Use Case 1
Assuming i have 2dArray Object of
let arr = [{'getName':'Report1'},{'getName':'User'},{'getName':'report 2'},{'getName':'User'},{'getName':'User'}]
let _NotRequiredSheet = ['User','Report 254',...]
Im trying to optimise my script with functional programming which will return me an array of
['report1','report2']
The current Method im using which does not have any error is :
for(let i =0;i < arr.length;i++){
if(arr[i].getName != _NotRequiredSheet[0]){
console.log(arr[i].getName)
}
}
But this will impact if _notrequiredSheet have a big list of what is not required
I tried using this approach which is using filter but since its 2dObject Array, im unsure how should this be implemented. What i did on my poc is
//Approach 1 : Not Working
let result = arr.filter(function (arr) {
return arr.getName != _NotRequiredSheet.values();
})
//Output should be as 1dArray['report1','report2'] , not (5) [{…}, {…}, {…}, {…}, {…}]
console.log(result)
//Approach 2 : Will output as 2D array with filtered value
// Will require to hardcord the index which is not advisable
let result = arr.filter(function (arr) {
return arr.getName != _NotRequiredSheet[0];
})
console.log(result)
i wanted to check if there is any way i could pass on using for loop with filter function. Result should return as 1D array which is
['Report1','Report2']
Use case 1 is Solved
Use Case 2 : 2D Object Array Assuming data is declared as
let arr2 = [
{$0:{'Name':'Report1'}},
{$0:{'Name':'Report2'}},
{$0:{'Name':'User'}}
]
Result should show this on console.log (2) [{…}, {…}] , filter function will remove 'User' as its reflected in _NotRequiredSheet.
Using the syntax i wrote
let result = arr2.map(item => item.$0.Name).filter(Name => !_NotRequiredSheet.includes(Name))
This will return as a single array
You could filter your data with looking for unwanted values and map only the wanted property.
const
data = [{ getName: 'Report1' }, { getName: 'User' }, { getName: 'report 2' }, { getName: 'User' }, { getName: 'User' }],
_NotRequiredSheet = ['User', 'Report 254'],
result = data
.filter(({ getName }) => !_NotRequiredSheet.includes(getName))
.map(({ getName }) => getName);
console.log(result);With a Set
const
data = [{ getName: 'Report1' }, { getName: 'User' }, { getName: 'report 2' }, { getName: 'User' }, { getName: 'User' }],
_NotRequiredSheet = ['User', 'Report 254'],
take = k => o => o[k],
hasNot = s => v => !s.has(v),
comp = f => g => o => f(g(o)),
result = data
.filter(
comp(hasNot(new Set(_NotRequiredSheet)))(take('getName'))
)
.map(({ getName }) => getName);
console.log(result);I'd recommend using reduce()
so you can return something based on _NotRequiredSheet.includes(cur.getName)
let arr = [{'getName':'Report1'},{'getName':'User'},{'getName':'report 2'},{'getName':'User'},{'getName':'User'}]
let _NotRequiredSheet = ['User','Report 254' ];
let res = arr.reduce((prev, cur) => {
if (_NotRequiredSheet.includes(cur.getName)) {
return prev;
} else {
return [ ...prev, cur.getName ];
}
}, []);
console.log(res);