How to order array of objects by the number of the keys the same as the given ones as a parameter?












1















I have a basic object with some keys, and the list of objects with similar, but not necessarily the exact same, keys.



const o1 = {k1: "", k2: "", k3: ""} // the basic object

const o2 = {k1: "", k4: ""}         // 1 the same key

const o3 = {k1: "", k3: ""}         // 2 the same keys

const o4 = {k3: "", k1: "", k2: ""} // 3 the same keys

const o5 = {k5: ""}                 // 0 the same keys


I want to write a function which takes the base object and an array of other objects, and it will sort/order them by the number of similar keys in the given object.



function order(obj, arr) {

  // ...

}


The result should be the ordered array of the given objects based on the number of the same keys as the base object.



order(o1, [o2, o3, o4, o5])

// result: [o4, o3, o2, o5]


What would you use for that?



I was thinking about sorting by the length of intersection on objects keys.






javascript ecmascript-6 functional-programming lodash







share|improve this question

























  • Object.keys(), Array.sort() and any form of loop to check/count the matching properties

    – Andreas
    Nov 19 '18 at 16:31
















1















I have a basic object with some keys, and the list of objects with similar, but not necessarily the exact same, keys.



const o1 = {k1: "", k2: "", k3: ""} // the basic object

const o2 = {k1: "", k4: ""}         // 1 the same key

const o3 = {k1: "", k3: ""}         // 2 the same keys

const o4 = {k3: "", k1: "", k2: ""} // 3 the same keys

const o5 = {k5: ""}                 // 0 the same keys


I want to write a function which takes the base object and an array of other objects, and it will sort/order them by the number of similar keys in the given object.



function order(obj, arr) {

  // ...

}


The result should be the ordered array of the given objects based on the number of the same keys as the base object.



order(o1, [o2, o3, o4, o5])

// result: [o4, o3, o2, o5]


What would you use for that?



I was thinking about sorting by the length of intersection on objects keys.






javascript ecmascript-6 functional-programming lodash







share|improve this question

























  • Object.keys(), Array.sort() and any form of loop to check/count the matching properties

    – Andreas
    Nov 19 '18 at 16:31














1












1








1








I have a basic object with some keys, and the list of objects with similar, but not necessarily the exact same, keys.



const o1 = {k1: "", k2: "", k3: ""} // the basic object

const o2 = {k1: "", k4: ""}         // 1 the same key

const o3 = {k1: "", k3: ""}         // 2 the same keys

const o4 = {k3: "", k1: "", k2: ""} // 3 the same keys

const o5 = {k5: ""}                 // 0 the same keys


I want to write a function which takes the base object and an array of other objects, and it will sort/order them by the number of similar keys in the given object.



function order(obj, arr) {

  // ...

}


The result should be the ordered array of the given objects based on the number of the same keys as the base object.



order(o1, [o2, o3, o4, o5])

// result: [o4, o3, o2, o5]


What would you use for that?



I was thinking about sorting by the length of intersection on objects keys.






javascript ecmascript-6 functional-programming lodash







share|improve this question
















I have a basic object with some keys, and the list of objects with similar, but not necessarily the exact same, keys.



const o1 = {k1: "", k2: "", k3: ""} // the basic object

const o2 = {k1: "", k4: ""}         // 1 the same key

const o3 = {k1: "", k3: ""}         // 2 the same keys

const o4 = {k3: "", k1: "", k2: ""} // 3 the same keys

const o5 = {k5: ""}                 // 0 the same keys


I want to write a function which takes the base object and an array of other objects, and it will sort/order them by the number of similar keys in the given object.



function order(obj, arr) {

  // ...

}


The result should be the ordered array of the given objects based on the number of the same keys as the base object.



order(o1, [o2, o3, o4, o5])

// result: [o4, o3, o2, o5]


What would you use for that?



