Use JavaScript or Lodash to create multi dimensional array

I want to create a multi-dimensional array like this using Lodash or vanilla JS:

[

  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],

  [11, 12, 13, 14, 15, 16, 17, 18, 19, 20],

  etc

]

This is a simplistic example since I want this pattern to continue up to 1,000,000 but for demonstration 1 to 20 is fine.

Any ideas? I've tried _.range(20) so far but I need this Array to be multi-dimensional. Thanks

asked Nov 20 '18 at 23:49

danday74

18.8k1491130

What happened when you tried _.range(20)? Perhaps you'd then be interested in _.chunk()

– Phil
Nov 20 '18 at 23:54

1

Pure JS: Array.from(new Array(30)).map((_, i) => i + 1).reduce((acc, v) => v % 10 === 1 ? [...acc, [v]] : (acc[Math.floor((v - 1) / 10)].push(v), acc), )

– zerkms
Nov 20 '18 at 23:58

add a comment |

I want to create a multi-dimensional array like this using Lodash or vanilla JS:

[

  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],

  [11, 12, 13, 14, 15, 16, 17, 18, 19, 20],

  etc

]

This is a simplistic example since I want this pattern to continue up to 1,000,000 but for demonstration 1 to 20 is fine.

Any ideas? I've tried _.range(20) so far but I need this Array to be multi-dimensional. Thanks

asked Nov 20 '18 at 23:49

danday74

18.8k1491130

What happened when you tried _.range(20)? Perhaps you'd then be interested in _.chunk()

– Phil
Nov 20 '18 at 23:54

1

Pure JS: Array.from(new Array(30)).map((_, i) => i + 1).reduce((acc, v) => v % 10 === 1 ? [...acc, [v]] : (acc[Math.floor((v - 1) / 10)].push(v), acc), )

– zerkms
Nov 20 '18 at 23:58

add a comment |

I want to create a multi-dimensional array like this using Lodash or vanilla JS:

[

  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],

  [11, 12, 13, 14, 15, 16, 17, 18, 19, 20],

  etc

]

This is a simplistic example since I want this pattern to continue up to 1,000,000 but for demonstration 1 to 20 is fine.

Any ideas? I've tried _.range(20) so far but I need this Array to be multi-dimensional. Thanks

asked Nov 20 '18 at 23:49

danday74

18.8k1491130

I want to create a multi-dimensional array like this using Lodash or vanilla JS:

[

  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],

  [11, 12, 13, 14, 15, 16, 17, 18, 19, 20],

  etc

]

This is a simplistic example since I want this pattern to continue up to 1,000,000 but for demonstration 1 to 20 is fine.

Any ideas? I've tried _.range(20) so far but I need this Array to be multi-dimensional. Thanks

javascript lodash

asked Nov 20 '18 at 23:49

danday74

18.8k1491130

asked Nov 20 '18 at 23:49

danday74

18.8k1491130

asked Nov 20 '18 at 23:49

danday74

18.8k1491130

asked Nov 20 '18 at 23:49

danday74

18.8k1491130

asked Nov 20 '18 at 23:49

danday74

18.8k1491130

What happened when you tried _.range(20)? Perhaps you'd then be interested in _.chunk()

– Phil
Nov 20 '18 at 23:54

1

Pure JS: Array.from(new Array(30)).map((_, i) => i + 1).reduce((acc, v) => v % 10 === 1 ? [...acc, [v]] : (acc[Math.floor((v - 1) / 10)].push(v), acc), )

– zerkms
Nov 20 '18 at 23:58

add a comment |

What happened when you tried _.range(20)? Perhaps you'd then be interested in _.chunk()

– Phil
Nov 20 '18 at 23:54

1

Pure JS: Array.from(new Array(30)).map((_, i) => i + 1).reduce((acc, v) => v % 10 === 1 ? [...acc, [v]] : (acc[Math.floor((v - 1) / 10)].push(v), acc), )

– zerkms
Nov 20 '18 at 23:58

What happened when you tried _.range(20)? Perhaps you'd then be interested in _.chunk()

