Mixing
Filtering array with objects inside an object
Clash Royale CLAN TAG#URR8PPP
up vote
7
down vote
favorite
1
I have a problem with filtering an array with nested objects.
[
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 25,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
]
I want to get the objects where the sum of all expenses is > 35. How to get inside expenses? Or maybe filter is not a proper way here.
javascript arrays
edited 1 hour ago
Felix Kling
535k121835885
asked 2 hours ago
Kamil Staszewski
1009
 |Â
show 3 more comments
up vote
7
down vote
favorite
1
I have a problem with filtering an array with nested objects.
[
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 25,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
]
I want to get the objects where the sum of all expenses is > 35. How to get inside expenses? Or maybe filter is not a proper way here.
javascript arrays
edited 1 hour ago
Felix Kling
535k121835885
asked 2 hours ago
Kamil Staszewski
1009
This should give you an idea: Sum all properties in object. "Or maybe filter is not a proper way here." It's absolutely the right way. You just need to sum the values ofexpensesinside the filter callback.
– Felix Kling
2 hours ago
const result = arr.filter(obj => Object.values(obj.expenses).every(expense => expense > 45))
– Pavlo
2 hours ago
What code are you using to evaluate?
– Mister Positive
2 hours ago
Change it to 25. How to iterate and get values of all expenses and filter that object?
– Kamil Staszewski
1 hour ago
1
I want to receive objects where the sum of values inside a expenses object is above 25. That the example. Your code doesn't work : /
– Kamil Staszewski
1 hour ago
 |Â
show 3 more comments
up vote
7
down vote
favorite
1
up vote
7
down vote
favorite
1
1
I have a problem with filtering an array with nested objects.
[
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 25,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
]
I want to get the objects where the sum of all expenses is > 35. How to get inside expenses? Or maybe filter is not a proper way here.
javascript arrays
edited 1 hour ago
Felix Kling
535k121835885
asked 2 hours ago
Kamil Staszewski
1009
I have a problem with filtering an array with nested objects.
[
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 25,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
]
I want to get the objects where the sum of all expenses is > 35. How to get inside expenses? Or maybe filter is not a proper way here.
javascript arrays
javascript arrays
edited 1 hour ago
Felix Kling
535k121835885
asked 2 hours ago
Kamil Staszewski
1009
edited 1 hour ago
Felix Kling
535k121835885
asked 2 hours ago
Kamil Staszewski
1009
edited 1 hour ago
Felix Kling
535k121835885
edited 1 hour ago
Felix Kling
535k121835885
edited 1 hour ago
Felix Kling
535k121835885
535k121835885
asked 2 hours ago
Kamil Staszewski
1009
asked 2 hours ago
Kamil Staszewski
1009
asked 2 hours ago
Kamil Staszewski
1009
1009
This should give you an idea: Sum all properties in object. "Or maybe filter is not a proper way here." It's absolutely the right way. You just need to sum the values ofexpensesinside the filter callback.
– Felix Kling
2 hours ago
const result = arr.filter(obj => Object.values(obj.expenses).every(expense => expense > 45))
– Pavlo
2 hours ago
What code are you using to evaluate?
– Mister Positive
2 hours ago
Change it to 25. How to iterate and get values of all expenses and filter that object?
– Kamil Staszewski
1 hour ago
1
I want to receive objects where the sum of values inside a expenses object is above 25. That the example. Your code doesn't work : /
– Kamil Staszewski
1 hour ago
 |Â
show 3 more comments
This should give you an idea: Sum all properties in object. "Or maybe filter is not a proper way here." It's absolutely the right way. You just need to sum the values ofexpensesinside the filter callback.
– Felix Kling
2 hours ago
const result = arr.filter(obj => Object.values(obj.expenses).every(expense => expense > 45))
– Pavlo
2 hours ago
What code are you using to evaluate?
– Mister Positive
2 hours ago
Change it to 25. How to iterate and get values of all expenses and filter that object?
