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looping through javascript object without sorting based on keys

What is the most efficient way to deep clone an object in JavaScript?How do I remove a property from a JavaScript object?How do I check if an array includes an object in JavaScript?How do I loop through or enumerate a JavaScript object?How do I correctly clone a JavaScript object?JavaScript closure inside loops – simple practical exampleHow to loop through a plain JavaScript object with the objects as members?Checking if a key exists in a JavaScript object?Loop through an array in JavaScriptIterate through object properties

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;

this is the problem description:
Given an array of integers, calculate the fractions of its elements that are positive, negative, and are zeros. Print the decimal value of each fraction on a new line.
for example given the array arr=[1,1,0,-1,-1] output should be:

0.400000
0.400000
0.200000

I know there is more more simple solution for it ,and i am sorry for my silly simple question but i wanna make my code work, my code sorts the output based on the key and removes duplicates. for this arr, my code output is:

0.200000
0.400000

thank you so much in advance for any help.

function plusMinus(arr) 
 var freq = ;
 for (var i = 0; i < arr.length; i++)
 if (freq[arr[i]]) 
 freq[arr[i]]++;
 else 
 freq[arr[i]] = 1;
 
 for(var key in freq)

 console.log((freq[key]/arr.length).toFixed(6));

edited Mar 24 at 10:30

asked Mar 24 at 8:55

eng

255

add a comment |

0.400000
0.400000
0.200000

0.200000
0.400000

thank you so much in advance for any help.

function plusMinus(arr) 
 var freq = ;
 for (var i = 0; i < arr.length; i++)
 if (freq[arr[i]]) 
 freq[arr[i]]++;
 else 
 freq[arr[i]] = 1;
 
 for(var key in freq)

 console.log((freq[key]/arr.length).toFixed(6));

edited Mar 24 at 10:30

asked Mar 24 at 8:55

eng

255

add a comment |

0.400000
0.400000
0.200000

0.200000
0.400000

thank you so much in advance for any help.

function plusMinus(arr) 
 var freq = ;
 for (var i = 0; i < arr.length; i++)
 if (freq[arr[i]]) 
 freq[arr[i]]++;
 else 
 freq[arr[i]] = 1;
 
 for(var key in freq)

 console.log((freq[key]/arr.length).toFixed(6));

edited Mar 24 at 10:30

asked Mar 24 at 8:55

eng

255

0.400000
0.400000
0.200000

0.200000
0.400000

thank you so much in advance for any help.

function plusMinus(arr) 
 var freq = ;
 for (var i = 0; i < arr.length; i++)
 if (freq[arr[i]]) 
 freq[arr[i]]++;
 else 
 freq[arr[i]] = 1;
 
 for(var key in freq)

 console.log((freq[key]/arr.length).toFixed(6));

javascript object

edited Mar 24 at 10:30

asked Mar 24 at 8:55

eng

255

edited Mar 24 at 10:30

asked Mar 24 at 8:55

eng

255

edited Mar 24 at 10:30

asked Mar 24 at 8:55

eng

255

asked Mar 24 at 8:55

eng

255

asked Mar 24 at 8:55

eng

255

add a comment |

7 Answers
7

active

oldest

votes

Let's make sure the order of key in the map by defining it first.

function plusMinus(arr) 
 var freq = 
 posetive: 0,
 negative: 0,
 zero: 0
 ;
 for (var i = 0; i < arr.length; i++)
 if( arr[i] < 0) 
 freq.negative++;
 else if(arr[i] > 0) 
 freq.posetive++;
 else 
 freq.zero++;
 
 
 for(var key in freq)
 console.log((freq[key]/arr.length).toFixed(6));
 

plusMinus([1,1,0,-1,-1]);

answered Mar 24 at 9:11

Dipak

3,18144064

add a comment |

You could take an object with predifined properties, this prevents each loop for checking the existence and take an array of keys for getting the result in a wanted order.

function plusMinus(arr) 
 var freq = 1: 0, '-1': 0, 0: 0 ,
 i, key;

 for (i = 0; i < arr.length; i++) 
 freq[arr[i]]++;
 

 for (key of [1, -1, 0]) 
 console.log((freq[key] / arr.length).toFixed(6));
 


plusMinus([1, 1, 0, -1, -1]);

answered Mar 24 at 9:04

Nina Scholz

207k16117190

1

The number can be any integer according the question. You might have to do: freq[Math.sign(arr[i])]++

– adiga
Mar 24 at 9:47

@adiga, right, for any integer, the sign sould be sufficient.