I was thinking about sorting by the length of intersection on objects keys.






javascript ecmascript-6 functional-programming lodash




javascript ecmascript-6 functional-programming lodash






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 19 '18 at 16:39







Kamil Lelonek

















asked Nov 19 '18 at 16:28









Kamil LelonekKamil Lelonek

9,45594876




9,45594876













  • Object.keys(), Array.sort() and any form of loop to check/count the matching properties

    – Andreas
    Nov 19 '18 at 16:31



















  • Object.keys(), Array.sort() and any form of loop to check/count the matching properties

    – Andreas
    Nov 19 '18 at 16:31

















Object.keys(), Array.sort() and any form of loop to check/count the matching properties

– Andreas
Nov 19 '18 at 16:31





Object.keys(), Array.sort() and any form of loop to check/count the matching properties

– Andreas
Nov 19 '18 at 16:31












3 Answers
3






active

oldest

votes


















1














You can use a combination of Object.keys and filter to determine the "real" length of the array and sort appropriately:






const o1 = {k1: "", k2: "", k3: ""} // the basic object

const o2 = {k1: "", k4: ""}         // 1 the same key

const o3 = {k1: "", k3: ""}         // 2 the same keys

const o4 = {k3: "", k1: "", k2: ""} // 3 the same keys

const o5 = {k5: ""}                 // 0 the same keys



function order(obj, arr) {

  return arr.sort((a, b) => 

              Object.keys(b).filter(k => k in obj).length - 

              Object.keys(a).filter(k => k in obj).length);

}



console.log(order(o1, [o2, o3, o4, o5]));








share|improve this answer
























  • Just a caveat: Array.sort will mutate arr, which may be a problem.

    – tokland
    Nov 19 '18 at 22:36



















2














You could count same keys and take this value for sorting.






const

    o1 = { k1: "", k2: "", k3: "" }, // the basic object

    o2 = { k1: "", k4: "" },         // 1 the same key

    o3 = { k1: "", k3: "" },         // 2 the same keys

    o4 = { k3: "", k1: "", k2: "" }, // 3 the same keys

    o5 = { k5: "" };                 // 0 the same keys



function order(object, array) {

    function getCount(o) {

        return Object.keys(object).reduce((s, k) => s + (k in o), 0);

    }

    return array.sort((a, b) => getCount(b) - getCount(a));

}



console.log(order(o1, [o2, o3, o4, o5])); // [o4, o3, o2, o5]








share|improve this answer































    1














    Using lodash:



    function order(reference, objs) {
    
  const referenceKeys = _.keys(reference);
    
  return _.sortBy(objs, obj => -_(referenceKeys).intersection(_.keys(obj)).size());
    
}


    Another way:



    function order(reference, objs) {
    
  const referenceKeys = new Set(_.keys(reference));
    
  return _.sortBy(objs, obj => _(obj).keys().sumBy(key => referenceKeys.has(key) ? -1 : 0))
    
}





    share|improve this answer


























    • I really like the first example!

      – Kamil Lelonek
      Nov 19 '18 at 23:45











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    You can use a combination of Object.keys and filter to determine the "real" length of the array and sort appropriately:






    const o1 = {k1: "", k2: "", k3: ""} // the basic object
    
const o2 = {k1: "", k4: ""}         // 1 the same key
    
const o3 = {k1: "", k3: ""}         // 2 the same keys
    
const o4 = {k3: "", k1: "", k2: ""} // 3 the same keys
    
const o5 = {k5: ""}                 // 0 the same keys
    

    
function order(obj, arr) {
    
  return arr.sort((a, b) => 
    
              Object.keys(b).filter(k => k in obj).length - 
    
              Object.keys(a).filter(k => k in obj).length);
    
}
    

    
console.log(order(o1, [o2, o3, o4, o5]));








    share|improve this answer
























    • Just a caveat: Array.sort will mutate arr, which may be a problem.

