Extracting data in ES6












1















I have this array const idArray = ["12", "231", "73", "4"] and an object 



const blueprints = {

  12: {color: red, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  231: {color: white, views: [{name: "front}, {name: "back}]},

  73: {color: black, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  4: {color: silver, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

}


How can I return an array of the following objects that have all front, back, top, and bottom using ES6 map/filter/some and etc?:



result =[

    {colorId: "12", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "73", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "4", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

]


I did it here but I feel like it is too messy and hard to read. Anybody could recommend on how to shorten it and make it easier to read using the ES6 functions (map, filter ...)?



 const result = idArray.map(id => {

      const bluePrint = bluePrints[id];

      const exists = blurPrint.views.some(view => view.name === 'top' || view.name === 'bottom');



      if (exists) {

        return {

          colorId: id,

          views: bluePrint.views

        }

      }

    }).filter(bluePrint => !bluePrint);





javascript ecmascript-6







share|improve this question


















  • 2





    Questions on how to optimize working code are better suited to be asked on codereview.stackexchange.com . You could use one reduce() instead of chaining map() and filter()

    – charlietfl
    Nov 15 '18 at 18:05











  • You have lots of missing quotes in your original array and desired result.

    – Barmar
    Nov 15 '18 at 18:14











  • Another typo: blurPrint should be bluePrint

    – Barmar
    Nov 15 '18 at 18:19











  • bluePrint => !bluePrint should be bluePrint => bluePrint, otherwise you'll just get an array of undefined.

    – Barmar
    Nov 15 '18 at 18:20


















1















I have this array const idArray = ["12", "231", "73", "4"] and an object 



const blueprints = {

  12: {color: red, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  231: {color: white, views: [{name: "front}, {name: "back}]},

  73: {color: black, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  4: {color: silver, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

}


How can I return an array of the following objects that have all front, back, top, and bottom using ES6 map/filter/some and etc?:



result =[

    {colorId: "12", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "73", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "4", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

]


I did it here but I feel like it is too messy and hard to read. Anybody could recommend on how to shorten it and make it easier to read using the ES6 functions (map, filter ...)?



 const result = idArray.map(id => {

      const bluePrint = bluePrints[id];

      const exists = blurPrint.views.some(view => view.name === 'top' || view.name === 'bottom');



      if (exists) {

        return {

          colorId: id,

          views: bluePrint.views

        }

      }

    }).filter(bluePrint => !bluePrint);





javascript ecmascript-6







share|improve this question


















  • 2





    Questions on how to optimize working code are better suited to be asked on codereview.stackexchange.com . You could use one reduce() instead of chaining map() and filter()

    – charlietfl
    Nov 15 '18 at 18:05











  • You have lots of missing quotes in your original array and desired result.

    – Barmar
    Nov 15 '18 at 18:14











  • Another typo: blurPrint should be bluePrint

    – Barmar
    Nov 15 '18 at 18:19











  • bluePrint => !bluePrint should be bluePrint => bluePrint, otherwise you'll just get an array of undefined.

    – Barmar
    Nov 15 '18 at 18:20
















1












1








1








I have this array const idArray = ["12", "231", "73", "4"] and an object 



const blueprints = {

  12: {color: red, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  231: {color: white, views: [{name: "front}, {name: "back}]},

  73: {color: black, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  4: {color: silver, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

}


How can I return an array of the following objects that have all front, back, top, and bottom using ES6 map/filter/some and etc?:



result =[

    {colorId: "12", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "73", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "4", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

]


I did it here but I feel like it is too messy and hard to read. Anybody could recommend on how to shorten it and make it easier to read using the ES6 functions (map, filter ...)?



 const result = idArray.map(id => {

      const bluePrint = bluePrints[id];

      const exists = blurPrint.views.some(view => view.name === 'top' || view.name === 'bottom');



      if (exists) {

        return {

          colorId: id,

          views: bluePrint.views

        }

      }

    }).filter(bluePrint => !bluePrint);





javascript ecmascript-6







share|improve this question














I have this array const idArray = ["12", "231", "73", "4"] and an object 



const blueprints = {

  12: {color: red, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  231: {color: white, views: [{name: "front}, {name: "back}]},

