Find time average of hh:mm:ss2019 Community Moderator ElectionFind sum of time in an array of time format hh:mm:ssCalculate relative time in C#How to get the current time in PythonWhat do 'real', 'user' and 'sys' mean in the output of time(1)?Convert a Unix timestamp to time in JavaScriptHow do I get time of a Python program's execution?How to calculate the time interval between two time stringsGet current time and date on AndroidFind object by id in an array of JavaScript objectsHow to calculate average call time in Microsoft Accesscalculating average for time in hh:mm format

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Find time average of hh:mm:ss

2019 Community Moderator ElectionFind sum of time in an array of time format hh:mm:ssCalculate relative time in C#How to get the current time in PythonWhat do 'real', 'user' and 'sys' mean in the output of time(1)?Convert a Unix timestamp to time in JavaScriptHow do I get time of a Python program's execution?How to calculate the time interval between two time stringsGet current time and date on AndroidFind object by id in an array of JavaScript objectsHow to calculate average call time in Microsoft Accesscalculating average for time in hh:mm format

I have this fiddle that calculates average time with miliseconds. However, my DB stores data in format of hh:mm:ss.
fiddle

var times= [ '00:00:03.00', '00:00:05.00', '00:00:02.00', '00:00:06.00'],
 date = 0,
 result = '';
function offsetify(t)
 return t < 10 ? '0' + t : t;

for(var x = 0; x < times.length; x++ ) 
 var tarr = times[x].split(':');
 date += new Date(0, 0, 0, tarr[0], tarr[1], tarr[2].split('.')[0], tarr[2].split('.')[1]).getTime(); 

var avg = new Date(date/times.length);
result = offsetify(avg.getHours()) + ':' + offsetify(avg.getMinutes()) + ':' + offsetify(avg.getSeconds()) + '.' + offsetify(avg.getMilliseconds());
document.write(result);

I need to make sure when seconds are averaged for example to 8.75 then average is returned as 00:00:9
using array like this:

var times= [ '00:00:03', '00:00:05', '00:00:020', '00:00:07']

can someone please help me modify this to properly round off hh:mm:ss format. thanks.

asked Mar 7 at 17:30

coder

819

1

I think Nina's is a great answer, but want to ask -- when you say "my DB stores data in format of hh:mm:ss" do you mean that you have a table which you are not allowed to alter that uses a datetime or timestamp field? ...because your DB certainly could store average milliseconds as an int.

– Stephen P
Mar 7 at 19:59

add a comment |

I have this fiddle that calculates average time with miliseconds. However, my DB stores data in format of hh:mm:ss.
fiddle

var times= [ '00:00:03.00', '00:00:05.00', '00:00:02.00', '00:00:06.00'],
 date = 0,
 result = '';
function offsetify(t)
 return t < 10 ? '0' + t : t;

for(var x = 0; x < times.length; x++ ) 
 var tarr = times[x].split(':');
 date += new Date(0, 0, 0, tarr[0], tarr[1], tarr[2].split('.')[0], tarr[2].split('.')[1]).getTime(); 

var avg = new Date(date/times.length);
result = offsetify(avg.getHours()) + ':' + offsetify(avg.getMinutes()) + ':' + offsetify(avg.getSeconds()) + '.' + offsetify(avg.getMilliseconds());
document.write(result);

I need to make sure when seconds are averaged for example to 8.75 then average is returned as 00:00:9
using array like this:

var times= [ '00:00:03', '00:00:05', '00:00:020', '00:00:07']

can someone please help me modify this to properly round off hh:mm:ss format. thanks.

asked Mar 7 at 17:30

coder

819

1

I think Nina's is a great answer, but want to ask -- when you say "my DB stores data in format of hh:mm:ss" do you mean that you have a table which you are not allowed to alter that uses a datetime or timestamp field? ...because your DB certainly could store average milliseconds as an int.

