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Multiply nested array lengths and return a single value

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0

1

How can i multiply the length of the nested values array using this array:

const options = [
 
 name: "Colors",
 values: [
 label: "Blue"
 ,
 
 label: "Red"
 ]
 ,
 
 name: "Sizes",
 values: [
 label: "Small"
 ,
 
 label: "Large"
 ]
 
]

My code:

const options = [
 
 name: "Colors",
 values: [
 
 label: "Blue"
 ,
 
 label: "Red"
 
 ]
 ,
 
 name: "Sizes",
 values: [
 
 label: "Small"
 ,
 
 label: "Large"
 
 ]
 
];

const multiply = options.reduce((a, b) => a * b.values.length, 0);

console.log(multiply);

I've tried with reduce but always return 0. It should return ex. 4

playground: https://codesandbox.io/s/08jrnr72w

edit: thanks everyone for the answers, I missed initial position...

javascript arrays

share|improve this question

edited Mar 7 at 11:40

Nick Parsons

10.2k2926

asked Mar 7 at 11:33

Stathis NtonasStathis Ntonas

453920

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  You are multiplying with an initial value of 0. Any number multiplied by 0 will return 0. Just replace 0 with 1 codesandbox.io/s/l9p345r3x7
  
  – briosheje
  Mar 7 at 11:38
  
  

  
  
  
  

add a comment | 

0

1

How can i multiply the length of the nested values array using this array:

const options = [
 
 name: "Colors",
 values: [
 label: "Blue"
 ,
 
 label: "Red"
 ]
 ,
 
 name: "Sizes",
 values: [
 label: "Small"
 ,
 
 label: "Large"
 ]
 
]

My code:

const options = [
 
 name: "Colors",
 values: [
 
 label: "Blue"
 ,
 
 label: "Red"
 
 ]
 ,
 
 name: "Sizes",
 values: [
 
 label: "Small"
 ,
 
 label: "Large"
 
 ]
 
];

const multiply = options.reduce((a, b) => a * b.values.length, 0);

console.log(multiply);

I've tried with reduce but always return 0. It should return ex. 4

playground: https://codesandbox.io/s/08jrnr72w

edit: thanks everyone for the answers, I missed initial position...

javascript arrays

share|improve this question

edited Mar 7 at 11:40

Nick Parsons

10.2k2926

asked Mar 7 at 11:33

Stathis NtonasStathis Ntonas

453920

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  You are multiplying with an initial value of 0. Any number multiplied by 0 will return 0. Just replace 0 with 1 codesandbox.io/s/l9p345r3x7
  
  – briosheje
  Mar 7 at 11:38
  
  

  
  
  
  

add a comment | 

How can i multiply the length of the nested values array using this array:

const options = [
 
 name: "Colors",
 values: [
 label: "Blue"
 ,
 
 label: "Red"
 ]
 ,
 
 name: "Sizes",
 values: [
 label: "Small"
 ,
 
 label: "Large"
 ]
 
]

My code:

const options = [
 
 name: "Colors",
 values: [
 
 label: "Blue"
 ,
 
 label: "Red"
 
 ]
 ,
 
 name: "Sizes",
 values: [
 
 label: "Small"
 ,
 
 label: "Large"
 
 ]
 
];

const multiply = options.reduce((a, b) => a * b.values.length, 0);

console.log(multiply);

I've tried with reduce but always return 0. It should return ex. 4

playground: https://codesandbox.io/s/08jrnr72w

edit: thanks everyone for the answers, I missed initial position...

javascript arrays

share|improve this question

edited Mar 7 at 11:40

Nick Parsons

10.2k2926

asked Mar 7 at 11:33

Stathis NtonasStathis Ntonas

453920

const options = [
 
 name: "Colors",
 values: [
 label: "Blue"
 ,
 
 label: "Red"
 ]
 ,
 
 name: "Sizes",
 values: [
 label: "Small"
 ,
 
 label: "Large"
 ]
 
]

const options = [
 
 name: "Colors",
 values: [
 
 label: "Blue"
 ,
 
 label: "Red"
 
 ]
 ,
 
 name: "Sizes",
 values: [
 
 label: "Small"
 ,
 
 label: "Large"
 
 ]
 
];

const multiply = options.reduce((a, b) => a * b.values.length, 0);

console.log(multiply);

const options = [
 
 name: "Colors",
 values: [
 
 label: "Blue"
 ,
 
 label: "Red"
 
