In javascript, need to perform sum of dynamic array
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2
I have a dynamic array and I need to perform a sum of it.
{ClientName: "A", Alex: 2, Da: 0, Cal: 0}
{ClientName: "b", Alex: 0, Da: 0, Cal: 4}
{ClientName: "c", Alex: 1, Da: 0, Cal: 5}
{ClientName: "d", Alex: 2, Da: 0, Cal: 0}
in this array, client name is fixed but other columns like Alex, Da, Cal are dynamic and will vary. i need to create a generic function in typescript where output will be like this.
{ClientName: "Total", Alex: 5, Da: 0, Cal: 9}
javascript
angular edited Jan 4 at 13:24
Nurbol Alpysbayev
4,9091634
asked Jan 4 at 12:49
RajeshRajesh
4017
add a comment |
2
I have a dynamic array and I need to perform a sum of it.
{ClientName: "A", Alex: 2, Da: 0, Cal: 0}
{ClientName: "b", Alex: 0, Da: 0, Cal: 4}
{ClientName: "c", Alex: 1, Da: 0, Cal: 5}
{ClientName: "d", Alex: 2, Da: 0, Cal: 0}
in this array, client name is fixed but other columns like Alex, Da, Cal are dynamic and will vary. i need to create a generic function in typescript where output will be like this.
{ClientName: "Total", Alex: 5, Da: 0, Cal: 9}
javascript
angular edited Jan 4 at 13:24
Nurbol Alpysbayev
4,9091634
asked Jan 4 at 12:49
RajeshRajesh
4017
8
please add an attempt.
– Nina Scholz
Jan 4 at 12:51
1
“Columns are dynamic” means the property names are arbitrary and could be anything? Or is there a certain set of possible column names?
– deceze♦
Jan 4 at 12:57
add a comment |
2
2
2
I have a dynamic array and I need to perform a sum of it.
{ClientName: "A", Alex: 2, Da: 0, Cal: 0}
{ClientName: "b", Alex: 0, Da: 0, Cal: 4}
{ClientName: "c", Alex: 1, Da: 0, Cal: 5}
{ClientName: "d", Alex: 2, Da: 0, Cal: 0}
in this array, client name is fixed but other columns like Alex, Da, Cal are dynamic and will vary. i need to create a generic function in typescript where output will be like this.
{ClientName: "Total", Alex: 5, Da: 0, Cal: 9}
javascript
angular edited Jan 4 at 13:24
Nurbol Alpysbayev
4,9091634
asked Jan 4 at 12:49
RajeshRajesh
4017
I have a dynamic array and I need to perform a sum of it.
{ClientName: "A", Alex: 2, Da: 0, Cal: 0}
{ClientName: "b", Alex: 0, Da: 0, Cal: 4}
{ClientName: "c", Alex: 1, Da: 0, Cal: 5}
{ClientName: "d", Alex: 2, Da: 0, Cal: 0}
in this array, client name is fixed but other columns like Alex, Da, Cal are dynamic and will vary. i need to create a generic function in typescript where output will be like this.
{ClientName: "Total", Alex: 5, Da: 0, Cal: 9}
javascript
angular javascript
angular edited Jan 4 at 13:24
Nurbol Alpysbayev
4,9091634
asked Jan 4 at 12:49
RajeshRajesh
4017
edited Jan 4 at 13:24
Nurbol Alpysbayev
4,9091634
asked Jan 4 at 12:49
RajeshRajesh
4017
edited Jan 4 at 13:24
Nurbol Alpysbayev
4,9091634
edited Jan 4 at 13:24
Nurbol Alpysbayev
4,9091634
edited Jan 4 at 13:24
Nurbol Alpysbayev
4,9091634
4,9091634
asked Jan 4 at 12:49
RajeshRajesh
4017
asked Jan 4 at 12:49
RajeshRajesh
4017
asked Jan 4 at 12:49
RajeshRajesh
4017
4017
8
please add an attempt.
– Nina Scholz
Jan 4 at 12:51
1
“Columns are dynamic” means the property names are arbitrary and could be anything? Or is there a certain set of possible column names?
– deceze♦
Jan 4 at 12:57
add a comment |
8
please add an attempt.
– Nina Scholz
Jan 4 at 12:51
1
“Columns are dynamic” means the property names are arbitrary and could be anything? Or is there a certain set of possible column names?