– Stephen P
Mar 7 at 19:59

add a comment |

I have this fiddle that calculates average time with miliseconds. However, my DB stores data in format of hh:mm:ss.
fiddle

var times= [ '00:00:03.00', '00:00:05.00', '00:00:02.00', '00:00:06.00'],
 date = 0,
 result = '';
function offsetify(t)
 return t < 10 ? '0' + t : t;

for(var x = 0; x < times.length; x++ ) 
 var tarr = times[x].split(':');
 date += new Date(0, 0, 0, tarr[0], tarr[1], tarr[2].split('.')[0], tarr[2].split('.')[1]).getTime(); 

var avg = new Date(date/times.length);
result = offsetify(avg.getHours()) + ':' + offsetify(avg.getMinutes()) + ':' + offsetify(avg.getSeconds()) + '.' + offsetify(avg.getMilliseconds());
document.write(result);

I need to make sure when seconds are averaged for example to 8.75 then average is returned as 00:00:9
using array like this:

var times= [ '00:00:03', '00:00:05', '00:00:020', '00:00:07']

can someone please help me modify this to properly round off hh:mm:ss format. thanks.

asked Mar 7 at 17:30

coder

819

I have this fiddle that calculates average time with miliseconds. However, my DB stores data in format of hh:mm:ss.
fiddle

var times= [ '00:00:03.00', '00:00:05.00', '00:00:02.00', '00:00:06.00'],
 date = 0,
 result = '';
function offsetify(t)
 return t < 10 ? '0' + t : t;

for(var x = 0; x < times.length; x++ ) 
 var tarr = times[x].split(':');
 date += new Date(0, 0, 0, tarr[0], tarr[1], tarr[2].split('.')[0], tarr[2].split('.')[1]).getTime(); 

var avg = new Date(date/times.length);
result = offsetify(avg.getHours()) + ':' + offsetify(avg.getMinutes()) + ':' + offsetify(avg.getSeconds()) + '.' + offsetify(avg.getMilliseconds());
document.write(result);

I need to make sure when seconds are averaged for example to 8.75 then average is returned as 00:00:9
using array like this:

var times= [ '00:00:03', '00:00:05', '00:00:020', '00:00:07']

can someone please help me modify this to properly round off hh:mm:ss format. thanks.

javascript time

asked Mar 7 at 17:30

coder

819

asked Mar 7 at 17:30

coder

819

asked Mar 7 at 17:30

coder

819

asked Mar 7 at 17:30

coder

819

asked Mar 7 at 17:30

coder

819

1

I think Nina's is a great answer, but want to ask -- when you say "my DB stores data in format of hh:mm:ss" do you mean that you have a table which you are not allowed to alter that uses a datetime or timestamp field? ...because your DB certainly could store average milliseconds as an int.

– Stephen P
Mar 7 at 19:59

add a comment |

1

I think Nina's is a great answer, but want to ask -- when you say "my DB stores data in format of hh:mm:ss" do you mean that you have a table which you are not allowed to alter that uses a datetime or timestamp field? ...because your DB certainly could store average milliseconds as an int.

– Stephen P
Mar 7 at 19:59

I think Nina's is a great answer, but want to ask -- when you say "my DB stores data in format of hh:mm:ss" do you mean that you have a table which you are not allowed to alter that uses a datetime or timestamp field? ...because your DB certainly could store average milliseconds as an int.

– Stephen P
Mar 7 at 19:59

add a comment |

4 Answers
4

active

oldest

votes

You could get the seconds, round the value and build a new string of the average time.

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(s => s.split(':').reduce((s, v, i) => s + times[i] * v, 0)),
 avg = Math.round(parts.reduce((a, b) => a + b, 0) / parts.length);

 return times
 .map(t => [Math.floor(avg / t), avg %= t][0])
 .map(v => v.toString().padStart(2, 0))
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

ES5

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(function (s) 
 return s.split(':').reduce(function (s, v, i) 
 return s + times[i] * v;
 , 0);
 ),
 avg = Math.round(parts.reduce(function (a, b) 
 return a + b;
 , 0) / parts.length);

 return times
 .map(function (t) 
 var value = Math.floor(avg / t);
 avg %= t;
 return value;
 )
 .map(function (v) 
 return v.toString().padStart(2, 0);
 )
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

edited Mar 7 at 19:13

answered Mar 7 at 17:50

Nina Scholz

191k15104176

interesting solution. will this work properly once there are minutes more than 0 to make sure properly rounds to hours... etc

– coder
Mar 7 at 17:57

do you have an example?