– Kamil Staszewski
1 hour ago
1
I want to receive objects where the sum of values inside a expenses object is above 25. That the example. Your code doesn't work : /
– Kamil Staszewski
1 hour ago
This should give you an idea: Sum all properties in object. "Or maybe filter is not a proper way here." It's absolutely the right way. You just need to sum the values of
expenses inside the filter callback.– Felix Kling
2 hours ago
This should give you an idea: Sum all properties in object. "Or maybe filter is not a proper way here." It's absolutely the right way. You just need to sum the values of
expenses inside the filter callback.– Felix Kling
2 hours ago
const result = arr.filter(obj => Object.values(obj.expenses).every(expense => expense > 45))– Pavlo
2 hours ago
const result = arr.filter(obj => Object.values(obj.expenses).every(expense => expense > 45))– Pavlo
2 hours ago
What code are you using to evaluate?
– Mister Positive
2 hours ago
What code are you using to evaluate?
– Mister Positive
2 hours ago
Change it to 25. How to iterate and get values of all expenses and filter that object?
– Kamil Staszewski
1 hour ago
Change it to 25. How to iterate and get values of all expenses and filter that object?
– Kamil Staszewski
1 hour ago
1
1
I want to receive objects where the sum of values inside a expenses object is above 25. That the example. Your code doesn't work : /
– Kamil Staszewski
1 hour ago
I want to receive objects where the sum of values inside a expenses object is above 25. That the example. Your code doesn't work : /
– Kamil Staszewski
1 hour ago
 |Â
show 3 more comments
5 Answers
5
active
oldest
votes
up vote
9
down vote
Just filter it, with a condition using reduce to sum the expenses! Pretty straight forward :)
const input = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
const output = input.filter(user => Object.values(user.expenses).reduce((acc, expense) => acc + expense) > 45);
console.log(output);answered 1 hour ago
sjahan
2,8041722
We have a winner. Thank you!
– Kamil Staszewski
1 hour ago
Glad it helps ;) Don't forget to mark the answer as accepted so that it can help the rest of the community :)
– sjahan
1 hour ago
I didn't knew about Object.values. Nice!
– Donny Verduijn
1 hour ago
add a comment |Â
up vote
0
down vote
You can try something like this
var data = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 25,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
]
var filtered = data.filter(c => Object.keys(c.expenses)
.reduce(function(p, e)
return p + c.expenses[e]
, 0) >= 35
);
console.log(filtered);answered 1 hour ago
Alfredo A.
81911232
add a comment |Â
up vote
0
down vote
You can access object at index i inside array arr with expression arr[i]
What you need to do is to loop over your array. Inside your loop you access each object with expression i have mentioned: arr[i] and then on this object you can access expenses following way arr[i].expenses after this - if i am understanding correctly you sum contents of your arr[i].expenses object and select those objects which satisfy your condition.
Please, See code below:
var expensesAbove35Arr = ;
var yourArray = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
for(var i=0; i<yourArray.length; i++)
if(yourArray[i].expenses.coffe + yourArray[i].expenses.cake > 35 )
expensesAbove35Arr.push(yourArray[i]);
You have got your result inside array expensesAbove35Arr
answered 1 hour ago
Tornike Shavishvili
5331922
add a comment |Â
up vote
0
down vote
I'm assuming that you need to keep you array of users the way it is, but with the expenses filtered.
(I assumed wrong as pointed in the comment, keeping this answer just in case someone sees any value on it)
Probably this can be optmized or simplified, but here it goes:
arr.reduce((acc, user) => [...acc,
Object.keys(user).reduce((userResult, key) =>
if (key === 'expenses')
return
...userResult,
expenses: Object.entries(elem.expenses)
.filter(([product, value]) => value > 35)
// now "reversing" the object.entries
.reduce((acc, [product, value]) => ( [product]: value ), )
return
...userResult,
[key]: elem[key]
, user) // starts with the user
], ) //starts with empty array
answered 1 hour ago
graciano
1289
1
The OP want's to get all user objects whose sum of the expenses is larger than N. They don't want to filter out expenses.