– deceze♦
Jan 4 at 12:57
8
8
please add an attempt.
– Nina Scholz
Jan 4 at 12:51
please add an attempt.
– Nina Scholz
Jan 4 at 12:51
1
1
“Columns are dynamic” means the property names are arbitrary and could be anything? Or is there a certain set of possible column names?
– deceze♦
Jan 4 at 12:57
“Columns are dynamic” means the property names are arbitrary and could be anything? Or is there a certain set of possible column names?
– deceze♦
Jan 4 at 12:57
add a comment |
5 Answers
5
active
oldest
votes
3
You can try something like this:
Idea:
- Uses
Object.keysto get all enumerable keys. - Loop over them ans set them to default if not found. Then add current object's value.
- Since you need hardcoded value for
ClientName, you can set it either in loop or set it after loop.
const data = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0},
]
const output = data.reduce((acc, item) => {
Object
.keys(item)
.forEach((key) => {
acc[key] = (acc[key] || 0) + item[key]
});
acc.ClientName = 'Total'
return acc;
}, {})
console.log(output)answered Jan 4 at 12:58
RajeshRajesh
16.7k52254
add a comment |
2
You could reduce the wanted values.
var data = [{ ClientName: "A", Alex: 2, Da: 0, Cal: 0 }, { ClientName: "b", Alex: 0, Da: 0, Cal: 4 }, { ClientName: "c", Alex: 1, Da: 0, Cal: 5 }, { ClientName: "d", Alex: 2, Da: 0, Cal: 0 }],
total = data.reduce((a, { ClientName, ...b }) => Object.assign(
{},
a,
{ ClientName: 'total' },
...Object.entries(b).map(([k, v]) => ({ [k]: (a[k] || 0) + v }))
));
console.log(total);answered Jan 4 at 13:13
Nina ScholzNina Scholz
199k15112182
add a comment |
0
Something along these lines should work as long as all keys besides ClientName are defined as numbers
arr.reduce((sumObject, item) => {
Object.keys(item).forEach(key => {
if(key === 'ClientName'){
return false
}
if(!sumObject[key]) {
sumObject[key] = item[key]
} else {
sumObject[key] += item[key]
}
})
return sumObject
}, {ClientName: 'Total'})
answered Jan 4 at 12:59
Dennis RuiterDennis Ruiter
18528
add a comment |
0
You can iterate with reduce to generate single output and to merge the key count you can use Object.keys and map will allow you to iterate over your object keys. Also added one dynamic key Abc: 9 just to verify the result.
const object = [{
ClientName: "A",
Alex: 2,
Da: 0,
Cal: 0
},
{
ClientName: "b",
Alex: 0,
Da: 0,
Cal: 4
},
{
ClientName: "c",
Alex: 1,
Da: 0,
Cal: 5
},
{
ClientName: "d",
Alex: 2,
Da: 0,
Cal: 0,
Abc: 9
},
]
const result = object.reduce((accumulator, currentValue) => {
Object.keys(currentValue).map((indexKey) => {
accumulator[indexKey] = (accumulator[indexKey] || 0) + currentValue[indexKey]
});
accumulator.ClientName = "Total";
return accumulator;
})
console.log(result);answered Jan 4 at 13:29
Rohit.007Rohit.007
1,7892521
Your solution is exactly same as mine. Please check other answers before posting
– Rajesh
Jan 4 at 13:59
Yes I do agree, but It seems similar but not exactly, See in you code there isforEachwhich is less optimized than.mapwhich is being used in my solution. Anyways I have already voted up for your solution.
– Rohit.007
Jan 4 at 14:11
using a map without return statement is so much worse. Also i don't think map is more optimised compared to forEach but that discussion can be done later
– Rajesh
Jan 4 at 14:13
add a comment |
-1
let arr = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0}
];
let output = arr.reduce((curr, next) => {
return {ClientName: "Total", Alex: (curr.Alex + next.Alex ), Da: (curr.Da + next.Da), Cal: (curr.Cal + next.Cal)};
});
console.log(output);
Check this can get the answer simply
answered Jan 4 at 13:07
loghi ahaloghi aha
114
OP is looking for a solutions that handles Dynamic columns
– Rajesh
Jan 4 at 13:09
yes you are right , it shoud not work for dynamic columns.thank you,
– loghi aha
Jan 4 at 13:23
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
3
You can try something like this:
Idea:
- Uses
Object.keysto get all enumerable keys. - Loop over them ans set them to default if not found. Then add current object's value.