– Phil
Nov 20 '18 at 23:54

Pure JS:

Array.from(new Array(30)).map((_, i) => i + 1).reduce((acc, v) => v % 10 === 1 ? [...acc, [v]] : (acc[Math.floor((v - 1) / 10)].push(v), acc), )

– zerkms
Nov 20 '18 at 23:58

Pure JS:

Array.from(new Array(30)).map((_, i) => i + 1).reduce((acc, v) => v % 10 === 1 ? [...acc, [v]] : (acc[Math.floor((v - 1) / 10)].push(v), acc), )

– zerkms
Nov 20 '18 at 23:58

add a comment |

4 Answers
4

active

oldest

votes

With lodash you can use chunk:

const result = _.chunk(_.range(1, 21), 10);

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

edited Nov 20 '18 at 23:58

answered Nov 20 '18 at 23:56

slider

8,25811130

1

Just a minor correction... OP's arrays start at 1

– Phil
Nov 20 '18 at 23:57

@Phil thanks, corrected.

– slider
Nov 20 '18 at 23:58

1

Upvote! First thing I though about when I read the question and then realized you already wrote it :)

– Akrion
Nov 21 '18 at 0:48

add a comment |

Using nested native Array#from()

const 

   limit = 100,

   fr = Array.from;



const res = fr({ length:limit/10 }, (_,i) => fr({ length:10 }, (_,j) => i*10 + j+1 ));



console.log(res)

edited Nov 21 '18 at 1:20

answered Nov 21 '18 at 0:03

charlietfl

139k1388120

2

That's really really really really cool

– zerkms
Nov 21 '18 at 0:04

1

Array#from() 👈 spot the Java dev ;-)

– Phil
Nov 21 '18 at 0:05

I know OP doesn't mention, but what if the limit isn't a multiple of 10, say 25? Slider's answer handles that case via chunking

– Phil
Nov 21 '18 at 0:07

@Phil right...I assumed even 10's based on 20 and 1,000,000

– charlietfl
Nov 21 '18 at 0:09

2

It's not as pretty, but you could have something like {length:Math.min(10, limit-(i*10))} in the inside loop to handle non-even groupings.

– Mark Meyer
Nov 21 '18 at 0:16

|
show 2 more comments

With vanilla javascript:

x = 

for (i=1; i<=20; i=i+10) {

    y = 

    for (j=0; j<10; j++) {

        y.push(i + j)

    }

    x.push(y)

}

console.log(x)

For anyone who's interested, I time clocked these answers and this is the fastest. Using 1 million entries, this clocks at .100 seconds. Lodash chunk clocks close behind at .110s. Array.from lags behind at .254s.

edited Nov 21 '18 at 0:32

answered Nov 21 '18 at 0:01

Conner

23.3k84568

Old school cool :)

– slider
Nov 21 '18 at 0:15

1

@Phil hah, fixed... and time-clocked

– Conner
Nov 21 '18 at 0:33

add a comment |

Looking at the ES6 Array.from documentation you can see a range generator function:

const range = (start, stop, step) => Array.from({ length: (stop -

start) / step }, (_, i) => start + (i * step));

So with that in mind you could generate the range via:

range(1, 21, 1)  // account for the 0 with the 21 otherwise you get 1-19

Then all that is missing is the chunk function:

const range = (start, stop, step) => Array.from({ length: (stop - start) / step }, (_, i) => start + (i * step));

const chunkBy = (arr, by=2) => arr.reduce((r,c,i) => (i%by==0 ? r.push([c]) : r[r.length-1] = [...r[r.length-1], c], r), )



console.log(chunkBy(range(1,21,1), 10))

The chunkBy uses Array.reduce & the % operator to chunk the data.

edited Nov 21 '18 at 0:47

answered Nov 21 '18 at 0:16

Akrion

9,42711224

1

And you wrote a mini lodash :D

– slider
Nov 21 '18 at 0:50

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

With lodash you can use chunk:

const result = _.chunk(_.range(1, 21), 10);

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

edited Nov 20 '18 at 23:58

answered Nov 20 '18 at 23:56

slider

8,25811130

1

Just a minor correction... OP's arrays start at 1

– Phil
Nov 20 '18 at 23:57

@Phil thanks, corrected.