– Nina Scholz
Mar 7 at 18:41

I am having trouble adding your code to my code.. it doesn't like =>

– coder
Mar 7 at 18:55

maybe the ES5 version is working for you.

– Nina Scholz
Mar 7 at 19:01

add a comment |

var seconds = avg.getMilliseconds() > 500 ? avg.getSeconds() + 1 : avg.getSeconds();

result = offsetify(avg.getHours()) + ':' + offsetify(avg.getMinutes()) + ':' + 
offsetify(seconds);

Basically just checking to see if ms is more than 50, if it is can manually add one to seconds, if not leave seconds as is. Not sure if you wanted seconds to be offset as well.

edited Mar 7 at 19:24

answered Mar 7 at 17:43

Carlos Reyes

988

thanks it worked well with seconds. I wonder if it will work properly with minutes..or if i need something similar for minutes,, can you plz review this and see if it works properly fiddle

– coder
Mar 7 at 17:54

2

there is 1000 ms in 1 s, so wouldn't you want to test if it's greater than 500?

– Steven Stark
Mar 7 at 18:44

actually can you please revisit this... I tried to average these values: 00:01:20, 00:00:40, 00:01:00, 00:00:09 result was given to me 00:00:48 instead of 00:00:47

– coder
Mar 7 at 18:47

1

Bingo Steven. thank you for that comment. You answered my question before I asked it. checking it >500 did the trick and i did get 00:00:47

– coder
Mar 7 at 18:50

1

nice one Steven edited my answer accordingly

– Carlos Reyes
Mar 7 at 19:25

add a comment |

Date can be used to convert them to milliseconds :

function averageTime(arr) 
 var sum = arr.reduce(function(a, b) return a + +new Date('1970T' + b + 'Z'); , 0);
 return new Date(sum / arr.length + 500).toJSON().slice(11, 19);


console.log( averageTime(['00:00:03.00', '00:00:05.00', '00:00:02.00', '00:00:06.00']) );
console.log( averageTime(['00:00:03', '00:00:05', '00:00:20', '00:00:07']) );

answered Mar 7 at 22:23

Slai

15.5k32436

add a comment |

First, convert all times into their UNIX timestamps using Date.getTime(), producing an array, then average the results. example: ]

let times = [
1551984787316,
1551984789662,
1551984790162,
1551984790730,
1551984791310
]


let average = ( values ) => 
 let sum = 0;
 for( let i = 0; i < values.length; i++ )
 sum += parseInt( values[i], 10 ); // base 10
 

 return Math.floor( sum/values.length );


let avgTime = average( times );
let avgDate = new Date( avgTime );
console.log( avgDate );

edited Mar 7 at 22:56

answered Mar 7 at 18:57

Steven Stark

666618

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

You could get the seconds, round the value and build a new string of the average time.

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(s => s.split(':').reduce((s, v, i) => s + times[i] * v, 0)),
 avg = Math.round(parts.reduce((a, b) => a + b, 0) / parts.length);

 return times
 .map(t => [Math.floor(avg / t), avg %= t][0])
 .map(v => v.toString().padStart(2, 0))
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

ES5

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(function (s) 
 return s.split(':').reduce(function (s, v, i) 
 return s + times[i] * v;
 , 0);
 ),
 avg = Math.round(parts.reduce(function (a, b) 
 return a + b;
 , 0) / parts.length);

 return times
 .map(function (t) 
 var value = Math.floor(avg / t);
 avg %= t;
 return value;
 )
 .map(function (v) 
 return v.toString().padStart(2, 0);
 )
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

edited Mar 7 at 19:13

answered Mar 7 at 17:50

Nina Scholz

191k15104176

interesting solution. will this work properly once there are minutes more than 0 to make sure properly rounds to hours... etc

– coder
Mar 7 at 17:57

do you have an example?

