Incorrect value returns

Question

I use this function:

function search (){
  v = [[2.0], [3.0], [4.0], [5.0]]
  if(v.indexOf(2.0) != -1){
    Logger.log ('if')
  }else{
    Logger.log('else')
  }
}

editing

My real need is in the values of words like this I was answered about numbers so I did not change and only added

function search (){
  v = [["One"], ["two"], ["three"], ["four"]]
  if(v.indexOf("One") != -1){
    Logger.log ('if')
  }else{
    Logger.log('else')
  }
}

And instead of "if" returning to me, "else" returns to me. I apologize for the broken English

user_16046521 · Answer

You input array is a multi/two dimensional array. You are searching for 2.0 which is actually embedded inside another smaller array. Try changing the function to below.

function search() {
  v = [
    ["One"],
    ["two"],
    ["three"],
    ["four"]
  ]
  if (v.findIndex(e => e[0] === 'One') != -1) {
    console.log('if')
  } else {
    console.log('else')
  }
}

search();

user_4830196 · Answer

You have an array of arrays, so trying to use indexOf on the array of arrays will not work for your case. One option is to use flat and some in conjunction to flatten the array of arrays to one array and then check if the value is found inside the flattened array:

function search (){
  v = [[2.0], [3.0], [4.0], [5.0]];
  if(v.flat().some((a) => a === 2.0)){
    console.log('if');
  }else{
    console.log('else');
  }
}

search();

function search (){
  v = [[2.0], [3.0], [4.0], [5.0]];
  if(v.flat().some((a) => a === 2.0)){
    console.log('if');
  }else{
    console.log('else');
  }
}

search();

user_7015157 · Answer

Did you mean to write v = [2.0, 3.0, 4.0, 5.0];?

Because that will print out 'if'. Reason is that in your example, each element of v array is an array of it's own.

If you really want it to be an array of arrays, then to make it work you can do this in the if statement:

if ( v.flat().includes(2.0) )

flat converts it from an array of array that contain a number, to an array of numbers.
includes is a nicer method to check if the array contains a value - no need to check for index :)

user_10850045 · Answer

array.indexOf() is not applicable for nested array. If your array is primitive types then it's working, but if array is non-primitive types then through indexOf you can't match the value.

Use this instated of your function:

v.findIndex(ele => [...new Set(ele)][0] == 2.0) >=0  ? 'if' : 'else'

Incorrect value returns

4 Answers