– slider
Nov 20 '18 at 23:58

1

Upvote! First thing I though about when I read the question and then realized you already wrote it :)

– Akrion
Nov 21 '18 at 0:48

add a comment |

With lodash you can use chunk:

const result = _.chunk(_.range(1, 21), 10);

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

edited Nov 20 '18 at 23:58

answered Nov 20 '18 at 23:56

slider

8,25811130

1

Just a minor correction... OP's arrays start at 1

– Phil
Nov 20 '18 at 23:57

@Phil thanks, corrected.

– slider
Nov 20 '18 at 23:58

1

Upvote! First thing I though about when I read the question and then realized you already wrote it :)

– Akrion
Nov 21 '18 at 0:48

add a comment |

With lodash you can use chunk:

const result = _.chunk(_.range(1, 21), 10);

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

edited Nov 20 '18 at 23:58

answered Nov 20 '18 at 23:56

slider

8,25811130

With lodash you can use chunk:

const result = _.chunk(_.range(1, 21), 10);

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

const result = _.chunk(_.range(1, 21), 10);

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

const result = _.chunk(_.range(1, 21), 10);

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

edited Nov 20 '18 at 23:58

answered Nov 20 '18 at 23:56

slider

8,25811130

edited Nov 20 '18 at 23:58

answered Nov 20 '18 at 23:56

slider

8,25811130

answered Nov 20 '18 at 23:56

slider

8,25811130

answered Nov 20 '18 at 23:56

slider

8,25811130

1

Just a minor correction... OP's arrays start at 1

– Phil
Nov 20 '18 at 23:57

@Phil thanks, corrected.

– slider
Nov 20 '18 at 23:58

1

Upvote! First thing I though about when I read the question and then realized you already wrote it :)

– Akrion
Nov 21 '18 at 0:48

add a comment |

1

Just a minor correction... OP's arrays start at 1

– Phil
Nov 20 '18 at 23:57

@Phil thanks, corrected.

– slider
Nov 20 '18 at 23:58

1

Upvote! First thing I though about when I read the question and then realized you already wrote it :)

– Akrion
Nov 21 '18 at 0:48

Just a minor correction... OP's arrays start at 1

– Phil
Nov 20 '18 at 23:57

@Phil thanks, corrected.

– slider
Nov 20 '18 at 23:58

Upvote! First thing I though about when I read the question and then realized you already wrote it :)

– Akrion
Nov 21 '18 at 0:48

add a comment |

Using nested native Array#from()

const 

   limit = 100,

   fr = Array.from;



const res = fr({ length:limit/10 }, (_,i) => fr({ length:10 }, (_,j) => i*10 + j+1 ));



console.log(res)

edited Nov 21 '18 at 1:20

answered Nov 21 '18 at 0:03

charlietfl

139k1388120

2

That's really really really really cool

– zerkms
Nov 21 '18 at 0:04

1

Array#from() 👈 spot the Java dev ;-)

– Phil
Nov 21 '18 at 0:05

I know OP doesn't mention, but what if the limit isn't a multiple of 10, say 25? Slider's answer handles that case via chunking

– Phil
Nov 21 '18 at 0:07

@Phil right...I assumed even 10's based on 20 and 1,000,000

– charlietfl
Nov 21 '18 at 0:09

2

It's not as pretty, but you could have something like {length:Math.min(10, limit-(i*10))} in the inside loop to handle non-even groupings.