- Since you need hardcoded value for
ClientName, you can set it either in loop or set it after loop.
const data = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0},
]
const output = data.reduce((acc, item) => {
Object
.keys(item)
.forEach((key) => {
acc[key] = (acc[key] || 0) + item[key]
});
acc.ClientName = 'Total'
return acc;
}, {})
console.log(output)answered Jan 4 at 12:58
RajeshRajesh
16.7k52254
add a comment |
3
You can try something like this:
Idea:
- Uses
Object.keysto get all enumerable keys. - Loop over them ans set them to default if not found. Then add current object's value.
- Since you need hardcoded value for
ClientName, you can set it either in loop or set it after loop.
const data = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0},
]
const output = data.reduce((acc, item) => {
Object
.keys(item)
.forEach((key) => {
acc[key] = (acc[key] || 0) + item[key]
});
acc.ClientName = 'Total'
return acc;
}, {})
console.log(output)answered Jan 4 at 12:58
RajeshRajesh
16.7k52254
add a comment |
3
3
3
You can try something like this:
Idea:
- Uses
Object.keysto get all enumerable keys. - Loop over them ans set them to default if not found. Then add current object's value.
- Since you need hardcoded value for
ClientName, you can set it either in loop or set it after loop.
const data = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0},
]
const output = data.reduce((acc, item) => {
Object
.keys(item)
.forEach((key) => {
acc[key] = (acc[key] || 0) + item[key]
});
acc.ClientName = 'Total'
return acc;
}, {})
console.log(output)answered Jan 4 at 12:58
RajeshRajesh
16.7k52254
You can try something like this:
Idea:
- Uses
Object.keysto get all enumerable keys. - Loop over them ans set them to default if not found. Then add current object's value.
- Since you need hardcoded value for
ClientName, you can set it either in loop or set it after loop.
const data = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0},
]
const output = data.reduce((acc, item) => {
Object
.keys(item)
.forEach((key) => {
acc[key] = (acc[key] || 0) + item[key]
});
acc.ClientName = 'Total'
return acc;
}, {})
console.log(output)const data = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0},
]
const output = data.reduce((acc, item) => {
Object
.keys(item)
.forEach((key) => {
acc[key] = (acc[key] || 0) + item[key]
});
acc.ClientName = 'Total'
return acc;
}, {})
console.log(output)const data = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0},
]
const output = data.reduce((acc, item) => {
Object
.keys(item)
.forEach((key) => {
acc[key] = (acc[key] || 0) + item[key]
});
acc.ClientName = 'Total'
return acc;
}, {})
console.log(output)answered Jan 4 at 12:58
RajeshRajesh
16.7k52254
answered Jan 4 at 12:58
RajeshRajesh
16.7k52254
answered Jan 4 at 12:58
RajeshRajesh
16.7k52254
answered Jan 4 at 12:58
RajeshRajesh
16.7k52254
16.7k52254
add a comment |
add a comment |
2
You could reduce the wanted values.
var data = [{ ClientName: "A", Alex: 2, Da: 0, Cal: 0 }, { ClientName: "b", Alex: 0, Da: 0, Cal: 4 }, { ClientName: "c", Alex: 1, Da: 0, Cal: 5 }, { ClientName: "d", Alex: 2, Da: 0, Cal: 0 }],
total = data.reduce((a, { ClientName, ...b }) => Object.assign(
{},
a,
{ ClientName: 'total' },
...Object.entries(b).map(([k, v]) => ({ [k]: (a[k] || 0) + v }))
));
console.log(total);answered Jan 4 at 13:13
Nina ScholzNina Scholz
199k15112182
add a comment |
2
You could reduce the wanted values.
