Get the index of the first element in an array with value greater than x

I have this array:

var array = [400, 4000, 400, 400, 4000];

How can I get the index of the first element with value greater than 400?

Note: While making sure this question was unique, I came across questions asking this same problem- but for different programming languages.

If there are duplicates of my question that apply to JS, I'd really want to see them.

asked Jan 12 '17 at 0:10

someGuy

325

add a comment |

I have this array:

var array = [400, 4000, 400, 400, 4000];

How can I get the index of the first element with value greater than 400?

asked Jan 12 '17 at 0:10

someGuy

325

add a comment |

I have this array:

var array = [400, 4000, 400, 400, 4000];

How can I get the index of the first element with value greater than 400?

asked Jan 12 '17 at 0:10

someGuy

325

I have this array:

var array = [400, 4000, 400, 400, 4000];

How can I get the index of the first element with value greater than 400?

javascript arrays

asked Jan 12 '17 at 0:10

someGuy

325

asked Jan 12 '17 at 0:10

someGuy

325

asked Jan 12 '17 at 0:10

someGuy

325

asked Jan 12 '17 at 0:10

someGuy

325

asked Jan 12 '17 at 0:10

someGuy

325

add a comment |

2 Answers
2

active

oldest

votes

You can use findIndex here

check this snippet

var array = [400, 4000, 400, 400, 4000];
var index=array.findIndex(function(number) 
 return number > 400;
);
console.log(index);

answered Jan 12 '17 at 0:17

Geeky

6,00411237

3

or in ES6 var index = array.findIndex(n => n > 400);

– Alnitak
Jan 12 '17 at 0:19

@RobG so it is - I misread the MDN page as saying it was in ES5

– Alnitak
Jan 12 '17 at 0:25

1

It's a little confusing that the edition numbers are one off the year number. Perhaps they should skip a year so they align. ;-)

– RobG
Jan 12 '17 at 0:27

add a comment |

You can use a simple for loop and check each element.

var array = [400, 4000, 400, 400, 4000];

var result;

for(var i=0, l=array.length; i<l; i++)
 if(array[i] > 400)
 result = i;
 break;
 


if(typeof result !== 'undefined')
 console.log('number greater than 400 found at array index: ' + result);
 else 
 console.log('no number greater than 400 found in the given arrry.');

Read up: for - JavaScript | MDN

edited Nov 15 '18 at 18:22

answered Jan 12 '17 at 0:14

Rahul Desai

11k1064106

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

You can use findIndex here

check this snippet

var array = [400, 4000, 400, 400, 4000];
var index=array.findIndex(function(number) 
 return number > 400;
);
console.log(index);

answered Jan 12 '17 at 0:17

Geeky

6,00411237

3

or in ES6 var index = array.findIndex(n => n > 400);

– Alnitak
Jan 12 '17 at 0:19

@RobG so it is - I misread the MDN page as saying it was in ES5

– Alnitak
Jan 12 '17 at 0:25

1

It's a little confusing that the edition numbers are one off the year number. Perhaps they should skip a year so they align. ;-)

– RobG
Jan 12 '17 at 0:27

add a comment |

You can use findIndex here

check this snippet

var array = [400, 4000, 400, 400, 4000];
var index=array.findIndex(function(number) 
 return number > 400;
);
console.log(index);

answered Jan 12 '17 at 0:17

Geeky

6,00411237

3

or in ES6 var index = array.findIndex(n => n > 400);

– Alnitak
Jan 12 '17 at 0:19

@RobG so it is - I misread the MDN page as saying it was in ES5

– Alnitak
Jan 12 '17 at 0:25

1

It's a little confusing that the edition numbers are one off the year number. Perhaps they should skip a year so they align. ;-)

– RobG
Jan 12 '17 at 0:27

add a comment |

You can use findIndex here

check this snippet

var array = [400, 4000, 400, 400, 4000];
var index=array.findIndex(function(number) 
 return number > 400;
);
console.log(index);

answered Jan 12 '17 at 0:17

Geeky

6,00411237

You can use findIndex here

check this snippet

var array = [400, 4000, 400, 400, 4000];
var index=array.findIndex(function(number) 
 return number > 400;
);
console.log(index);

var array = [400, 4000, 400, 400, 4000];
var index=array.findIndex(function(number) 
 return number > 400;
);
console.log(index);

var array = [400, 4000, 400, 400, 4000];
var index=array.findIndex(function(number) 
 return number > 400;
);
console.log(index);

answered Jan 12 '17 at 0:17

Geeky

6,00411237

answered Jan 12 '17 at 0:17

Geeky

6,00411237

answered Jan 12 '17 at 0:17

Geeky

6,00411237

answered Jan 12 '17 at 0:17

Geeky

6,00411237

3

or in ES6 var index = array.findIndex(n => n > 400);