      – tokland
      Nov 19 '18 at 22:36
















    1














    You can use a combination of Object.keys and filter to determine the "real" length of the array and sort appropriately:






    const o1 = {k1: "", k2: "", k3: ""} // the basic object
    
const o2 = {k1: "", k4: ""}         // 1 the same key
    
const o3 = {k1: "", k3: ""}         // 2 the same keys
    
const o4 = {k3: "", k1: "", k2: ""} // 3 the same keys
    
const o5 = {k5: ""}                 // 0 the same keys
    

    
function order(obj, arr) {
    
  return arr.sort((a, b) => 
    
              Object.keys(b).filter(k => k in obj).length - 
    
              Object.keys(a).filter(k => k in obj).length);
    
}
    

    
console.log(order(o1, [o2, o3, o4, o5]));








    share|improve this answer
























    • Just a caveat: Array.sort will mutate arr, which may be a problem.

      – tokland
      Nov 19 '18 at 22:36














    1












    1








    1







    You can use a combination of Object.keys and filter to determine the "real" length of the array and sort appropriately:






    const o1 = {k1: "", k2: "", k3: ""} // the basic object
    
const o2 = {k1: "", k4: ""}         // 1 the same key
    
const o3 = {k1: "", k3: ""}         // 2 the same keys
    
const o4 = {k3: "", k1: "", k2: ""} // 3 the same keys
    
const o5 = {k5: ""}                 // 0 the same keys
    

    
function order(obj, arr) {
    
  return arr.sort((a, b) => 
    
              Object.keys(b).filter(k => k in obj).length - 
    
              Object.keys(a).filter(k => k in obj).length);
    
}
    

    
console.log(order(o1, [o2, o3, o4, o5]));








    share|improve this answer













    You can use a combination of Object.keys and filter to determine the "real" length of the array and sort appropriately:






    const o1 = {k1: "", k2: "", k3: ""} // the basic object
    
const o2 = {k1: "", k4: ""}         // 1 the same key
    
const o3 = {k1: "", k3: ""}         // 2 the same keys
    
const o4 = {k3: "", k1: "", k2: ""} // 3 the same keys
    
const o5 = {k5: ""}                 // 0 the same keys
    

    
function order(obj, arr) {
    
  return arr.sort((a, b) => 
    
              Object.keys(b).filter(k => k in obj).length - 
    
              Object.keys(a).filter(k => k in obj).length);
    
}
    

    
console.log(order(o1, [o2, o3, o4, o5]));








    const o1 = {k1: "", k2: "", k3: ""} // the basic object
    
const o2 = {k1: "", k4: ""}         // 1 the same key
    
const o3 = {k1: "", k3: ""}         // 2 the same keys
    
const o4 = {k3: "", k1: "", k2: ""} // 3 the same keys
    
const o5 = {k5: ""}                 // 0 the same keys
    

    
function order(obj, arr) {
    
  return arr.sort((a, b) => 
    
              Object.keys(b).filter(k => k in obj).length - 
    
              Object.keys(a).filter(k => k in obj).length);
    
}
    

    
console.log(order(o1, [o2, o3, o4, o5]));





    const o1 = {k1: "", k2: "", k3: ""} // the basic object
    
const o2 = {k1: "", k4: ""}         // 1 the same key
    
const o3 = {k1: "", k3: ""}         // 2 the same keys
    
const o4 = {k3: "", k1: "", k2: ""} // 3 the same keys
    
const o5 = {k5: ""}                 // 0 the same keys
    

    
function order(obj, arr) {
    
  return arr.sort((a, b) => 
    
              Object.keys(b).filter(k => k in obj).length - 
    
              Object.keys(a).filter(k => k in obj).length);
    
}
    

    
console.log(order(o1, [o2, o3, o4, o5]));






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 19 '18 at 16:36









    sliderslider

    8,16511129




    8,16511129













    • Just a caveat: Array.sort will mutate arr, which may be a problem.

      – tokland
      Nov 19 '18 at 22:36



















    • Just a caveat: Array.sort will mutate arr, which may be a problem.