  73: {color: black, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  4: {color: silver, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

}


How can I return an array of the following objects that have all front, back, top, and bottom using ES6 map/filter/some and etc?:



result =[

    {colorId: "12", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "73", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "4", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

]


I did it here but I feel like it is too messy and hard to read. Anybody could recommend on how to shorten it and make it easier to read using the ES6 functions (map, filter ...)?



 const result = idArray.map(id => {

      const bluePrint = bluePrints[id];

      const exists = blurPrint.views.some(view => view.name === 'top' || view.name === 'bottom');



      if (exists) {

        return {

          colorId: id,

          views: bluePrint.views

        }

      }

    }).filter(bluePrint => !bluePrint);





javascript ecmascript-6




javascript ecmascript-6






share|improve this question













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asked Nov 15 '18 at 17:57









j doej doe

63




63








  • 2





    Questions on how to optimize working code are better suited to be asked on codereview.stackexchange.com . You could use one reduce() instead of chaining map() and filter()

    – charlietfl
    Nov 15 '18 at 18:05











  • You have lots of missing quotes in your original array and desired result.

    – Barmar
    Nov 15 '18 at 18:14











  • Another typo: blurPrint should be bluePrint

    – Barmar
    Nov 15 '18 at 18:19











  • bluePrint => !bluePrint should be bluePrint => bluePrint, otherwise you'll just get an array of undefined.

    – Barmar
    Nov 15 '18 at 18:20
















  • 2





    Questions on how to optimize working code are better suited to be asked on codereview.stackexchange.com . You could use one reduce() instead of chaining map() and filter()

    – charlietfl
    Nov 15 '18 at 18:05











  • You have lots of missing quotes in your original array and desired result.

    – Barmar
    Nov 15 '18 at 18:14











  • Another typo: blurPrint should be bluePrint

    – Barmar
    Nov 15 '18 at 18:19











  • bluePrint => !bluePrint should be bluePrint => bluePrint, otherwise you'll just get an array of undefined.

    – Barmar
    Nov 15 '18 at 18:20










2




2





Questions on how to optimize working code are better suited to be asked on codereview.stackexchange.com . You could use one reduce() instead of chaining map() and filter()

– charlietfl
Nov 15 '18 at 18:05





Questions on how to optimize working code are better suited to be asked on codereview.stackexchange.com . You could use one reduce() instead of chaining map() and filter()

– charlietfl
Nov 15 '18 at 18:05













You have lots of missing quotes in your original array and desired result.

– Barmar
Nov 15 '18 at 18:14





You have lots of missing quotes in your original array and desired result.

– Barmar
Nov 15 '18 at 18:14













Another typo: blurPrint should be bluePrint

– Barmar
Nov 15 '18 at 18:19





Another typo: blurPrint should be bluePrint

– Barmar
Nov 15 '18 at 18:19













bluePrint => !bluePrint should be bluePrint => bluePrint, otherwise you'll just get an array of undefined.

– Barmar
Nov 15 '18 at 18:20







bluePrint => !bluePrint should be bluePrint => bluePrint, otherwise you'll just get an array of undefined.

– Barmar
Nov 15 '18 at 18:20














3 Answers
3






active

oldest

votes


















0














You can filter ids so that every color in your target set of colors is in that id's blueprint.views and then map those ids to your desired result object:






const idArray = ["12", "231", "73", "4"];

const blueprints = {

  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},

  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

};



const result = idArray

                .filter(id => {

                  const colors = blueprints[id].views.map(e => e.name);

                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));

                })

                .map(id => ({colorId: id, views: blueprints[id].views}))



console.log(result);








share|improve this answer

































    0














    You can map() over your idArray to create the array in the desired format, then use filter() to test if every() required string is in the view array and filter out the incomplete items:






    const blueprints = {
    
    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
    
    231: {color: "white", views: [{name: "front"}, {name: "back"}]},
    
    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
    
    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
    
  }
    

    
const idArray = ["12", "231", "73", "4"] 
    
const required = ['front', 'back', 'top', 'bottom']
    

    
let newArry = idArray
    
        .map(colorID => ({colorID, views: blueprints[colorID].views }) )
    
