18 at 21:37
im confused by the result. you say you wantato map tobon the basis of their ids, but{id: 1, price: 40}inamaps to "apple" but its not in the apple object in your result?
– Shawn Andrews
Nov 24 '18 at 21:45
Why would you want to repeat the id with each price? Certainly the id will be the same for the prices that appear in the same array. It seems more appropriate to avoid this repetition, and put the id at the name level in your output structure.
– trincot
Nov 24 '18 at 22:12
Any feed-back on my comment, and on the answers below?
– trincot
Nov 25 '18 at 19:29
add a comment |
-3
-3
-3
0
I have two arrays like this:
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
Now I want a result like this:
result = [ {name: "apple", prices: [{id: 1, price: 50}, {id: 1, price: 40}]}, {name: "orange", prices: [{id: 2, price: 30}]}, {name: "others", prices: [{id: null, price: 80}]}]
I want to map the elements of the array a to the name of the second array b on the basis of their ids.
javascript arrays json reactjs
asked Nov 24 '18 at 21:30
Nitesh RanjanNitesh Ranjan
10116
I have two arrays like this:
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
Now I want a result like this:
result = [ {name: "apple", prices: [{id: 1, price: 50}, {id: 1, price: 40}]}, {name: "orange", prices: [{id: 2, price: 30}]}, {name: "others", prices: [{id: null, price: 80}]}]
I want to map the elements of the array a to the name of the second array b on the basis of their ids.
javascript arrays json reactjs
javascript arrays json reactjs
asked Nov 24 '18 at 21:30
Nitesh RanjanNitesh Ranjan
10116
asked Nov 24 '18 at 21:30
Nitesh RanjanNitesh Ranjan
10116
edited Nov 25 '18 at 8:22
Nitesh Ranjan
asked Nov 24 '18 at 21:30
Nitesh RanjanNitesh Ranjan
10116
asked Nov 24 '18 at 21:30
Nitesh RanjanNitesh Ranjan
10116
asked Nov 24 '18 at 21:30
Nitesh RanjanNitesh Ranjan
10116
10116
is the result for apple's prices supposed to be[{id: 1, price: 50}, {id: 1, price: 40}]instead?
– Shawn Andrews
Nov 24 '18 at 21:36
this is just an example of the real problem i have. i did give a bad example but yes i want exactly what i wrote.
– Nitesh Ranjan
Nov 24 '18 at 21:37
im confused by the result. you say you wantato map tobon the basis of their ids, but{id: 1, price: 40}inamaps to "apple" but its not in the apple object in your result?
– Shawn Andrews
Nov 24 '18 at 21:45
Why would you want to repeat the id with each price? Certainly the id will be the same for the prices that appear in the same array. It seems more appropriate to avoid this repetition, and put the id at the name level in your output structure.
– trincot
Nov 24 '18 at 22:12
Any feed-back on my comment, and on the answers below?
– trincot
Nov 25 '18 at 19:29
add a comment |
is the result for apple's prices supposed to be[{id: 1, price: 50}, {id: 1, price: 40}]instead?
– Shawn Andrews
Nov 24 '18 at 21:36
this is just an example of the real problem i have. i did give a bad example but yes i want exactly what i wrote.
– Nitesh Ranjan
Nov 24 '18 at 21:37
im confused by the result. you say you wantato map tobon the basis of their ids, but{id: 1, price: 40}inamaps to "apple" but its not in the apple object in your result?
– Shawn Andrews
Nov 24 '18 at 21:45
Why would you want to repeat the id with each price? Certainly the id will be the same for the prices that appear in the same array. It seems more appropriate to avoid this repetition, and put the id at the name level in your output structure.
– trincot
Nov 24 '18 at 22:12
Any feed-back on my comment, and on the answers below?
– trincot
Nov 25 '18 at 19:29
is the result for apple's prices supposed to be
[{id: 1, price: 50}, {id: 1, price: 40}] instead?– Shawn Andrews
Nov 24 '18 at 21:36
is the result for apple's prices supposed to be
[{id: 1, price: 50}, {id: 1, price: 40}] instead?– Shawn Andrews
Nov 24 '18 at 21:36
this is just an example of the real problem i have. i did give a bad example but yes i want exactly what i wrote.