– Nina Scholz
Mar 24 at 9:49

@NinaScholz the order is right now but still deletes duplicates, the output should be 0.4,0.4,0.2 but its giving 0.4,0.2 , why is that?

– eng
Mar 24 at 10:32

i don't understand the last output. what do you get?

– Nina Scholz
Mar 24 at 10:37

@NinaScholz Sorry, it was my mistake. thats right

– eng
Mar 24 at 10:52

add a comment |

You can use reduce.

Here idea is

First loop through original array and check for the value.

If value is zero we increment count of zero key.

If value is positive we increment count of pos key.

If value is negative we increment count of neg key.

Finally we divide each count by length of array.

let arr = [1,1,0,-1,-1]

let op = arr.reduce((op,inp)=>
 if(inp === 0)
 op.zero.count++
 else if (inp > 0)
 op.pos.count++;
 else 
 op.neg.count++;
 
 return op
,zero:count:0,pos:count:0,neg:count:0)


let final = Object.entries(op).map(([key,value])=>(
 [key] : value.count / arr.length
))

console.log(final)

answered Mar 24 at 9:03

Code Maniac

14k21034

add a comment |

Use reduce, map and filter:

const arr = [1, 1, 0, -1, -1];

const counts = arr.reduce((acc, curr) => 
 if (!curr) acc[0]++;
 else if (curr > 0) acc[1]++;
 else acc[2]++;
 return acc
, [0, 0, 0]);

const result = counts.map(e => e / arr.length).filter((e, i, a) => a.indexOf(e) == i);

console.log(result);

answered Mar 24 at 9:11

Jack Bashford

23k52251

add a comment |

You can try using Array.reduce and the resulting array will have the fraction of positive number at the '0'th index, negative at '1'st and zero at the '2'nd index.

Now if you want to control the count of the number of elements after decimal point, use Array.map at the end to transform it.

const array = [1,1,0,-1,-1];
function plusMinus(arr)
 const output = arr.reduce((acc, ele) => 
 if(ele > 0)
 if(ele < 0)
 acc[1] = ((acc[1] 
 if(ele === 0) 0 ) + 1 / arr.length);
 
 return acc;
 , []).map(ele => ele.toFixed(6));
 console.log(...output);

plusMinus(array);

edited Mar 24 at 9:20

answered Mar 24 at 9:07

Amardeep Bhowmick

6,53621231

add a comment |

Math.sign is your friend here.
Math.sign

Also lodash would really help this snippet to be cleaner, I highly recommend _.countBy. Lodash .countBy

Here's the code.

const plusMinus = (numbers) => 
 // Count by Sign (-1, 0 1)
 const countSign = _.countBy(numbers, Math.sign);
 // _.countBy return object, of counted '1': 2, '0': 1, '-1': 2 
 
 // Print them in orders
 const printOrder = [1, -1, 0];
 printOrder.forEach(sign => 
 console.log((countSign[sign] / numbers.length).toFixed(6));
 );


const testArr = [1,1,0,-1,-1];
plusMinus(testArr);

<script src="https://cdn.jsdelivr.net/npm/lodash@4.17.11/lodash.min.js"></script>

answered Mar 24 at 9:28

Doppio

942611

add a comment |

Here is another one-line solution using Array.reduce() and Array.forEach() functions:

const plusMinus = arr => arr
 .reduce((res, curr) => ++res[!curr ? 2 : curr < 0 ? 1 : 0] && res, [0, 0, 0])
 .forEach(i => console.log((i / arr.length).toFixed(6)));

plusMinus([1, 1, 0, -1, -1]);

answered Mar 24 at 9:35

Andriy

8,84022135

add a comment |

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7 Answers
7

active

oldest

votes

7 Answers
7

active

oldest

votes

Let's make sure the order of key in the map by defining it first.

function plusMinus(arr) 
 var freq = 
 posetive: 0,
 negative: 0,
 zero: 0
 ;
 for (var i = 0; i < arr.length; i++)
 if( arr[i] < 0) 
 freq.negative++;
 else if(arr[i] > 0) 
 freq.posetive++;
 else 
 freq.zero++;
 
 
 for(var key in freq)
 console.log((freq[key]/arr.length).toFixed(6));
 

plusMinus([1,1,0,-1,-1]);

answered Mar 24 at 9:11

Dipak

3,18144064

add a comment |

Let's make sure the order of key in the map by defining it first.