– Nina Scholz
Mar 7 at 18:41

I am having trouble adding your code to my code.. it doesn't like =>

– coder
Mar 7 at 18:55

maybe the ES5 version is working for you.

– Nina Scholz
Mar 7 at 19:01

add a comment |

You could get the seconds, round the value and build a new string of the average time.

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(s => s.split(':').reduce((s, v, i) => s + times[i] * v, 0)),
 avg = Math.round(parts.reduce((a, b) => a + b, 0) / parts.length);

 return times
 .map(t => [Math.floor(avg / t), avg %= t][0])
 .map(v => v.toString().padStart(2, 0))
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

ES5

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(function (s) 
 return s.split(':').reduce(function (s, v, i) 
 return s + times[i] * v;
 , 0);
 ),
 avg = Math.round(parts.reduce(function (a, b) 
 return a + b;
 , 0) / parts.length);

 return times
 .map(function (t) 
 var value = Math.floor(avg / t);
 avg %= t;
 return value;
 )
 .map(function (v) 
 return v.toString().padStart(2, 0);
 )
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

edited Mar 7 at 19:13

answered Mar 7 at 17:50

Nina Scholz

191k15104176

interesting solution. will this work properly once there are minutes more than 0 to make sure properly rounds to hours... etc

– coder
Mar 7 at 17:57

do you have an example?

– Nina Scholz
Mar 7 at 18:41

I am having trouble adding your code to my code.. it doesn't like =>

– coder
Mar 7 at 18:55

maybe the ES5 version is working for you.

– Nina Scholz
Mar 7 at 19:01

add a comment |

You could get the seconds, round the value and build a new string of the average time.

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(s => s.split(':').reduce((s, v, i) => s + times[i] * v, 0)),
 avg = Math.round(parts.reduce((a, b) => a + b, 0) / parts.length);

 return times
 .map(t => [Math.floor(avg / t), avg %= t][0])
 .map(v => v.toString().padStart(2, 0))
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

ES5

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(function (s) 
 return s.split(':').reduce(function (s, v, i) 
 return s + times[i] * v;
 , 0);
 ),
 avg = Math.round(parts.reduce(function (a, b) 
 return a + b;
 , 0) / parts.length);

 return times
 .map(function (t) 
 var value = Math.floor(avg / t);
 avg %= t;
 return value;
 )
 .map(function (v) 
 return v.toString().padStart(2, 0);
 )
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

edited Mar 7 at 19:13

answered Mar 7 at 17:50

Nina Scholz

191k15104176

You could get the seconds, round the value and build a new string of the average time.

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(s => s.split(':').reduce((s, v, i) => s + times[i] * v, 0)),
 avg = Math.round(parts.reduce((a, b) => a + b, 0) / parts.length);

 return times
 .map(t => [Math.floor(avg / t), avg %= t][0])
 .map(v => v.toString().padStart(2, 0))
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

ES5

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(function (s) 
 return s.split(':').reduce(function (s, v, i) 
 return s + times[i] * v;
 , 0);
 ),
 avg = Math.round(parts.reduce(function (a, b) 
 return a + b;
 , 0) / parts.length);

 return times
 .map(function (t) 
 var value = Math.floor(avg / t);
 avg %= t;
 return value;
 )
 .map(function (v) 
 return v.toString().padStart(2, 0);
 )
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(s => s.split(':').reduce((s, v, i) => s + times[i] * v, 0)),
 avg = Math.round(parts.reduce((a, b) => a + b, 0) / parts.length);

 return times
 .map(t => [Math.floor(avg / t), avg %= t][0])
 .map(v => v.toString().padStart(2, 0))
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(s => s.split(':').reduce((s, v, i) => s + times[i] * v, 0)),
 avg = Math.round(parts.reduce((a, b) => a + b, 0) / parts.length);