– Nitesh Ranjan
Nov 24 '18 at 21:37
this is just an example of the real problem i have. i did give a bad example but yes i want exactly what i wrote.
– Nitesh Ranjan
Nov 24 '18 at 21:37
im confused by the result. you say you want
a to map to b on the basis of their ids, but {id: 1, price: 40} in a maps to "apple" but its not in the apple object in your result?– Shawn Andrews
Nov 24 '18 at 21:45
im confused by the result. you say you want
a to map to b on the basis of their ids, but {id: 1, price: 40} in a maps to "apple" but its not in the apple object in your result?– Shawn Andrews
Nov 24 '18 at 21:45
Why would you want to repeat the id with each price? Certainly the id will be the same for the prices that appear in the same array. It seems more appropriate to avoid this repetition, and put the id at the name level in your output structure.
– trincot
Nov 24 '18 at 22:12
Why would you want to repeat the id with each price? Certainly the id will be the same for the prices that appear in the same array. It seems more appropriate to avoid this repetition, and put the id at the name level in your output structure.
– trincot
Nov 24 '18 at 22:12
Any feed-back on my comment, and on the answers below?
– trincot
Nov 25 '18 at 19:29
Any feed-back on my comment, and on the answers below?
– trincot
Nov 25 '18 at 19:29
add a comment |
4 Answers
4
active
oldest
votes
1
Here's an approach that using reduce to build a lookup set and avoid repeated searches in b. Another reduction pass builds the result arrays by name using the lookup table. Lastly, map is used to format the result.
Time complexity is linear (three passes with a lot of constant time object lookups).
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const lookup = b.reduce((a, e) => {
a[e.id] = e.name;
return a;
}, {});
const result = Object.entries(
a.reduce((a, e) => {
const key = lookup[e.id] || "others";
if (!(key in a)) {
a[key] = ;
}
a[key].push(e);
return a;
}, {})
).map(e => ({name: e[0], prices: e[1]}));
console.log(result);answered Nov 24 '18 at 21:52
ggorlenggorlen
7,1883825
add a comment |
1
It would be more logical to not repeat the id in the prices part of the result, since the id belongs with the name.
I would suggest using a temporary map (for efficiency):
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const map = new Map(b.map(o => [o.id, Object.assign(o, { prices: })]))
.set(null, {id: null, name: "others", prices: });
a.forEach(o => map.get(o.id).prices.push(o.price));
const result = [...map.values()];
console.log(result);answered Nov 24 '18 at 22:11
trincottrincot
125k1588121
add a comment |
1
Yes, you can do it simply with a map and a filter
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
b.map(({ name, id }) => ({
name,
id,
prices: a.filter(item => item.id === id).map(({ price }) => price)
}));
answered Nov 24 '18 at 22:12
Andy RayAndy Ray
17.7k76298
1
map*filter= O(n²).
– trincot
Nov 24 '18 at 22:13
add a comment |
1
You can do this with a single Array.reduce and Object.values if you start by combining the 2 arrays together:
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const result = Object.values([...b, ...a].reduce((r, c) => {
if ('name' in c || c.id == null)
r[c.id || 'others'] = ({name: c.name || 'others', prices: })
if ('name' in c)
return r
else if (c.id != null)
r[c.id].prices.push(c)
else
r['others'].prices.push(c)
return r
}, {}))
console.log(result)The idea is to start with the one containing the names so the grouping creates first the object groupings and then just fill the group arrays.
answered Nov 25 '18 at 0:21
AkrionAkrion
9,48511224
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53462526%2fmerge-two-arrays-of-objects-by-id%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Here's an approach that using reduce to build a lookup set and avoid repeated searches in b. Another reduction pass builds the result arrays by name using the lookup table. Lastly, map is used to format the result.
Time complexity is linear (three passes with a lot of constant time object lookups).