– Felix Kling
1 hour ago
add a comment |Â
up vote
-2
down vote
A possible solution might be:
arr.filter(function(v)
for(expense in v.expenses)
if(v.expenses[expense] > 10)
return true;
else
return false;
)
answered 2 hours ago
Velimir Tchatchevsky
1,91511117
1
Why should the OP try this? How does this solve their problem?
– Felix Kling
2 hours ago
1
On the contrary OP's problem is getting to the "expenses" array, "How to get inside expenses", I don't think SO is a coding service
– Velimir Tchatchevsky
2 hours ago
1
I want iterate over all values inside expenses object
– Kamil Staszewski
2 hours ago
3
So, if they are asking how to sum all property values, you are showing how to access a single property, because actually answering the question would be a too much? O_o
– Felix Kling
1 hour ago
1
I agree. But if you only say "try this" then that seems to imply that you are providing a full fledged solution to their problem. But you are not. Your code example is somewhat related but doesn't produce the result they want. So at the very least you should explain what you are doing and how this helps them get to solving their problem.
– Felix Kling
1 hour ago
 |Â
show 2 more comments
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
Just filter it, with a condition using reduce to sum the expenses! Pretty straight forward :)
const input = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
const output = input.filter(user => Object.values(user.expenses).reduce((acc, expense) => acc + expense) > 45);
console.log(output);answered 1 hour ago
sjahan
2,8041722
We have a winner. Thank you!
– Kamil Staszewski
1 hour ago
Glad it helps ;) Don't forget to mark the answer as accepted so that it can help the rest of the community :)
– sjahan
1 hour ago
I didn't knew about Object.values. Nice!
– Donny Verduijn
1 hour ago
add a comment |Â
up vote
9
down vote
Just filter it, with a condition using reduce to sum the expenses! Pretty straight forward :)
const input = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
const output = input.filter(user => Object.values(user.expenses).reduce((acc, expense) => acc + expense) > 45);
console.log(output);answered 1 hour ago
sjahan
2,8041722
We have a winner. Thank you!
– Kamil Staszewski
1 hour ago
Glad it helps ;) Don't forget to mark the answer as accepted so that it can help the rest of the community :)
– sjahan
1 hour ago
I didn't knew about Object.values. Nice!
– Donny Verduijn
1 hour ago
add a comment |Â
up vote
9
down vote
up vote
9
down vote
Just filter it, with a condition using reduce to sum the expenses! Pretty straight forward :)
const input = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
const output = input.filter(user => Object.values(user.expenses).reduce((acc, expense) => acc + expense) > 45);
console.log(output);answered 1 hour ago
sjahan
2,8041722
Just filter it, with a condition using reduce to sum the expenses! Pretty straight forward :)
const input = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
const output = input.filter(user => Object.values(user.expenses).reduce((acc, expense) => acc + expense) > 45);
console.log(output);const input = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
const output = input.filter(user => Object.values(user.expenses).reduce((acc, expense) => acc + expense) > 45);
console.log(output);const input = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
const output = input.filter(user => Object.values(user.expenses).reduce((acc, expense) => acc + expense) > 45);
console.log(output);answered 1 hour ago
sjahan
2,8041722
answered 1 hour ago
sjahan
2,8041722
answered 1 hour ago
sjahan
2,8041722
answered 1 hour ago
sjahan
2,8041722
2,8041722
We have a winner. Thank you!
– Kamil Staszewski
1 hour ago
Glad it helps ;) Don't forget to mark the answer as accepted so that it can help the rest of the community :)
– sjahan
1 hour ago
I didn't knew about Object.values. Nice!
– Donny Verduijn
1 hour ago
add a comment |Â
We have a winner. Thank you!
– Kamil Staszewski
1 hour ago
Glad it helps ;) Don't forget to mark the answer as accepted so that it can help the rest of the community :)
– sjahan
1 hour ago
I didn't knew about Object.values. Nice!
– Donny Verduijn
1 hour ago
We have a winner. Thank you!
– Kamil Staszewski
1 hour ago
We have a winner. Thank you!