var data = [{ ClientName: "A", Alex: 2, Da: 0, Cal: 0 }, { ClientName: "b", Alex: 0, Da: 0, Cal: 4 }, { ClientName: "c", Alex: 1, Da: 0, Cal: 5 }, { ClientName: "d", Alex: 2, Da: 0, Cal: 0 }],
total = data.reduce((a, { ClientName, ...b }) => Object.assign(
{},
a,
{ ClientName: 'total' },
...Object.entries(b).map(([k, v]) => ({ [k]: (a[k] || 0) + v }))
));
console.log(total);answered Jan 4 at 13:13
Nina ScholzNina Scholz
199k15112182
add a comment |
2
2
2
You could reduce the wanted values.
var data = [{ ClientName: "A", Alex: 2, Da: 0, Cal: 0 }, { ClientName: "b", Alex: 0, Da: 0, Cal: 4 }, { ClientName: "c", Alex: 1, Da: 0, Cal: 5 }, { ClientName: "d", Alex: 2, Da: 0, Cal: 0 }],
total = data.reduce((a, { ClientName, ...b }) => Object.assign(
{},
a,
{ ClientName: 'total' },
...Object.entries(b).map(([k, v]) => ({ [k]: (a[k] || 0) + v }))
));
console.log(total);answered Jan 4 at 13:13
Nina ScholzNina Scholz
199k15112182
You could reduce the wanted values.
var data = [{ ClientName: "A", Alex: 2, Da: 0, Cal: 0 }, { ClientName: "b", Alex: 0, Da: 0, Cal: 4 }, { ClientName: "c", Alex: 1, Da: 0, Cal: 5 }, { ClientName: "d", Alex: 2, Da: 0, Cal: 0 }],
total = data.reduce((a, { ClientName, ...b }) => Object.assign(
{},
a,
{ ClientName: 'total' },
...Object.entries(b).map(([k, v]) => ({ [k]: (a[k] || 0) + v }))
));
console.log(total);var data = [{ ClientName: "A", Alex: 2, Da: 0, Cal: 0 }, { ClientName: "b", Alex: 0, Da: 0, Cal: 4 }, { ClientName: "c", Alex: 1, Da: 0, Cal: 5 }, { ClientName: "d", Alex: 2, Da: 0, Cal: 0 }],
total = data.reduce((a, { ClientName, ...b }) => Object.assign(
{},
a,
{ ClientName: 'total' },
...Object.entries(b).map(([k, v]) => ({ [k]: (a[k] || 0) + v }))
));
console.log(total);var data = [{ ClientName: "A", Alex: 2, Da: 0, Cal: 0 }, { ClientName: "b", Alex: 0, Da: 0, Cal: 4 }, { ClientName: "c", Alex: 1, Da: 0, Cal: 5 }, { ClientName: "d", Alex: 2, Da: 0, Cal: 0 }],
total = data.reduce((a, { ClientName, ...b }) => Object.assign(
{},
a,
{ ClientName: 'total' },
...Object.entries(b).map(([k, v]) => ({ [k]: (a[k] || 0) + v }))
));
console.log(total);answered Jan 4 at 13:13
Nina ScholzNina Scholz
199k15112182
answered Jan 4 at 13:13
Nina ScholzNina Scholz
199k15112182
answered Jan 4 at 13:13
Nina ScholzNina Scholz
199k15112182
answered Jan 4 at 13:13
Nina ScholzNina Scholz
199k15112182
199k15112182
add a comment |
add a comment |
0
Something along these lines should work as long as all keys besides ClientName are defined as numbers
arr.reduce((sumObject, item) => {
Object.keys(item).forEach(key => {
if(key === 'ClientName'){
return false
}
if(!sumObject[key]) {
sumObject[key] = item[key]
} else {
sumObject[key] += item[key]
}
})
return sumObject
}, {ClientName: 'Total'})
answered Jan 4 at 12:59
Dennis RuiterDennis Ruiter
18528
add a comment |
0
Something along these lines should work as long as all keys besides ClientName are defined as numbers
arr.reduce((sumObject, item) => {
Object.keys(item).forEach(key => {
if(key === 'ClientName'){
return false
}
if(!sumObject[key]) {
sumObject[key] = item[key]
} else {
sumObject[key] += item[key]
}
})
return sumObject
}, {ClientName: 'Total'})
answered Jan 4 at 12:59
Dennis RuiterDennis Ruiter
18528
add a comment |
0
0
0
Something along these lines should work as long as all keys besides ClientName are defined as numbers
arr.reduce((sumObject, item) => {
Object.keys(item).forEach(key => {
if(key === 'ClientName'){
return false
}
if(!sumObject[key]) {
sumObject[key] = item[key]
} else {
sumObject[key] += item[key]
}
})
return sumObject
}, {ClientName: 'Total'})
answered Jan 4 at 12:59
Dennis RuiterDennis Ruiter
18528
Something along these lines should work as long as all keys besides ClientName are defined as numbers
arr.reduce((sumObject, item) => {
Object.keys(item).forEach(key => {
if(key === 'ClientName'){
return false
}
if(!sumObject[key]) {
sumObject[key] = item[key]
} else {
sumObject[key] += item[key]
}
})
return sumObject
}, {ClientName: 'Total'})
answered Jan 4 at 12:59
Dennis RuiterDennis Ruiter
18528
answered Jan 4 at 12:59
Dennis RuiterDennis Ruiter
18528
answered Jan 4 at 12:59
Dennis RuiterDennis Ruiter
18528
answered Jan 4 at 12:59
Dennis RuiterDennis Ruiter
18528
18528
add a comment |
add a comment |
0
You can iterate with reduce to generate single output and to merge the key count you can use Object.keys and map will allow you to iterate over your object keys. Also added one dynamic key Abc: 9 just to verify the result.
