refactoring function with different methods

refactoring function with different methods - javascript

This function current takes in an array, and outputs it a character with its count if the element's length is greater than one

Is there a way I can do this same thing, perhaps with a different javascript array method without having to use a new array or a result variable?

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  let newArray = ;



  let result = arr.forEach(e => e.length > 1 ? newArray.push(`${e[0]}${e.length}`) : newArray.push(e))



  return newArray.join("");

}



console.log(compressCharacters(a));

edited Nov 13 '18 at 1:16

Ele

22.8k42044

asked Nov 13 '18 at 1:11

totalnoob

3261631

add a comment |

This function current takes in an array, and outputs it a character with its count if the element's length is greater than one

Is there a way I can do this same thing, perhaps with a different javascript array method without having to use a new array or a result variable?

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  let newArray = ;



  let result = arr.forEach(e => e.length > 1 ? newArray.push(`${e[0]}${e.length}`) : newArray.push(e))



  return newArray.join("");

}



console.log(compressCharacters(a));

edited Nov 13 '18 at 1:16

Ele

22.8k42044

asked Nov 13 '18 at 1:11

totalnoob

3261631

add a comment |

This function current takes in an array, and outputs it a character with its count if the element's length is greater than one

Is there a way I can do this same thing, perhaps with a different javascript array method without having to use a new array or a result variable?

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  let newArray = ;



  let result = arr.forEach(e => e.length > 1 ? newArray.push(`${e[0]}${e.length}`) : newArray.push(e))



  return newArray.join("");

}



console.log(compressCharacters(a));

edited Nov 13 '18 at 1:16

Ele

22.8k42044

asked Nov 13 '18 at 1:11

totalnoob

3261631

This function current takes in an array, and outputs it a character with its count if the element's length is greater than one

Is there a way I can do this same thing, perhaps with a different javascript array method without having to use a new array or a result variable?

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  let newArray = ;



  let result = arr.forEach(e => e.length > 1 ? newArray.push(`${e[0]}${e.length}`) : newArray.push(e))



  return newArray.join("");

}



console.log(compressCharacters(a));

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  let newArray = ;



  let result = arr.forEach(e => e.length > 1 ? newArray.push(`${e[0]}${e.length}`) : newArray.push(e))



  return newArray.join("");

}



console.log(compressCharacters(a));

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  let newArray = ;



  let result = arr.forEach(e => e.length > 1 ? newArray.push(`${e[0]}${e.length}`) : newArray.push(e))



  return newArray.join("");

}



console.log(compressCharacters(a));

javascript

edited Nov 13 '18 at 1:16

Ele

22.8k42044

asked Nov 13 '18 at 1:11

totalnoob

3261631

edited Nov 13 '18 at 1:16

Ele

22.8k42044

asked Nov 13 '18 at 1:11

totalnoob

3261631

edited Nov 13 '18 at 1:16

Ele

22.8k42044

edited Nov 13 '18 at 1:16

Ele

22.8k42044

edited Nov 13 '18 at 1:16

Ele

22.8k42044

asked Nov 13 '18 at 1:11

totalnoob

3261631

asked Nov 13 '18 at 1:11

totalnoob

3261631

asked Nov 13 '18 at 1:11

totalnoob

3261631

add a comment |

3 Answers
3

active

oldest

votes

You can just use map() instead of forEach() and return the values you want. There's no need for the extra array. It's often the case than forEach + push() == map()

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  return arr.map(e => e.length > 1

           ? `${e[0]}${e.length}` 

           : e)

           .join("");

}



console.log(compressCharacters(a));

answered Nov 13 '18 at 1:17

Mark Meyer

38.1k33159

add a comment |

You can construct and immediately return a mapped array, that's joined by the empty string:

const a = ['aaa', 'z', 'eeeee'];

const compressCharacters = arr => (

  arr.map(str => str[0] + (str.length > 1 ? str.length : ''))

    .join('')

)



console.log(compressCharacters(a));

When you're trying to construct a new array by transforming all elements from some other array, .map is the appropriate method to use. forEach should be reserved for generic iteration with side effects, when you aren't trying to construct a new object. (when you are trying to construct a new object, but it's not an array, or the new array isn't one-to-one with the old array, usually you can use .reduce)

answered Nov 13 '18 at 1:18

CertainPerformance

81.3k143865

add a comment |

This is an alternative using the function reduce to build the desired output.

const a = ['aaa', 'z', 'eeeee'],

      result = a.reduce((a, [m], i, arr) => {

        let {length} = arr[i];

        return a + m + (length <= 1 ? "" : length);

      }, "");



console.log(result);

edited Nov 13 '18 at 2:13

answered Nov 13 '18 at 1:29

Ele

22.8k42044

...............

