Test for an array being subset of another master array
$begingroup$
I was trying to build a small utility function to check if an array is part of other array. It's testing if an array is a subset of another master array.
const masterArray = [1,2,3,4,5,6];
const candidateArray = [2,5,6];
//Test for subset.
//create a set from the two.
const s1 = new Set(masterArray.concat(candidateArray));
//Compare the sizes of the master array and the created set
//If the sizes are same, no new elements are added that means
//the candidate is complete subset.
s1.size === masterArray.length;
Can this be handled in a better way?
typescript
edited Dec 25 '18 at 20:42
Jamal♦
30.3k11116226
asked Jul 16 '18 at 6:50
HarshalHarshal
1064
$endgroup$
bumped to the homepage by Community♦ 18 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
I was trying to build a small utility function to check if an array is part of other array. It's testing if an array is a subset of another master array.
const masterArray = [1,2,3,4,5,6];
const candidateArray = [2,5,6];
//Test for subset.
//create a set from the two.
const s1 = new Set(masterArray.concat(candidateArray));
//Compare the sizes of the master array and the created set
//If the sizes are same, no new elements are added that means
//the candidate is complete subset.
s1.size === masterArray.length;
Can this be handled in a better way?
typescript
edited Dec 25 '18 at 20:42
Jamal♦
30.3k11116226
asked Jul 16 '18 at 6:50
HarshalHarshal
1064
$endgroup$
bumped to the homepage by Community♦ 18 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
2
$begingroup$
Wouldn't that fail ifmasterArrayhas duplicate elements?
$endgroup$
– Martin R
Jul 16 '18 at 7:15
$begingroup$
Good point. In my case the master array is always unique.
$endgroup$
– Harshal
Jul 16 '18 at 7:22
$begingroup$
I can enhance it by comparing with a new set created from master array(new Set(masterArray)).size
$endgroup$
– Harshal
Jul 16 '18 at 8:46
add a comment |
$begingroup$
I was trying to build a small utility function to check if an array is part of other array. It's testing if an array is a subset of another master array.
const masterArray = [1,2,3,4,5,6];
const candidateArray = [2,5,6];
//Test for subset.
//create a set from the two.
const s1 = new Set(masterArray.concat(candidateArray));
//Compare the sizes of the master array and the created set
//If the sizes are same, no new elements are added that means
//the candidate is complete subset.
s1.size === masterArray.length;
Can this be handled in a better way?
typescript
edited Dec 25 '18 at 20:42
Jamal♦
30.3k11116226
asked Jul 16 '18 at 6:50
HarshalHarshal
1064
$endgroup$
I was trying to build a small utility function to check if an array is part of other array. It's testing if an array is a subset of another master array.
const masterArray = [1,2,3,4,5,6];
const candidateArray = [2,5,6];
//Test for subset.
//create a set from the two.
const s1 = new Set(masterArray.concat(candidateArray));
//Compare the sizes of the master array and the created set
//If the sizes are same, no new elements are added that means
//the candidate is complete subset.
s1.size === masterArray.length;
Can this be handled in a better way?
typescript
typescript
edited Dec 25 '18 at 20:42
Jamal♦
30.3k11116226
asked Jul 16 '18 at 6:50
HarshalHarshal
1064
edited Dec 25 '18 at 20:42
Jamal♦
30.3k11116226
asked Jul 16 '18 at 6:50
HarshalHarshal
1064
edited Dec 25 '18 at 20:42
Jamal♦
30.3k11116226
edited Dec 25 '18 at 20:42
Jamal♦
30.3k11116226
edited Dec 25 '18 at 20:42
Jamal♦
30.3k11116226
30.3k11116226
asked Jul 16 '18 at 6:50
HarshalHarshal
1064
asked Jul 16 '18 at 6:50
HarshalHarshal
1064
asked Jul 16 '18 at 6:50
HarshalHarshal
1064
1064
bumped to the homepage by Community♦ 18 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 18 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
2
$begingroup$
Wouldn't that fail ifmasterArrayhas duplicate elements?
$endgroup$
– Martin R
Jul 16 '18 at 7:15
$begingroup$
Good point. In my case the master array is always unique.
