Iterating through an array of objects and extracting values from unspecified keys

Question

Iterating through an array of objects and extracting values from unspecified keys

Calculating which color holds a higher value in each array element of the data. Then adding the color with the higher value to an empty object, or increasing the count if already present. Finally, sorting the totals object from highest to lowest based on the total values and returning the color with the highest value.

Facing challenges mapping over this structured array as the property keys are not consistent. Should I consider restructuring it?

*I am open to redesigning the data structure if necessary, so please suggest if there's an easier solution with a different design!

data = [
   { orange: 4, green: 4},
   { green: 0, yellow: 0},
   { yellow: 1, orange: 4 },
   { blue: 2, green: 1 }, 
   { blue: 2, yellow: 1 }, 
   { green: 3, yellow: 2 },
   { green: 1, blue: 3},
   { green: 5, yellow: 2 }, 
 ]


totals = {
  blue: 3,
  green: 2,
  orange: 1,
}

Solution: highValueColor = blue

// PSEUDOCODE //map over the array => data.map() //identify highest value between two elements => propA - propB //check to see if the color's (key) in the element has already been added to totals object //IF the key does not yet exist, create a property in the tally object with the color(key) and set its value to 1 //IF the key is already listed in tally object, increment its property value by 1 => ++ //sort totals object => Math.max() //return highest value color

javascript arrays maps

Answer 1

Answer №1

It's worth noting that there may be a small issue with @hopzebordah's answer, as it seems to count a color even when two colors have the same value. For example, { orange: 4, green: 4} would be counted as orange.

In the comments, I provided an alternative version using map since you appeared to show interest in that method. However, I might have misinterpreted your intended goal.

If you only require the highest value and not necessarily a sorted object, then skipping the sorting step might be more efficient. You can see this in action with highest_value_unsort.

const data = [
  { orange: 4, green: 4},
  { green: 0, yellow: 0},
  { yellow: 1, orange: 4 },
  { blue: 2, green: 1 }, 
  { blue: 2, yellow: 1 }, 
  { green: 3, yellow: 2 },
  { green: 1, blue: 3},
  { green: 5, yellow: 2 }, 
];

const pick_color = (color_obj) => {
  const [[color1, val1], [color2, val2]] = Object.entries(color_obj);
  
  return val1 === val2 ?
    null :
    val1 > val2 ?
        color1 :
        color2;
};


const unsorted = {};
for(const color_obj of data) {
  const color = pick_color(color_obj);
  
  if(color) {
    unsorted[color] = (unsorted[color] ?? 0) + 1;
  }
}

// version of the above using reduce:
// const unsorted = data.reduce((acc, val) => {
//   const color = pick_color(val);
//   
//   return !color ?
//     acc :
//     { ...acc, [color]: (acc[color] ?? 0) + 1 };
// }, {});

// version of the above using map then reduce:
// const unsorted = data
//   .map(pick_color)
//   .reduce(
//     (acc, color) => !color ?
//       acc :
//       { ...acc, [color]: (acc[color] ?? 0) + 1 },
//     {}
//   );

const sorted = Object.fromEntries(
  Object.entries(unsorted)
    .sort(([, a_val], [, b_val]) => b_val - a_val)
);

const highest_value = Object.entries(sorted)[0][0];
const highest_value_unsort = Object.entries(unsorted)
  .reduce(
    (acc, entry) => entry[1] > acc[1] ? entry : acc,
    ['none', 0]
  )[0];

console.log(sorted);
console.log(highest_value);
console.log(highest_value_unsort);

If you're unfamiliar with some of the techniques used above, here are some handy reference links:

Answer 2

It's worth noting that there may be a small issue with @hopzebordah's answer, as it seems to count a color even when two colors have the same value. For example, { orange: 4, green: 4} would be counted as orange.

In the comments, I provided an alternative version using map since you appeared to show interest in that method. However, I might have misinterpreted your intended goal.

