Mixing
Performing delete operations on an immutable JS Map object
0
I have an immutable JS map with a structure like the below
{
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
I want to filter this object to only show objects which are tracked - i.e
{
"a": [
{"name": "bar"}
],
"b": [
{"name": "baz"},
]
}
Is there a way for me to do this with immutable map operations? i tried the below but it doesnt seem to work
object.map((lis) => lis.filter((li) => li.untracked !== true)).toJS()
object.toList().map((lis) => lis.filter((li) => li.untracked !== true))
javascript immutable.js
asked Nov 20 '18 at 16:30
KannajKannaj
2,04521641
add a comment |
0
I have an immutable JS map with a structure like the below
{
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
I want to filter this object to only show objects which are tracked - i.e
{
"a": [
{"name": "bar"}
],
"b": [
{"name": "baz"},
]
}
Is there a way for me to do this with immutable map operations? i tried the below but it doesnt seem to work
object.map((lis) => lis.filter((li) => li.untracked !== true)).toJS()
object.toList().map((lis) => lis.filter((li) => li.untracked !== true))
javascript immutable.js
asked Nov 20 '18 at 16:30
KannajKannaj
2,04521641
add a comment |
0
0
0
1
I have an immutable JS map with a structure like the below
{
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
I want to filter this object to only show objects which are tracked - i.e
{
"a": [
{"name": "bar"}
],
"b": [
{"name": "baz"},
]
}
Is there a way for me to do this with immutable map operations? i tried the below but it doesnt seem to work
object.map((lis) => lis.filter((li) => li.untracked !== true)).toJS()
object.toList().map((lis) => lis.filter((li) => li.untracked !== true))
javascript immutable.js
asked Nov 20 '18 at 16:30
KannajKannaj
2,04521641
I have an immutable JS map with a structure like the below
{
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
I want to filter this object to only show objects which are tracked - i.e
{
"a": [
{"name": "bar"}
],
"b": [
{"name": "baz"},
]
}
Is there a way for me to do this with immutable map operations? i tried the below but it doesnt seem to work
object.map((lis) => lis.filter((li) => li.untracked !== true)).toJS()
object.toList().map((lis) => lis.filter((li) => li.untracked !== true))
javascript immutable.js
javascript immutable.js
asked Nov 20 '18 at 16:30
KannajKannaj
2,04521641
asked Nov 20 '18 at 16:30
KannajKannaj
2,04521641
asked Nov 20 '18 at 16:30
KannajKannaj
2,04521641
asked Nov 20 '18 at 16:30
KannajKannaj
2,04521641
asked Nov 20 '18 at 16:30
KannajKannaj
2,04521641
2,04521641
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
1
let data={
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
Object.entries(data).map(([key, value]) => {
data[key]=value.filter(ele=>ele.untracked!=true)
});
console.log("result is : ",data)answered Nov 20 '18 at 16:59
BittuSBittuS
587216
add a comment |
0
As per ImmutableJS docs:
filter() returns a new Map with only the entries for which the predicate function returns true.
example:
function someFilterFunction(entry) {
return !entry.untracked;
}
myMapOfStuff = myMapOfStuff.filter(someFilterFunction);
Link to documentation: https://facebook.github.io/immutable-js/docs/#/Map/filter
EDIT: Remember, the method on an immutable collection returns a COPY of the collection you are operating on. A reassignment must happen if you want to preserve the most up to date reference.
