Is there a way to return the missing values of list_a from list_b in TypeScrpit?
For example:
var a1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
var a2 = ['a', 'b', 'c', 'd', 'z'];
The result value is
['e', 'f', 'g'].
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1So you would like to compare a1 and a2 and check which is missing in a2 and print them or add them in a2? I'm not sure if I understood it right..webta.st.ic– webta.st.ic2016-07-21 07:59:26 +00:00Commented Jul 21, 2016 at 7:59
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I need element missing in a1 from a2 but I don't need element missing in a2 from a1.Galma88– Galma882016-07-21 08:18:48 +00:00Commented Jul 21, 2016 at 8:18
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6 Answers
There are probably a lot of ways, for example using the Array.prototype.filter():
var a1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
var a2 = ['a', 'b', 'c', 'd'];
let missing = a1.filter(item => a2.indexOf(item) < 0);
console.log(missing); // ["e", "f", "g"]
Edit
The filter function runs over the elements of a1 and it reduce it (but in a new array) to elements who are in a1 (because we're iterating over it's elements) and are missing in a2.
Elements in a2 which are missing in a1 won't be included in the result array (missing) as the filter function doesn't iterate over the a2 elements:
var a1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
var a2 = ['a', 'b', 'c', 'd', 'z', 'hey', 'there'];
let missing = a1.filter(item => a2.indexOf(item) < 0);
console.log(missing); // still ["e", "f", "g"]
7 Comments
Galma88
it seems good but ii shows also the values that I have in B and I don't have in A. but I don't need them.
Nitzan Tomer
It returns
["e", "f", "g"] just like you wanted... So I'm not sure what you meanGalma88
I edited my question. If I have a 'z' in a2 your script return also the 'z' but I don't need it
Nitzan Tomer
No, my script will still return
["e", "f", "g"] even with the added z to a2. even var a2 = ['a', 'b', 'c', 'd', 'z', 'hey', 'there']; it will still return the requested missing values, check it here.
oiawo
That's perfect, thant you. I just wrote this little "array diff" function for anyone interested: const arrayDiffs = (oldArray: any[] = [], newArray: any[] = []) => { const added = newArray.filter((item) => oldArray.indexOf(item) < 0); const removed = oldArray.filter((item) => newArray.indexOf(item) < 0); return { added, removed }; };
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Typescript only provides design / compile time help, it doesn't add JavaScript features. So the solution that works in JavaScript will work in Typescript.
Plenty of ways to solve this, my goto choice would be lodash: https://lodash.com/docs#difference
_.difference(['a', 'b', 'c', 'd', 'e', 'f', 'g'],['a', 'b', 'c', 'd']);
Comments
You can only use this method. With the larger table first.
const a1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
const a2 = ['a', 'b', 'c', 'd', 'x', 'y', 'z'];
const difference_a1 = a1.filter(x => !a2.includes(x));
const difference_a2 = a2.filter(x => !a1.includes(x));
console.log(difference_a1);
/* ["e","f","g"] */
console.log(difference_a2);
/* ["x","y","z"] */
Comments
You can look for both values:
const removed = before.filter((x: any) => !after.includes(x));
const added = after.filter((x: any) => !before.includes(x));
Comments
Why don't leverage the power of Javascript :-)
var a1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
var a2 = ['a', 'b', 'c', 'd', 'k', 'l', 'M', 'j'];
let difference: string[] = [];
difference = difference.concat(a1.filter(item => a2.indexOf(item) < 0))
.concat(a2.filter(item => a1.indexOf(item) < 0));
or
difference = difference.concat(a1.filter(item => !a2.include(item)))
.concat(a2.filter(item => !a1.include(item)));
console.log(difference); // ["e", "f", "g", "k", "l", "M", "j"]
Comments
const a1 = ['a', 'b', 'c', 'd', 'e', 'f'];
const a2 = ['a', 'b', 'c'];
let bigArray = null;
let smallArray = null;
if(a1.length >= a2.length)
{
bigArray = a1;
smallArray =a2;
} else {
bigArray= a2;
smallArray =a1;
}
const diff = bigArray.filter(item => smallArray.indexOf(item) < 0);
console.log(diff);