 ]
 ,
 
 name: "Sizes",
 values: [
 
 label: "Small"
 ,
 
 label: "Large"
 
 ]
 
];

const multiply = options.reduce((a, b) => a * b.values.length, 0);

console.log(multiply);

const options = [
 
 name: "Colors",
 values: [
 
 label: "Blue"
 ,
 
 label: "Red"
 
 ]
 ,
 
 name: "Sizes",
 values: [
 
 label: "Small"
 ,
 
 label: "Large"
 
 ]
 
];

const multiply = options.reduce((a, b) => a * b.values.length, 0);

console.log(multiply);

share|improve this question

edited Mar 7 at 11:40

Nick Parsons

10.2k2926

asked Mar 7 at 11:33

Stathis NtonasStathis Ntonas

453920

share|improve this question

edited Mar 7 at 11:40

Nick Parsons

10.2k2926

asked Mar 7 at 11:33

Stathis NtonasStathis Ntonas

453920

edited Mar 7 at 11:40

Nick Parsons

10.2k2926

edited Mar 7 at 11:40

Nick Parsons

10.2k2926

asked Mar 7 at 11:33

Stathis NtonasStathis Ntonas

453920

asked Mar 7 at 11:33

Stathis NtonasStathis Ntonas

453920

4 Answers
4

active

oldest

votes

3

The reduce function expects an initial value. The initial value you provided is 0. If you multiply 0 with anything, you'll get 0.

You should initialize your accumulator with 1. That should fix it.

But you might have to do more, i.e. handle the edge condition. For example, at one point you may get 0 as length for one particular object and in that case, your accumulator will evaluate to 0.

share|improve this answer

answered Mar 7 at 11:38

Jayendra SharanJayendra Sharan

34927

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This is the correct answer, since it also takes care of mentioning that there might be an element with 0 length in the array. Moreover, you may also take care of elements that do not have a values property.
  
  – briosheje
  Mar 7 at 11:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks for pointing it out, in my case values will always have values, @jayendra-sharan would you mind posting an edge condition? It would be great for reference.
  
  – Stathis Ntonas
  Mar 7 at 11:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I have created a fork, you can play around it and let me know if something doesn't work. codesandbox.io/s/zx90ww42rl Please mark as accepted if it resolved your issue.
  
  – Jayendra Sharan
  Mar 7 at 11:48
  
  

  
  
  
  

add a comment | 

2

You need to initialize Array.prototype.reduce with 1 for multiplication (multiplying by zero is always going to end up as zero):

const multiply = options.reduce((a, b) => a * b.values.length, 1);

share|improve this answer

answered Mar 7 at 11:38

Rob_MRob_M

11016

add a comment | 

1

Map them to lengths and then reduce the results.

const options = [
 
 name: "Colors",
 values: [
 label: "Blue"
 ,
 
 label: "Red"
 ]
 ,
 
 name: "Sizes",
 values: [
 label: "Small"
 ,
 
 label: "Large"
 ]
 
]

const multValues = options.map(opt => opt.values.length).reduce((a,b) => a*b);
console.log(multValues)

share|improve this answer

answered Mar 7 at 11:36

dporechnydporechny

50229

add a comment | 

1

Change your starting value to 1 as 0 times anything is 0 (thus your accumulator (a) will always be 0):

const options=[name:"Colors",values:[label:"Blue",label:"Red"],name:"Sizes",values:[label:"Small",label:"Large"]];

// change starting value -------------------------------------/
const multiply = options.reduce((a, b) => a * b.values.length, 1);

console.log(multiply);

share|improve this answer

answered Mar 7 at 11:37

Nick ParsonsNick Parsons

10.2k2926

add a comment | 

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3

The reduce function expects an initial value. The initial value you provided is 0. If you multiply 0 with anything, you'll get 0.

You should initialize your accumulator with 1. That should fix it.

But you might have to do more, i.e. handle the edge condition. For example, at one point you may get 0 as length for one particular object and in that case, your accumulator will evaluate to 0.

share|improve this answer

answered Mar 7 at 11:38

Jayendra SharanJayendra Sharan

34927

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This is the correct answer, since it also takes care of mentioning that there might be an element with 0 length in the array. Moreover, you may also take care of elements that do not have a values property.
  
