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How to calculate average using jq from JSON output


By : John McTigue
Date : July 30 2020, 11:00 PM
wish helps you You can create a single array instead of an array per entry, then pipe it to the filter add and the filter length, which will add the numbers in the array and compute the length of the array. FInally, you divide the results of both filters and... voila!
code :
jq '[.Datapoints[] | select(.ExtendedStatistics.p80>=0) | .ExtendedStatistics.p80] | add/length' file.json


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Calculate and output average from jQuery UI sliders


By : lucifer_rj
Date : March 29 2020, 07:55 AM
help you fix your problem Check this out I made a class so it would be easier to loop through
http://jsfiddle.net/6SftR/
code :
var sum = 0 ; 
 var nums = 0;
$('.slider').bind({

    slidechange : function(event,ui) {
      sum = 0;
      nums  = 0;
    $('.slider').each(function(){

        var value = $( this ).slider( "option", "value" );
        if(value > 0 ) 
         {
            sum += value ; 
            nums++;
          }

        }); 
       var avg = sum / nums ; 
       $('.average').val(Math.floor(avg));
       console.log(avg); 
    }
}).slider();
<div class="slider"></div>
<div class="slider"></div>
<div class="slider"></div>
<div class="slider"></div>
<div class="slider"></div>
<div class="slider"></div>


    <input type="text" class="average" name="pm_overall_score" id="pm_overall_score" value="" size="30">

Calculate average from JSON column


By : Lư Gia Lâm
Date : March 29 2020, 07:55 AM
I hope this helps . Your json array is messed up, like @posz commented. Would have to be:
code :
CREATE TABLE tbl (id int, speed_data json);

INSERT INTO tbl VALUES
  (391982, '{"speed":[1.3,1.3,1.4,1.5]}')
, (391983, '{"speed":[0.9,0.8,0.8,1.0]}');
SELECT t.id, avg(x::text::numeric) AS avg_speed
FROM   tbl t
     , json_array_elements(speed_data->'speed') x
GROUP  BY t.id;
SELECT t.id, avg(x::numeric) AS avg_speed
FROM   tbl t
     , json_array_elements_text(speed_data->'speed') x
GROUP  BY t.id;

Calculate Average for all columns and then output difference


By : S.salunkhe
Date : March 29 2020, 07:55 AM
like below fixes the issue There are two problems in your query. The first one is that since you're selecting the event_name, your average is calcluated at that level; the other problem is that you're using the Event_Cost column both as is and aggregated at the same time, from the same table, and that's not possible.
You should get the aggregated value from an inner query and cross join it with your initial table, like this
code :
SELECT  t1.Event_Name AS 'Event Name',
        ROUND(Event_Cost) AS 'Event Cost',
        ROUND(t2.aver) AS 'Average Cost Of Events',
        ROUND(Event_Cost - t2.aver) AS 'Amount Over Average £'
FROM    Event t1
CROSS JOIN
        (
            SELECT  AVG(Event_cost) as aver
            FROM    Event
        ) t2
WHERE   t1.Event_Cost > t2.aver;

Calculate Average of JSON Values


By : user2098761
Date : March 29 2020, 07:55 AM
I hope this helps . You could reduce the data by taking the nth (length) of the value and add it to the average in a single loop.
code :
var data = { a: { values: 9892308 }, b: { values: 9894522 }, c: { values: 3298714 }, d: { values: 3300055 } },
    average = Object
        .values(data)
        .reduce((avg, { values }, _, { length }) => avg + values / length, 0);
        
console.log(average);

output blank--Java program to calculate average of array


By : Truong Pham
Date : March 29 2020, 07:55 AM
this will help to answer the main question...
However, my problem is that when I run my program the output window is completely blank. I assume this is because I did not call in my method to the main method. However, I am not sure how to write that code.
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