function plusMinus(arr) 
 var freq = 
 posetive: 0,
 negative: 0,
 zero: 0
 ;
 for (var i = 0; i < arr.length; i++)
 if( arr[i] < 0) 
 freq.negative++;
 else if(arr[i] > 0) 
 freq.posetive++;
 else 
 freq.zero++;
 
 
 for(var key in freq)
 console.log((freq[key]/arr.length).toFixed(6));
 

plusMinus([1,1,0,-1,-1]);

answered Mar 24 at 9:11

Dipak

3,18144064

add a comment |

Let's make sure the order of key in the map by defining it first.

function plusMinus(arr) 
 var freq = 
 posetive: 0,
 negative: 0,
 zero: 0
 ;
 for (var i = 0; i < arr.length; i++)
 if( arr[i] < 0) 
 freq.negative++;
 else if(arr[i] > 0) 
 freq.posetive++;
 else 
 freq.zero++;
 
 
 for(var key in freq)
 console.log((freq[key]/arr.length).toFixed(6));
 

plusMinus([1,1,0,-1,-1]);

answered Mar 24 at 9:11

Dipak

3,18144064

Let's make sure the order of key in the map by defining it first.

function plusMinus(arr) 
 var freq = 
 posetive: 0,
 negative: 0,
 zero: 0
 ;
 for (var i = 0; i < arr.length; i++)
 if( arr[i] < 0) 
 freq.negative++;
 else if(arr[i] > 0) 
 freq.posetive++;
 else 
 freq.zero++;
 
 
 for(var key in freq)
 console.log((freq[key]/arr.length).toFixed(6));
 

plusMinus([1,1,0,-1,-1]);

function plusMinus(arr) 
 var freq = 
 posetive: 0,
 negative: 0,
 zero: 0
 ;
 for (var i = 0; i < arr.length; i++)
 if( arr[i] < 0) 
 freq.negative++;
 else if(arr[i] > 0) 
 freq.posetive++;
 else 
 freq.zero++;
 
 
 for(var key in freq)
 console.log((freq[key]/arr.length).toFixed(6));
 

plusMinus([1,1,0,-1,-1]);

function plusMinus(arr) 
 var freq = 
 posetive: 0,
 negative: 0,
 zero: 0
 ;
 for (var i = 0; i < arr.length; i++)
 if( arr[i] < 0) 
 freq.negative++;
 else if(arr[i] > 0) 
 freq.posetive++;
 else 
 freq.zero++;
 
 
 for(var key in freq)
 console.log((freq[key]/arr.length).toFixed(6));
 

plusMinus([1,1,0,-1,-1]);

answered Mar 24 at 9:11

Dipak

3,18144064

answered Mar 24 at 9:11

Dipak

3,18144064

answered Mar 24 at 9:11

Dipak

3,18144064

answered Mar 24 at 9:11

Dipak

3,18144064

add a comment |

You could take an object with predifined properties, this prevents each loop for checking the existence and take an array of keys for getting the result in a wanted order.

function plusMinus(arr) 
 var freq = 1: 0, '-1': 0, 0: 0 ,
 i, key;

 for (i = 0; i < arr.length; i++) 
 freq[arr[i]]++;
 

 for (key of [1, -1, 0]) 
 console.log((freq[key] / arr.length).toFixed(6));
 


plusMinus([1, 1, 0, -1, -1]);

answered Mar 24 at 9:04

Nina Scholz

207k16117190

1

The number can be any integer according the question. You might have to do: freq[Math.sign(arr[i])]++

– adiga
Mar 24 at 9:47

@adiga, right, for any integer, the sign sould be sufficient.

– Nina Scholz
Mar 24 at 9:49

@NinaScholz the order is right now but still deletes duplicates, the output should be 0.4,0.4,0.2 but its giving 0.4,0.2 , why is that?

– eng
Mar 24 at 10:32

i don't understand the last output. what do you get?

– Nina Scholz
Mar 24 at 10:37

@NinaScholz Sorry, it was my mistake. thats right

– eng
Mar 24 at 10:52

add a comment |

You could take an object with predifined properties, this prevents each loop for checking the existence and take an array of keys for getting the result in a wanted order.

function plusMinus(arr) 
 var freq = 1: 0, '-1': 0, 0: 0 ,
 i, key;

 for (i = 0; i < arr.length; i++) 
 freq[arr[i]]++;
 

 for (key of [1, -1, 0]) 
 console.log((freq[key] / arr.length).toFixed(6));
 


plusMinus([1, 1, 0, -1, -1]);

answered Mar 24 at 9:04

Nina Scholz

207k16117190

1

The number can be any integer according the question. You might have to do: freq[Math.sign(arr[i])]++

– adiga
Mar 24 at 9:47

@adiga, right, for any integer, the sign sould be sufficient.