  – briosheje
  Mar 7 at 11:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks for pointing it out, in my case values will always have values, @jayendra-sharan would you mind posting an edge condition? It would be great for reference.
  
  – Stathis Ntonas
  Mar 7 at 11:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I have created a fork, you can play around it and let me know if something doesn't work. codesandbox.io/s/zx90ww42rl Please mark as accepted if it resolved your issue.
  
  – Jayendra Sharan
  Mar 7 at 11:48
  
  

  
  
  
  

add a comment | 

3

The reduce function expects an initial value. The initial value you provided is 0. If you multiply 0 with anything, you'll get 0.

You should initialize your accumulator with 1. That should fix it.

But you might have to do more, i.e. handle the edge condition. For example, at one point you may get 0 as length for one particular object and in that case, your accumulator will evaluate to 0.

share|improve this answer

answered Mar 7 at 11:38

Jayendra SharanJayendra Sharan

34927

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This is the correct answer, since it also takes care of mentioning that there might be an element with 0 length in the array. Moreover, you may also take care of elements that do not have a values property.
  
  – briosheje
  Mar 7 at 11:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks for pointing it out, in my case values will always have values, @jayendra-sharan would you mind posting an edge condition? It would be great for reference.
  
  – Stathis Ntonas
  Mar 7 at 11:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I have created a fork, you can play around it and let me know if something doesn't work. codesandbox.io/s/zx90ww42rl Please mark as accepted if it resolved your issue.
  
  – Jayendra Sharan
  Mar 7 at 11:48
  
  

  
  
  
  

add a comment | 

The reduce function expects an initial value. The initial value you provided is 0. If you multiply 0 with anything, you'll get 0.

You should initialize your accumulator with 1. That should fix it.

But you might have to do more, i.e. handle the edge condition. For example, at one point you may get 0 as length for one particular object and in that case, your accumulator will evaluate to 0.

share|improve this answer

answered Mar 7 at 11:38

Jayendra SharanJayendra Sharan

34927

share|improve this answer

answered Mar 7 at 11:38

Jayendra SharanJayendra Sharan

34927

answered Mar 7 at 11:38

Jayendra SharanJayendra Sharan

34927

answered Mar 7 at 11:38

Jayendra SharanJayendra Sharan

34927

2

You need to initialize Array.prototype.reduce with 1 for multiplication (multiplying by zero is always going to end up as zero):

const multiply = options.reduce((a, b) => a * b.values.length, 1);

share|improve this answer

answered Mar 7 at 11:38

Rob_MRob_M

11016

add a comment | 

2

You need to initialize Array.prototype.reduce with 1 for multiplication (multiplying by zero is always going to end up as zero):

const multiply = options.reduce((a, b) => a * b.values.length, 1);

share|improve this answer

answered Mar 7 at 11:38

Rob_MRob_M

11016

add a comment | 

You need to initialize Array.prototype.reduce with 1 for multiplication (multiplying by zero is always going to end up as zero):

const multiply = options.reduce((a, b) => a * b.values.length, 1);

share|improve this answer

answered Mar 7 at 11:38

Rob_MRob_M

11016

const multiply = options.reduce((a, b) => a * b.values.length, 1);

share|improve this answer

answered Mar 7 at 11:38

Rob_MRob_M

11016

answered Mar 7 at 11:38

Rob_MRob_M

11016

answered Mar 7 at 11:38

Rob_MRob_M

11016

1

Map them to lengths and then reduce the results.

const options = [
 
 name: "Colors",
 values: [
 label: "Blue"
 ,
 
 label: "Red"
 ]
 ,
 
 name: "Sizes",
 values: [
 label: "Small"
 ,
 
 label: "Large"
 ]
 
]

const multValues = options.map(opt => opt.values.length).reduce((a,b) => a*b);
console.log(multValues)

share|improve this answer

answered Mar 7 at 11:36

dporechnydporechny

50229

add a comment | 

1

Map them to lengths and then reduce the results.

const options = [
 
 name: "Colors",
 values: [
 label: "Blue"
 ,
 
 label: "Red"
 ]
 ,
 
 name: "Sizes",
 values: [
 label: "Small"
 ,
 
 label: "Large"
 ]
 
]

const multValues = options.map(opt => opt.values.length).reduce((a,b) => a*b);
console.log(multValues)

share|improve this answer

answered Mar 7 at 11:36

dporechnydporechny

50229

add a comment | 

Map them to lengths and then reduce the results.