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const lookup = b.reduce((a, e) => {
a[e.id] = e.name;
return a;
}, {});
const result = Object.entries(
a.reduce((a, e) => {
const key = lookup[e.id] || "others";
if (!(key in a)) {
a[key] = ;
}
a[key].push(e);
return a;
}, {})
).map(e => ({name: e[0], prices: e[1]}));
console.log(result);answered Nov 24 '18 at 21:52
ggorlenggorlen
7,1883825
add a comment |
1
Here's an approach that using reduce to build a lookup set and avoid repeated searches in b. Another reduction pass builds the result arrays by name using the lookup table. Lastly, map is used to format the result.
Time complexity is linear (three passes with a lot of constant time object lookups).
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const lookup = b.reduce((a, e) => {
a[e.id] = e.name;
return a;
}, {});
const result = Object.entries(
a.reduce((a, e) => {
const key = lookup[e.id] || "others";
if (!(key in a)) {
a[key] = ;
}
a[key].push(e);
return a;
}, {})
).map(e => ({name: e[0], prices: e[1]}));
console.log(result);answered Nov 24 '18 at 21:52
ggorlenggorlen
7,1883825
add a comment |
1
1
1
Here's an approach that using reduce to build a lookup set and avoid repeated searches in b. Another reduction pass builds the result arrays by name using the lookup table. Lastly, map is used to format the result.
Time complexity is linear (three passes with a lot of constant time object lookups).
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const lookup = b.reduce((a, e) => {
a[e.id] = e.name;
return a;
}, {});
const result = Object.entries(
a.reduce((a, e) => {
const key = lookup[e.id] || "others";
if (!(key in a)) {
a[key] = ;
}
a[key].push(e);
return a;
}, {})
).map(e => ({name: e[0], prices: e[1]}));
console.log(result);answered Nov 24 '18 at 21:52
ggorlenggorlen
7,1883825
Here's an approach that using reduce to build a lookup set and avoid repeated searches in b. Another reduction pass builds the result arrays by name using the lookup table. Lastly, map is used to format the result.
Time complexity is linear (three passes with a lot of constant time object lookups).
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const lookup = b.reduce((a, e) => {
a[e.id] = e.name;
return a;
}, {});
const result = Object.entries(
a.reduce((a, e) => {
const key = lookup[e.id] || "others";
if (!(key in a)) {
a[key] = ;
}
a[key].push(e);
return a;
}, {})
).map(e => ({name: e[0], prices: e[1]}));
console.log(result);let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const lookup = b.reduce((a, e) => {
a[e.id] = e.name;
return a;
}, {});
const result = Object.entries(
a.reduce((a, e) => {
const key = lookup[e.id] || "others";
if (!(key in a)) {
a[key] = ;
}
a[key].push(e);
return a;
}, {})
).map(e => ({name: e[0], prices: e[1]}));
console.log(result);let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const lookup = b.reduce((a, e) => {
a[e.id] = e.name;
return a;
}, {});
const result = Object.entries(
a.reduce((a, e) => {
const key = lookup[e.id] || "others";
if (!(key in a)) {
a[key] = ;
}
a[key].push(e);
return a;
}, {})
).map(e => ({name: e[0], prices: e[1]}));
console.log(result);answered Nov 24 '18 at 21:52
ggorlenggorlen
7,1883825
edited Nov 24 '18 at 22:03
answered Nov 24 '18 at 21:52
ggorlenggorlen
7,1883825
answered Nov 24 '18 at 21:52
ggorlenggorlen
7,1883825
answered Nov 24 '18 at 21:52
ggorlenggorlen
7,1883825
7,1883825
add a comment |
add a comment |
1
It would be more logical to not repeat the id in the prices part of the result, since the id belongs with the name.
I would suggest using a temporary map (for efficiency):
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const map = new Map(b.map(o => [o.id, Object.assign(o, { prices: })]))
.set(null, {id: null, name: "others", prices: });
a.forEach(o => map.get(o.id).prices.push(o.price));
const result = [...map.values()];
console.log(result);answered Nov 24 '18 at 22:11
trincottrincot
125k1588121
add a comment |
1
It would be more logical to not repeat the id in the prices part of the result, since the id belongs with the name.
I would suggest using a temporary map (for efficiency):
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const map = new Map(b.map(o => [o.id, Object.assign(o, { prices: })]))
.set(null, {id: null, name: "others", prices: });
a.forEach(o => map.get(o.id).prices.push(o.price));
const result = [...map.values()];
console.log(result);answered Nov 24 '18 at 22:11
trincottrincot
125k1588121
add a comment |
1
1
1
It would be more logical to not repeat the id in the prices part of the result, since the id belongs with the name.
