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Using a map function on a 'Map' to change values

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11

I can use a Map and then set values:

const a = new Map([["a", "b"], ["c", "d"]])

Now if I want to apply a function to all values in a functional way (without for ... of or .forEach), I thought I could have done something like this:

const b = a.map(([k, v]) => [k, doSomethingWith(v)]);

But there is no map function on a Map. Is there a built-in/elegant way to map a Map?

javascript dictionary ecmascript-6

share|improve this question

edited Apr 24 at 10:26

Peter Mortensen

14k1987114

asked Apr 24 at 8:56

rap-2-hrap-2-h

10.4k1176141

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Use Map#entries then .map on them a.entries().map(...)
  
  – ponury-kostek
  Apr 24 at 9:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ponury-kostek It won't work because Map.entries() return an Iterator (not Array); and Iterators haven't .map method...
  
  – FZs
  Apr 24 at 9:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Unfortunately, there is no elegant way. You have to process the entries and then produce a new Map out of the result.
  
  – VLAZ
  Apr 24 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of Using map/filter behavior on Map instance
  
  – Bergi
  Apr 24 at 13:57
  

  
  
  
  

add a comment | 

11

I can use a Map and then set values:

const a = new Map([["a", "b"], ["c", "d"]])

Now if I want to apply a function to all values in a functional way (without for ... of or .forEach), I thought I could have done something like this:

const b = a.map(([k, v]) => [k, doSomethingWith(v)]);

But there is no map function on a Map. Is there a built-in/elegant way to map a Map?

javascript dictionary ecmascript-6

share|improve this question

edited Apr 24 at 10:26

Peter Mortensen

14k1987114

asked Apr 24 at 8:56

rap-2-hrap-2-h

10.4k1176141

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Use Map#entries then .map on them a.entries().map(...)
  
  – ponury-kostek
  Apr 24 at 9:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ponury-kostek It won't work because Map.entries() return an Iterator (not Array); and Iterators haven't .map method...
  
  – FZs
  Apr 24 at 9:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Unfortunately, there is no elegant way. You have to process the entries and then produce a new Map out of the result.
  
  – VLAZ
  Apr 24 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of Using map/filter behavior on Map instance
  
  – Bergi
  Apr 24 at 13:57
  

  
  
  
  

add a comment | 

I can use a Map and then set values:

const a = new Map([["a", "b"], ["c", "d"]])

Now if I want to apply a function to all values in a functional way (without for ... of or .forEach), I thought I could have done something like this:

const b = a.map(([k, v]) => [k, doSomethingWith(v)]);

But there is no map function on a Map. Is there a built-in/elegant way to map a Map?

javascript dictionary ecmascript-6

share|improve this question

edited Apr 24 at 10:26

Peter Mortensen

14k1987114

asked Apr 24 at 8:56

rap-2-hrap-2-h

10.4k1176141

const a = new Map([["a", "b"], ["c", "d"]])

const b = a.map(([k, v]) => [k, doSomethingWith(v)]);

share|improve this question

edited Apr 24 at 10:26

Peter Mortensen

14k1987114

asked Apr 24 at 8:56

rap-2-hrap-2-h

10.4k1176141

share|improve this question

edited Apr 24 at 10:26

Peter Mortensen

14k1987114

asked Apr 24 at 8:56

rap-2-hrap-2-h

10.4k1176141

edited Apr 24 at 10:26

Peter Mortensen

14k1987114

edited Apr 24 at 10:26

Peter Mortensen

14k1987114

asked Apr 24 at 8:56

rap-2-hrap-2-h

10.4k1176141

asked Apr 24 at 8:56

rap-2-hrap-2-h

10.4k1176141

5 Answers
5

active

oldest

votes

12

You could use Array.from for getting entries and mapping new values and take this result for a new Map.