 return times
 .map(t => [Math.floor(avg / t), avg %= t][0])
 .map(v => v.toString().padStart(2, 0))
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(function (s) 
 return s.split(':').reduce(function (s, v, i) 
 return s + times[i] * v;
 , 0);
 ),
 avg = Math.round(parts.reduce(function (a, b) 
 return a + b;
 , 0) / parts.length);

 return times
 .map(function (t) 
 var value = Math.floor(avg / t);
 avg %= t;
 return value;
 )
 .map(function (v) 
 return v.toString().padStart(2, 0);
 )
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

function getAverageTime(array) 
 var times = [3600, 60, 1],
 parts = array.map(function (s) 
 return s.split(':').reduce(function (s, v, i) 
 return s + times[i] * v;
 , 0);
 ),
 avg = Math.round(parts.reduce(function (a, b) 
 return a + b;
 , 0) / parts.length);

 return times
 .map(function (t) 
 var value = Math.floor(avg / t);
 avg %= t;
 return value;
 )
 .map(function (v) 
 return v.toString().padStart(2, 0);
 )
 .join(':');


console.log(getAverageTime(['00:00:03', '00:00:05', '00:00:020', '00:00:07']));
console.log(getAverageTime(['00:00:03', '00:30:05', '00:30:020', '03:00:07']));

edited Mar 7 at 19:13

answered Mar 7 at 17:50

Nina Scholz

191k15104176

edited Mar 7 at 19:13

answered Mar 7 at 17:50

Nina Scholz

191k15104176

answered Mar 7 at 17:50

Nina Scholz

191k15104176

answered Mar 7 at 17:50

Nina Scholz

191k15104176

interesting solution. will this work properly once there are minutes more than 0 to make sure properly rounds to hours... etc

– coder
Mar 7 at 17:57

do you have an example?

– Nina Scholz
Mar 7 at 18:41

I am having trouble adding your code to my code.. it doesn't like =>

– coder
Mar 7 at 18:55

maybe the ES5 version is working for you.

– Nina Scholz
Mar 7 at 19:01

add a comment |

interesting solution. will this work properly once there are minutes more than 0 to make sure properly rounds to hours... etc

– coder
Mar 7 at 17:57

do you have an example?

– Nina Scholz
Mar 7 at 18:41

I am having trouble adding your code to my code.. it doesn't like =>

– coder
Mar 7 at 18:55

maybe the ES5 version is working for you.

– Nina Scholz
Mar 7 at 19:01

interesting solution. will this work properly once there are minutes more than 0 to make sure properly rounds to hours... etc

– coder
Mar 7 at 17:57

do you have an example?

– Nina Scholz
Mar 7 at 18:41

I am having trouble adding your code to my code.. it doesn't like =>

– coder
Mar 7 at 18:55

maybe the ES5 version is working for you.

– Nina Scholz
Mar 7 at 19:01

add a comment |

var seconds = avg.getMilliseconds() > 500 ? avg.getSeconds() + 1 : avg.getSeconds();

result = offsetify(avg.getHours()) + ':' + offsetify(avg.getMinutes()) + ':' + 
offsetify(seconds);

Basically just checking to see if ms is more than 50, if it is can manually add one to seconds, if not leave seconds as is. Not sure if you wanted seconds to be offset as well.

edited Mar 7 at 19:24

answered Mar 7 at 17:43

Carlos Reyes

988

thanks it worked well with seconds. I wonder if it will work properly with minutes..or if i need something similar for minutes,, can you plz review this and see if it works properly fiddle

– coder
Mar 7 at 17:54

2

there is 1000 ms in 1 s, so wouldn't you want to test if it's greater than 500?