I would suggest using a temporary map (for efficiency):
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const map = new Map(b.map(o => [o.id, Object.assign(o, { prices: })]))
.set(null, {id: null, name: "others", prices: });
a.forEach(o => map.get(o.id).prices.push(o.price));
const result = [...map.values()];
console.log(result);answered Nov 24 '18 at 22:11
trincottrincot
125k1588121
It would be more logical to not repeat the id in the prices part of the result, since the id belongs with the name.
I would suggest using a temporary map (for efficiency):
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const map = new Map(b.map(o => [o.id, Object.assign(o, { prices: })]))
.set(null, {id: null, name: "others", prices: });
a.forEach(o => map.get(o.id).prices.push(o.price));
const result = [...map.values()];
console.log(result);let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const map = new Map(b.map(o => [o.id, Object.assign(o, { prices: })]))
.set(null, {id: null, name: "others", prices: });
a.forEach(o => map.get(o.id).prices.push(o.price));
const result = [...map.values()];
console.log(result);let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const map = new Map(b.map(o => [o.id, Object.assign(o, { prices: })]))
.set(null, {id: null, name: "others", prices: });
a.forEach(o => map.get(o.id).prices.push(o.price));
const result = [...map.values()];
console.log(result);answered Nov 24 '18 at 22:11
trincottrincot
125k1588121
answered Nov 24 '18 at 22:11
trincottrincot
125k1588121
answered Nov 24 '18 at 22:11
trincottrincot
125k1588121
answered Nov 24 '18 at 22:11
trincottrincot
125k1588121
125k1588121
add a comment |
add a comment |
1
Yes, you can do it simply with a map and a filter
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
b.map(({ name, id }) => ({
name,
id,
prices: a.filter(item => item.id === id).map(({ price }) => price)
}));
answered Nov 24 '18 at 22:12
Andy RayAndy Ray
17.7k76298
1
map*filter= O(n²).
– trincot
Nov 24 '18 at 22:13
add a comment |
1
Yes, you can do it simply with a map and a filter
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
b.map(({ name, id }) => ({
name,
id,
prices: a.filter(item => item.id === id).map(({ price }) => price)
}));
answered Nov 24 '18 at 22:12
Andy RayAndy Ray
17.7k76298
1
map*filter= O(n²).
– trincot
Nov 24 '18 at 22:13
add a comment |
1
1
1
Yes, you can do it simply with a map and a filter
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
b.map(({ name, id }) => ({
name,
id,
prices: a.filter(item => item.id === id).map(({ price }) => price)
}));
answered Nov 24 '18 at 22:12
Andy RayAndy Ray
17.7k76298
Yes, you can do it simply with a map and a filter
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
b.map(({ name, id }) => ({
name,
id,
prices: a.filter(item => item.id === id).map(({ price }) => price)
}));
answered Nov 24 '18 at 22:12
Andy RayAndy Ray
17.7k76298
answered Nov 24 '18 at 22:12
Andy RayAndy Ray
17.7k76298
answered Nov 24 '18 at 22:12
Andy RayAndy Ray
17.7k76298
answered Nov 24 '18 at 22:12
Andy RayAndy Ray
17.7k76298
17.7k76298
1
map*filter= O(n²).
– trincot
Nov 24 '18 at 22:13
add a comment |
1
map*filter= O(n²).
– trincot
Nov 24 '18 at 22:13
1
1
map*filter = O(n²).– trincot
Nov 24 '18 at 22:13
map*filter = O(n²).– trincot
Nov 24 '18 at 22:13
add a comment |
1
You can do this with a single Array.reduce and Object.values if you start by combining the 2 arrays together:
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const result = Object.values([...b, ...a].reduce((r, c) => {
if ('name' in c || c.id == null)
r[c.id || 'others'] = ({name: c.name || 'others', prices: })
if ('name' in c)
return r
else if (c.id != null)
r[c.id].prices.push(c)
else
r['others'].prices.push(c)
return r
}, {}))
console.log(result)The idea is to start with the one containing the names so the grouping creates first the object groupings and then just fill the group arrays.