– Nina Scholz
Mar 24 at 9:49

@NinaScholz the order is right now but still deletes duplicates, the output should be 0.4,0.4,0.2 but its giving 0.4,0.2 , why is that?

– eng
Mar 24 at 10:32

i don't understand the last output. what do you get?

– Nina Scholz
Mar 24 at 10:37

@NinaScholz Sorry, it was my mistake. thats right

– eng
Mar 24 at 10:52

add a comment |

You could take an object with predifined properties, this prevents each loop for checking the existence and take an array of keys for getting the result in a wanted order.

function plusMinus(arr) 
 var freq = 1: 0, '-1': 0, 0: 0 ,
 i, key;

 for (i = 0; i < arr.length; i++) 
 freq[arr[i]]++;
 

 for (key of [1, -1, 0]) 
 console.log((freq[key] / arr.length).toFixed(6));
 


plusMinus([1, 1, 0, -1, -1]);

answered Mar 24 at 9:04

Nina Scholz

207k16117190

You could take an object with predifined properties, this prevents each loop for checking the existence and take an array of keys for getting the result in a wanted order.

function plusMinus(arr) 
 var freq = 1: 0, '-1': 0, 0: 0 ,
 i, key;

 for (i = 0; i < arr.length; i++) 
 freq[arr[i]]++;
 

 for (key of [1, -1, 0]) 
 console.log((freq[key] / arr.length).toFixed(6));
 


plusMinus([1, 1, 0, -1, -1]);

function plusMinus(arr) 
 var freq = 1: 0, '-1': 0, 0: 0 ,
 i, key;

 for (i = 0; i < arr.length; i++) 
 freq[arr[i]]++;
 

 for (key of [1, -1, 0]) 
 console.log((freq[key] / arr.length).toFixed(6));
 


plusMinus([1, 1, 0, -1, -1]);

function plusMinus(arr) 
 var freq = 1: 0, '-1': 0, 0: 0 ,
 i, key;

 for (i = 0; i < arr.length; i++) 
 freq[arr[i]]++;
 

 for (key of [1, -1, 0]) 
 console.log((freq[key] / arr.length).toFixed(6));
 


plusMinus([1, 1, 0, -1, -1]);

answered Mar 24 at 9:04

Nina Scholz

207k16117190

answered Mar 24 at 9:04

Nina Scholz

207k16117190

answered Mar 24 at 9:04

Nina Scholz

207k16117190

answered Mar 24 at 9:04

Nina Scholz

207k16117190

1

The number can be any integer according the question. You might have to do: freq[Math.sign(arr[i])]++

– adiga
Mar 24 at 9:47

@adiga, right, for any integer, the sign sould be sufficient.

– Nina Scholz
Mar 24 at 9:49

@NinaScholz the order is right now but still deletes duplicates, the output should be 0.4,0.4,0.2 but its giving 0.4,0.2 , why is that?

– eng
Mar 24 at 10:32

i don't understand the last output. what do you get?

– Nina Scholz
Mar 24 at 10:37

@NinaScholz Sorry, it was my mistake. thats right

– eng
Mar 24 at 10:52

add a comment |

1

The number can be any integer according the question. You might have to do: freq[Math.sign(arr[i])]++

– adiga
Mar 24 at 9:47

@adiga, right, for any integer, the sign sould be sufficient.

– Nina Scholz
Mar 24 at 9:49

@NinaScholz the order is right now but still deletes duplicates, the output should be 0.4,0.4,0.2 but its giving 0.4,0.2 , why is that?

– eng
Mar 24 at 10:32

i don't understand the last output. what do you get?

– Nina Scholz
Mar 24 at 10:37

@NinaScholz Sorry, it was my mistake. thats right

– eng
Mar 24 at 10:52

The number can be any integer according the question. You might have to do: freq[Math.sign(arr[i])]++

– adiga
Mar 24 at 9:47

@adiga, right, for any integer, the sign sould be sufficient.

– Nina Scholz
Mar 24 at 9:49

@NinaScholz the order is right now but still deletes duplicates, the output should be 0.4,0.4,0.2 but its giving 0.4,0.2 , why is that?

– eng
Mar 24 at 10:32

i don't understand the last output. what do you get?