$endgroup$
– Harshal
Jul 16 '18 at 7:22
$begingroup$
I can enhance it by comparing with a new set created from master array(new Set(masterArray)).size
$endgroup$
– Harshal
Jul 16 '18 at 8:46
add a comment |
2
$begingroup$
Wouldn't that fail ifmasterArrayhas duplicate elements?
$endgroup$
– Martin R
Jul 16 '18 at 7:15
$begingroup$
Good point. In my case the master array is always unique.
$endgroup$
– Harshal
Jul 16 '18 at 7:22
$begingroup$
I can enhance it by comparing with a new set created from master array(new Set(masterArray)).size
$endgroup$
– Harshal
Jul 16 '18 at 8:46
2
2
$begingroup$
Wouldn't that fail if
masterArray has duplicate elements?$endgroup$
– Martin R
Jul 16 '18 at 7:15
$begingroup$
Wouldn't that fail if
masterArray has duplicate elements?$endgroup$
– Martin R
Jul 16 '18 at 7:15
$begingroup$
Good point. In my case the master array is always unique.
$endgroup$
– Harshal
Jul 16 '18 at 7:22
$begingroup$
Good point. In my case the master array is always unique.
$endgroup$
– Harshal
Jul 16 '18 at 7:22
$begingroup$
I can enhance it by comparing with a new set created from master array
(new Set(masterArray)).size$endgroup$
– Harshal
Jul 16 '18 at 8:46
$begingroup$
I can enhance it by comparing with a new set created from master array
(new Set(masterArray)).size$endgroup$
– Harshal
Jul 16 '18 at 8:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So the key here is that I value code that's obvious to others.
Someone else is likely to try to stuff duplicate elements in unless it requires a type refactor; making const masterSet: Set<number> and using that throughout the code is a Good Idea. That "someone else" can often be you two months down the line, too.
In this case I would not say that the point of this length comparison is obvious; it will work but it takes some thinking -- that is why it has bugs with duplicate elements. In this case I would write something which I find more straightforward like,
function isSubsetOf<x>(sub: Iterable<x>, sup: Set<x>): boolean {
for (const x of sub) {
if (!sup.has(x)) {
return false;
}
}
return true;
}
I am also hinting through the type system how this function works by insisting that the subset is any iterable -- so I must be iterating through it.
With a restriction to an array and a set, one can get a little swankier by using the reduce function,
function isSubsetOf<x>(sub: Array<x>, sup: Set<x>): boolean {
return sub.reduce((acc, x) => acc && sup.has(x), true);
}
but probably not all other readers will find that as intuitive as I do, and the former has slightly better runtime characteristics on very large lists that are obviously not subsets.
answered Dec 26 '18 at 3:44
CR DrostCR Drost
22113
$endgroup$
add a comment |
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$begingroup$
So the key here is that I value code that's obvious to others.
Someone else is likely to try to stuff duplicate elements in unless it requires a type refactor; making const masterSet: Set<number> and using that throughout the code is a Good Idea. That "someone else" can often be you two months down the line, too.
In this case I would not say that the point of this length comparison is obvious; it will work but it takes some thinking -- that is why it has bugs with duplicate elements. In this case I would write something which I find more straightforward like,
function isSubsetOf<x>(sub: Iterable<x>, sup: Set<x>): boolean {
for (const x of sub) {
if (!sup.has(x)) {
return false;
}
}
return true;
}
I am also hinting through the type system how this function works by insisting that the subset is any iterable -- so I must be iterating through it.
With a restriction to an array and a set, one can get a little swankier by using the reduce function,
function isSubsetOf<x>(sub: Array<x>, sup: Set<x>): boolean {
return sub.reduce((acc, x) => acc && sup.has(x), true);
}
but probably not all other readers will find that as intuitive as I do, and the former has slightly better runtime characteristics on very large lists that are obviously not subsets.
answered Dec 26 '18 at 3:44
CR DrostCR Drost
22113
$endgroup$
add a comment |
$begingroup$
So the key here is that I value code that's obvious to others.
