JavaScript inline function to array map
up vote
1
down vote
favorite
Is it possibile to use an inline return with javascript map function?
Instead of doing
array.map(token => { var x=new Object(); x[token]=words[token]; return x;} )
I would like to do it inline as doing
array.map(token => token )
so applying a inline method like
array.map(token => inline_function(token) )
I have tried like
Object.keys(chart).sort((a,b) => words[b]-words[a]).map(token => ( (token) => (new Object())[token]=words[token] )(token) )
but cannot get the return using a anonymous call with ().
Here is an example using the non inline case:
text = "Lorem ipsum dolor sit amet,n consectetur adipiscing elit,nsed do eiusmod tempor incididuntnut labore et dolore magna aliqua.nUt enim ad minim veniam,nquis nostrud exercitation ullamco laboris nisinut aliquip ex ea commodo consequat.nDuis aute irure dolor in reprehenderit in voluptate velit essencillum dolore eu fugiat nulla pariatur.nExcepteur sint occaecat cupidatat non proident,nLorem ipsum dolor sit amet etwas,nsunt in culpa qui officia deserunt mollit anim id est laborum"
words = text.split(/s+/g)
count = words.reduce(function(m, v) {
m[v] = m[v] ? m[v] + 1 : 1;
return m;
}, {})
sorted = Object.keys(count).sort((a, b) => count[b] - count[a]).map(token => {
var x = new Object();
x[token] = count[token];
return x;
})
console.log(sorted)javascript arrays
asked Nov 12 at 14:26
loretoparisi
7,55754771
add a comment |
up vote
1
down vote
favorite
Is it possibile to use an inline return with javascript map function?
Instead of doing
array.map(token => { var x=new Object(); x[token]=words[token]; return x;} )
I would like to do it inline as doing
array.map(token => token )
so applying a inline method like
array.map(token => inline_function(token) )
I have tried like
Object.keys(chart).sort((a,b) => words[b]-words[a]).map(token => ( (token) => (new Object())[token]=words[token] )(token) )
but cannot get the return using a anonymous call with ().
Here is an example using the non inline case:
text = "Lorem ipsum dolor sit amet,n consectetur adipiscing elit,nsed do eiusmod tempor incididuntnut labore et dolore magna aliqua.nUt enim ad minim veniam,nquis nostrud exercitation ullamco laboris nisinut aliquip ex ea commodo consequat.nDuis aute irure dolor in reprehenderit in voluptate velit essencillum dolore eu fugiat nulla pariatur.nExcepteur sint occaecat cupidatat non proident,nLorem ipsum dolor sit amet etwas,nsunt in culpa qui officia deserunt mollit anim id est laborum"
words = text.split(/s+/g)
count = words.reduce(function(m, v) {
m[v] = m[v] ? m[v] + 1 : 1;
return m;
}, {})
sorted = Object.keys(count).sort((a, b) => count[b] - count[a]).map(token => {
var x = new Object();
x[token] = count[token];
return x;
})
console.log(sorted)javascript arrays
asked Nov 12 at 14:26
loretoparisi
7,55754771
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is it possibile to use an inline return with javascript map function?
Instead of doing
array.map(token => { var x=new Object(); x[token]=words[token]; return x;} )
I would like to do it inline as doing
array.map(token => token )
so applying a inline method like
array.map(token => inline_function(token) )
I have tried like
Object.keys(chart).sort((a,b) => words[b]-words[a]).map(token => ( (token) => (new Object())[token]=words[token] )(token) )
but cannot get the return using a anonymous call with ().
