We Keep Coding

Why is this attempt to retrieve the minimum number of an array not working?

Max should be = 9.99, and min should be = 6.88
let arr = [["2019","00","01", 9.99], ["2018","00","01", 9.32], ["2017","00","01", 6.88]]
let max = Math.max(Number(...arr.map((o) => { return o[3] }))); //9.99
let min = Math.min(Number(...arr.map((o) => { return o[3] }))); //9.99
console.log(min);
console.log(max);

let arr = [["2019","00","01", 9.99], ["2018","00","01", 9.32], ["2017","00","01", 6.88]];
let max = Math.max(...arr.map((o) => { return o[3] })); //9.99
let min = Math.min(...arr.map((o) => { return o[3] })); //6.88
console.log({
max , min
})

Earlier, you had put Number around the mapped array. This converts the array to a number. However, you only wanted the individual elements to be numbers so move it inside the map function.
let arr = [
["2019","00","01", 9.99],
["2018","00","01", 9.32],
["2017","00","01", 6.88]
];
let max = Math.max(...arr.map((o) => { return Number(o[3]) })); //9.99
let min = Math.min(...arr.map((o) => { return Number(o[3]) })); //6.88
// this can be rewritten like so:
// Math.min(...arr.map((o) => Number(o[3])));
//
// https://www.w3schools.com/js/js_arrow_function.asp
console.log(max);
console.log(min);
So,
before:
let min = Math.min(Number(...arr.map((o) => { return o[3] }))); //9.99
after
let min = Math.min(...arr.map((o) => { return Number(o[3]) })); //6.88

How to populate an array with integers

Please, how do you populate an array say ‘num’ with numbers not in a second array say ‘fig’? I’m trying to use a loop to have the values of the already populated array ‘fig’ compared to ‘num’ which is to be populated with integers not found in ‘fig’. I’m a bit confused.

If you need to do an array with n numbers you can use this two ways.
const arrayLength = 100;
const numberArray = [...new Array(arrayLength).keys()]
const anotherWay = new Array(arrayLength).fill().map((_, idx) => idx + 1);
console.log(numberArray, anotherWay)

so to do this we have to do a few things:
1) define an existing array with numbers to avoid
2) define length on new array
3) generate a random number and make it an integer
4) check to see if we need to avoid
5) if it's a new value add it to the second array
var first=[55,45,35,1,2,3,4,5];
var second = [];
var i = 7;
var x;
while (i != 0){
x = ~~(Math.random()*100);
var check = false;
for(let j=0; j<first.length;j++){
if(x == first[j]){
check = true;
}
}
if(!check){
second.push(x);
i--;
}
}
console.log(second);

const fig = [-21, 0, 3, 6, 7, 42]
const min = Math.min(...fig) // or fig[0] if the the array is already sorted
const max = Math.max(...fig) // or fig[fig.length - 1]
const num = Array.from({ length: max - min }, (_, i) => i + min)
.filter(el => !fig.includes(el))
or, saving one loop
const num = Array.from({ length: max - min }).reduce((acc, _, i) => {
const curr = i + min
if (!fig.includes(curr)) {
return acc.concat(curr)
}
return acc
}, [])
This is assuming your range is from the smallest number in fig to the largest in fig.

Javascript: Convert string to array of objects?

Tell me, how can I optimally convert a string like
const data = "1350, 1351, 1390-1391, 1401, 1402 - 1407";
to an array like
const range = [
{
min: 1350,
max: 1350,
},
{
min: 1351,
max: 1351,
},
{
min: 1390,
max: 1391,
},
{
min: 1401,
max: 1401,
},
{
min: 1402,
max: 1407,
},
];
?
In other words, you need to create an array of number ranges using a string in which these numbers ranges are explicitly specified.
The most obvious of the possible algorithms is:
1) split the string using a delimiter,
2) the resulting parts are cleaned of spaces using the command trim
3) check whether the part is a number
4) if not, then split the part using the delimiter -
5) the parts obtained are cleaned of spaces using the command trim,
6) check that the amount of component eq 2 and it's a number
But is it possible to make it more optimal, more beautiful, more effective?

You can use .split() and .map():
const data = "1350, 1351, 1390-1391, 1401, 1402 - 1407";
const range = data.split(",").map(s => {
let [min, max = min] = s.split("-");
return {min: Number(min), max: Number(max)}
});
console.log(range);
.as-console-wrapper { max-height: 100% !important; top: 0; }

Try something like this:
const data = "1350, 1351, 1390-1391, 1401, 1402 - 1407"
const result = data
.split(/\s*,\s*/)
.map(pair => {
const [min, max = min] = pair
.split(/\s*-\s*/)
.map(Number)
return {
min,
max
}
})
console.log(result)

You can use split method in combination with reduce method.
The reduce() method applies a function against an accumulator and each
element in the array (from left to right) to reduce it to a single
value.
Also, use + operator in order to force result to Number.
const data = "1350, 1351, 1390-1391, 1401, 1402 - 1407";
const array = data.split(', ').reduce(function(arr, elem){
var [min, max] = elem.split('-');
arr.push({
min : + min,
max: + (max || min)
});
return arr;
},[]);
console.log(array);

