Comparing numerical values of an array against eachother javascript-CodePudding

let arrayOfNumbers = [1,2,3,4,5]

what would be the best way to compare the numbers against each other? For instance, comparing 1 to 2 then 2 to 3 then 3 to four, and so on?

function t(a) {
 let t = 0
 for (let i = 0; i < a.length; i  ) {
 if (a[t] > a[t   1]) {
  console.log('down')
 } else if (a[t] < a[t   1]) {
   console.log('up')
 } else if (a[t] === a[t   1]) {
  console.log('no change')
}
t

} }

CodePudding user response：

You could start from index one and check the previous value.

function t(a) {
  for (let i = 1; i < a.length; i  ) {
    if (a[i - 1] > a[i]) console.log('down');
    else if (a[i - 1] < a[i]) console.log('up');
    else console.log('no change');
  }
}

t([0, 1, 3, 2, 4, 4, 2]);

CodePudding user response：

I would do a for loop(since you don't describe why you need it I will have it to see if they are all the same, also I may have misunderstood the question, if I did it should be changeable to fit what you need)

    var Numbers = [1,1,1,1,1,0,1,1];
//I am using this for the exsample

for (i=0;i<Numbers.length-1;i  ){
  //the minus one is so it dosn't do 1 isn't the same as undefined

  if (Numbers[i]==Numbers[i 1]){

console.log(String(Numbers[i]) ' is the same as ' String(Numbers[i 1]));

    //this is for when the numbers are the same

  } else{
    console.log(String(Numbers[i]) ' is different than ' String(Numbers[i 1]));

    //the is for when the numbers are different
  }
}

sorry if this is a weird format, this is my first time answering a question on StackOverflow