– Nina Scholz
Mar 24 at 10:37

@NinaScholz Sorry, it was my mistake. thats right

– eng
Mar 24 at 10:52

add a comment |

You can use reduce.

Here idea is

First loop through original array and check for the value.

If value is zero we increment count of zero key.

If value is positive we increment count of pos key.

If value is negative we increment count of neg key.

Finally we divide each count by length of array.

let arr = [1,1,0,-1,-1]

let op = arr.reduce((op,inp)=>
 if(inp === 0)
 op.zero.count++
 else if (inp > 0)
 op.pos.count++;
 else 
 op.neg.count++;
 
 return op
,zero:count:0,pos:count:0,neg:count:0)


let final = Object.entries(op).map(([key,value])=>(
 [key] : value.count / arr.length
))

console.log(final)

answered Mar 24 at 9:03

Code Maniac

14k21034

add a comment |

You can use reduce.

Here idea is

First loop through original array and check for the value.

If value is zero we increment count of zero key.

If value is positive we increment count of pos key.

If value is negative we increment count of neg key.

Finally we divide each count by length of array.

let arr = [1,1,0,-1,-1]

let op = arr.reduce((op,inp)=>
 if(inp === 0)
 op.zero.count++
 else if (inp > 0)
 op.pos.count++;
 else 
 op.neg.count++;
 
 return op
,zero:count:0,pos:count:0,neg:count:0)


let final = Object.entries(op).map(([key,value])=>(
 [key] : value.count / arr.length
))

console.log(final)

answered Mar 24 at 9:03

Code Maniac

14k21034

add a comment |

You can use reduce.

Here idea is

First loop through original array and check for the value.

If value is zero we increment count of zero key.

If value is positive we increment count of pos key.

If value is negative we increment count of neg key.

Finally we divide each count by length of array.

let arr = [1,1,0,-1,-1]

let op = arr.reduce((op,inp)=>
 if(inp === 0)
 op.zero.count++
 else if (inp > 0)
 op.pos.count++;
 else 
 op.neg.count++;
 
 return op
,zero:count:0,pos:count:0,neg:count:0)


let final = Object.entries(op).map(([key,value])=>(
 [key] : value.count / arr.length
))

console.log(final)

answered Mar 24 at 9:03

Code Maniac

14k21034

You can use reduce.

Here idea is

First loop through original array and check for the value.

If value is zero we increment count of zero key.

If value is positive we increment count of pos key.

If value is negative we increment count of neg key.

Finally we divide each count by length of array.

let arr = [1,1,0,-1,-1]

let op = arr.reduce((op,inp)=>
 if(inp === 0)
 op.zero.count++
 else if (inp > 0)
 op.pos.count++;
 else 
 op.neg.count++;
 
 return op
,zero:count:0,pos:count:0,neg:count:0)


let final = Object.entries(op).map(([key,value])=>(
 [key] : value.count / arr.length
))

console.log(final)

let arr = [1,1,0,-1,-1]

let op = arr.reduce((op,inp)=>
 if(inp === 0)
 op.zero.count++
 else if (inp > 0)
 op.pos.count++;
 else 
 op.neg.count++;
 
 return op
,zero:count:0,pos:count:0,neg:count:0)


let final = Object.entries(op).map(([key,value])=>(
 [key] : value.count / arr.length
))

console.log(final)

let arr = [1,1,0,-1,-1]

let op = arr.reduce((op,inp)=>
 if(inp === 0)
 op.zero.count++
 else if (inp > 0)
 op.pos.count++;
 else 
 op.neg.count++;
 
 return op
,zero:count:0,pos:count:0,neg:count:0)


let final = Object.entries(op).map(([key,value])=>(
 [key] : value.count / arr.length
))

console.log(final)

answered Mar 24 at 9:03

Code Maniac

14k21034

answered Mar 24 at 9:03

Code Maniac

14k21034

answered Mar 24 at 9:03

Code Maniac

14k21034

answered Mar 24 at 9:03

Code Maniac

14k21034

add a comment |

Use reduce, map and filter:

const arr = [1, 1, 0, -1, -1];

const counts = arr.reduce((acc, curr) => 
 if (!curr) acc[0]++;
 else if (curr > 0) acc[1]++;
 else acc[2]++;
 return acc
, [0, 0, 0]);

const result = counts.map(e => e / arr.length).filter((e, i, a) => a.indexOf(e) == i);

console.log(result);

answered Mar 24 at 9:11

Jack Bashford

23k52251

add a comment |

Use reduce, map and filter:

const arr = [1, 1, 0, -1, -1];