const b = new Map(Array.from(
 a, 
 ([k, v]) => [k, doSomethingWith(v)]
));

share|improve this answer

answered Apr 24 at 9:04

Nina ScholzNina Scholz

202k16115184

add a comment | 

4

The most elegant/concise way I am aware of is by converting the map to an array with the spread operator (...), applying .map to that array and then creating a Map from it again:

const a = new Map([["a", "b"], ["c", "d"]])
const b = new Map([...a].map(([k,v])=>[k, v.toUpperCase()]))
// b: Map(2) "a" => "B", "c" => "D"

share|improve this answer

edited Apr 24 at 10:20

answered Apr 24 at 10:14

nitzelnitzel

31127

add a comment | 

2

There are no builtin methods for this (yet!). The most elegant way currently is to use generators:

const b = new Map((function*() 
 for (const [k, v] of a)
 yield [k, doSomethingWith(v)];
)());

I would however recommend to write helper functions for this that work with arbitrary iterables:

function* mapValue(iterable, callback) 
 for (const [k, v] of a)
 yield [k, callback(v)];

const b = new Map(mapValue(a, doSomethingWith));

share|improve this answer

answered Apr 24 at 14:10

BergiBergi

385k64596926

add a comment | 

0

You could use Symbol.iterator for changing or creating a new Map.

function f (iterator) 
 for (let item of iterator1) 
 item[0] = 1;
 item[1] = 'hello world';
 console.log(item);
 


var map1 = new Map();
map1.set('0', 'foo');
map1.set(1, 'bar');
var iterator1 = map1[Symbol.iterator]();

f(iterator1);

share|improve this answer

answered Apr 24 at 9:12

koλzarkoλzar

5761518

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  for (let item of iterator1) that should be iterator, no?
  
  – VLAZ
  Apr 24 at 9:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And the question was: 'in a functional way (without for ... of or .forEach)'. This example uses for...of
  
  – FZs
  Apr 24 at 9:15
  

  
  
  
  

add a comment | 

0

You could do that like this:

let b = new Map(a)
b.forEach((value,key,myMap) => myMap.set(key, dosomething(value)))

or:

let b
(b = new Map(a)).forEach((value,key,myMap) => myMap.set(key, dosomething(value)))

share|improve this answer

edited Apr 24 at 9:35

answered Apr 24 at 9:26

欧阳维杰欧阳维杰

431512

add a comment | 

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12

You could use Array.from for getting entries and mapping new values and take this result for a new Map.

const b = new Map(Array.from(
 a, 
 ([k, v]) => [k, doSomethingWith(v)]
));

share|improve this answer

answered Apr 24 at 9:04

Nina ScholzNina Scholz

202k16115184

add a comment | 

12

You could use Array.from for getting entries and mapping new values and take this result for a new Map.

const b = new Map(Array.from(
 a, 
 ([k, v]) => [k, doSomethingWith(v)]
));

share|improve this answer

answered Apr 24 at 9:04

Nina ScholzNina Scholz

202k16115184

add a comment | 

You could use Array.from for getting entries and mapping new values and take this result for a new Map.

const b = new Map(Array.from(
 a, 
 ([k, v]) => [k, doSomethingWith(v)]
));

share|improve this answer

answered Apr 24 at 9:04

Nina ScholzNina Scholz

202k16115184

const b = new Map(Array.from(
 a, 
 ([k, v]) => [k, doSomethingWith(v)]
));

share|improve this answer

answered Apr 24 at 9:04

Nina ScholzNina Scholz

202k16115184

answered Apr 24 at 9:04

Nina ScholzNina Scholz

202k16115184

answered Apr 24 at 9:04

Nina ScholzNina Scholz

202k16115184

4

The most elegant/concise way I am aware of is by converting the map to an array with the spread operator (...), applying .map to that array and then creating a Map from it again:

const a = new Map([["a", "b"], ["c", "d"]])
const b = new Map([...a].map(([k,v])=>[k, v.toUpperCase()]))
// b: Map(2) "a" => "B", "c" => "D"

share|improve this answer

edited Apr 24 at 10:20

answered Apr 24 at 10:14

nitzelnitzel

31127

add a comment | 

4

The most elegant/concise way I am aware of is by converting the map to an array with the spread operator (...), applying .map to that array and then creating a Map from it again:

