Grouping elements in array by multiple properties underscore
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0
down vote
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During work, I was given this task : to group elements with similar properties in the array.
In general, the problem is as follows :
var order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
]
If I group this elements by 'ru', 'area', 'unit', 'tot', 'equipment' and 'parameter', I will get this result :
var result = [
{
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"tot":2,
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"tot":1,
"equipment":"37 P 552 A",
"parameter":"Discharge Pressure"
}, {
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"tot":1,
"equipment":"37 P 552 B",
"parameter":"Discharge Pressure"
}];
After some experimentation, I came to the following code :
var groups = _.groupBy(order, function(value) {
return value.ru + "#" + value.area + "#" + value.unit + "#" + value.equipment + "#" + value.parameter + "#";
});
groups = _.map(groups, function(group) {
return _.extend(group[0], {tot: group.length});
});
Now I have problem how to group the similar result elements. If there is someone around here, please help me. Thanks in advance.
javascript arrays json underscore.js
asked Nov 9 at 3:53
Hamzah Aznageel
198
add a comment |
up vote
0
down vote
favorite
1
During work, I was given this task : to group elements with similar properties in the array.
In general, the problem is as follows :
var order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
]
If I group this elements by 'ru', 'area', 'unit', 'tot', 'equipment' and 'parameter', I will get this result :
var result = [
{
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"tot":2,
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"tot":1,
"equipment":"37 P 552 A",
"parameter":"Discharge Pressure"
}, {
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"tot":1,
"equipment":"37 P 552 B",
"parameter":"Discharge Pressure"
}];
After some experimentation, I came to the following code :
var groups = _.groupBy(order, function(value) {
return value.ru + "#" + value.area + "#" + value.unit + "#" + value.equipment + "#" + value.parameter + "#";
});
groups = _.map(groups, function(group) {
return _.extend(group[0], {tot: group.length});
});
Now I have problem how to group the similar result elements. If there is someone around here, please help me. Thanks in advance.
javascript arrays json underscore.js
asked Nov 9 at 3:53
Hamzah Aznageel
198
What do you mean by similar result elements?
– Andreas
Nov 9 at 3:59
@Andreas how to group the result data above, because the result have some similar elements.
– Hamzah Aznageel
Nov 9 at 4:01
Then, what is the problem with your current code?
– Andreas
Nov 9 at 4:05
This solution works, but is this a right and best way? It still looks a little ugly to me.
– Hamzah Aznageel
Nov 9 at 4:11
1
Please edit your question accordingly, because currently you are saying that you have a problem with the grouping, not with the optimizing
– Andreas
Nov 9 at 4:15
add a comment |
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
During work, I was given this task : to group elements with similar properties in the array.
In general, the problem is as follows :
var order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
]
If I group this elements by 'ru', 'area', 'unit', 'tot', 'equipment' and 'parameter', I will get this result :
var result = [
{
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"tot":2,
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"tot":1,
"equipment":"37 P 552 A",
"parameter":"Discharge Pressure"
}, {
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"tot":1,
"equipment":"37 P 552 B",
"parameter":"Discharge Pressure"
}];
After some experimentation, I came to the following code :
var groups = _.groupBy(order, function(value) {
return value.ru + "#" + value.area + "#" + value.unit + "#" + value.equipment + "#" + value.parameter + "#";
});
groups = _.map(groups, function(group) {
return _.extend(group[0], {tot: group.length});
});
Now I have problem how to group the similar result elements. If there is someone around here, please help me. Thanks in advance.
javascript arrays json underscore.js
asked Nov 9 at 3:53
Hamzah Aznageel
198
During work, I was given this task : to group elements with similar properties in the array.