answered Nov 20 '18 at 16:35
Francisco GarciaFrancisco Garcia
704
add a comment |
0
Loop through that object and create a new object with keys from old object and then filter the values from the old object where untracked is not equal to false
let obj = {
"a": [{
"name": "foo",
"untracked": true
},
{
"name": "bar"
}
],
"b": [{
"name": "baz"
},
{
"name": "bar",
"untracked": true
}
]
}
let newObj = {};
function filterObject(obj) {
for (let keys in obj) {
newObj[keys] = obj[keys].filter(item => {
return item.untracked !== true
})
}
}
filterObject(obj)
console.log(newObj)answered Nov 20 '18 at 16:37
brkbrk
26.5k31940
add a comment |
0
data = {
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
let result = Object.entries(data).map(([key, values])=>(
{ key: values.filter(({untracked})=> !untracked) }
))
console.log(result)answered Nov 20 '18 at 17:14
Mohammed AshfaqMohammed Ashfaq
1,6672612
Uncaught ReferenceError: result is not defined. Also you are changing the structure of the given data as output will be=> [ { "key": [ { "name": "bar" } ] }, { "key": [ { "name": "baz" } ] } ]
– Monica Acha
Nov 20 '18 at 17:40
I forgot to declare the result variable.Thanks @Monic
– Mohammed Ashfaq
Nov 21 '18 at 1:23
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
let data={
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
Object.entries(data).map(([key, value]) => {
data[key]=value.filter(ele=>ele.untracked!=true)
});
console.log("result is : ",data)answered Nov 20 '18 at 16:59
BittuSBittuS
587216
add a comment |
1
let data={
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
Object.entries(data).map(([key, value]) => {
data[key]=value.filter(ele=>ele.untracked!=true)
});
console.log("result is : ",data)answered Nov 20 '18 at 16:59
BittuSBittuS
587216
add a comment |
1
1
1
let data={
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
Object.entries(data).map(([key, value]) => {
data[key]=value.filter(ele=>ele.untracked!=true)
});
console.log("result is : ",data)answered Nov 20 '18 at 16:59
BittuSBittuS
587216
let data={
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
Object.entries(data).map(([key, value]) => {
data[key]=value.filter(ele=>ele.untracked!=true)
});
console.log("result is : ",data)let data={
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
Object.entries(data).map(([key, value]) => {
data[key]=value.filter(ele=>ele.untracked!=true)
});
console.log("result is : ",data)let data={
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
Object.entries(data).map(([key, value]) => {
data[key]=value.filter(ele=>ele.untracked!=true)
});
console.log("result is : ",data)answered Nov 20 '18 at 16:59
BittuSBittuS
587216
edited Nov 20 '18 at 17:05
answered Nov 20 '18 at 16:59
BittuSBittuS
587216
answered Nov 20 '18 at 16:59
BittuSBittuS
587216
answered Nov 20 '18 at 16:59
BittuSBittuS
587216
587216
add a comment |
add a comment |
0
As per ImmutableJS docs:
filter() returns a new Map with only the entries for which the predicate function returns true.
example:
function someFilterFunction(entry) {
return !entry.untracked;
}
myMapOfStuff = myMapOfStuff.filter(someFilterFunction);
Link to documentation: https://facebook.github.io/immutable-js/docs/#/Map/filter
EDIT: Remember, the method on an immutable collection returns a COPY of the collection you are operating on. A reassignment must happen if you want to preserve the most up to date reference.
answered Nov 20 '18 at 16:35
Francisco GarciaFrancisco Garcia
704
add a comment |
0
As per ImmutableJS docs:
filter() returns a new Map with only the entries for which the predicate function returns true.
example:
function someFilterFunction(entry) {
return !entry.untracked;
}
myMapOfStuff = myMapOfStuff.filter(someFilterFunction);
Link to documentation: https://facebook.github.io/immutable-js/docs/#/Map/filter
EDIT: Remember, the method on an immutable collection returns a COPY of the collection you are operating on. A reassignment must happen if you want to preserve the most up to date reference.
answered Nov 20 '18 at 16:35
Francisco GarciaFrancisco Garcia
704
add a comment |
0
0
0
As per ImmutableJS docs:
filter() returns a new Map with only the entries for which the predicate function returns true.
example:
function someFilterFunction(entry) {
return !entry.untracked;
}
myMapOfStuff = myMapOfStuff.filter(someFilterFunction);
Link to documentation: https://facebook.github.io/immutable-js/docs/#/Map/filter
EDIT: Remember, the method on an immutable collection returns a COPY of the collection you are operating on. A reassignment must happen if you want to preserve the most up to date reference.
answered Nov 20 '18 at 16:35
Francisco GarciaFrancisco Garcia
704
As per ImmutableJS docs:
filter() returns a new Map with only the entries for which the predicate function returns true.