const counts = arr.reduce((acc, curr) => 
 if (!curr) acc[0]++;
 else if (curr > 0) acc[1]++;
 else acc[2]++;
 return acc
, [0, 0, 0]);

const result = counts.map(e => e / arr.length).filter((e, i, a) => a.indexOf(e) == i);

console.log(result);

answered Mar 24 at 9:11

Jack Bashford

23k52251

add a comment |

Use reduce, map and filter:

const arr = [1, 1, 0, -1, -1];

const counts = arr.reduce((acc, curr) => 
 if (!curr) acc[0]++;
 else if (curr > 0) acc[1]++;
 else acc[2]++;
 return acc
, [0, 0, 0]);

const result = counts.map(e => e / arr.length).filter((e, i, a) => a.indexOf(e) == i);

console.log(result);

answered Mar 24 at 9:11

Jack Bashford

23k52251

Use reduce, map and filter:

const arr = [1, 1, 0, -1, -1];

const counts = arr.reduce((acc, curr) => 
 if (!curr) acc[0]++;
 else if (curr > 0) acc[1]++;
 else acc[2]++;
 return acc
, [0, 0, 0]);

const result = counts.map(e => e / arr.length).filter((e, i, a) => a.indexOf(e) == i);

console.log(result);

const arr = [1, 1, 0, -1, -1];

const counts = arr.reduce((acc, curr) => 
 if (!curr) acc[0]++;
 else if (curr > 0) acc[1]++;
 else acc[2]++;
 return acc
, [0, 0, 0]);

const result = counts.map(e => e / arr.length).filter((e, i, a) => a.indexOf(e) == i);

console.log(result);

const arr = [1, 1, 0, -1, -1];

const counts = arr.reduce((acc, curr) => 
 if (!curr) acc[0]++;
 else if (curr > 0) acc[1]++;
 else acc[2]++;
 return acc
, [0, 0, 0]);

const result = counts.map(e => e / arr.length).filter((e, i, a) => a.indexOf(e) == i);

console.log(result);

answered Mar 24 at 9:11

Jack Bashford

23k52251

answered Mar 24 at 9:11

Jack Bashford

23k52251

answered Mar 24 at 9:11

Jack Bashford

23k52251

answered Mar 24 at 9:11

Jack Bashford

23k52251

add a comment |

You can try using Array.reduce and the resulting array will have the fraction of positive number at the '0'th index, negative at '1'st and zero at the '2'nd index.

Now if you want to control the count of the number of elements after decimal point, use Array.map at the end to transform it.

const array = [1,1,0,-1,-1];
function plusMinus(arr)
 const output = arr.reduce((acc, ele) => 
 if(ele > 0)
 if(ele < 0)
 acc[1] = ((acc[1] 
 if(ele === 0) 0 ) + 1 / arr.length);
 
 return acc;
 , []).map(ele => ele.toFixed(6));
 console.log(...output);

plusMinus(array);

edited Mar 24 at 9:20

answered Mar 24 at 9:07

Amardeep Bhowmick

6,53621231

add a comment |

You can try using Array.reduce and the resulting array will have the fraction of positive number at the '0'th index, negative at '1'st and zero at the '2'nd index.

Now if you want to control the count of the number of elements after decimal point, use Array.map at the end to transform it.

const array = [1,1,0,-1,-1];
function plusMinus(arr)
 const output = arr.reduce((acc, ele) => 
 if(ele > 0)
 if(ele < 0)
 acc[1] = ((acc[1] 
 if(ele === 0) 0 ) + 1 / arr.length);
 
 return acc;
 , []).map(ele => ele.toFixed(6));
 console.log(...output);

plusMinus(array);

edited Mar 24 at 9:20

answered Mar 24 at 9:07

Amardeep Bhowmick

6,53621231

add a comment |

You can try using Array.reduce and the resulting array will have the fraction of positive number at the '0'th index, negative at '1'st and zero at the '2'nd index.

Now if you want to control the count of the number of elements after decimal point, use Array.map at the end to transform it.

const array = [1,1,0,-1,-1];
function plusMinus(arr)
 const output = arr.reduce((acc, ele) => 
 if(ele > 0)
 if(ele < 0)
 acc[1] = ((acc[1] 
 if(ele === 0) 0 ) + 1 / arr.length);
 
 return acc;
 , []).map(ele => ele.toFixed(6));
 console.log(...output);

plusMinus(array);

edited Mar 24 at 9:20

answered Mar 24 at 9:07

Amardeep Bhowmick

6,53621231

You can try using Array.reduce and the resulting array will have the fraction of positive number at the '0'th index, negative at '1'st and zero at the '2'nd index.