In general, the problem is as follows :
var order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
]
If I group this elements by 'ru', 'area', 'unit', 'tot', 'equipment' and 'parameter', I will get this result :
var result = [
{
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"tot":2,
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"tot":1,
"equipment":"37 P 552 A",
"parameter":"Discharge Pressure"
}, {
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"tot":1,
"equipment":"37 P 552 B",
"parameter":"Discharge Pressure"
}];
After some experimentation, I came to the following code :
var groups = _.groupBy(order, function(value) {
return value.ru + "#" + value.area + "#" + value.unit + "#" + value.equipment + "#" + value.parameter + "#";
});
groups = _.map(groups, function(group) {
return _.extend(group[0], {tot: group.length});
});
Now I have problem how to group the similar result elements. If there is someone around here, please help me. Thanks in advance.
javascript arrays json underscore.js
javascript arrays json underscore.js
asked Nov 9 at 3:53
Hamzah Aznageel
198
asked Nov 9 at 3:53
Hamzah Aznageel
198
edited Nov 9 at 4:01
asked Nov 9 at 3:53
Hamzah Aznageel
198
asked Nov 9 at 3:53
Hamzah Aznageel
198
asked Nov 9 at 3:53
Hamzah Aznageel
198
198
What do you mean by similar result elements?
– Andreas
Nov 9 at 3:59
@Andreas how to group the result data above, because the result have some similar elements.
– Hamzah Aznageel
Nov 9 at 4:01
Then, what is the problem with your current code?
– Andreas
Nov 9 at 4:05
This solution works, but is this a right and best way? It still looks a little ugly to me.
– Hamzah Aznageel
Nov 9 at 4:11
1
Please edit your question accordingly, because currently you are saying that you have a problem with the grouping, not with the optimizing
– Andreas
Nov 9 at 4:15
add a comment |
What do you mean by similar result elements?
– Andreas
Nov 9 at 3:59
@Andreas how to group the result data above, because the result have some similar elements.
– Hamzah Aznageel
Nov 9 at 4:01
Then, what is the problem with your current code?
– Andreas
Nov 9 at 4:05
This solution works, but is this a right and best way? It still looks a little ugly to me.
– Hamzah Aznageel
Nov 9 at 4:11
1
Please edit your question accordingly, because currently you are saying that you have a problem with the grouping, not with the optimizing
– Andreas
Nov 9 at 4:15
What do you mean by similar result elements?
– Andreas
Nov 9 at 3:59
What do you mean by similar result elements?
– Andreas
Nov 9 at 3:59
@Andreas how to group the result data above, because the result have some similar elements.
– Hamzah Aznageel
Nov 9 at 4:01
@Andreas how to group the result data above, because the result have some similar elements.
– Hamzah Aznageel
Nov 9 at 4:01
Then, what is the problem with your current code?
– Andreas
Nov 9 at 4:05
Then, what is the problem with your current code?
– Andreas
Nov 9 at 4:05
This solution works, but is this a right and best way? It still looks a little ugly to me.
– Hamzah Aznageel
Nov 9 at 4:11
This solution works, but is this a right and best way? It still looks a little ugly to me.
– Hamzah Aznageel
Nov 9 at 4:11
1
1
Please edit your question accordingly, because currently you are saying that you have a problem with the grouping, not with the optimizing
– Andreas
Nov 9 at 4:15
Please edit your question accordingly, because currently you are saying that you have a problem with the grouping, not with the optimizing
– Andreas
Nov 9 at 4:15
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
How about this:
var order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
];
var result = ;
order.forEach(current=>{
let index = -1;
result.forEach((c,i)=>{
if (c.ru==current.ru && c.area==current.area && c.unit==current.unit && c.tot==current.tot && c.equipment==current.equipment && c.parameter==current.parameter){
index = i;
}
});
// console.log(index);
if(index==-1){
result.push(current);
}else{
result[index]["tot"] = result[index]["tot"]+current["tot"];
}
});
console.log(result);answered Nov 9 at 4:29
protoproto
90069
how to get specific object for example 'ru'?
– Hamzah Aznageel
Nov 9 at 4:41
Get result just has only "ru"?
– protoproto
Nov 9 at 4:44
yes, get result only "ru", I think just console.log(result.ru), but the result is undefined
– Hamzah Aznageel
Nov 9 at 4:55
You can use:console.log(order.map(current=>current["ru"]));--> get only column 'ru'!
– protoproto
Nov 9 at 5:03
add a comment |
up vote
0
down vote
Maybe you can avoid grouping and directly update the tot property of the merged object in your "index" map/object if you encounter similar elements:
Here is a solution that does that with reduce:
const order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
]
const merged = Object.values(order.reduce((acc, curr) => {
const key = `${curr.ru}#${curr.area}#${curr.unit}#${curr.equipment}#${curr.parameter}`;
if (key in acc) acc[key].tot += 1;
else acc[key] = Object.assign({}, curr);
return acc;
}, {}));
console.log(merged);
console.log(merged.map(d => d.ru));answered Nov 9 at 4:41
slider
6,5051129
how to get all "ru" objects in array?