example:
function someFilterFunction(entry) {
return !entry.untracked;
}
myMapOfStuff = myMapOfStuff.filter(someFilterFunction);
Link to documentation: https://facebook.github.io/immutable-js/docs/#/Map/filter
EDIT: Remember, the method on an immutable collection returns a COPY of the collection you are operating on. A reassignment must happen if you want to preserve the most up to date reference.
answered Nov 20 '18 at 16:35
Francisco GarciaFrancisco Garcia
704
answered Nov 20 '18 at 16:35
Francisco GarciaFrancisco Garcia
704
answered Nov 20 '18 at 16:35
Francisco GarciaFrancisco Garcia
704
answered Nov 20 '18 at 16:35
Francisco GarciaFrancisco Garcia
704
704
add a comment |
add a comment |
0
Loop through that object and create a new object with keys from old object and then filter the values from the old object where untracked is not equal to false
let obj = {
"a": [{
"name": "foo",
"untracked": true
},
{
"name": "bar"
}
],
"b": [{
"name": "baz"
},
{
"name": "bar",
"untracked": true
}
]
}
let newObj = {};
function filterObject(obj) {
for (let keys in obj) {
newObj[keys] = obj[keys].filter(item => {
return item.untracked !== true
})
}
}
filterObject(obj)
console.log(newObj)answered Nov 20 '18 at 16:37
brkbrk
26.5k31940
add a comment |
0
Loop through that object and create a new object with keys from old object and then filter the values from the old object where untracked is not equal to false
let obj = {
"a": [{
"name": "foo",
"untracked": true
},
{
"name": "bar"
}
],
"b": [{
"name": "baz"
},
{
"name": "bar",
"untracked": true
}
]
}
let newObj = {};
function filterObject(obj) {
for (let keys in obj) {
newObj[keys] = obj[keys].filter(item => {
return item.untracked !== true
})
}
}
filterObject(obj)
console.log(newObj)answered Nov 20 '18 at 16:37
brkbrk
26.5k31940
add a comment |
0
0
0
Loop through that object and create a new object with keys from old object and then filter the values from the old object where untracked is not equal to false
let obj = {
"a": [{
"name": "foo",
"untracked": true
},
{
"name": "bar"
}
],
"b": [{
"name": "baz"
},
{
"name": "bar",
"untracked": true
}
]
}
let newObj = {};
function filterObject(obj) {
for (let keys in obj) {
newObj[keys] = obj[keys].filter(item => {
return item.untracked !== true
})
}
}
filterObject(obj)
console.log(newObj)answered Nov 20 '18 at 16:37
brkbrk
26.5k31940
Loop through that object and create a new object with keys from old object and then filter the values from the old object where untracked is not equal to false
let obj = {
"a": [{
"name": "foo",
"untracked": true
},
{
"name": "bar"
}
],
"b": [{
"name": "baz"
},
{
"name": "bar",
"untracked": true
}
]
}
let newObj = {};
function filterObject(obj) {
for (let keys in obj) {
newObj[keys] = obj[keys].filter(item => {
return item.untracked !== true
})
}
}
filterObject(obj)
console.log(newObj)let obj = {
"a": [{
"name": "foo",
"untracked": true
},
{
"name": "bar"
}
],
"b": [{
"name": "baz"
},
{
"name": "bar",
"untracked": true
}
]
}
let newObj = {};
function filterObject(obj) {
for (let keys in obj) {
newObj[keys] = obj[keys].filter(item => {
return item.untracked !== true
})
}
}
filterObject(obj)
console.log(newObj)let obj = {
"a": [{
"name": "foo",
"untracked": true
},
{
"name": "bar"
}
],
"b": [{
"name": "baz"
},
{
"name": "bar",
"untracked": true
}
]
}
let newObj = {};
function filterObject(obj) {
for (let keys in obj) {
newObj[keys] = obj[keys].filter(item => {
return item.untracked !== true
})
}
}
filterObject(obj)
console.log(newObj)answered Nov 20 '18 at 16:37
brkbrk
26.5k31940
answered Nov 20 '18 at 16:37
brkbrk
26.5k31940
answered Nov 20 '18 at 16:37
brkbrk
26.5k31940
answered Nov 20 '18 at 16:37
brkbrk
26.5k31940
26.5k31940
add a comment |
add a comment |
0
data = {