Here is an example using the non inline case:
text = "Lorem ipsum dolor sit amet,n consectetur adipiscing elit,nsed do eiusmod tempor incididuntnut labore et dolore magna aliqua.nUt enim ad minim veniam,nquis nostrud exercitation ullamco laboris nisinut aliquip ex ea commodo consequat.nDuis aute irure dolor in reprehenderit in voluptate velit essencillum dolore eu fugiat nulla pariatur.nExcepteur sint occaecat cupidatat non proident,nLorem ipsum dolor sit amet etwas,nsunt in culpa qui officia deserunt mollit anim id est laborum"
words = text.split(/s+/g)
count = words.reduce(function(m, v) {
m[v] = m[v] ? m[v] + 1 : 1;
return m;
}, {})
sorted = Object.keys(count).sort((a, b) => count[b] - count[a]).map(token => {
var x = new Object();
x[token] = count[token];
return x;
})
console.log(sorted)javascript arrays
asked Nov 12 at 14:26
loretoparisi
7,55754771
Is it possibile to use an inline return with javascript map function?
Instead of doing
array.map(token => { var x=new Object(); x[token]=words[token]; return x;} )
I would like to do it inline as doing
array.map(token => token )
so applying a inline method like
array.map(token => inline_function(token) )
I have tried like
Object.keys(chart).sort((a,b) => words[b]-words[a]).map(token => ( (token) => (new Object())[token]=words[token] )(token) )
but cannot get the return using a anonymous call with ().
Here is an example using the non inline case:
text = "Lorem ipsum dolor sit amet,n consectetur adipiscing elit,nsed do eiusmod tempor incididuntnut labore et dolore magna aliqua.nUt enim ad minim veniam,nquis nostrud exercitation ullamco laboris nisinut aliquip ex ea commodo consequat.nDuis aute irure dolor in reprehenderit in voluptate velit essencillum dolore eu fugiat nulla pariatur.nExcepteur sint occaecat cupidatat non proident,nLorem ipsum dolor sit amet etwas,nsunt in culpa qui officia deserunt mollit anim id est laborum"
words = text.split(/s+/g)
count = words.reduce(function(m, v) {
m[v] = m[v] ? m[v] + 1 : 1;
return m;
}, {})
sorted = Object.keys(count).sort((a, b) => count[b] - count[a]).map(token => {
var x = new Object();
x[token] = count[token];
return x;
})
console.log(sorted)text = "Lorem ipsum dolor sit amet,n consectetur adipiscing elit,nsed do eiusmod tempor incididuntnut labore et dolore magna aliqua.nUt enim ad minim veniam,nquis nostrud exercitation ullamco laboris nisinut aliquip ex ea commodo consequat.nDuis aute irure dolor in reprehenderit in voluptate velit essencillum dolore eu fugiat nulla pariatur.nExcepteur sint occaecat cupidatat non proident,nLorem ipsum dolor sit amet etwas,nsunt in culpa qui officia deserunt mollit anim id est laborum"
words = text.split(/s+/g)
count = words.reduce(function(m, v) {
m[v] = m[v] ? m[v] + 1 : 1;
return m;
}, {})
sorted = Object.keys(count).sort((a, b) => count[b] - count[a]).map(token => {
var x = new Object();
x[token] = count[token];
return x;
})
console.log(sorted)text = "Lorem ipsum dolor sit amet,n consectetur adipiscing elit,nsed do eiusmod tempor incididuntnut labore et dolore magna aliqua.nUt enim ad minim veniam,nquis nostrud exercitation ullamco laboris nisinut aliquip ex ea commodo consequat.nDuis aute irure dolor in reprehenderit in voluptate velit essencillum dolore eu fugiat nulla pariatur.nExcepteur sint occaecat cupidatat non proident,nLorem ipsum dolor sit amet etwas,nsunt in culpa qui officia deserunt mollit anim id est laborum"
words = text.split(/s+/g)
count = words.reduce(function(m, v) {
m[v] = m[v] ? m[v] + 1 : 1;
return m;
}, {})
sorted = Object.keys(count).sort((a, b) => count[b] - count[a]).map(token => {
var x = new Object();
x[token] = count[token];
return x;
})
console.log(sorted)javascript arrays
javascript arrays
asked Nov 12 at 14:26
loretoparisi
7,55754771
asked Nov 12 at 14:26
loretoparisi
7,55754771
asked Nov 12 at 14:26
loretoparisi
7,55754771
asked Nov 12 at 14:26
loretoparisi
7,55754771
asked Nov 12 at 14:26
loretoparisi
7,55754771
7,55754771
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
All of those are inline functions. Fundamentally, you're asking how to write this callback:
token => {
var x = new Object();
x[token] = count[token];
return x;
}
with a concise body rather than a function body. In this specific case, you can do that with a computed property name:
token => ({[token]: count[token]})
// ^^^^^^^---- computed property name, new in ES2015
Note we need the parens (()) around the concise body because the first character of it would be { (the beginning of the object initializer) otherwise, which would make the parser think it had a function body rather than a concise one.