– Hamzah Aznageel
Nov 9 at 5:01
@HamzahAznageel it's not that complicated. As @protoproto said, you can usemap(see updated answer).merged.map(d => d.ru)
– slider
Nov 9 at 5:08
well, okay thank you so much
– Hamzah Aznageel
Nov 9 at 6:11
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
How about this:
var order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
];
var result = ;
order.forEach(current=>{
let index = -1;
result.forEach((c,i)=>{
if (c.ru==current.ru && c.area==current.area && c.unit==current.unit && c.tot==current.tot && c.equipment==current.equipment && c.parameter==current.parameter){
index = i;
}
});
// console.log(index);
if(index==-1){
result.push(current);
}else{
result[index]["tot"] = result[index]["tot"]+current["tot"];
}
});
console.log(result);answered Nov 9 at 4:29
protoproto
90069
how to get specific object for example 'ru'?
– Hamzah Aznageel
Nov 9 at 4:41
Get result just has only "ru"?
– protoproto
Nov 9 at 4:44
yes, get result only "ru", I think just console.log(result.ru), but the result is undefined
– Hamzah Aznageel
Nov 9 at 4:55
You can use:console.log(order.map(current=>current["ru"]));--> get only column 'ru'!
– protoproto
Nov 9 at 5:03
add a comment |
up vote
0
down vote
accepted
How about this:
var order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
];
var result = ;
order.forEach(current=>{
let index = -1;
result.forEach((c,i)=>{
if (c.ru==current.ru && c.area==current.area && c.unit==current.unit && c.tot==current.tot && c.equipment==current.equipment && c.parameter==current.parameter){
index = i;
}
});
// console.log(index);
if(index==-1){
result.push(current);
}else{
result[index]["tot"] = result[index]["tot"]+current["tot"];
}
});
console.log(result);answered Nov 9 at 4:29
protoproto
90069
how to get specific object for example 'ru'?
– Hamzah Aznageel
Nov 9 at 4:41
Get result just has only "ru"?
– protoproto
Nov 9 at 4:44
yes, get result only "ru", I think just console.log(result.ru), but the result is undefined
– Hamzah Aznageel
Nov 9 at 4:55
You can use:console.log(order.map(current=>current["ru"]));--> get only column 'ru'!
– protoproto
Nov 9 at 5:03
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
How about this:
var order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
];
var result = ;
order.forEach(current=>{
let index = -1;
result.forEach((c,i)=>{
if (c.ru==current.ru && c.area==current.area && c.unit==current.unit && c.tot==current.tot && c.equipment==current.equipment && c.parameter==current.parameter){
index = i;
}
});
// console.log(index);
if(index==-1){
result.push(current);
}else{
result[index]["tot"] = result[index]["tot"]+current["tot"];
}
});
console.log(result);answered Nov 9 at 4:29
protoproto
90069
How about this:
var order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
];
var result = ;
order.forEach(current=>{
let index = -1;
result.forEach((c,i)=>{
if (c.ru==current.ru && c.area==current.area && c.unit==current.unit && c.tot==current.tot && c.equipment==current.equipment && c.parameter==current.parameter){
index = i;
}
});
// console.log(index);
if(index==-1){
result.push(current);
}else{
result[index]["tot"] = result[index]["tot"]+current["tot"];
}
});
console.log(result);var order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
];
var result = ;
order.forEach(current=>{
let index = -1;
result.forEach((c,i)=>{
if (c.ru==current.ru && c.area==current.area && c.unit==current.unit && c.tot==current.tot && c.equipment==current.equipment && c.parameter==current.parameter){
index = i;
}
});
// console.log(index);
if(index==-1){
result.push(current);
}else{
result[index]["tot"] = result[index]["tot"]+current["tot"];
}
});
console.log(result);var order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
];
var result = ;
order.forEach(current=>{
let index = -1;
result.forEach((c,i)=>{
if (c.ru==current.ru && c.area==current.area && c.unit==current.unit && c.tot==current.tot && c.equipment==current.equipment && c.parameter==current.parameter){
index = i;
}
});
// console.log(index);
if(index==-1){
result.push(current);
}else{
result[index]["tot"] = result[index]["tot"]+current["tot"];
}
});
console.log(result);answered Nov 9 at 4:29
protoproto
90069
answered Nov 9 at 4:29
protoproto
90069
answered Nov 9 at 4:29
protoproto
90069
answered Nov 9 at 4:29
protoproto
90069
90069
how to get specific object for example 'ru'?