– Steven Stark
Mar 7 at 18:44

actually can you please revisit this... I tried to average these values: 00:01:20, 00:00:40, 00:01:00, 00:00:09 result was given to me 00:00:48 instead of 00:00:47

– coder
Mar 7 at 18:47

1

Bingo Steven. thank you for that comment. You answered my question before I asked it. checking it >500 did the trick and i did get 00:00:47

– coder
Mar 7 at 18:50

1

nice one Steven edited my answer accordingly

– Carlos Reyes
Mar 7 at 19:25

add a comment |

var seconds = avg.getMilliseconds() > 500 ? avg.getSeconds() + 1 : avg.getSeconds();

result = offsetify(avg.getHours()) + ':' + offsetify(avg.getMinutes()) + ':' + 
offsetify(seconds);

Basically just checking to see if ms is more than 50, if it is can manually add one to seconds, if not leave seconds as is. Not sure if you wanted seconds to be offset as well.

edited Mar 7 at 19:24

answered Mar 7 at 17:43

Carlos Reyes

988

thanks it worked well with seconds. I wonder if it will work properly with minutes..or if i need something similar for minutes,, can you plz review this and see if it works properly fiddle

– coder
Mar 7 at 17:54

2

there is 1000 ms in 1 s, so wouldn't you want to test if it's greater than 500?

– Steven Stark
Mar 7 at 18:44

actually can you please revisit this... I tried to average these values: 00:01:20, 00:00:40, 00:01:00, 00:00:09 result was given to me 00:00:48 instead of 00:00:47

– coder
Mar 7 at 18:47

1

Bingo Steven. thank you for that comment. You answered my question before I asked it. checking it >500 did the trick and i did get 00:00:47

– coder
Mar 7 at 18:50

1

nice one Steven edited my answer accordingly

– Carlos Reyes
Mar 7 at 19:25

add a comment |

var seconds = avg.getMilliseconds() > 500 ? avg.getSeconds() + 1 : avg.getSeconds();

result = offsetify(avg.getHours()) + ':' + offsetify(avg.getMinutes()) + ':' + 
offsetify(seconds);

Basically just checking to see if ms is more than 50, if it is can manually add one to seconds, if not leave seconds as is. Not sure if you wanted seconds to be offset as well.

edited Mar 7 at 19:24

answered Mar 7 at 17:43

Carlos Reyes

988

var seconds = avg.getMilliseconds() > 500 ? avg.getSeconds() + 1 : avg.getSeconds();

result = offsetify(avg.getHours()) + ':' + offsetify(avg.getMinutes()) + ':' + 
offsetify(seconds);

Basically just checking to see if ms is more than 50, if it is can manually add one to seconds, if not leave seconds as is. Not sure if you wanted seconds to be offset as well.

edited Mar 7 at 19:24

answered Mar 7 at 17:43

Carlos Reyes

988

edited Mar 7 at 19:24

answered Mar 7 at 17:43

Carlos Reyes

988

answered Mar 7 at 17:43

Carlos Reyes

988

answered Mar 7 at 17:43

Carlos Reyes

988

thanks it worked well with seconds. I wonder if it will work properly with minutes..or if i need something similar for minutes,, can you plz review this and see if it works properly fiddle

– coder
Mar 7 at 17:54

2

there is 1000 ms in 1 s, so wouldn't you want to test if it's greater than 500?

– Steven Stark
Mar 7 at 18:44

actually can you please revisit this... I tried to average these values: 00:01:20, 00:00:40, 00:01:00, 00:00:09 result was given to me 00:00:48 instead of 00:00:47

– coder
Mar 7 at 18:47

1

Bingo Steven. thank you for that comment. You answered my question before I asked it. checking it >500 did the trick and i did get 00:00:47

– coder
Mar 7 at 18:50

1

nice one Steven edited my answer accordingly

– Carlos Reyes
Mar 7 at 19:25

add a comment |

thanks it worked well with seconds. I wonder if it will work properly with minutes..or if i need something similar for minutes,, can you plz review this and see if it works properly fiddle

– coder
Mar 7 at 17:54

2

there is 1000 ms in 1 s, so wouldn't you want to test if it's greater than 500?