– Kamil Staszewski
1 hour ago
Glad it helps ;) Don't forget to mark the answer as accepted so that it can help the rest of the community :)
– sjahan
1 hour ago
Glad it helps ;) Don't forget to mark the answer as accepted so that it can help the rest of the community :)
– sjahan
1 hour ago
I didn't knew about Object.values. Nice!
– Donny Verduijn
1 hour ago
I didn't knew about Object.values. Nice!
– Donny Verduijn
1 hour ago
add a comment |Â
up vote
0
down vote
You can try something like this
var data = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 25,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
]
var filtered = data.filter(c => Object.keys(c.expenses)
.reduce(function(p, e)
return p + c.expenses[e]
, 0) >= 35
);
console.log(filtered);answered 1 hour ago
Alfredo A.
81911232
add a comment |Â
up vote
0
down vote
You can try something like this
var data = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 25,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
]
var filtered = data.filter(c => Object.keys(c.expenses)
.reduce(function(p, e)
return p + c.expenses[e]
, 0) >= 35
);
console.log(filtered);answered 1 hour ago
Alfredo A.
81911232
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can try something like this
var data = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 25,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
]
var filtered = data.filter(c => Object.keys(c.expenses)
.reduce(function(p, e)
return p + c.expenses[e]
, 0) >= 35
);
console.log(filtered);answered 1 hour ago
Alfredo A.
81911232
You can try something like this
var data = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 25,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
]
var filtered = data.filter(c => Object.keys(c.expenses)
.reduce(function(p, e)
return p + c.expenses[e]
, 0) >= 35
);
console.log(filtered);var data = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 25,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
]
var filtered = data.filter(c => Object.keys(c.expenses)
.reduce(function(p, e)
return p + c.expenses[e]
, 0) >= 35
);
console.log(filtered);var data = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 25,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
]
var filtered = data.filter(c => Object.keys(c.expenses)
.reduce(function(p, e)
return p + c.expenses[e]
, 0) >= 35
);
console.log(filtered);answered 1 hour ago
Alfredo A.
81911232
edited 1 hour ago
answered 1 hour ago
Alfredo A.
81911232
answered 1 hour ago
Alfredo A.
81911232
answered 1 hour ago
Alfredo A.
81911232
81911232
add a comment |Â
add a comment |Â
up vote
0
down vote
You can access object at index i inside array arr with expression arr[i]
What you need to do is to loop over your array. Inside your loop you access each object with expression i have mentioned: arr[i] and then on this object you can access expenses following way arr[i].expenses after this - if i am understanding correctly you sum contents of your arr[i].expenses object and select those objects which satisfy your condition.
Please, See code below:
var expensesAbove35Arr = ;
var yourArray = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
for(var i=0; i<yourArray.length; i++)
if(yourArray[i].expenses.coffe + yourArray[i].expenses.cake > 35 )
expensesAbove35Arr.push(yourArray[i]);
You have got your result inside array expensesAbove35Arr
answered 1 hour ago
Tornike Shavishvili
5331922
add a comment |Â
up vote
0
down vote
You can access object at index i inside array arr with expression arr[i]
What you need to do is to loop over your array. Inside your loop you access each object with expression i have mentioned: arr[i] and then on this object you can access expenses following way arr[i].expenses after this - if i am understanding correctly you sum contents of your arr[i].expenses object and select those objects which satisfy your condition.
Please, See code below:
var expensesAbove35Arr = ;
var yourArray = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
for(var i=0; i<yourArray.length; i++)
if(yourArray[i].expenses.coffe + yourArray[i].expenses.cake > 35 )
expensesAbove35Arr.push(yourArray[i]);
You have got your result inside array expensesAbove35Arr
answered 1 hour ago
Tornike Shavishvili
5331922
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can access object at index i inside array arr with expression arr[i]
What you need to do is to loop over your array. Inside your loop you access each object with expression i have mentioned: arr[i] and then on this object you can access expenses following way arr[i].expenses after this - if i am understanding correctly you sum contents of your arr[i].expenses object and select those objects which satisfy your condition.