In the more general case, you can't turn a function body into a concise body unless everything fits within a single expression, such as the object initializer above. Sometimes you can use (or arguably abuse) the comma operator for that, with the left-hand operand to , having side-effects, but in such cases it's usually clearer to just leave the function body instead.
As an example of that (ab)use, this logs the values prior to doubling them:
doubled = values.map(v => (console.log(v), v * 2));
I don't recommend it, but you'll see it done. :-)
answered Nov 12 at 14:31
T.J. Crowder
674k11811911287
1
Beat me to it by 2 seconds! :)
– ibrahim mahrir
Nov 12 at 14:32
2
@ibrahimmahrir - But perhaps you wouldn't have had the silly typo (=instead of:). Fixed now. :-)
– T.J. Crowder
Nov 12 at 14:34
amazing I missed the computed property. Thanks T.J. ;)
– loretoparisi
Nov 12 at 14:35
add a comment |
up vote
2
down vote
Why not pass a function to .map?
function myFunction(token){
var x = new Object();
x[token] = count[token];
return x;
}
sorted = Object.keys(count).sort((a, b) => count[b] - count[a]).map(myFunction);
answered Nov 12 at 14:30
Adriani6
4,05021123
Yes that is correct. I was trying to understand if I can infer()to inline function call like like(() => something))()
– loretoparisi
Nov 12 at 14:31
1
@loretoparisi Ah, apologies. I must of misunderstood what you're asking.
– Adriani6
Nov 12 at 14:32
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
All of those are inline functions. Fundamentally, you're asking how to write this callback:
token => {
var x = new Object();
x[token] = count[token];
return x;
}
with a concise body rather than a function body. In this specific case, you can do that with a computed property name:
token => ({[token]: count[token]})
// ^^^^^^^---- computed property name, new in ES2015
Note we need the parens (()) around the concise body because the first character of it would be { (the beginning of the object initializer) otherwise, which would make the parser think it had a function body rather than a concise one.
In the more general case, you can't turn a function body into a concise body unless everything fits within a single expression, such as the object initializer above. Sometimes you can use (or arguably abuse) the comma operator for that, with the left-hand operand to , having side-effects, but in such cases it's usually clearer to just leave the function body instead.
As an example of that (ab)use, this logs the values prior to doubling them:
doubled = values.map(v => (console.log(v), v * 2));
I don't recommend it, but you'll see it done. :-)
answered Nov 12 at 14:31
T.J. Crowder
674k11811911287
1
Beat me to it by 2 seconds! :)
– ibrahim mahrir
Nov 12 at 14:32
2
@ibrahimmahrir - But perhaps you wouldn't have had the silly typo (=instead of:). Fixed now. :-)
– T.J. Crowder
Nov 12 at 14:34
amazing I missed the computed property. Thanks T.J. ;)
– loretoparisi
Nov 12 at 14:35
add a comment |
up vote
3
down vote
accepted
All of those are inline functions. Fundamentally, you're asking how to write this callback:
token => {
var x = new Object();
x[token] = count[token];
return x;
}
with a concise body rather than a function body. In this specific case, you can do that with a computed property name:
token => ({[token]: count[token]})
// ^^^^^^^---- computed property name, new in ES2015
Note we need the parens (()) around the concise body because the first character of it would be { (the beginning of the object initializer) otherwise, which would make the parser think it had a function body rather than a concise one.