– totalnoob
Nov 13 '18 at 4:38

can you explain how this works? I've never used an array in the reduce method before or deconstructed the length of an array like that before

– totalnoob
Nov 13 '18 at 4:42

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

You can just use map() instead of forEach() and return the values you want. There's no need for the extra array. It's often the case than forEach + push() == map()

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  return arr.map(e => e.length > 1

           ? `${e[0]}${e.length}` 

           : e)

           .join("");

}



console.log(compressCharacters(a));

answered Nov 13 '18 at 1:17

Mark Meyer

38.1k33159

add a comment |

You can just use map() instead of forEach() and return the values you want. There's no need for the extra array. It's often the case than forEach + push() == map()

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  return arr.map(e => e.length > 1

           ? `${e[0]}${e.length}` 

           : e)

           .join("");

}



console.log(compressCharacters(a));

answered Nov 13 '18 at 1:17

Mark Meyer

38.1k33159

add a comment |

You can just use map() instead of forEach() and return the values you want. There's no need for the extra array. It's often the case than forEach + push() == map()

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  return arr.map(e => e.length > 1

           ? `${e[0]}${e.length}` 

           : e)

           .join("");

}



console.log(compressCharacters(a));

answered Nov 13 '18 at 1:17

Mark Meyer

38.1k33159

You can just use map() instead of forEach() and return the values you want. There's no need for the extra array. It's often the case than forEach + push() == map()

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  return arr.map(e => e.length > 1

           ? `${e[0]}${e.length}` 

           : e)

           .join("");

}



console.log(compressCharacters(a));

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  return arr.map(e => e.length > 1

           ? `${e[0]}${e.length}` 

           : e)

           .join("");

}



console.log(compressCharacters(a));

const a = ['aaa', 'z', 'eeeee'];



const compressCharacters = arr => {

  return arr.map(e => e.length > 1

           ? `${e[0]}${e.length}` 

           : e)

           .join("");

}



console.log(compressCharacters(a));

answered Nov 13 '18 at 1:17

Mark Meyer

38.1k33159

answered Nov 13 '18 at 1:17

Mark Meyer

38.1k33159

answered Nov 13 '18 at 1:17

Mark Meyer

38.1k33159

answered Nov 13 '18 at 1:17

Mark Meyer

38.1k33159

add a comment |

You can construct and immediately return a mapped array, that's joined by the empty string:

const a = ['aaa', 'z', 'eeeee'];

const compressCharacters = arr => (

  arr.map(str => str[0] + (str.length > 1 ? str.length : ''))

    .join('')

)



console.log(compressCharacters(a));

answered Nov 13 '18 at 1:18

CertainPerformance

81.3k143865

add a comment |

You can construct and immediately return a mapped array, that's joined by the empty string:

const a = ['aaa', 'z', 'eeeee'];

const compressCharacters = arr => (

  arr.map(str => str[0] + (str.length > 1 ? str.length : ''))

    .join('')

)



console.log(compressCharacters(a));

answered Nov 13 '18 at 1:18

CertainPerformance

81.3k143865

add a comment |

You can construct and immediately return a mapped array, that's joined by the empty string:

const a = ['aaa', 'z', 'eeeee'];

const compressCharacters = arr => (

  arr.map(str => str[0] + (str.length > 1 ? str.length : ''))

    .join('')

)



console.log(compressCharacters(a));

answered Nov 13 '18 at 1:18

CertainPerformance

81.3k143865

You can construct and immediately return a mapped array, that's joined by the empty string:

const a = ['aaa', 'z', 'eeeee'];

const compressCharacters = arr => (

  arr.map(str => str[0] + (str.length > 1 ? str.length : ''))