– Mark Meyer
Nov 21 '18 at 0:16

|
show 2 more comments

Using nested native Array#from()

const 

   limit = 100,

   fr = Array.from;



const res = fr({ length:limit/10 }, (_,i) => fr({ length:10 }, (_,j) => i*10 + j+1 ));



console.log(res)

edited Nov 21 '18 at 1:20

answered Nov 21 '18 at 0:03

charlietfl

139k1388120

2

That's really really really really cool

– zerkms
Nov 21 '18 at 0:04

1

Array#from() 👈 spot the Java dev ;-)

– Phil
Nov 21 '18 at 0:05

I know OP doesn't mention, but what if the limit isn't a multiple of 10, say 25? Slider's answer handles that case via chunking

– Phil
Nov 21 '18 at 0:07

@Phil right...I assumed even 10's based on 20 and 1,000,000

– charlietfl
Nov 21 '18 at 0:09

2

It's not as pretty, but you could have something like {length:Math.min(10, limit-(i*10))} in the inside loop to handle non-even groupings.

– Mark Meyer
Nov 21 '18 at 0:16

|
show 2 more comments

Using nested native Array#from()

const 

   limit = 100,

   fr = Array.from;



const res = fr({ length:limit/10 }, (_,i) => fr({ length:10 }, (_,j) => i*10 + j+1 ));



console.log(res)

edited Nov 21 '18 at 1:20

answered Nov 21 '18 at 0:03

charlietfl

139k1388120

Using nested native Array#from()

const 

   limit = 100,

   fr = Array.from;



const res = fr({ length:limit/10 }, (_,i) => fr({ length:10 }, (_,j) => i*10 + j+1 ));



console.log(res)

const 

   limit = 100,

   fr = Array.from;



const res = fr({ length:limit/10 }, (_,i) => fr({ length:10 }, (_,j) => i*10 + j+1 ));



console.log(res)

const 

   limit = 100,

   fr = Array.from;



const res = fr({ length:limit/10 }, (_,i) => fr({ length:10 }, (_,j) => i*10 + j+1 ));



console.log(res)

edited Nov 21 '18 at 1:20

answered Nov 21 '18 at 0:03

charlietfl

139k1388120

edited Nov 21 '18 at 1:20

answered Nov 21 '18 at 0:03

charlietfl

139k1388120

answered Nov 21 '18 at 0:03

charlietfl

139k1388120

answered Nov 21 '18 at 0:03

charlietfl

139k1388120

2

That's really really really really cool

– zerkms
Nov 21 '18 at 0:04

1

Array#from() 👈 spot the Java dev ;-)

– Phil
Nov 21 '18 at 0:05

I know OP doesn't mention, but what if the limit isn't a multiple of 10, say 25? Slider's answer handles that case via chunking

– Phil
Nov 21 '18 at 0:07

@Phil right...I assumed even 10's based on 20 and 1,000,000

– charlietfl
Nov 21 '18 at 0:09

2

It's not as pretty, but you could have something like {length:Math.min(10, limit-(i*10))} in the inside loop to handle non-even groupings.

– Mark Meyer
Nov 21 '18 at 0:16

|
show 2 more comments

2

That's really really really really cool

– zerkms
Nov 21 '18 at 0:04

1

Array#from() 👈 spot the Java dev ;-)

– Phil
Nov 21 '18 at 0:05

I know OP doesn't mention, but what if the limit isn't a multiple of 10, say 25? Slider's answer handles that case via chunking

– Phil
Nov 21 '18 at 0:07

@Phil right...I assumed even 10's based on 20 and 1,000,000

– charlietfl
Nov 21 '18 at 0:09

2

It's not as pretty, but you could have something like {length:Math.min(10, limit-(i*10))} in the inside loop to handle non-even groupings.

– Mark Meyer
Nov 21 '18 at 0:16

That's really really really really cool

– zerkms
Nov 21 '18 at 0:04

Array#from() 👈 spot the Java dev ;-)

– Phil
Nov 21 '18 at 0:05

I know OP doesn't mention, but what if the limit isn't a multiple of 10, say 25? Slider's answer handles that case via chunking

– Phil
Nov 21 '18 at 0:07

@Phil right...I assumed even 10's based on 20 and 1,000,000

– charlietfl
Nov 21 '18 at 0:09

It's not as pretty, but you could have something like {length:Math.min(10, limit-(i*10))} in the inside loop to handle non-even groupings.