– Alnitak
Jan 12 '17 at 0:19

@RobG so it is - I misread the MDN page as saying it was in ES5

– Alnitak
Jan 12 '17 at 0:25

1

It's a little confusing that the edition numbers are one off the year number. Perhaps they should skip a year so they align. ;-)

– RobG
Jan 12 '17 at 0:27

add a comment |

3

or in ES6 var index = array.findIndex(n => n > 400);

– Alnitak
Jan 12 '17 at 0:19

@RobG so it is - I misread the MDN page as saying it was in ES5

– Alnitak
Jan 12 '17 at 0:25

1

It's a little confusing that the edition numbers are one off the year number. Perhaps they should skip a year so they align. ;-)

– RobG
Jan 12 '17 at 0:27

or in ES6 var index = array.findIndex(n => n > 400);

– Alnitak
Jan 12 '17 at 0:19

@RobG so it is - I misread the MDN page as saying it was in ES5

– Alnitak
Jan 12 '17 at 0:25

It's a little confusing that the edition numbers are one off the year number. Perhaps they should skip a year so they align. ;-)

– RobG
Jan 12 '17 at 0:27

add a comment |

You can use a simple for loop and check each element.

var array = [400, 4000, 400, 400, 4000];

var result;

for(var i=0, l=array.length; i<l; i++)
 if(array[i] > 400)
 result = i;
 break;
 


if(typeof result !== 'undefined')
 console.log('number greater than 400 found at array index: ' + result);
 else 
 console.log('no number greater than 400 found in the given arrry.');

Read up: for - JavaScript | MDN

edited Nov 15 '18 at 18:22

answered Jan 12 '17 at 0:14

Rahul Desai

11k1064106

add a comment |

You can use a simple for loop and check each element.

var array = [400, 4000, 400, 400, 4000];

var result;

for(var i=0, l=array.length; i<l; i++)
 if(array[i] > 400)
 result = i;
 break;
 


if(typeof result !== 'undefined')
 console.log('number greater than 400 found at array index: ' + result);
 else 
 console.log('no number greater than 400 found in the given arrry.');

Read up: for - JavaScript | MDN

edited Nov 15 '18 at 18:22

answered Jan 12 '17 at 0:14

Rahul Desai

11k1064106

add a comment |

You can use a simple for loop and check each element.

var array = [400, 4000, 400, 400, 4000];

var result;

for(var i=0, l=array.length; i<l; i++)
 if(array[i] > 400)
 result = i;
 break;
 


if(typeof result !== 'undefined')
 console.log('number greater than 400 found at array index: ' + result);
 else 
 console.log('no number greater than 400 found in the given arrry.');

Read up: for - JavaScript | MDN

edited Nov 15 '18 at 18:22

answered Jan 12 '17 at 0:14

Rahul Desai

11k1064106

You can use a simple for loop and check each element.

var array = [400, 4000, 400, 400, 4000];

var result;

for(var i=0, l=array.length; i<l; i++)
 if(array[i] > 400)
 result = i;
 break;
 


if(typeof result !== 'undefined')
 console.log('number greater than 400 found at array index: ' + result);
 else 
 console.log('no number greater than 400 found in the given arrry.');

Read up: for - JavaScript | MDN

var array = [400, 4000, 400, 400, 4000];

var result;

for(var i=0, l=array.length; i<l; i++)
 if(array[i] > 400)
 result = i;
 break;
 


if(typeof result !== 'undefined')
 console.log('number greater than 400 found at array index: ' + result);
 else 
 console.log('no number greater than 400 found in the given arrry.');

var array = [400, 4000, 400, 400, 4000];

var result;

for(var i=0, l=array.length; i<l; i++)
 if(array[i] > 400)
 result = i;
 break;
 


if(typeof result !== 'undefined')
 console.log('number greater than 400 found at array index: ' + result);
 else 
 console.log('no number greater than 400 found in the given arrry.');

edited Nov 15 '18 at 18:22

answered Jan 12 '17 at 0:14

Rahul Desai

11k1064106

edited Nov 15 '18 at 18:22

answered Jan 12 '17 at 0:14

Rahul Desai

11k1064106

answered Jan 12 '17 at 0:14

Rahul Desai

11k1064106

answered Jan 12 '17 at 0:14

Rahul Desai

11k1064106

add a comment |

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