const a = new Map([["a", "b"], ["c", "d"]])
const b = new Map([...a].map(([k,v])=>[k, v.toUpperCase()]))
// b: Map(2) "a" => "B", "c" => "D"

share|improve this answer

edited Apr 24 at 10:20

answered Apr 24 at 10:14

nitzelnitzel

31127

add a comment | 

The most elegant/concise way I am aware of is by converting the map to an array with the spread operator (...), applying .map to that array and then creating a Map from it again:

const a = new Map([["a", "b"], ["c", "d"]])
const b = new Map([...a].map(([k,v])=>[k, v.toUpperCase()]))
// b: Map(2) "a" => "B", "c" => "D"

share|improve this answer

edited Apr 24 at 10:20

answered Apr 24 at 10:14

nitzelnitzel

31127

const a = new Map([["a", "b"], ["c", "d"]])
const b = new Map([...a].map(([k,v])=>[k, v.toUpperCase()]))
// b: Map(2) "a" => "B", "c" => "D"

share|improve this answer

edited Apr 24 at 10:20

answered Apr 24 at 10:14

nitzelnitzel

31127

answered Apr 24 at 10:14

nitzelnitzel

31127

answered Apr 24 at 10:14

nitzelnitzel

31127

2

There are no builtin methods for this (yet!). The most elegant way currently is to use generators:

const b = new Map((function*() 
 for (const [k, v] of a)
 yield [k, doSomethingWith(v)];
)());

I would however recommend to write helper functions for this that work with arbitrary iterables:

function* mapValue(iterable, callback) 
 for (const [k, v] of a)
 yield [k, callback(v)];

const b = new Map(mapValue(a, doSomethingWith));

share|improve this answer

answered Apr 24 at 14:10

BergiBergi

385k64596926

add a comment | 

2

There are no builtin methods for this (yet!). The most elegant way currently is to use generators:

const b = new Map((function*() 
 for (const [k, v] of a)
 yield [k, doSomethingWith(v)];
)());

I would however recommend to write helper functions for this that work with arbitrary iterables:

function* mapValue(iterable, callback) 
 for (const [k, v] of a)
 yield [k, callback(v)];

const b = new Map(mapValue(a, doSomethingWith));

share|improve this answer

answered Apr 24 at 14:10

BergiBergi

385k64596926

add a comment | 

There are no builtin methods for this (yet!). The most elegant way currently is to use generators:

const b = new Map((function*() 
 for (const [k, v] of a)
 yield [k, doSomethingWith(v)];
)());

I would however recommend to write helper functions for this that work with arbitrary iterables:

function* mapValue(iterable, callback) 
 for (const [k, v] of a)
 yield [k, callback(v)];

const b = new Map(mapValue(a, doSomethingWith));

share|improve this answer

answered Apr 24 at 14:10

BergiBergi

385k64596926

const b = new Map((function*() 
 for (const [k, v] of a)
 yield [k, doSomethingWith(v)];
)());

function* mapValue(iterable, callback) 
 for (const [k, v] of a)
 yield [k, callback(v)];

const b = new Map(mapValue(a, doSomethingWith));

share|improve this answer

answered Apr 24 at 14:10

BergiBergi

385k64596926

answered Apr 24 at 14:10

BergiBergi

385k64596926

answered Apr 24 at 14:10

BergiBergi

385k64596926

0

You could use Symbol.iterator for changing or creating a new Map.

function f (iterator) 
 for (let item of iterator1) 
 item[0] = 1;
 item[1] = 'hello world';
 console.log(item);
 


var map1 = new Map();
map1.set('0', 'foo');
map1.set(1, 'bar');
var iterator1 = map1[Symbol.iterator]();

f(iterator1);

share|improve this answer

answered Apr 24 at 9:12

koλzarkoλzar

5761518

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  for (let item of iterator1) that should be iterator, no?
  