– Mark Meyer
Nov 21 '18 at 0:16

|
show 2 more comments

With vanilla javascript:

x = 

for (i=1; i<=20; i=i+10) {

    y = 

    for (j=0; j<10; j++) {

        y.push(i + j)

    }

    x.push(y)

}

console.log(x)

edited Nov 21 '18 at 0:32

answered Nov 21 '18 at 0:01

Conner

23.3k84568

Old school cool :)

– slider
Nov 21 '18 at 0:15

1

@Phil hah, fixed... and time-clocked

– Conner
Nov 21 '18 at 0:33

add a comment |

With vanilla javascript:

x = 

for (i=1; i<=20; i=i+10) {

    y = 

    for (j=0; j<10; j++) {

        y.push(i + j)

    }

    x.push(y)

}

console.log(x)

edited Nov 21 '18 at 0:32

answered Nov 21 '18 at 0:01

Conner

23.3k84568

Old school cool :)

– slider
Nov 21 '18 at 0:15

1

@Phil hah, fixed... and time-clocked

– Conner
Nov 21 '18 at 0:33

add a comment |

With vanilla javascript:

x = 

for (i=1; i<=20; i=i+10) {

    y = 

    for (j=0; j<10; j++) {

        y.push(i + j)

    }

    x.push(y)

}

console.log(x)

edited Nov 21 '18 at 0:32

answered Nov 21 '18 at 0:01

Conner

23.3k84568

With vanilla javascript:

x = 

for (i=1; i<=20; i=i+10) {

    y = 

    for (j=0; j<10; j++) {

        y.push(i + j)

    }

    x.push(y)

}

console.log(x)

x = 

for (i=1; i<=20; i=i+10) {

    y = 

    for (j=0; j<10; j++) {

        y.push(i + j)

    }

    x.push(y)

}

console.log(x)

x = 

for (i=1; i<=20; i=i+10) {

    y = 

    for (j=0; j<10; j++) {

        y.push(i + j)

    }

    x.push(y)

}

console.log(x)

edited Nov 21 '18 at 0:32

answered Nov 21 '18 at 0:01

Conner

23.3k84568

edited Nov 21 '18 at 0:32

answered Nov 21 '18 at 0:01

Conner

23.3k84568

answered Nov 21 '18 at 0:01

Conner

23.3k84568

answered Nov 21 '18 at 0:01

Conner

23.3k84568

Old school cool :)

– slider
Nov 21 '18 at 0:15

1

@Phil hah, fixed... and time-clocked

– Conner
Nov 21 '18 at 0:33

add a comment |

Old school cool :)

– slider
Nov 21 '18 at 0:15

1

@Phil hah, fixed... and time-clocked

– Conner
Nov 21 '18 at 0:33

Old school cool :)

– slider
Nov 21 '18 at 0:15

@Phil hah, fixed... and time-clocked

– Conner
Nov 21 '18 at 0:33

add a comment |

Looking at the ES6 Array.from documentation you can see a range generator function:

const range = (start, stop, step) => Array.from({ length: (stop -

start) / step }, (_, i) => start + (i * step));

So with that in mind you could generate the range via:

range(1, 21, 1)  // account for the 0 with the 21 otherwise you get 1-19

Then all that is missing is the chunk function:

const range = (start, stop, step) => Array.from({ length: (stop - start) / step }, (_, i) => start + (i * step));

const chunkBy = (arr, by=2) => arr.reduce((r,c,i) => (i%by==0 ? r.push([c]) : r[r.length-1] = [...r[r.length-1], c], r), )



console.log(chunkBy(range(1,21,1), 10))

The chunkBy uses Array.reduce & the % operator to chunk the data.