Now if you want to control the count of the number of elements after decimal point, use Array.map at the end to transform it.

const array = [1,1,0,-1,-1];
function plusMinus(arr)
 const output = arr.reduce((acc, ele) => 
 if(ele > 0)
 if(ele < 0)
 acc[1] = ((acc[1] 
 if(ele === 0) 0 ) + 1 / arr.length);
 
 return acc;
 , []).map(ele => ele.toFixed(6));
 console.log(...output);

plusMinus(array);

const array = [1,1,0,-1,-1];
function plusMinus(arr)
 const output = arr.reduce((acc, ele) => 
 if(ele > 0)
 if(ele < 0)
 acc[1] = ((acc[1] 
 if(ele === 0) 0 ) + 1 / arr.length);
 
 return acc;
 , []).map(ele => ele.toFixed(6));
 console.log(...output);

plusMinus(array);

const array = [1,1,0,-1,-1];
function plusMinus(arr)
 const output = arr.reduce((acc, ele) => 
 if(ele > 0)
 if(ele < 0)
 acc[1] = ((acc[1] 
 if(ele === 0) 0 ) + 1 / arr.length);
 
 return acc;
 , []).map(ele => ele.toFixed(6));
 console.log(...output);

plusMinus(array);

edited Mar 24 at 9:20

answered Mar 24 at 9:07

Amardeep Bhowmick

6,53621231

edited Mar 24 at 9:20

answered Mar 24 at 9:07

Amardeep Bhowmick

6,53621231

answered Mar 24 at 9:07

Amardeep Bhowmick

6,53621231

answered Mar 24 at 9:07

Amardeep Bhowmick

6,53621231

add a comment |

Math.sign is your friend here.
Math.sign

Also lodash would really help this snippet to be cleaner, I highly recommend _.countBy. Lodash .countBy

Here's the code.

const plusMinus = (numbers) => 
 // Count by Sign (-1, 0 1)
 const countSign = _.countBy(numbers, Math.sign);
 // _.countBy return object, of counted '1': 2, '0': 1, '-1': 2 
 
 // Print them in orders
 const printOrder = [1, -1, 0];
 printOrder.forEach(sign => 
 console.log((countSign[sign] / numbers.length).toFixed(6));
 );


const testArr = [1,1,0,-1,-1];
plusMinus(testArr);

<script src="https://cdn.jsdelivr.net/npm/lodash@4.17.11/lodash.min.js"></script>

answered Mar 24 at 9:28

Doppio

942611

add a comment |

Math.sign is your friend here.
Math.sign

Also lodash would really help this snippet to be cleaner, I highly recommend _.countBy. Lodash .countBy

Here's the code.

const plusMinus = (numbers) => 
 // Count by Sign (-1, 0 1)
 const countSign = _.countBy(numbers, Math.sign);
 // _.countBy return object, of counted '1': 2, '0': 1, '-1': 2 
 
 // Print them in orders
 const printOrder = [1, -1, 0];
 printOrder.forEach(sign => 
 console.log((countSign[sign] / numbers.length).toFixed(6));
 );


const testArr = [1,1,0,-1,-1];
plusMinus(testArr);

<script src="https://cdn.jsdelivr.net/npm/lodash@4.17.11/lodash.min.js"></script>

answered Mar 24 at 9:28

Doppio

942611

add a comment |

Math.sign is your friend here.
Math.sign

Also lodash would really help this snippet to be cleaner, I highly recommend _.countBy. Lodash .countBy

Here's the code.

const plusMinus = (numbers) => 
 // Count by Sign (-1, 0 1)
 const countSign = _.countBy(numbers, Math.sign);
 // _.countBy return object, of counted '1': 2, '0': 1, '-1': 2 
 
 // Print them in orders
 const printOrder = [1, -1, 0];
 printOrder.forEach(sign => 
 console.log((countSign[sign] / numbers.length).toFixed(6));
 );


const testArr = [1,1,0,-1,-1];
plusMinus(testArr);

<script src="https://cdn.jsdelivr.net/npm/lodash@4.17.11/lodash.min.js"></script>

answered Mar 24 at 9:28

Doppio

942611

Math.sign is your friend here.
Math.sign

Also lodash would really help this snippet to be cleaner, I highly recommend _.countBy. Lodash .countBy

Here's the code.