Please, See code below:
var expensesAbove35Arr = ;
var yourArray = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
for(var i=0; i<yourArray.length; i++)
if(yourArray[i].expenses.coffe + yourArray[i].expenses.cake > 35 )
expensesAbove35Arr.push(yourArray[i]);
You have got your result inside array expensesAbove35Arr
answered 1 hour ago
Tornike Shavishvili
5331922
You can access object at index i inside array arr with expression arr[i]
What you need to do is to loop over your array. Inside your loop you access each object with expression i have mentioned: arr[i] and then on this object you can access expenses following way arr[i].expenses after this - if i am understanding correctly you sum contents of your arr[i].expenses object and select those objects which satisfy your condition.
Please, See code below:
var expensesAbove35Arr = ;
var yourArray = [
"firstName": "Kevin",
"lastName": "Smith",
"expenses":
"drink1": 26,
"drink2": 20
,
"firstName": "John",
"lastName": "Rambo",
"expenses":
"coffe": 10,
"cake": 20
];
for(var i=0; i<yourArray.length; i++)
if(yourArray[i].expenses.coffe + yourArray[i].expenses.cake > 35 )
expensesAbove35Arr.push(yourArray[i]);
You have got your result inside array expensesAbove35Arr
answered 1 hour ago
Tornike Shavishvili
5331922
answered 1 hour ago
Tornike Shavishvili
5331922
answered 1 hour ago
Tornike Shavishvili
5331922
answered 1 hour ago
Tornike Shavishvili
5331922
5331922
add a comment |Â
add a comment |Â
up vote
0
down vote
I'm assuming that you need to keep you array of users the way it is, but with the expenses filtered.
(I assumed wrong as pointed in the comment, keeping this answer just in case someone sees any value on it)
Probably this can be optmized or simplified, but here it goes:
arr.reduce((acc, user) => [...acc,
Object.keys(user).reduce((userResult, key) =>
if (key === 'expenses')
return
...userResult,
expenses: Object.entries(elem.expenses)
.filter(([product, value]) => value > 35)
// now "reversing" the object.entries
.reduce((acc, [product, value]) => ( [product]: value ), )
return
...userResult,
[key]: elem[key]
, user) // starts with the user
], ) //starts with empty array
answered 1 hour ago
graciano
1289
1
The OP want's to get all user objects whose sum of the expenses is larger than N. They don't want to filter out expenses.
– Felix Kling
1 hour ago
add a comment |Â
up vote
0
down vote
I'm assuming that you need to keep you array of users the way it is, but with the expenses filtered.
(I assumed wrong as pointed in the comment, keeping this answer just in case someone sees any value on it)
Probably this can be optmized or simplified, but here it goes:
arr.reduce((acc, user) => [...acc,
Object.keys(user).reduce((userResult, key) =>
if (key === 'expenses')
return
...userResult,
expenses: Object.entries(elem.expenses)
.filter(([product, value]) => value > 35)
// now "reversing" the object.entries
.reduce((acc, [product, value]) => ( [product]: value ), )
return
...userResult,
[key]: elem[key]
, user) // starts with the user
], ) //starts with empty array
answered 1 hour ago
graciano
1289
1
The OP want's to get all user objects whose sum of the expenses is larger than N. They don't want to filter out expenses.
– Felix Kling
1 hour ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I'm assuming that you need to keep you array of users the way it is, but with the expenses filtered.
(I assumed wrong as pointed in the comment, keeping this answer just in case someone sees any value on it)
Probably this can be optmized or simplified, but here it goes:
arr.reduce((acc, user) => [...acc,
Object.keys(user).reduce((userResult, key) =>
if (key === 'expenses')
return
...userResult,
expenses: Object.entries(elem.expenses)
.filter(([product, value]) => value > 35)
// now "reversing" the object.entries
.reduce((acc, [product, value]) => ( [product]: value ), )
return
...userResult,
[key]: elem[key]
, user) // starts with the user
], ) //starts with empty array
answered 1 hour ago
graciano
1289
I'm assuming that you need to keep you array of users the way it is, but with the expenses filtered.