const object = [{
ClientName: "A",
Alex: 2,
Da: 0,
Cal: 0
},
{
ClientName: "b",
Alex: 0,
Da: 0,
Cal: 4
},
{
ClientName: "c",
Alex: 1,
Da: 0,
Cal: 5
},
{
ClientName: "d",
Alex: 2,
Da: 0,
Cal: 0,
Abc: 9
},
]
const result = object.reduce((accumulator, currentValue) => {
Object.keys(currentValue).map((indexKey) => {
accumulator[indexKey] = (accumulator[indexKey] || 0) + currentValue[indexKey]
});
accumulator.ClientName = "Total";
return accumulator;
})
console.log(result);answered Jan 4 at 13:29
Rohit.007Rohit.007
1,7892521
Your solution is exactly same as mine. Please check other answers before posting
– Rajesh
Jan 4 at 13:59
Yes I do agree, but It seems similar but not exactly, See in you code there isforEachwhich is less optimized than.mapwhich is being used in my solution. Anyways I have already voted up for your solution.
– Rohit.007
Jan 4 at 14:11
using a map without return statement is so much worse. Also i don't think map is more optimised compared to forEach but that discussion can be done later
– Rajesh
Jan 4 at 14:13
add a comment |
0
You can iterate with reduce to generate single output and to merge the key count you can use Object.keys and map will allow you to iterate over your object keys. Also added one dynamic key Abc: 9 just to verify the result.
const object = [{
ClientName: "A",
Alex: 2,
Da: 0,
Cal: 0
},
{
ClientName: "b",
Alex: 0,
Da: 0,
Cal: 4
},
{
ClientName: "c",
Alex: 1,
Da: 0,
Cal: 5
},
{
ClientName: "d",
Alex: 2,
Da: 0,
Cal: 0,
Abc: 9
},
]
const result = object.reduce((accumulator, currentValue) => {
Object.keys(currentValue).map((indexKey) => {
accumulator[indexKey] = (accumulator[indexKey] || 0) + currentValue[indexKey]
});
accumulator.ClientName = "Total";
return accumulator;
})
console.log(result);answered Jan 4 at 13:29
Rohit.007Rohit.007
1,7892521
Your solution is exactly same as mine. Please check other answers before posting
– Rajesh
Jan 4 at 13:59
Yes I do agree, but It seems similar but not exactly, See in you code there isforEachwhich is less optimized than.mapwhich is being used in my solution. Anyways I have already voted up for your solution.