answered Nov 25 '18 at 0:21
AkrionAkrion
9,48511224
add a comment |
1
You can do this with a single Array.reduce and Object.values if you start by combining the 2 arrays together:
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const result = Object.values([...b, ...a].reduce((r, c) => {
if ('name' in c || c.id == null)
r[c.id || 'others'] = ({name: c.name || 'others', prices: })
if ('name' in c)
return r
else if (c.id != null)
r[c.id].prices.push(c)
else
r['others'].prices.push(c)
return r
}, {}))
console.log(result)The idea is to start with the one containing the names so the grouping creates first the object groupings and then just fill the group arrays.
answered Nov 25 '18 at 0:21
AkrionAkrion
9,48511224
add a comment |
1
1
1
You can do this with a single Array.reduce and Object.values if you start by combining the 2 arrays together:
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const result = Object.values([...b, ...a].reduce((r, c) => {
if ('name' in c || c.id == null)
r[c.id || 'others'] = ({name: c.name || 'others', prices: })
if ('name' in c)
return r
else if (c.id != null)
r[c.id].prices.push(c)
else
r['others'].prices.push(c)
return r
}, {}))
console.log(result)The idea is to start with the one containing the names so the grouping creates first the object groupings and then just fill the group arrays.
answered Nov 25 '18 at 0:21
AkrionAkrion
9,48511224
You can do this with a single Array.reduce and Object.values if you start by combining the 2 arrays together:
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const result = Object.values([...b, ...a].reduce((r, c) => {
if ('name' in c || c.id == null)
r[c.id || 'others'] = ({name: c.name || 'others', prices: })
if ('name' in c)
return r
else if (c.id != null)
r[c.id].prices.push(c)
else
r['others'].prices.push(c)
return r
}, {}))
console.log(result)The idea is to start with the one containing the names so the grouping creates first the object groupings and then just fill the group arrays.
let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const result = Object.values([...b, ...a].reduce((r, c) => {
if ('name' in c || c.id == null)
r[c.id || 'others'] = ({name: c.name || 'others', prices: })
if ('name' in c)
return r
else if (c.id != null)
r[c.id].prices.push(c)
else
r['others'].prices.push(c)
return r
}, {}))
console.log(result)let a = [{id: 1, price: 50}, {id: 2, price: 30}, {id: 1, price: 40}, {id: null, price: 80}];
let b = [{id: 1, name: "apple"}, {id: 2, name: "orange"}];
const result = Object.values([...b, ...a].reduce((r, c) => {
if ('name' in c || c.id == null)
r[c.id || 'others'] = ({name: c.name || 'others', prices: })
if ('name' in c)
return r
else if (c.id != null)
r[c.id].prices.push(c)
else
r['others'].prices.push(c)
return r
}, {}))
console.log(result)answered Nov 25 '18 at 0:21
AkrionAkrion
9,48511224
edited Nov 25 '18 at 2:11
answered Nov 25 '18 at 0:21
AkrionAkrion
9,48511224
answered Nov 25 '18 at 0:21
AkrionAkrion
9,48511224
answered Nov 25 '18 at 0:21
AkrionAkrion
9,48511224
9,48511224
add a comment |
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53462526%2fmerge-two-arrays-of-objects-by-id%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
is the result for apple's prices supposed to be
[{id: 1, price: 50}, {id: 1, price: 40}]instead?– Shawn Andrews
Nov 24 '18 at 21:36
this is just an example of the real problem i have. i did give a bad example but yes i want exactly what i wrote.
– Nitesh Ranjan
Nov 24 '18 at 21:37
im confused by the result. you say you want
ato map tobon the basis of their ids, but{id: 1, price: 40}inamaps to "apple" but its not in the apple object in your result?– Shawn Andrews
Nov 24 '18 at 21:45
Why would you want to repeat the id with each price? Certainly the id will be the same for the prices that appear in the same array. It seems more appropriate to avoid this repetition, and put the id at the name level in your output structure.
– trincot
Nov 24 '18 at 22:12
Any feed-back on my comment, and on the answers below?
– trincot
Nov 25 '18 at 19:29