– Steven Stark
Mar 7 at 18:44

actually can you please revisit this... I tried to average these values: 00:01:20, 00:00:40, 00:01:00, 00:00:09 result was given to me 00:00:48 instead of 00:00:47

– coder
Mar 7 at 18:47

1

Bingo Steven. thank you for that comment. You answered my question before I asked it. checking it >500 did the trick and i did get 00:00:47

– coder
Mar 7 at 18:50

1

nice one Steven edited my answer accordingly

– Carlos Reyes
Mar 7 at 19:25

thanks it worked well with seconds. I wonder if it will work properly with minutes..or if i need something similar for minutes,, can you plz review this and see if it works properly fiddle

– coder
Mar 7 at 17:54

there is 1000 ms in 1 s, so wouldn't you want to test if it's greater than 500?

– Steven Stark
Mar 7 at 18:44

actually can you please revisit this... I tried to average these values: 00:01:20, 00:00:40, 00:01:00, 00:00:09 result was given to me 00:00:48 instead of 00:00:47

– coder
Mar 7 at 18:47

Bingo Steven. thank you for that comment. You answered my question before I asked it. checking it >500 did the trick and i did get 00:00:47

– coder
Mar 7 at 18:50

nice one Steven edited my answer accordingly

– Carlos Reyes
Mar 7 at 19:25

add a comment |

Date can be used to convert them to milliseconds :

function averageTime(arr) 
 var sum = arr.reduce(function(a, b) return a + +new Date('1970T' + b + 'Z'); , 0);
 return new Date(sum / arr.length + 500).toJSON().slice(11, 19);


console.log( averageTime(['00:00:03.00', '00:00:05.00', '00:00:02.00', '00:00:06.00']) );
console.log( averageTime(['00:00:03', '00:00:05', '00:00:20', '00:00:07']) );

answered Mar 7 at 22:23

Slai

15.5k32436

add a comment |

Date can be used to convert them to milliseconds :

function averageTime(arr) 
 var sum = arr.reduce(function(a, b) return a + +new Date('1970T' + b + 'Z'); , 0);
 return new Date(sum / arr.length + 500).toJSON().slice(11, 19);


console.log( averageTime(['00:00:03.00', '00:00:05.00', '00:00:02.00', '00:00:06.00']) );
console.log( averageTime(['00:00:03', '00:00:05', '00:00:20', '00:00:07']) );

answered Mar 7 at 22:23

Slai

15.5k32436

add a comment |

Date can be used to convert them to milliseconds :

function averageTime(arr) 
 var sum = arr.reduce(function(a, b) return a + +new Date('1970T' + b + 'Z'); , 0);
 return new Date(sum / arr.length + 500).toJSON().slice(11, 19);


console.log( averageTime(['00:00:03.00', '00:00:05.00', '00:00:02.00', '00:00:06.00']) );
console.log( averageTime(['00:00:03', '00:00:05', '00:00:20', '00:00:07']) );

answered Mar 7 at 22:23

Slai

15.5k32436

Date can be used to convert them to milliseconds :

function averageTime(arr) 
 var sum = arr.reduce(function(a, b) return a + +new Date('1970T' + b + 'Z'); , 0);
 return new Date(sum / arr.length + 500).toJSON().slice(11, 19);


console.log( averageTime(['00:00:03.00', '00:00:05.00', '00:00:02.00', '00:00:06.00']) );
console.log( averageTime(['00:00:03', '00:00:05', '00:00:20', '00:00:07']) );

function averageTime(arr) 
 var sum = arr.reduce(function(a, b) return a + +new Date('1970T' + b + 'Z'); , 0);
 return new Date(sum / arr.length + 500).toJSON().slice(11, 19);


console.log( averageTime(['00:00:03.00', '00:00:05.00', '00:00:02.00', '00:00:06.00']) );
console.log( averageTime(['00:00:03', '00:00:05', '00:00:20', '00:00:07']) );

function averageTime(arr) 
 var sum = arr.reduce(function(a, b) return a + +new Date('1970T' + b + 'Z'); , 0);
 return new Date(sum / arr.length + 500).toJSON().slice(11, 19);