      – tokland
      Nov 19 '18 at 22:36

















    Just a caveat: Array.sort will mutate arr, which may be a problem.

    – tokland
    Nov 19 '18 at 22:36





    Just a caveat: Array.sort will mutate arr, which may be a problem.

    – tokland
    Nov 19 '18 at 22:36













    2














    You could count same keys and take this value for sorting.






    const
    
    o1 = { k1: "", k2: "", k3: "" }, // the basic object
    
    o2 = { k1: "", k4: "" },         // 1 the same key
    
    o3 = { k1: "", k3: "" },         // 2 the same keys
    
    o4 = { k3: "", k1: "", k2: "" }, // 3 the same keys
    
    o5 = { k5: "" };                 // 0 the same keys
    

    
function order(object, array) {
    
    function getCount(o) {
    
        return Object.keys(object).reduce((s, k) => s + (k in o), 0);
    
    }
    
    return array.sort((a, b) => getCount(b) - getCount(a));
    
}
    

    
console.log(order(o1, [o2, o3, o4, o5])); // [o4, o3, o2, o5]








    share|improve this answer




























      2














      You could count same keys and take this value for sorting.






      const
      
    o1 = { k1: "", k2: "", k3: "" }, // the basic object
      
    o2 = { k1: "", k4: "" },         // 1 the same key
      
    o3 = { k1: "", k3: "" },         // 2 the same keys
      
    o4 = { k3: "", k1: "", k2: "" }, // 3 the same keys
      
    o5 = { k5: "" };                 // 0 the same keys
      

      
function order(object, array) {
      
    function getCount(o) {
      
        return Object.keys(object).reduce((s, k) => s + (k in o), 0);
      
    }
      
    return array.sort((a, b) => getCount(b) - getCount(a));
      
}
      

      
console.log(order(o1, [o2, o3, o4, o5])); // [o4, o3, o2, o5]








      share|improve this answer


























        2












        2








        2







        You could count same keys and take this value for sorting.






        const
        
    o1 = { k1: "", k2: "", k3: "" }, // the basic object
        
    o2 = { k1: "", k4: "" },         // 1 the same key
        
    o3 = { k1: "", k3: "" },         // 2 the same keys
        
    o4 = { k3: "", k1: "", k2: "" }, // 3 the same keys
        
    o5 = { k5: "" };                 // 0 the same keys
        

        
function order(object, array) {
        
    function getCount(o) {
        
        return Object.keys(object).reduce((s, k) => s + (k in o), 0);
        
    }
        
    return array.sort((a, b) => getCount(b) - getCount(a));
        
}
        

        
console.log(order(o1, [o2, o3, o4, o5])); // [o4, o3, o2, o5]








        share|improve this answer













        You could count same keys and take this value for sorting.






        const
        
    o1 = { k1: "", k2: "", k3: "" }, // the basic object
        
    o2 = { k1: "", k4: "" },         // 1 the same key
        
    o3 = { k1: "", k3: "" },         // 2 the same keys
        
    o4 = { k3: "", k1: "", k2: "" }, // 3 the same keys
        
    o5 = { k5: "" };                 // 0 the same keys
        

        
function order(object, array) {
        
    function getCount(o) {
        
        return Object.keys(object).reduce((s, k) => s + (k in o), 0);
        
    }
        
    return array.sort((a, b) => getCount(b) - getCount(a));
        
}
        

        
console.log(order(o1, [o2, o3, o4, o5])); // [o4, o3, o2, o5]








        const
        
    o1 = { k1: "", k2: "", k3: "" }, // the basic object
        
    o2 = { k1: "", k4: "" },         // 1 the same key
        
    o3 = { k1: "", k3: "" },         // 2 the same keys
        
    o4 = { k3: "", k1: "", k2: "" }, // 3 the same keys
        
    o5 = { k5: "" };                 // 0 the same keys
        

        
function order(object, array) {
        
    function getCount(o) {
        
        return Object.keys(object).reduce((s, k) => s + (k in o), 0);
        