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
let result = Object.entries(data).map(([key, values])=>(
{ key: values.filter(({untracked})=> !untracked) }
))
console.log(result)answered Nov 20 '18 at 17:14
Mohammed AshfaqMohammed Ashfaq
1,6672612
Uncaught ReferenceError: result is not defined. Also you are changing the structure of the given data as output will be=> [ { "key": [ { "name": "bar" } ] }, { "key": [ { "name": "baz" } ] } ]
– Monica Acha
Nov 20 '18 at 17:40
I forgot to declare the result variable.Thanks @Monic
– Mohammed Ashfaq
Nov 21 '18 at 1:23
add a comment |
0
data = {
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
let result = Object.entries(data).map(([key, values])=>(
{ key: values.filter(({untracked})=> !untracked) }
))
console.log(result)answered Nov 20 '18 at 17:14
Mohammed AshfaqMohammed Ashfaq
1,6672612
Uncaught ReferenceError: result is not defined. Also you are changing the structure of the given data as output will be=> [ { "key": [ { "name": "bar" } ] }, { "key": [ { "name": "baz" } ] } ]
– Monica Acha
Nov 20 '18 at 17:40
I forgot to declare the result variable.Thanks @Monic
– Mohammed Ashfaq
Nov 21 '18 at 1:23
add a comment |
0
0
0
data = {
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
let result = Object.entries(data).map(([key, values])=>(
{ key: values.filter(({untracked})=> !untracked) }
))
console.log(result)answered Nov 20 '18 at 17:14
Mohammed AshfaqMohammed Ashfaq
1,6672612
data = {
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
let result = Object.entries(data).map(([key, values])=>(
{ key: values.filter(({untracked})=> !untracked) }
))
console.log(result)data = {
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
let result = Object.entries(data).map(([key, values])=>(
{ key: values.filter(({untracked})=> !untracked) }
))
console.log(result)data = {
"a": [
{"name": "foo", "untracked": true},
{"name": "bar"}
],
"b": [
{"name": "baz"},
{"name": "bar", "untracked": true}
]
}
let result = Object.entries(data).map(([key, values])=>(
{ key: values.filter(({untracked})=> !untracked) }
))
console.log(result)answered Nov 20 '18 at 17:14
Mohammed AshfaqMohammed Ashfaq
1,6672612
edited Nov 21 '18 at 1:21
answered Nov 20 '18 at 17:14
Mohammed AshfaqMohammed Ashfaq
1,6672612
answered Nov 20 '18 at 17:14
Mohammed AshfaqMohammed Ashfaq
1,6672612
answered Nov 20 '18 at 17:14
Mohammed AshfaqMohammed Ashfaq
1,6672612
1,6672612
Uncaught ReferenceError: result is not defined. Also you are changing the structure of the given data as output will be=> [ { "key": [ { "name": "bar" } ] }, { "key": [ { "name": "baz" } ] } ]
– Monica Acha
Nov 20 '18 at 17:40
I forgot to declare the result variable.Thanks @Monic
– Mohammed Ashfaq
Nov 21 '18 at 1:23
add a comment |
Uncaught ReferenceError: result is not defined. Also you are changing the structure of the given data as output will be=> [ { "key": [ { "name": "bar" } ] }, { "key": [ { "name": "baz" } ] } ]
– Monica Acha
Nov 20 '18 at 17:40
I forgot to declare the result variable.Thanks @Monic
– Mohammed Ashfaq
Nov 21 '18 at 1:23
Uncaught ReferenceError: result is not defined. Also you are changing the structure of the given data as output will be=> [ { "key": [ { "name": "bar" } ] }, { "key": [ { "name": "baz" } ] } ]
– Monica Acha
Nov 20 '18 at 17:40
Uncaught ReferenceError: result is not defined. Also you are changing the structure of the given data as output will be=> [ { "key": [ { "name": "bar" } ] }, { "key": [ { "name": "baz" } ] } ]
– Monica Acha
Nov 20 '18 at 17:40
I forgot to declare the result variable.Thanks @Monic
– Mohammed Ashfaq
Nov 21 '18 at 1:23
I forgot to declare the result variable.Thanks @Monic
– Mohammed Ashfaq
Nov 21 '18 at 1:23
add a comment |
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