– Hamzah Aznageel
Nov 9 at 4:41
Get result just has only "ru"?
– protoproto
Nov 9 at 4:44
yes, get result only "ru", I think just console.log(result.ru), but the result is undefined
– Hamzah Aznageel
Nov 9 at 4:55
You can use:console.log(order.map(current=>current["ru"]));--> get only column 'ru'!
– protoproto
Nov 9 at 5:03
add a comment |
how to get specific object for example 'ru'?
– Hamzah Aznageel
Nov 9 at 4:41
Get result just has only "ru"?
– protoproto
Nov 9 at 4:44
yes, get result only "ru", I think just console.log(result.ru), but the result is undefined
– Hamzah Aznageel
Nov 9 at 4:55
You can use:console.log(order.map(current=>current["ru"]));--> get only column 'ru'!
– protoproto
Nov 9 at 5:03
how to get specific object for example 'ru'?
– Hamzah Aznageel
Nov 9 at 4:41
how to get specific object for example 'ru'?
– Hamzah Aznageel
Nov 9 at 4:41
Get result just has only "ru"?
– protoproto
Nov 9 at 4:44
Get result just has only "ru"?
– protoproto
Nov 9 at 4:44
yes, get result only "ru", I think just console.log(result.ru), but the result is undefined
– Hamzah Aznageel
Nov 9 at 4:55
yes, get result only "ru", I think just console.log(result.ru), but the result is undefined
– Hamzah Aznageel
Nov 9 at 4:55
You can use:
console.log(order.map(current=>current["ru"])); --> get only column 'ru'!– protoproto
Nov 9 at 5:03
You can use:
console.log(order.map(current=>current["ru"])); --> get only column 'ru'!– protoproto
Nov 9 at 5:03
add a comment |
up vote
0
down vote
Maybe you can avoid grouping and directly update the tot property of the merged object in your "index" map/object if you encounter similar elements:
Here is a solution that does that with reduce:
const order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
]
const merged = Object.values(order.reduce((acc, curr) => {
const key = `${curr.ru}#${curr.area}#${curr.unit}#${curr.equipment}#${curr.parameter}`;
if (key in acc) acc[key].tot += 1;
else acc[key] = Object.assign({}, curr);
return acc;
}, {}));
console.log(merged);
console.log(merged.map(d => d.ru));answered Nov 9 at 4:41
slider
6,5051129
how to get all "ru" objects in array?
– Hamzah Aznageel
Nov 9 at 5:01
@HamzahAznageel it's not that complicated. As @protoproto said, you can usemap(see updated answer).merged.map(d => d.ru)
– slider
Nov 9 at 5:08
well, okay thank you so much
– Hamzah Aznageel
Nov 9 at 6:11
add a comment |
up vote
0
down vote
Maybe you can avoid grouping and directly update the tot property of the merged object in your "index" map/object if you encounter similar elements:
Here is a solution that does that with reduce:
const order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
]
const merged = Object.values(order.reduce((acc, curr) => {
const key = `${curr.ru}#${curr.area}#${curr.unit}#${curr.equipment}#${curr.parameter}`;
if (key in acc) acc[key].tot += 1;
else acc[key] = Object.assign({}, curr);
return acc;
}, {}));
console.log(merged);
console.log(merged.map(d => d.ru));answered Nov 9 at 4:41
slider
6,5051129
how to get all "ru" objects in array?