In the more general case, you can't turn a function body into a concise body unless everything fits within a single expression, such as the object initializer above. Sometimes you can use (or arguably abuse) the comma operator for that, with the left-hand operand to , having side-effects, but in such cases it's usually clearer to just leave the function body instead.
As an example of that (ab)use, this logs the values prior to doubling them:
doubled = values.map(v => (console.log(v), v * 2));
I don't recommend it, but you'll see it done. :-)
answered Nov 12 at 14:31
T.J. Crowder
674k11811911287
1
Beat me to it by 2 seconds! :)
– ibrahim mahrir
Nov 12 at 14:32
2
@ibrahimmahrir - But perhaps you wouldn't have had the silly typo (=instead of:). Fixed now. :-)
– T.J. Crowder
Nov 12 at 14:34
amazing I missed the computed property. Thanks T.J. ;)
– loretoparisi
Nov 12 at 14:35
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
All of those are inline functions. Fundamentally, you're asking how to write this callback:
token => {
var x = new Object();
x[token] = count[token];
return x;
}
with a concise body rather than a function body. In this specific case, you can do that with a computed property name:
token => ({[token]: count[token]})
// ^^^^^^^---- computed property name, new in ES2015
Note we need the parens (()) around the concise body because the first character of it would be { (the beginning of the object initializer) otherwise, which would make the parser think it had a function body rather than a concise one.
In the more general case, you can't turn a function body into a concise body unless everything fits within a single expression, such as the object initializer above. Sometimes you can use (or arguably abuse) the comma operator for that, with the left-hand operand to , having side-effects, but in such cases it's usually clearer to just leave the function body instead.
As an example of that (ab)use, this logs the values prior to doubling them:
doubled = values.map(v => (console.log(v), v * 2));
I don't recommend it, but you'll see it done. :-)
answered Nov 12 at 14:31
T.J. Crowder
674k11811911287
All of those are inline functions. Fundamentally, you're asking how to write this callback:
token => {
var x = new Object();
x[token] = count[token];
return x;
}
with a concise body rather than a function body. In this specific case, you can do that with a computed property name:
token => ({[token]: count[token]})
// ^^^^^^^---- computed property name, new in ES2015
Note we need the parens (()) around the concise body because the first character of it would be { (the beginning of the object initializer) otherwise, which would make the parser think it had a function body rather than a concise one.
In the more general case, you can't turn a function body into a concise body unless everything fits within a single expression, such as the object initializer above. Sometimes you can use (or arguably abuse) the comma operator for that, with the left-hand operand to , having side-effects, but in such cases it's usually clearer to just leave the function body instead.
As an example of that (ab)use, this logs the values prior to doubling them:
doubled = values.map(v => (console.log(v), v * 2));
I don't recommend it, but you'll see it done. :-)
answered Nov 12 at 14:31
T.J. Crowder
674k11811911287
edited Nov 12 at 14:36
answered Nov 12 at 14:31
T.J. Crowder
674k11811911287
answered Nov 12 at 14:31
T.J. Crowder
674k11811911287
answered Nov 12 at 14:31
T.J. Crowder
674k11811911287
674k11811911287
1
Beat me to it by 2 seconds! :)
– ibrahim mahrir
Nov 12 at 14:32
2
@ibrahimmahrir - But perhaps you wouldn't have had the silly typo (=instead of:). Fixed now. :-)
– T.J. Crowder
Nov 12 at 14:34
amazing I missed the computed property. Thanks T.J. ;)
– loretoparisi
Nov 12 at 14:35
add a comment |
1
Beat me to it by 2 seconds! :)
– ibrahim mahrir
Nov 12 at 14:32
2
@ibrahimmahrir - But perhaps you wouldn't have had the silly typo (=instead of:). Fixed now. :-)
– T.J. Crowder
Nov 12 at 14:34
amazing I missed the computed property. Thanks T.J. ;)
– loretoparisi
Nov 12 at 14:35
1
1
Beat me to it by 2 seconds! :)
– ibrahim mahrir
Nov 12 at 14:32
Beat me to it by 2 seconds! :)
– ibrahim mahrir
Nov 12 at 14:32
2
2
@ibrahimmahrir - But perhaps you wouldn't have had the silly typo (
= instead of :). Fixed now. :-)– T.J. Crowder
Nov 12 at 14:34
@ibrahimmahrir - But perhaps you wouldn't have had the silly typo (
= instead of :). Fixed now. :-)– T.J. Crowder
Nov 12 at 14:34
amazing I missed the computed property. Thanks T.J. ;)
– loretoparisi
Nov 12 at 14:35
amazing I missed the computed property. Thanks T.J. ;)
– loretoparisi
Nov 12 at 14:35
add a comment |
up vote
2
down vote
Why not pass a function to .map?