(I assumed wrong as pointed in the comment, keeping this answer just in case someone sees any value on it)
Probably this can be optmized or simplified, but here it goes:
arr.reduce((acc, user) => [...acc,
Object.keys(user).reduce((userResult, key) =>
if (key === 'expenses')
return
...userResult,
expenses: Object.entries(elem.expenses)
.filter(([product, value]) => value > 35)
// now "reversing" the object.entries
.reduce((acc, [product, value]) => ( [product]: value ), )
return
...userResult,
[key]: elem[key]
, user) // starts with the user
], ) //starts with empty array
answered 1 hour ago
graciano
1289
edited 1 hour ago
answered 1 hour ago
graciano
1289
answered 1 hour ago
graciano
1289
answered 1 hour ago
graciano
1289
1289
1
The OP want's to get all user objects whose sum of the expenses is larger than N. They don't want to filter out expenses.
– Felix Kling
1 hour ago
add a comment |Â
1
The OP want's to get all user objects whose sum of the expenses is larger than N. They don't want to filter out expenses.
– Felix Kling
1 hour ago
1
1
The OP want's to get all user objects whose sum of the expenses is larger than N. They don't want to filter out expenses.
– Felix Kling
1 hour ago
The OP want's to get all user objects whose sum of the expenses is larger than N. They don't want to filter out expenses.
– Felix Kling
1 hour ago
add a comment |Â
up vote
-2
down vote
A possible solution might be:
arr.filter(function(v)
for(expense in v.expenses)
if(v.expenses[expense] > 10)
return true;
else
return false;
)
answered 2 hours ago
Velimir Tchatchevsky
1,91511117
1
Why should the OP try this? How does this solve their problem?
– Felix Kling
2 hours ago
1
On the contrary OP's problem is getting to the "expenses" array, "How to get inside expenses", I don't think SO is a coding service
– Velimir Tchatchevsky
2 hours ago
1
I want iterate over all values inside expenses object
– Kamil Staszewski
2 hours ago
3
So, if they are asking how to sum all property values, you are showing how to access a single property, because actually answering the question would be a too much? O_o
– Felix Kling
1 hour ago
1
I agree. But if you only say "try this" then that seems to imply that you are providing a full fledged solution to their problem. But you are not. Your code example is somewhat related but doesn't produce the result they want. So at the very least you should explain what you are doing and how this helps them get to solving their problem.
– Felix Kling
1 hour ago
 |Â
show 2 more comments
up vote
-2
down vote
A possible solution might be:
arr.filter(function(v)
for(expense in v.expenses)
if(v.expenses[expense] > 10)
return true;
else
return false;
)
answered 2 hours ago
Velimir Tchatchevsky
1,91511117
1
Why should the OP try this? How does this solve their problem?
– Felix Kling
2 hours ago
1
On the contrary OP's problem is getting to the "expenses" array, "How to get inside expenses", I don't think SO is a coding service
– Velimir Tchatchevsky
2 hours ago
1
I want iterate over all values inside expenses object
– Kamil Staszewski
2 hours ago
3
So, if they are asking how to sum all property values, you are showing how to access a single property, because actually answering the question would be a too much? O_o
– Felix Kling
1 hour ago
1
I agree. But if you only say "try this" then that seems to imply that you are providing a full fledged solution to their problem. But you are not. Your code example is somewhat related but doesn't produce the result they want. So at the very least you should explain what you are doing and how this helps them get to solving their problem.
– Felix Kling
1 hour ago
 |Â
show 2 more comments
up vote
-2
down vote
up vote
-2
down vote
A possible solution might be:
arr.filter(function(v)
for(expense in v.expenses)
if(v.expenses[expense] > 10)
return true;
else
return false;
)
answered 2 hours ago
Velimir Tchatchevsky
1,91511117
A possible solution might be:
arr.filter(function(v)
for(expense in v.expenses)
if(v.expenses[expense] > 10)
return true;
else
return false;
)
answered 2 hours ago
Velimir Tchatchevsky
1,91511117
edited 1 hour ago
answered 2 hours ago
Velimir Tchatchevsky
1,91511117
answered 2 hours ago
Velimir Tchatchevsky
1,91511117
answered 2 hours ago
Velimir Tchatchevsky
1,91511117
1,91511117
1
Why should the OP try this? How does this solve their problem?