– Rohit.007
Jan 4 at 14:11
using a map without return statement is so much worse. Also i don't think map is more optimised compared to forEach but that discussion can be done later
– Rajesh
Jan 4 at 14:13
add a comment |
0
0
0
You can iterate with reduce to generate single output and to merge the key count you can use Object.keys and map will allow you to iterate over your object keys. Also added one dynamic key Abc: 9 just to verify the result.
const object = [{
ClientName: "A",
Alex: 2,
Da: 0,
Cal: 0
},
{
ClientName: "b",
Alex: 0,
Da: 0,
Cal: 4
},
{
ClientName: "c",
Alex: 1,
Da: 0,
Cal: 5
},
{
ClientName: "d",
Alex: 2,
Da: 0,
Cal: 0,
Abc: 9
},
]
const result = object.reduce((accumulator, currentValue) => {
Object.keys(currentValue).map((indexKey) => {
accumulator[indexKey] = (accumulator[indexKey] || 0) + currentValue[indexKey]
});
accumulator.ClientName = "Total";
return accumulator;
})
console.log(result);answered Jan 4 at 13:29
Rohit.007Rohit.007
1,7892521
You can iterate with reduce to generate single output and to merge the key count you can use Object.keys and map will allow you to iterate over your object keys. Also added one dynamic key Abc: 9 just to verify the result.
const object = [{
ClientName: "A",
Alex: 2,
Da: 0,
Cal: 0
},
{
ClientName: "b",
Alex: 0,
Da: 0,
Cal: 4
},
{
ClientName: "c",
Alex: 1,
Da: 0,
Cal: 5
},
{
ClientName: "d",
Alex: 2,
Da: 0,
Cal: 0,
Abc: 9
},
]
const result = object.reduce((accumulator, currentValue) => {
Object.keys(currentValue).map((indexKey) => {
accumulator[indexKey] = (accumulator[indexKey] || 0) + currentValue[indexKey]
});
accumulator.ClientName = "Total";
return accumulator;
})
console.log(result);const object = [{
ClientName: "A",
Alex: 2,
Da: 0,
Cal: 0
},
{
ClientName: "b",
Alex: 0,
Da: 0,
Cal: 4
},
{
ClientName: "c",
Alex: 1,
Da: 0,
Cal: 5
},
{
ClientName: "d",
Alex: 2,
Da: 0,
Cal: 0,
Abc: 9
},
]
const result = object.reduce((accumulator, currentValue) => {
Object.keys(currentValue).map((indexKey) => {
accumulator[indexKey] = (accumulator[indexKey] || 0) + currentValue[indexKey]
});
accumulator.ClientName = "Total";
return accumulator;
})
console.log(result);const object = [{
ClientName: "A",
Alex: 2,
Da: 0,
Cal: 0
},
{
ClientName: "b",
Alex: 0,
Da: 0,
Cal: 4
},
{
ClientName: "c",
Alex: 1,
Da: 0,
Cal: 5
},
{
ClientName: "d",
Alex: 2,
Da: 0,
Cal: 0,
Abc: 9
},
]
const result = object.reduce((accumulator, currentValue) => {
Object.keys(currentValue).map((indexKey) => {
accumulator[indexKey] = (accumulator[indexKey] || 0) + currentValue[indexKey]
});
accumulator.ClientName = "Total";
return accumulator;
})
console.log(result);answered Jan 4 at 13:29
Rohit.007Rohit.007
1,7892521
answered Jan 4 at 13:29
Rohit.007Rohit.007
1,7892521
answered Jan 4 at 13:29
Rohit.007Rohit.007
1,7892521
answered Jan 4 at 13:29
Rohit.007Rohit.007
1,7892521
1,7892521
Your solution is exactly same as mine. Please check other answers before posting
– Rajesh
Jan 4 at 13:59
Yes I do agree, but It seems similar but not exactly, See in you code there isforEachwhich is less optimized than.mapwhich is being used in my solution. Anyways I have already voted up for your solution.
– Rohit.007
Jan 4 at 14:11
using a map without return statement is so much worse. Also i don't think map is more optimised compared to forEach but that discussion can be done later
– Rajesh
Jan 4 at 14:13
add a comment |
Your solution is exactly same as mine. Please check other answers before posting
– Rajesh
Jan 4 at 13:59
Yes I do agree, but It seems similar but not exactly, See in you code there isforEachwhich is less optimized than.mapwhich is being used in my solution. Anyways I have already voted up for your solution.