edited Nov 21 '18 at 0:47

answered Nov 21 '18 at 0:16

Akrion

9,42711224

1

And you wrote a mini lodash :D

– slider
Nov 21 '18 at 0:50

add a comment |

Looking at the ES6 Array.from documentation you can see a range generator function:

const range = (start, stop, step) => Array.from({ length: (stop -

start) / step }, (_, i) => start + (i * step));

So with that in mind you could generate the range via:

range(1, 21, 1)  // account for the 0 with the 21 otherwise you get 1-19

Then all that is missing is the chunk function:

const range = (start, stop, step) => Array.from({ length: (stop - start) / step }, (_, i) => start + (i * step));

const chunkBy = (arr, by=2) => arr.reduce((r,c,i) => (i%by==0 ? r.push([c]) : r[r.length-1] = [...r[r.length-1], c], r), )



console.log(chunkBy(range(1,21,1), 10))

The chunkBy uses Array.reduce & the % operator to chunk the data.

edited Nov 21 '18 at 0:47

answered Nov 21 '18 at 0:16

Akrion

9,42711224

1

And you wrote a mini lodash :D

– slider
Nov 21 '18 at 0:50

add a comment |

Looking at the ES6 Array.from documentation you can see a range generator function:

const range = (start, stop, step) => Array.from({ length: (stop -

start) / step }, (_, i) => start + (i * step));

So with that in mind you could generate the range via:

range(1, 21, 1)  // account for the 0 with the 21 otherwise you get 1-19

Then all that is missing is the chunk function:

const range = (start, stop, step) => Array.from({ length: (stop - start) / step }, (_, i) => start + (i * step));

const chunkBy = (arr, by=2) => arr.reduce((r,c,i) => (i%by==0 ? r.push([c]) : r[r.length-1] = [...r[r.length-1], c], r), )



console.log(chunkBy(range(1,21,1), 10))

The chunkBy uses Array.reduce & the % operator to chunk the data.

edited Nov 21 '18 at 0:47

answered Nov 21 '18 at 0:16

Akrion

9,42711224

Looking at the ES6 Array.from documentation you can see a range generator function:

const range = (start, stop, step) => Array.from({ length: (stop -

start) / step }, (_, i) => start + (i * step));

So with that in mind you could generate the range via:

range(1, 21, 1)  // account for the 0 with the 21 otherwise you get 1-19

Then all that is missing is the chunk function:

const range = (start, stop, step) => Array.from({ length: (stop - start) / step }, (_, i) => start + (i * step));

const chunkBy = (arr, by=2) => arr.reduce((r,c,i) => (i%by==0 ? r.push([c]) : r[r.length-1] = [...r[r.length-1], c], r), )



console.log(chunkBy(range(1,21,1), 10))

The chunkBy uses Array.reduce & the % operator to chunk the data.

const range = (start, stop, step) => Array.from({ length: (stop - start) / step }, (_, i) => start + (i * step));

const chunkBy = (arr, by=2) => arr.reduce((r,c,i) => (i%by==0 ? r.push([c]) : r[r.length-1] = [...r[r.length-1], c], r), )



console.log(chunkBy(range(1,21,1), 10))

const range = (start, stop, step) => Array.from({ length: (stop - start) / step }, (_, i) => start + (i * step));

const chunkBy = (arr, by=2) => arr.reduce((r,c,i) => (i%by==0 ? r.push([c]) : r[r.length-1] = [...r[r.length-1], c], r), )



console.log(chunkBy(range(1,21,1), 10))

edited Nov 21 '18 at 0:47

answered Nov 21 '18 at 0:16

Akrion

9,42711224

edited Nov 21 '18 at 0:47

answered Nov 21 '18 at 0:16

Akrion

9,42711224

answered Nov 21 '18 at 0:16

Akrion

9,42711224

answered Nov 21 '18 at 0:16

Akrion

9,42711224

1

And you wrote a mini lodash :D

– slider
Nov 21 '18 at 0:50

add a comment |

1

And you wrote a mini lodash :D

– slider
Nov 21 '18 at 0:50

And you wrote a mini lodash :D

– slider
Nov 21 '18 at 0:50

add a comment |

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