const plusMinus = (numbers) => 
 // Count by Sign (-1, 0 1)
 const countSign = _.countBy(numbers, Math.sign);
 // _.countBy return object, of counted '1': 2, '0': 1, '-1': 2 
 
 // Print them in orders
 const printOrder = [1, -1, 0];
 printOrder.forEach(sign => 
 console.log((countSign[sign] / numbers.length).toFixed(6));
 );


const testArr = [1,1,0,-1,-1];
plusMinus(testArr);

<script src="https://cdn.jsdelivr.net/npm/lodash@4.17.11/lodash.min.js"></script>

const plusMinus = (numbers) => 
 // Count by Sign (-1, 0 1)
 const countSign = _.countBy(numbers, Math.sign);
 // _.countBy return object, of counted '1': 2, '0': 1, '-1': 2 
 
 // Print them in orders
 const printOrder = [1, -1, 0];
 printOrder.forEach(sign => 
 console.log((countSign[sign] / numbers.length).toFixed(6));
 );


const testArr = [1,1,0,-1,-1];
plusMinus(testArr);

<script src="https://cdn.jsdelivr.net/npm/lodash@4.17.11/lodash.min.js"></script>

const plusMinus = (numbers) => 
 // Count by Sign (-1, 0 1)
 const countSign = _.countBy(numbers, Math.sign);
 // _.countBy return object, of counted '1': 2, '0': 1, '-1': 2 
 
 // Print them in orders
 const printOrder = [1, -1, 0];
 printOrder.forEach(sign => 
 console.log((countSign[sign] / numbers.length).toFixed(6));
 );


const testArr = [1,1,0,-1,-1];
plusMinus(testArr);

<script src="https://cdn.jsdelivr.net/npm/lodash@4.17.11/lodash.min.js"></script>

answered Mar 24 at 9:28

Doppio

942611

answered Mar 24 at 9:28

Doppio

942611

answered Mar 24 at 9:28

Doppio

942611

answered Mar 24 at 9:28

Doppio

942611

add a comment |

Here is another one-line solution using Array.reduce() and Array.forEach() functions:

const plusMinus = arr => arr
 .reduce((res, curr) => ++res[!curr ? 2 : curr < 0 ? 1 : 0] && res, [0, 0, 0])
 .forEach(i => console.log((i / arr.length).toFixed(6)));

plusMinus([1, 1, 0, -1, -1]);

answered Mar 24 at 9:35

Andriy

8,84022135

add a comment |

Here is another one-line solution using Array.reduce() and Array.forEach() functions:

const plusMinus = arr => arr
 .reduce((res, curr) => ++res[!curr ? 2 : curr < 0 ? 1 : 0] && res, [0, 0, 0])
 .forEach(i => console.log((i / arr.length).toFixed(6)));

plusMinus([1, 1, 0, -1, -1]);

answered Mar 24 at 9:35

Andriy

8,84022135

add a comment |

Here is another one-line solution using Array.reduce() and Array.forEach() functions:

const plusMinus = arr => arr
 .reduce((res, curr) => ++res[!curr ? 2 : curr < 0 ? 1 : 0] && res, [0, 0, 0])
 .forEach(i => console.log((i / arr.length).toFixed(6)));

plusMinus([1, 1, 0, -1, -1]);

answered Mar 24 at 9:35

Andriy

8,84022135

Here is another one-line solution using Array.reduce() and Array.forEach() functions:

const plusMinus = arr => arr
 .reduce((res, curr) => ++res[!curr ? 2 : curr < 0 ? 1 : 0] && res, [0, 0, 0])
 .forEach(i => console.log((i / arr.length).toFixed(6)));

plusMinus([1, 1, 0, -1, -1]);

const plusMinus = arr => arr
 .reduce((res, curr) => ++res[!curr ? 2 : curr < 0 ? 1 : 0] && res, [0, 0, 0])
 .forEach(i => console.log((i / arr.length).toFixed(6)));

plusMinus([1, 1, 0, -1, -1]);

const plusMinus = arr => arr
 .reduce((res, curr) => ++res[!curr ? 2 : curr < 0 ? 1 : 0] && res, [0, 0, 0])
 .forEach(i => console.log((i / arr.length).toFixed(6)));

plusMinus([1, 1, 0, -1, -1]);

answered Mar 24 at 9:35

Andriy

8,84022135

answered Mar 24 at 9:35

Andriy

8,84022135

answered Mar 24 at 9:35

Andriy

8,84022135

answered Mar 24 at 9:35

Andriy

8,84022135

add a comment |

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