– Rohit.007
Jan 4 at 14:11
using a map without return statement is so much worse. Also i don't think map is more optimised compared to forEach but that discussion can be done later
– Rajesh
Jan 4 at 14:13
Your solution is exactly same as mine. Please check other answers before posting
– Rajesh
Jan 4 at 13:59
Your solution is exactly same as mine. Please check other answers before posting
– Rajesh
Jan 4 at 13:59
Yes I do agree, but It seems similar but not exactly, See in you code there is
forEach which is less optimized than .map which is being used in my solution. Anyways I have already voted up for your solution.– Rohit.007
Jan 4 at 14:11
Yes I do agree, but It seems similar but not exactly, See in you code there is
forEach which is less optimized than .map which is being used in my solution. Anyways I have already voted up for your solution.– Rohit.007
Jan 4 at 14:11
using a map without return statement is so much worse. Also i don't think map is more optimised compared to forEach but that discussion can be done later
– Rajesh
Jan 4 at 14:13
using a map without return statement is so much worse. Also i don't think map is more optimised compared to forEach but that discussion can be done later
– Rajesh
Jan 4 at 14:13
add a comment |
-1
let arr = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0}
];
let output = arr.reduce((curr, next) => {
return {ClientName: "Total", Alex: (curr.Alex + next.Alex ), Da: (curr.Da + next.Da), Cal: (curr.Cal + next.Cal)};
});
console.log(output);
Check this can get the answer simply
answered Jan 4 at 13:07
loghi ahaloghi aha
114
OP is looking for a solutions that handles Dynamic columns
– Rajesh
Jan 4 at 13:09
yes you are right , it shoud not work for dynamic columns.thank you,
– loghi aha
Jan 4 at 13:23
add a comment |
-1
let arr = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0}
];
let output = arr.reduce((curr, next) => {
return {ClientName: "Total", Alex: (curr.Alex + next.Alex ), Da: (curr.Da + next.Da), Cal: (curr.Cal + next.Cal)};
});
console.log(output);
Check this can get the answer simply
answered Jan 4 at 13:07
loghi ahaloghi aha
114
OP is looking for a solutions that handles Dynamic columns
– Rajesh
Jan 4 at 13:09
yes you are right , it shoud not work for dynamic columns.thank you,
– loghi aha
Jan 4 at 13:23
add a comment |
-1
-1
-1
let arr = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0}
];
let output = arr.reduce((curr, next) => {
return {ClientName: "Total", Alex: (curr.Alex + next.Alex ), Da: (curr.Da + next.Da), Cal: (curr.Cal + next.Cal)};
});
console.log(output);
Check this can get the answer simply
answered Jan 4 at 13:07
loghi ahaloghi aha
114
let arr = [
{ClientName: "A", Alex: 2, Da: 0, Cal: 0},
{ClientName: "b", Alex: 0, Da: 0, Cal: 4},
{ClientName: "c", Alex: 1, Da: 0, Cal: 5},
{ClientName: "d", Alex: 2, Da: 0, Cal: 0}
];
let output = arr.reduce((curr, next) => {
return {ClientName: "Total", Alex: (curr.Alex + next.Alex ), Da: (curr.Da + next.Da), Cal: (curr.Cal + next.Cal)};
});
console.log(output);
Check this can get the answer simply
answered Jan 4 at 13:07
loghi ahaloghi aha
114
answered Jan 4 at 13:07
loghi ahaloghi aha
114
answered Jan 4 at 13:07
loghi ahaloghi aha
114
answered Jan 4 at 13:07
loghi ahaloghi aha
114
114
OP is looking for a solutions that handles Dynamic columns
– Rajesh
Jan 4 at 13:09
yes you are right , it shoud not work for dynamic columns.thank you,
– loghi aha
Jan 4 at 13:23
add a comment |
OP is looking for a solutions that handles Dynamic columns
– Rajesh
Jan 4 at 13:09
yes you are right , it shoud not work for dynamic columns.thank you,
– loghi aha
Jan 4 at 13:23
OP is looking for a solutions that handles Dynamic columns
– Rajesh
Jan 4 at 13:09
OP is looking for a solutions that handles Dynamic columns
– Rajesh
Jan 4 at 13:09
yes you are right , it shoud not work for dynamic columns.thank you,
– loghi aha
Jan 4 at 13:23
yes you are right , it shoud not work for dynamic columns.thank you,
– loghi aha
Jan 4 at 13:23
add a comment |
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8
please add an attempt.
– Nina Scholz
Jan 4 at 12:51
1
“Columns are dynamic” means the property names are arbitrary and could be anything? Or is there a certain set of possible column names?
– deceze♦
Jan 4 at 12:57