    .join('')

)



console.log(compressCharacters(a));

const a = ['aaa', 'z', 'eeeee'];

const compressCharacters = arr => (

  arr.map(str => str[0] + (str.length > 1 ? str.length : ''))

    .join('')

)



console.log(compressCharacters(a));

const a = ['aaa', 'z', 'eeeee'];

const compressCharacters = arr => (

  arr.map(str => str[0] + (str.length > 1 ? str.length : ''))

    .join('')

)



console.log(compressCharacters(a));

answered Nov 13 '18 at 1:18

CertainPerformance

81.3k143865

answered Nov 13 '18 at 1:18

CertainPerformance

81.3k143865

answered Nov 13 '18 at 1:18

CertainPerformance

81.3k143865

answered Nov 13 '18 at 1:18

CertainPerformance

81.3k143865

add a comment |

This is an alternative using the function reduce to build the desired output.

const a = ['aaa', 'z', 'eeeee'],

      result = a.reduce((a, [m], i, arr) => {

        let {length} = arr[i];

        return a + m + (length <= 1 ? "" : length);

      }, "");



console.log(result);

edited Nov 13 '18 at 2:13

answered Nov 13 '18 at 1:29

Ele

22.8k42044

...............

– totalnoob
Nov 13 '18 at 4:38

can you explain how this works? I've never used an array in the reduce method before or deconstructed the length of an array like that before

– totalnoob
Nov 13 '18 at 4:42

add a comment |

This is an alternative using the function reduce to build the desired output.

const a = ['aaa', 'z', 'eeeee'],

      result = a.reduce((a, [m], i, arr) => {

        let {length} = arr[i];

        return a + m + (length <= 1 ? "" : length);

      }, "");



console.log(result);

edited Nov 13 '18 at 2:13

answered Nov 13 '18 at 1:29

Ele

22.8k42044

...............

– totalnoob
Nov 13 '18 at 4:38

can you explain how this works? I've never used an array in the reduce method before or deconstructed the length of an array like that before

– totalnoob
Nov 13 '18 at 4:42

add a comment |

This is an alternative using the function reduce to build the desired output.

const a = ['aaa', 'z', 'eeeee'],

      result = a.reduce((a, [m], i, arr) => {

        let {length} = arr[i];

        return a + m + (length <= 1 ? "" : length);

      }, "");



console.log(result);

edited Nov 13 '18 at 2:13

answered Nov 13 '18 at 1:29

Ele

22.8k42044

This is an alternative using the function reduce to build the desired output.

const a = ['aaa', 'z', 'eeeee'],

      result = a.reduce((a, [m], i, arr) => {

        let {length} = arr[i];

        return a + m + (length <= 1 ? "" : length);

      }, "");



console.log(result);

const a = ['aaa', 'z', 'eeeee'],

      result = a.reduce((a, [m], i, arr) => {

        let {length} = arr[i];

        return a + m + (length <= 1 ? "" : length);

      }, "");



console.log(result);

const a = ['aaa', 'z', 'eeeee'],

      result = a.reduce((a, [m], i, arr) => {

        let {length} = arr[i];

        return a + m + (length <= 1 ? "" : length);

      }, "");



console.log(result);

edited Nov 13 '18 at 2:13

answered Nov 13 '18 at 1:29

Ele

22.8k42044

edited Nov 13 '18 at 2:13

answered Nov 13 '18 at 1:29

Ele

22.8k42044

answered Nov 13 '18 at 1:29

Ele

22.8k42044

answered Nov 13 '18 at 1:29

Ele

22.8k42044

...............

– totalnoob
Nov 13 '18 at 4:38

can you explain how this works? I've never used an array in the reduce method before or deconstructed the length of an array like that before

– totalnoob
Nov 13 '18 at 4:42

add a comment |

...............

– totalnoob
Nov 13 '18 at 4:38

can you explain how this works? I've never used an array in the reduce method before or deconstructed the length of an array like that before

– totalnoob
Nov 13 '18 at 4:42

...............

– totalnoob
Nov 13 '18 at 4:38

can you explain how this works? I've never used an array in the reduce method before or deconstructed the length of an array like that before

– totalnoob
Nov 13 '18 at 4:42

add a comment |

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