console.log( averageTime(['00:00:03.00', '00:00:05.00', '00:00:02.00', '00:00:06.00']) );
console.log( averageTime(['00:00:03', '00:00:05', '00:00:20', '00:00:07']) );

answered Mar 7 at 22:23

Slai

15.5k32436

answered Mar 7 at 22:23

Slai

15.5k32436

answered Mar 7 at 22:23

Slai

15.5k32436

answered Mar 7 at 22:23

Slai

15.5k32436

add a comment |

First, convert all times into their UNIX timestamps using Date.getTime(), producing an array, then average the results. example: ]

let times = [
1551984787316,
1551984789662,
1551984790162,
1551984790730,
1551984791310
]


let average = ( values ) => 
 let sum = 0;
 for( let i = 0; i < values.length; i++ )
 sum += parseInt( values[i], 10 ); // base 10
 

 return Math.floor( sum/values.length );


let avgTime = average( times );
let avgDate = new Date( avgTime );
console.log( avgDate );

edited Mar 7 at 22:56

answered Mar 7 at 18:57

Steven Stark

666618

add a comment |

First, convert all times into their UNIX timestamps using Date.getTime(), producing an array, then average the results. example: ]

let times = [
1551984787316,
1551984789662,
1551984790162,
1551984790730,
1551984791310
]


let average = ( values ) => 
 let sum = 0;
 for( let i = 0; i < values.length; i++ )
 sum += parseInt( values[i], 10 ); // base 10
 

 return Math.floor( sum/values.length );


let avgTime = average( times );
let avgDate = new Date( avgTime );
console.log( avgDate );

edited Mar 7 at 22:56

answered Mar 7 at 18:57

Steven Stark

666618

add a comment |

First, convert all times into their UNIX timestamps using Date.getTime(), producing an array, then average the results. example: ]

let times = [
1551984787316,
1551984789662,
1551984790162,
1551984790730,
1551984791310
]


let average = ( values ) => 
 let sum = 0;
 for( let i = 0; i < values.length; i++ )
 sum += parseInt( values[i], 10 ); // base 10
 

 return Math.floor( sum/values.length );


let avgTime = average( times );
let avgDate = new Date( avgTime );
console.log( avgDate );

edited Mar 7 at 22:56

answered Mar 7 at 18:57

Steven Stark

666618

First, convert all times into their UNIX timestamps using Date.getTime(), producing an array, then average the results. example: ]

let times = [
1551984787316,
1551984789662,
1551984790162,
1551984790730,
1551984791310
]


let average = ( values ) => 
 let sum = 0;
 for( let i = 0; i < values.length; i++ )
 sum += parseInt( values[i], 10 ); // base 10
 

 return Math.floor( sum/values.length );


let avgTime = average( times );
let avgDate = new Date( avgTime );
console.log( avgDate );

let times = [
1551984787316,
1551984789662,
1551984790162,
1551984790730,
1551984791310
]


let average = ( values ) => 
 let sum = 0;
 for( let i = 0; i < values.length; i++ )
 sum += parseInt( values[i], 10 ); // base 10
 

 return Math.floor( sum/values.length );


let avgTime = average( times );
let avgDate = new Date( avgTime );
console.log( avgDate );

let times = [
1551984787316,
1551984789662,
1551984790162,
1551984790730,
1551984791310
]


let average = ( values ) => 
 let sum = 0;
 for( let i = 0; i < values.length; i++ )
 sum += parseInt( values[i], 10 ); // base 10
 

 return Math.floor( sum/values.length );


let avgTime = average( times );
let avgDate = new Date( avgTime );
console.log( avgDate );

edited Mar 7 at 22:56

answered Mar 7 at 18:57

Steven Stark

666618

edited Mar 7 at 22:56

answered Mar 7 at 18:57

Steven Stark

666618

answered Mar 7 at 18:57

Steven Stark

666618

answered Mar 7 at 18:57

Steven Stark

666618

add a comment |

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