  – VLAZ
  Apr 24 at 9:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And the question was: 'in a functional way (without for ... of or .forEach)'. This example uses for...of
  
  – FZs
  Apr 24 at 9:15
  

  
  
  
  

add a comment | 

0

You could use Symbol.iterator for changing or creating a new Map.

function f (iterator) 
 for (let item of iterator1) 
 item[0] = 1;
 item[1] = 'hello world';
 console.log(item);
 


var map1 = new Map();
map1.set('0', 'foo');
map1.set(1, 'bar');
var iterator1 = map1[Symbol.iterator]();

f(iterator1);

share|improve this answer

answered Apr 24 at 9:12

koλzarkoλzar

5761518

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  for (let item of iterator1) that should be iterator, no?
  
  – VLAZ
  Apr 24 at 9:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And the question was: 'in a functional way (without for ... of or .forEach)'. This example uses for...of
  
  – FZs
  Apr 24 at 9:15
  

  
  
  
  

add a comment | 

You could use Symbol.iterator for changing or creating a new Map.

function f (iterator) 
 for (let item of iterator1) 
 item[0] = 1;
 item[1] = 'hello world';
 console.log(item);
 


var map1 = new Map();
map1.set('0', 'foo');
map1.set(1, 'bar');
var iterator1 = map1[Symbol.iterator]();

f(iterator1);

share|improve this answer

answered Apr 24 at 9:12

koλzarkoλzar

5761518

function f (iterator) 
 for (let item of iterator1) 
 item[0] = 1;
 item[1] = 'hello world';
 console.log(item);
 


var map1 = new Map();
map1.set('0', 'foo');
map1.set(1, 'bar');
var iterator1 = map1[Symbol.iterator]();

f(iterator1);

function f (iterator) 
 for (let item of iterator1) 
 item[0] = 1;
 item[1] = 'hello world';
 console.log(item);
 


var map1 = new Map();
map1.set('0', 'foo');
map1.set(1, 'bar');
var iterator1 = map1[Symbol.iterator]();

f(iterator1);

function f (iterator) 
 for (let item of iterator1) 
 item[0] = 1;
 item[1] = 'hello world';
 console.log(item);
 


var map1 = new Map();
map1.set('0', 'foo');
map1.set(1, 'bar');
var iterator1 = map1[Symbol.iterator]();

f(iterator1);

share|improve this answer

answered Apr 24 at 9:12

koλzarkoλzar

5761518

answered Apr 24 at 9:12

koλzarkoλzar

5761518

answered Apr 24 at 9:12

koλzarkoλzar

5761518

0

You could do that like this:

let b = new Map(a)
b.forEach((value,key,myMap) => myMap.set(key, dosomething(value)))

or:

let b
(b = new Map(a)).forEach((value,key,myMap) => myMap.set(key, dosomething(value)))

share|improve this answer

edited Apr 24 at 9:35

answered Apr 24 at 9:26

欧阳维杰欧阳维杰

431512

add a comment | 

0

You could do that like this:

let b = new Map(a)
b.forEach((value,key,myMap) => myMap.set(key, dosomething(value)))

or:

let b
(b = new Map(a)).forEach((value,key,myMap) => myMap.set(key, dosomething(value)))

share|improve this answer

edited Apr 24 at 9:35

answered Apr 24 at 9:26

欧阳维杰欧阳维杰

431512

add a comment | 

You could do that like this:

let b = new Map(a)
b.forEach((value,key,myMap) => myMap.set(key, dosomething(value)))

or:

let b
(b = new Map(a)).forEach((value,key,myMap) => myMap.set(key, dosomething(value)))

share|improve this answer

edited Apr 24 at 9:35

answered Apr 24 at 9:26

欧阳维杰欧阳维杰

431512

let b = new Map(a)
b.forEach((value,key,myMap) => myMap.set(key, dosomething(value)))

let b
(b = new Map(a)).forEach((value,key,myMap) => myMap.set(key, dosomething(value)))

share|improve this answer

edited Apr 24 at 9:35

answered Apr 24 at 9:26

欧阳维杰欧阳维杰

431512

answered Apr 24 at 9:26

欧阳维杰欧阳维杰

431512

answered Apr 24 at 9:26

欧阳维杰欧阳维杰

431512