– Hamzah Aznageel
Nov 9 at 5:01
@HamzahAznageel it's not that complicated. As @protoproto said, you can usemap(see updated answer).merged.map(d => d.ru)
– slider
Nov 9 at 5:08
well, okay thank you so much
– Hamzah Aznageel
Nov 9 at 6:11
add a comment |
up vote
0
down vote
up vote
0
down vote
Maybe you can avoid grouping and directly update the tot property of the merged object in your "index" map/object if you encounter similar elements:
Here is a solution that does that with reduce:
const order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
]
const merged = Object.values(order.reduce((acc, curr) => {
const key = `${curr.ru}#${curr.area}#${curr.unit}#${curr.equipment}#${curr.parameter}`;
if (key in acc) acc[key].tot += 1;
else acc[key] = Object.assign({}, curr);
return acc;
}, {}));
console.log(merged);
console.log(merged.map(d => d.ru));answered Nov 9 at 4:41
slider
6,5051129
Maybe you can avoid grouping and directly update the tot property of the merged object in your "index" map/object if you encounter similar elements:
Here is a solution that does that with reduce:
const order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
]
const merged = Object.values(order.reduce((acc, curr) => {
const key = `${curr.ru}#${curr.area}#${curr.unit}#${curr.equipment}#${curr.parameter}`;
if (key in acc) acc[key].tot += 1;
else acc[key] = Object.assign({}, curr);
return acc;
}, {}));
console.log(merged);
console.log(merged.map(d => d.ru));const order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
]
const merged = Object.values(order.reduce((acc, curr) => {
const key = `${curr.ru}#${curr.area}#${curr.unit}#${curr.equipment}#${curr.parameter}`;
if (key in acc) acc[key].tot += 1;
else acc[key] = Object.assign({}, curr);
return acc;
}, {}));
console.log(merged);
console.log(merged.map(d => d.ru));const order = [
{
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Discharge pressure"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"OFFSITE",
"equipment":"37 P 552 A",
"parameter":"Speed"
}, {
"tot":1,
"ru":"R401",
"area":"RFCC",
"unit":"RCU",
"equipment":"37 P 552 B",
"parameter":"Discharge pressure"
}
]
const merged = Object.values(order.reduce((acc, curr) => {
const key = `${curr.ru}#${curr.area}#${curr.unit}#${curr.equipment}#${curr.parameter}`;
if (key in acc) acc[key].tot += 1;
else acc[key] = Object.assign({}, curr);
return acc;
}, {}));
console.log(merged);
console.log(merged.map(d => d.ru));answered Nov 9 at 4:41
slider
6,5051129
edited Nov 9 at 5:07
answered Nov 9 at 4:41
slider
6,5051129
answered Nov 9 at 4:41
slider
6,5051129
answered Nov 9 at 4:41
slider
6,5051129
6,5051129
how to get all "ru" objects in array?
– Hamzah Aznageel
Nov 9 at 5:01
@HamzahAznageel it's not that complicated. As @protoproto said, you can usemap(see updated answer).merged.map(d => d.ru)
– slider
Nov 9 at 5:08
well, okay thank you so much
– Hamzah Aznageel
Nov 9 at 6:11
add a comment |
how to get all "ru" objects in array?
– Hamzah Aznageel
Nov 9 at 5:01
@HamzahAznageel it's not that complicated. As @protoproto said, you can usemap(see updated answer).merged.map(d => d.ru)
– slider
Nov 9 at 5:08
well, okay thank you so much
– Hamzah Aznageel
Nov 9 at 6:11
how to get all "ru" objects in array?
– Hamzah Aznageel
Nov 9 at 5:01
how to get all "ru" objects in array?
– Hamzah Aznageel
Nov 9 at 5:01
@HamzahAznageel it's not that complicated. As @protoproto said, you can use
map (see updated answer). merged.map(d => d.ru)– slider
Nov 9 at 5:08
@HamzahAznageel it's not that complicated. As @protoproto said, you can use
map (see updated answer). merged.map(d => d.ru)– slider
Nov 9 at 5:08
well, okay thank you so much
– Hamzah Aznageel
Nov 9 at 6:11
well, okay thank you so much
– Hamzah Aznageel
Nov 9 at 6:11
add a comment |
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What do you mean by similar result elements?
– Andreas
Nov 9 at 3:59
@Andreas how to group the result data above, because the result have some similar elements.
– Hamzah Aznageel
Nov 9 at 4:01
Then, what is the problem with your current code?
– Andreas
Nov 9 at 4:05
This solution works, but is this a right and best way? It still looks a little ugly to me.
– Hamzah Aznageel
Nov 9 at 4:11
1
Please edit your question accordingly, because currently you are saying that you have a problem with the grouping, not with the optimizing
– Andreas
Nov 9 at 4:15