    }
        
    return array.sort((a, b) => getCount(b) - getCount(a));
        
}
        

        
console.log(order(o1, [o2, o3, o4, o5])); // [o4, o3, o2, o5]





        const
        
    o1 = { k1: "", k2: "", k3: "" }, // the basic object
        
    o2 = { k1: "", k4: "" },         // 1 the same key
        
    o3 = { k1: "", k3: "" },         // 2 the same keys
        
    o4 = { k3: "", k1: "", k2: "" }, // 3 the same keys
        
    o5 = { k5: "" };                 // 0 the same keys
        

        
function order(object, array) {
        
    function getCount(o) {
        
        return Object.keys(object).reduce((s, k) => s + (k in o), 0);
        
    }
        
    return array.sort((a, b) => getCount(b) - getCount(a));
        
}
        

        
console.log(order(o1, [o2, o3, o4, o5])); // [o4, o3, o2, o5]






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 19 '18 at 16:41









        Nina ScholzNina Scholz

        178k1491157




        178k1491157























            1














            Using lodash:



            function order(reference, objs) {
            
  const referenceKeys = _.keys(reference);
            
  return _.sortBy(objs, obj => -_(referenceKeys).intersection(_.keys(obj)).size());
            
}


            Another way:



            function order(reference, objs) {
            
  const referenceKeys = new Set(_.keys(reference));
            
  return _.sortBy(objs, obj => _(obj).keys().sumBy(key => referenceKeys.has(key) ? -1 : 0))
            
}





            share|improve this answer


























            • I really like the first example!

              – Kamil Lelonek
              Nov 19 '18 at 23:45
















            1














            Using lodash:



            function order(reference, objs) {
            
  const referenceKeys = _.keys(reference);
            
  return _.sortBy(objs, obj => -_(referenceKeys).intersection(_.keys(obj)).size());
            
}


            Another way:



            function order(reference, objs) {
            
  const referenceKeys = new Set(_.keys(reference));
            
  return _.sortBy(objs, obj => _(obj).keys().sumBy(key => referenceKeys.has(key) ? -1 : 0))
            
}





            share|improve this answer


























            • I really like the first example!

              – Kamil Lelonek
              Nov 19 '18 at 23:45














            1












            1








            1







            Using lodash:



            function order(reference, objs) {
            
  const referenceKeys = _.keys(reference);
            
  return _.sortBy(objs, obj => -_(referenceKeys).intersection(_.keys(obj)).size());
            
}


            Another way:



            function order(reference, objs) {
            
  const referenceKeys = new Set(_.keys(reference));
            
  return _.sortBy(objs, obj => _(obj).keys().sumBy(key => referenceKeys.has(key) ? -1 : 0))
            
}





            share|improve this answer















            Using lodash:



            function order(reference, objs) {
            
  const referenceKeys = _.keys(reference);
            
  return _.sortBy(objs, obj => -_(referenceKeys).intersection(_.keys(obj)).size());
            
}


            Another way:



            function order(reference, objs) {
            
  const referenceKeys = new Set(_.keys(reference));
            
  return _.sortBy(objs, obj => _(obj).keys().sumBy(key => referenceKeys.has(key) ? -1 : 0))
            
}






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 19 '18 at 22:46

























            answered Nov 19 '18 at 22:12









            toklandtokland

            51.7k11115143




            51.7k11115143













            • I really like the first example!

              – Kamil Lelonek
              Nov 19 '18 at 23:45



















            • I really like the first example!

              – Kamil Lelonek
              Nov 19 '18 at 23:45

















            I really like the first example!

            – Kamil Lelonek
            Nov 19 '18 at 23:45





            I really like the first example!

            – Kamil Lelonek
            Nov 19 '18 at 23:45


















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            MongoDB - Not Authorized To Execute Command

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
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            Npm cannot find a required file even through it is in the searched directory

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            up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
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            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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            1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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