If you only require the highest value and not necessarily a sorted object, then skipping the sorting step might be more efficient. You can see this in action with highest_value_unsort.

const data = [
  { orange: 4, green: 4},
  { green: 0, yellow: 0},
  { yellow: 1, orange: 4 },
  { blue: 2, green: 1 }, 
  { blue: 2, yellow: 1 }, 
  { green: 3, yellow: 2 },
  { green: 1, blue: 3},
  { green: 5, yellow: 2 }, 
];

const pick_color = (color_obj) => {
  const [[color1, val1], [color2, val2]] = Object.entries(color_obj);
  
  return val1 === val2 ?
    null :
    val1 > val2 ?
        color1 :
        color2;
};


const unsorted = {};
for(const color_obj of data) {
  const color = pick_color(color_obj);
  
  if(color) {
    unsorted[color] = (unsorted[color] ?? 0) + 1;
  }
}

// version of the above using reduce:
// const unsorted = data.reduce((acc, val) => {
//   const color = pick_color(val);
//   
//   return !color ?
//     acc :
//     { ...acc, [color]: (acc[color] ?? 0) + 1 };
// }, {});

// version of the above using map then reduce:
// const unsorted = data
//   .map(pick_color)
//   .reduce(
//     (acc, color) => !color ?
//       acc :
//       { ...acc, [color]: (acc[color] ?? 0) + 1 },
//     {}
//   );

const sorted = Object.fromEntries(
  Object.entries(unsorted)
    .sort(([, a_val], [, b_val]) => b_val - a_val)
);

const highest_value = Object.entries(sorted)[0][0];
const highest_value_unsort = Object.entries(unsorted)
  .reduce(
    (acc, entry) => entry[1] > acc[1] ? entry : acc,
    ['none', 0]
  )[0];

console.log(sorted);
console.log(highest_value);
console.log(highest_value_unsort);

If you're unfamiliar with some of the techniques used above, here are some handy reference links:

Answer 3

Answer №2

If you're using JS, consider yourself lucky!

Setting and getting data in JS objects is a breeze with the { [someVariable]: value } notation. You can also easily check for key existence using the in operator:

const obj = { red: 'foo' };
const red = 'red';
console.log(red in obj) // true
console.log('red' in obj) // true
console.log(blue in obj) // false

With some simple loops, you can get this result:

const data = [
   { orange: 4, green: 4},
   { green: 0, yellow: 0},
   { yellow: 1, orange: 4 },
   { blue: 2, green: 1 }, 
   { blue: 2, yellow: 1 }, 
   { green: 3, yellow: 2 },
   { green: 1, blue: 3},
   { green: 5, yellow: 2 }, 
 ];
 
const totals =  {};

for (const colors of data) {
    const [color1, color2] = Object.keys(colors);
    let color = color1;
    if (colors[color1] < colors[color2]) {
        color = color2
    }
    totals[color] = totals[color] ? totals[color] + 1 : 1;
}

console.log(totals) // { orange: 2, green: 3, blue: 3 }

This solution isn't optimized for performance due to the structure of your data requiring iteration over each value to check keys and values.

JS objects are incredibly versatile and can offer faster solutions depending on dataset size and potential performance constraints.

Answer 4

If you're using JS, consider yourself lucky!

Setting and getting data in JS objects is a breeze with the { [someVariable]: value } notation. You can also easily check for key existence using the in operator:

const obj = { red: 'foo' };
const red = 'red';
console.log(red in obj) // true
console.log('red' in obj) // true
console.log(blue in obj) // false

With some simple loops, you can get this result:

const data = [
   { orange: 4, green: 4},
   { green: 0, yellow: 0},
   { yellow: 1, orange: 4 },
   { blue: 2, green: 1 }, 
   { blue: 2, yellow: 1 }, 
   { green: 3, yellow: 2 },
   { green: 1, blue: 3},
   { green: 5, yellow: 2 }, 
 ];
 
const totals =  {};

for (const colors of data) {
    const [color1, color2] = Object.keys(colors);
    let color = color1;
    if (colors[color1] < colors[color2]) {
        color = color2
    }
    totals[color] = totals[color] ? totals[color] + 1 : 1;
}

console.log(totals) // { orange: 2, green: 3, blue: 3 }

This solution isn't optimized for performance due to the structure of your data requiring iteration over each value to check keys and values.

JS objects are incredibly versatile and can offer faster solutions depending on dataset size and potential performance constraints.

Iterating through an array of objects and extracting values from unspecified keys

Answer №1

Answer №2

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