function myFunction(token){
var x = new Object();
x[token] = count[token];
return x;
}
sorted = Object.keys(count).sort((a, b) => count[b] - count[a]).map(myFunction);
answered Nov 12 at 14:30
Adriani6
4,05021123
Yes that is correct. I was trying to understand if I can infer()to inline function call like like(() => something))()
– loretoparisi
Nov 12 at 14:31
1
@loretoparisi Ah, apologies. I must of misunderstood what you're asking.
– Adriani6
Nov 12 at 14:32
add a comment |
up vote
2
down vote
Why not pass a function to .map?
function myFunction(token){
var x = new Object();
x[token] = count[token];
return x;
}
sorted = Object.keys(count).sort((a, b) => count[b] - count[a]).map(myFunction);
answered Nov 12 at 14:30
Adriani6
4,05021123
Yes that is correct. I was trying to understand if I can infer()to inline function call like like(() => something))()
– loretoparisi
Nov 12 at 14:31
1
@loretoparisi Ah, apologies. I must of misunderstood what you're asking.
– Adriani6
Nov 12 at 14:32
add a comment |
up vote
2
down vote
up vote
2
down vote
Why not pass a function to .map?
function myFunction(token){
var x = new Object();
x[token] = count[token];
return x;
}
sorted = Object.keys(count).sort((a, b) => count[b] - count[a]).map(myFunction);
answered Nov 12 at 14:30
Adriani6
4,05021123
Why not pass a function to .map?
function myFunction(token){
var x = new Object();
x[token] = count[token];
return x;
}
sorted = Object.keys(count).sort((a, b) => count[b] - count[a]).map(myFunction);
answered Nov 12 at 14:30
Adriani6
4,05021123
answered Nov 12 at 14:30
Adriani6
4,05021123
answered Nov 12 at 14:30
Adriani6
4,05021123
answered Nov 12 at 14:30
Adriani6
4,05021123
4,05021123
Yes that is correct. I was trying to understand if I can infer()to inline function call like like(() => something))()
– loretoparisi
Nov 12 at 14:31
1
@loretoparisi Ah, apologies. I must of misunderstood what you're asking.
– Adriani6
Nov 12 at 14:32
add a comment |
Yes that is correct. I was trying to understand if I can infer()to inline function call like like(() => something))()
– loretoparisi
Nov 12 at 14:31
1
@loretoparisi Ah, apologies. I must of misunderstood what you're asking.
– Adriani6
Nov 12 at 14:32
Yes that is correct. I was trying to understand if I can infer
() to inline function call like like (() => something))()– loretoparisi
Nov 12 at 14:31
Yes that is correct. I was trying to understand if I can infer
() to inline function call like like (() => something))()– loretoparisi
Nov 12 at 14:31
1
1
@loretoparisi Ah, apologies. I must of misunderstood what you're asking.
– Adriani6
Nov 12 at 14:32
@loretoparisi Ah, apologies. I must of misunderstood what you're asking.
– Adriani6
Nov 12 at 14:32
add a comment |
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