Someone else is likely to try to stuff duplicate elements in unless it requires a type refactor; making const masterSet: Set<number> and using that throughout the code is a Good Idea. That "someone else" can often be you two months down the line, too.
In this case I would not say that the point of this length comparison is obvious; it will work but it takes some thinking -- that is why it has bugs with duplicate elements. In this case I would write something which I find more straightforward like,
function isSubsetOf<x>(sub: Iterable<x>, sup: Set<x>): boolean {
for (const x of sub) {
if (!sup.has(x)) {
return false;
}
}
return true;
}
I am also hinting through the type system how this function works by insisting that the subset is any iterable -- so I must be iterating through it.
With a restriction to an array and a set, one can get a little swankier by using the reduce function,
function isSubsetOf<x>(sub: Array<x>, sup: Set<x>): boolean {
return sub.reduce((acc, x) => acc && sup.has(x), true);
}
but probably not all other readers will find that as intuitive as I do, and the former has slightly better runtime characteristics on very large lists that are obviously not subsets.
answered Dec 26 '18 at 3:44
CR DrostCR Drost
22113
$endgroup$
add a comment |
$begingroup$
So the key here is that I value code that's obvious to others.
Someone else is likely to try to stuff duplicate elements in unless it requires a type refactor; making const masterSet: Set<number> and using that throughout the code is a Good Idea. That "someone else" can often be you two months down the line, too.
In this case I would not say that the point of this length comparison is obvious; it will work but it takes some thinking -- that is why it has bugs with duplicate elements. In this case I would write something which I find more straightforward like,
function isSubsetOf<x>(sub: Iterable<x>, sup: Set<x>): boolean {
for (const x of sub) {
if (!sup.has(x)) {
return false;
}
}
return true;
}
I am also hinting through the type system how this function works by insisting that the subset is any iterable -- so I must be iterating through it.
With a restriction to an array and a set, one can get a little swankier by using the reduce function,
function isSubsetOf<x>(sub: Array<x>, sup: Set<x>): boolean {
return sub.reduce((acc, x) => acc && sup.has(x), true);
}
but probably not all other readers will find that as intuitive as I do, and the former has slightly better runtime characteristics on very large lists that are obviously not subsets.
answered Dec 26 '18 at 3:44
CR DrostCR Drost
22113
$endgroup$
So the key here is that I value code that's obvious to others.
Someone else is likely to try to stuff duplicate elements in unless it requires a type refactor; making const masterSet: Set<number> and using that throughout the code is a Good Idea. That "someone else" can often be you two months down the line, too.
In this case I would not say that the point of this length comparison is obvious; it will work but it takes some thinking -- that is why it has bugs with duplicate elements. In this case I would write something which I find more straightforward like,
function isSubsetOf<x>(sub: Iterable<x>, sup: Set<x>): boolean {
for (const x of sub) {
if (!sup.has(x)) {
return false;
}
}
return true;
}
I am also hinting through the type system how this function works by insisting that the subset is any iterable -- so I must be iterating through it.
With a restriction to an array and a set, one can get a little swankier by using the reduce function,
function isSubsetOf<x>(sub: Array<x>, sup: Set<x>): boolean {
return sub.reduce((acc, x) => acc && sup.has(x), true);
}
but probably not all other readers will find that as intuitive as I do, and the former has slightly better runtime characteristics on very large lists that are obviously not subsets.
answered Dec 26 '18 at 3:44
CR DrostCR Drost
22113
answered Dec 26 '18 at 3:44
CR DrostCR Drost
22113
answered Dec 26 '18 at 3:44
CR DrostCR Drost
22113
answered Dec 26 '18 at 3:44
CR DrostCR Drost
22113
22113
add a comment |
add a comment |
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2
$begingroup$
Wouldn't that fail if
masterArrayhas duplicate elements?$endgroup$
– Martin R
Jul 16 '18 at 7:15
$begingroup$
Good point. In my case the master array is always unique.
$endgroup$
– Harshal
Jul 16 '18 at 7:22
$begingroup$
I can enhance it by comparing with a new set created from master array
(new Set(masterArray)).size$endgroup$
– Harshal
Jul 16 '18 at 8:46