        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))
    

    
console.log(newArry)








    share|improve this answer

































      0














      Object.keys(blueprints)
      
  .map(k => ({colorId:k, views:blueprints[k].views}))
      
  .filter(el =>
      
    ['front', 'back', 'top', 'bottom'].every(it => 
      
      el.views.some(s => s.name === it)
      
  )
      
)





      share|improve this answer

























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        3 Answers
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        3 Answers
        3






        active

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        active

        oldest

        votes






        active

        oldest

        votes









        0














        You can filter ids so that every color in your target set of colors is in that id's blueprint.views and then map those ids to your desired result object:






        const idArray = ["12", "231", "73", "4"];
        
const blueprints = {
        
  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
        
  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},
        
  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
        
  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
        
};
        

        
const result = idArray
        
                .filter(id => {
        
                  const colors = blueprints[id].views.map(e => e.name);
        
                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));
        
                })
        
                .map(id => ({colorId: id, views: blueprints[id].views}))
        

        
console.log(result);








        share|improve this answer






























          0














          You can filter ids so that every color in your target set of colors is in that id's blueprint.views and then map those ids to your desired result object:






          const idArray = ["12", "231", "73", "4"];
          
const blueprints = {
          
  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
          
  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},
          
  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
          
  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
          
};
          

          
const result = idArray
          
                .filter(id => {
          
                  const colors = blueprints[id].views.map(e => e.name);
          
                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));
          
                })
          
                .map(id => ({colorId: id, views: blueprints[id].views}))
          

          
console.log(result);








          share|improve this answer




























            0












            0








            0







            You can filter ids so that every color in your target set of colors is in that id's blueprint.views and then map those ids to your desired result object:






            const idArray = ["12", "231", "73", "4"];
            
const blueprints = {
            
  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},
            
  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
};
            

            
const result = idArray
            
                .filter(id => {
            
                  const colors = blueprints[id].views.map(e => e.name);
            
                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));
            
                })
            
                .map(id => ({colorId: id, views: blueprints[id].views}))
            

            
console.log(result);








            share|improve this answer















            You can filter ids so that every color in your target set of colors is in that id's blueprint.views and then map those ids to your desired result object:






            const idArray = ["12", "231", "73", "4"];
            
const blueprints = {
            
  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},
            
  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
};
            

            
const result = idArray
            
                .filter(id => {
            
                  const colors = blueprints[id].views.map(e => e.name);
            
                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));
            
                })
            
                .map(id => ({colorId: id, views: blueprints[id].views}))
            

            
console.log(result);








            const idArray = ["12", "231", "73", "4"];
            
const blueprints = {
            
  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},
            
  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
};
            

            
const result = idArray
            
                .filter(id => {
            
                  const colors = blueprints[id].views.map(e => e.name);
            
                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));
            
                })
            
                .map(id => ({colorId: id, views: blueprints[id].views}))
            

            
console.log(result);





            const idArray = ["12", "231", "73", "4"];
            
const blueprints = {
            
  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},
            
  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
            
};
            

            
const result = idArray
            
                .filter(id => {
            
                  const colors = blueprints[id].views.map(e => e.name);
            
                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));
            
                })
            
                .map(id => ({colorId: id, views: blueprints[id].views}))
            

            
console.log(result);






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 15 '18 at 18:30

























            answered Nov 15 '18 at 18:21









            sliderslider

            8,49811331




            8,49811331

























                0














                You can map() over your idArray to create the array in the desired format, then use filter() to test if every() required string is in the view array and filter out the incomplete items:






                const blueprints = {
                
    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                
    231: {color: "white", views: [{name: "front"}, {name: "back"}]},
                
    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                
    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                
  }
                

                
const idArray = ["12", "231", "73", "4"] 
                
const required = ['front', 'back', 'top', 'bottom']
                

                
let newArry = idArray
                
        .map(colorID => ({colorID, views: blueprints[colorID].views }) )
                
        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))
                

                
console.log(newArry)








                share|improve this answer






























                  0














                  You can map() over your idArray to create the array in the desired format, then use filter() to test if every() required string is in the view array and filter out the incomplete items:






                  const blueprints = {
                  
    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                  
    231: {color: "white", views: [{name: "front"}, {name: "back"}]},
                  