const options = [
 
 name: "Colors",
 values: [
 label: "Blue"
 ,
 
 label: "Red"
 ]
 ,
 
 name: "Sizes",
 values: [
 label: "Small"
 ,
 
 label: "Large"
 ]
 
]

const multValues = options.map(opt => opt.values.length).reduce((a,b) => a*b);
console.log(multValues)

share|improve this answer

answered Mar 7 at 11:36

dporechnydporechny

50229

const options = [
 
 name: "Colors",
 values: [
 label: "Blue"
 ,
 
 label: "Red"
 ]
 ,
 
 name: "Sizes",
 values: [
 label: "Small"
 ,
 
 label: "Large"
 ]
 
]

const multValues = options.map(opt => opt.values.length).reduce((a,b) => a*b);
console.log(multValues)

const options = [
 
 name: "Colors",
 values: [
 label: "Blue"
 ,
 
 label: "Red"
 ]
 ,
 
 name: "Sizes",
 values: [
 label: "Small"
 ,
 
 label: "Large"
 ]
 
]

const multValues = options.map(opt => opt.values.length).reduce((a,b) => a*b);
console.log(multValues)

const options = [
 
 name: "Colors",
 values: [
 label: "Blue"
 ,
 
 label: "Red"
 ]
 ,
 
 name: "Sizes",
 values: [
 label: "Small"
 ,
 
 label: "Large"
 ]
 
]

const multValues = options.map(opt => opt.values.length).reduce((a,b) => a*b);
console.log(multValues)

share|improve this answer

answered Mar 7 at 11:36

dporechnydporechny

50229

answered Mar 7 at 11:36

dporechnydporechny

50229

answered Mar 7 at 11:36

dporechnydporechny

50229

1

Change your starting value to 1 as 0 times anything is 0 (thus your accumulator (a) will always be 0):

const options=[name:"Colors",values:[label:"Blue",label:"Red"],name:"Sizes",values:[label:"Small",label:"Large"]];

// change starting value -------------------------------------/
const multiply = options.reduce((a, b) => a * b.values.length, 1);

console.log(multiply);

share|improve this answer

answered Mar 7 at 11:37

Nick ParsonsNick Parsons

10.2k2926

add a comment | 

1

Change your starting value to 1 as 0 times anything is 0 (thus your accumulator (a) will always be 0):

const options=[name:"Colors",values:[label:"Blue",label:"Red"],name:"Sizes",values:[label:"Small",label:"Large"]];

// change starting value -------------------------------------/
const multiply = options.reduce((a, b) => a * b.values.length, 1);

console.log(multiply);

share|improve this answer

answered Mar 7 at 11:37

Nick ParsonsNick Parsons

10.2k2926

add a comment | 

Change your starting value to 1 as 0 times anything is 0 (thus your accumulator (a) will always be 0):

const options=[name:"Colors",values:[label:"Blue",label:"Red"],name:"Sizes",values:[label:"Small",label:"Large"]];

// change starting value -------------------------------------/
const multiply = options.reduce((a, b) => a * b.values.length, 1);

console.log(multiply);

share|improve this answer

answered Mar 7 at 11:37

Nick ParsonsNick Parsons

10.2k2926

const options=[name:"Colors",values:[label:"Blue",label:"Red"],name:"Sizes",values:[label:"Small",label:"Large"]];

// change starting value -------------------------------------/
const multiply = options.reduce((a, b) => a * b.values.length, 1);

console.log(multiply);

const options=[name:"Colors",values:[label:"Blue",label:"Red"],name:"Sizes",values:[label:"Small",label:"Large"]];

// change starting value -------------------------------------/
const multiply = options.reduce((a, b) => a * b.values.length, 1);

console.log(multiply);

const options=[name:"Colors",values:[label:"Blue",label:"Red"],name:"Sizes",values:[label:"Small",label:"Large"]];

// change starting value -------------------------------------/
const multiply = options.reduce((a, b) => a * b.values.length, 1);

console.log(multiply);

share|improve this answer

answered Mar 7 at 11:37

Nick ParsonsNick Parsons

10.2k2926

answered Mar 7 at 11:37

Nick ParsonsNick Parsons

10.2k2926

answered Mar 7 at 11:37

Nick ParsonsNick Parsons

10.2k2926