– Felix Kling
2 hours ago
1
On the contrary OP's problem is getting to the "expenses" array, "How to get inside expenses", I don't think SO is a coding service
– Velimir Tchatchevsky
2 hours ago
1
I want iterate over all values inside expenses object
– Kamil Staszewski
2 hours ago
3
So, if they are asking how to sum all property values, you are showing how to access a single property, because actually answering the question would be a too much? O_o
– Felix Kling
1 hour ago
1
I agree. But if you only say "try this" then that seems to imply that you are providing a full fledged solution to their problem. But you are not. Your code example is somewhat related but doesn't produce the result they want. So at the very least you should explain what you are doing and how this helps them get to solving their problem.
– Felix Kling
1 hour ago
 |Â
show 2 more comments
1
Why should the OP try this? How does this solve their problem?
– Felix Kling
2 hours ago
1
On the contrary OP's problem is getting to the "expenses" array, "How to get inside expenses", I don't think SO is a coding service
– Velimir Tchatchevsky
2 hours ago
1
I want iterate over all values inside expenses object
– Kamil Staszewski
2 hours ago
3
So, if they are asking how to sum all property values, you are showing how to access a single property, because actually answering the question would be a too much? O_o
– Felix Kling
1 hour ago
1
I agree. But if you only say "try this" then that seems to imply that you are providing a full fledged solution to their problem. But you are not. Your code example is somewhat related but doesn't produce the result they want. So at the very least you should explain what you are doing and how this helps them get to solving their problem.
– Felix Kling
1 hour ago
1
1
Why should the OP try this? How does this solve their problem?
– Felix Kling
2 hours ago
Why should the OP try this? How does this solve their problem?
– Felix Kling
2 hours ago
1
1
On the contrary OP's problem is getting to the "expenses" array, "How to get inside expenses", I don't think SO is a coding service
– Velimir Tchatchevsky
2 hours ago
On the contrary OP's problem is getting to the "expenses" array, "How to get inside expenses", I don't think SO is a coding service
– Velimir Tchatchevsky
2 hours ago
1
1
I want iterate over all values inside expenses object
– Kamil Staszewski
2 hours ago
I want iterate over all values inside expenses object
– Kamil Staszewski
2 hours ago
3
3
So, if they are asking how to sum all property values, you are showing how to access a single property, because actually answering the question would be a too much? O_o
– Felix Kling
1 hour ago
So, if they are asking how to sum all property values, you are showing how to access a single property, because actually answering the question would be a too much? O_o
– Felix Kling
1 hour ago
1
1
I agree. But if you only say "try this" then that seems to imply that you are providing a full fledged solution to their problem. But you are not. Your code example is somewhat related but doesn't produce the result they want. So at the very least you should explain what you are doing and how this helps them get to solving their problem.
– Felix Kling
1 hour ago
I agree. But if you only say "try this" then that seems to imply that you are providing a full fledged solution to their problem. But you are not. Your code example is somewhat related but doesn't produce the result they want. So at the very least you should explain what you are doing and how this helps them get to solving their problem.
– Felix Kling
1 hour ago
 |Â
show 2 more comments
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This should give you an idea: Sum all properties in object. "Or maybe filter is not a proper way here." It's absolutely the right way. You just need to sum the values of
expensesinside the filter callback.– Felix Kling
2 hours ago
const result = arr.filter(obj => Object.values(obj.expenses).every(expense => expense > 45))– Pavlo
2 hours ago
What code are you using to evaluate?
– Mister Positive
2 hours ago
Change it to 25. How to iterate and get values of all expenses and filter that object?
– Kamil Staszewski
1 hour ago
1
I want to receive objects where the sum of values inside a expenses object is above 25. That the example. Your code doesn't work : /
– Kamil Staszewski
1 hour ago