    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                  
    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                  
  }
                  

                  
const idArray = ["12", "231", "73", "4"] 
                  
const required = ['front', 'back', 'top', 'bottom']
                  

                  
let newArry = idArray
                  
        .map(colorID => ({colorID, views: blueprints[colorID].views }) )
                  
        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))
                  

                  
console.log(newArry)








                  share|improve this answer




























                    0












                    0








                    0







                    You can map() over your idArray to create the array in the desired format, then use filter() to test if every() required string is in the view array and filter out the incomplete items:






                    const blueprints = {
                    
    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
    231: {color: "white", views: [{name: "front"}, {name: "back"}]},
                    
    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
  }
                    

                    
const idArray = ["12", "231", "73", "4"] 
                    
const required = ['front', 'back', 'top', 'bottom']
                    

                    
let newArry = idArray
                    
        .map(colorID => ({colorID, views: blueprints[colorID].views }) )
                    
        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))
                    

                    
console.log(newArry)








                    share|improve this answer















                    You can map() over your idArray to create the array in the desired format, then use filter() to test if every() required string is in the view array and filter out the incomplete items:






                    const blueprints = {
                    
    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
    231: {color: "white", views: [{name: "front"}, {name: "back"}]},
                    
    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
  }
                    

                    
const idArray = ["12", "231", "73", "4"] 
                    
const required = ['front', 'back', 'top', 'bottom']
                    

                    
let newArry = idArray
                    
        .map(colorID => ({colorID, views: blueprints[colorID].views }) )
                    
        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))
                    

                    
console.log(newArry)








                    const blueprints = {
                    
    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
    231: {color: "white", views: [{name: "front"}, {name: "back"}]},
                    
    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
  }
                    

                    
const idArray = ["12", "231", "73", "4"] 
                    
const required = ['front', 'back', 'top', 'bottom']
                    

                    
let newArry = idArray
                    
        .map(colorID => ({colorID, views: blueprints[colorID].views }) )
                    
        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))
                    

                    
console.log(newArry)





                    const blueprints = {
                    
    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
    231: {color: "white", views: [{name: "front"}, {name: "back"}]},
                    
    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},
                    
  }
                    

                    
const idArray = ["12", "231", "73", "4"] 
                    
const required = ['front', 'back', 'top', 'bottom']
                    

                    
let newArry = idArray
                    
        .map(colorID => ({colorID, views: blueprints[colorID].views }) )
                    
        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))
                    

                    
console.log(newArry)






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Nov 15 '18 at 18:34

























                    answered Nov 15 '18 at 18:24









                    Mark MeyerMark Meyer

                    39.2k33162




                    39.2k33162























                        0














                        Object.keys(blueprints)
                        
  .map(k => ({colorId:k, views:blueprints[k].views}))
                        
  .filter(el =>
                        
    ['front', 'back', 'top', 'bottom'].every(it => 
                        
      el.views.some(s => s.name === it)
                        
  )
                        
)





                        share|improve this answer






























                          0














                          Object.keys(blueprints)
                          
  .map(k => ({colorId:k, views:blueprints[k].views}))
                          
  .filter(el =>
                          
    ['front', 'back', 'top', 'bottom'].every(it => 
                          
      el.views.some(s => s.name === it)
                          
  )
                          
)





                          share|improve this answer




























                            0












                            0








                            0







                            Object.keys(blueprints)
                            
  .map(k => ({colorId:k, views:blueprints[k].views}))
                            
  .filter(el =>
                            
    ['front', 'back', 'top', 'bottom'].every(it => 
                            
      el.views.some(s => s.name === it)
                            
  )
                            
)





                            share|improve this answer















                            Object.keys(blueprints)
                            
  .map(k => ({colorId:k, views:blueprints[k].views}))
                            
  .filter(el =>
                            
    ['front', 'back', 'top', 'bottom'].every(it => 
                            
      el.views.some(s => s.name === it)
                            
  )
                            
)






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Nov 16 '18 at 6:17

























                            answered Nov 15 '18 at 18:39









                            dee zgdee zg

                            4,80231437




                            4,80231437






























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