I think the simple and much understandable way would be to loop through the values and check if they have a range value (with hyphen) and create the object accordingly for the range array.
const data = "1350, 1351, 1390-1391, 1401, 1402 - 1407";
var range = [];
data.split(',').forEach((val)=>{
var obj = {};
if(val.indexOf('-') === -1 ){
obj.min = val;
obj.max = val;
} else {
obj.min = val.split('-')[0].trim();
obj.max = val.split('-')[1].trim();
}
range.push(obj);
});
console.log(range);

This code will help you.
const data = "1350, 1351, 1390-1391, 1401, 1402 - 1407";
var arr = data.split(',');
const range = [];
for(var i=0; i< arr.length; i++){
var b = arr[i].split('-');
if(!b[1])b[1]=b[0];
var obj = new Object();
obj.min = parseInt(b[0]);
obj.max = parseInt(b[1]);
range.push(obj);
}
console.log(range);

how to filter a set of data by variance in typescript?

Let's say I have a set of data like this with a row for each minute in the last 4 hours:
[
{ X:1000, Y:2000, Z:3000, DateTime: 12/15/2018 12:00 },
{ X:998, Y:2011, Z:3020, DateTime: 12/15/2018 12:01 }
]
I need an array of property names whose values are within a 20% variance for all rows. So if Y and Z above meet this criteria but X does not then the output should look like this:
[Y, Z]
What typescript code could I use to do this?

I don't know exactly what "variance" or "variance percentage" mean in your question. I just used this formula to calculate variance: https://www.wikihow.com/Calculate-Variance
For the variance percentage, I simply divided the variance by the mean value and expressed it in percentage.
Feel free to replace my calculateVariancePercentage with a more correct implementation.
const ACCEPTABLE_VARIANCE_THRESHOLD = 20;
const dataset = [
{ X:1000, Y:2000, Z:3000, DateTime: '12/15/2018 12:00' },
{ X:998, Y:2011, Z:3020, DateTime: '12/15/2018 12:01' }
];
const calculateVariancePercentage = (data) => {
const meanValue = data.reduce((sum, element) => sum + element, 0) / data.length;
const sumOfDeviations = data.reduce((sod, element) => Math.pow(element - meanValue, 2), 0);
const variance = sumOfDeviations / (data.length - 1);
return variance / meanValue * 100;
}
const variables = Object.keys(dataset[0]).filter(key => key !== 'DateTime');
const result = variables.filter(variable => {
const varData = dataset.map(row => row[variable]);
const varianceInPercentage = calculateVariancePercentage(varData);
console.log(varianceInPercentage);
return calculateVariancePercentage(varData) <= ACCEPTABLE_VARIANCE_THRESHOLD;
});
console.log(result);

find min and max value using reduce es2015

If I have array of object like this
[{min:5,max:10,id:1}, {min:50,max:3,id:2}, {min:1,max:40,id:3}]
How to find min and max using reduce? I know I can use generic loop and compare but I would like to explore reduce in es2015

You can use reduce like this to get the min and max numbers from each object in the array.
const arr = [{min:5,max:10,id:1}, {min:50,max:3,id:2}, {min:1,max:40,id:3}]
console.log(
arr.reduce((acc, x) => {
acc.min = Math.min(acc.min, x.min)
acc.max = Math.max(acc.max, x.max)
return acc
}, { min: Infinity, max: -Infinity })
)
// as Bergi suggested, we could just return a new Object literal
// from the reduction
console.log(
arr.reduce((acc, x) => ({
min: Math.min(acc.min, x.min),
max: Math.max(acc.max, x.max)
}), { min: Infinity, max: -Infinity })
)

Assuming that you want, for the minimum, to get the id of the item with the lowed min value, then this will do it.
const items = [{min:5,max:10,id:1}, {min:50,max:3,id:2}, {min:1,max:40,id:3}];
// Reduce to find the item with the lowest min
const min = items.reduce((res, item) => {
// No result yet, so the item must be the lowest seen
if (!res) return item;
// Use the lowest of the current lowest and the current item
return item.min < res.min ? item : res;
}, undefined).id;

const arr = [{min:5,max:10,id:1}, {min:50,max:3,id:2}, {min:1,max:40,id:3}];
const min = arr.reduce((m, o) => m < o.min? m: o.min, +Infinity),
max = arr.reduce((M, o) => M > o.max? M: o.max, -Infinity);
console.log("Min:", min);
console.log("Max:", max);
Explanation of min:
const min = arr.reduce((m, o) => { // for each object o in the array arr
return m < o.min? // if o.min is bigger than the minimum m
m: // then the minimum is still m
o.min; // otherwise, the new minimum is o.min
}, +Infinity); // pass in +Ifinity as the minimum (the initial value of m so whatever is arr[0].min is it will be smaller than m). You can pass in arr[0].min if you want instead of +Infinity