How to create two arrays in the same loop with CoffeeScript?

Question

I want to create two arrays b and c at the same time. I know two methods which can be able to achieve it. The first method is

b = ([i, i * 2] for i in [0..10])
c = ([i, i * 3] for i in [0..10])

alert "b=#{b}"
alert "c=#{c}"

This method is very handy for creating only one array. I can not be the better way to get the better performance for computation.

The second method is

b = []
c = []
for i in [0..10]
  b.push [i, i*2]
  c.push [i, i*3]

alert "b=#{b}"
alert "c=#{c}"

This method seems good for computation efficiency but two lines b = [] c = [] have to be written first. I don't want to write this 2 lines but I have not find a good idea to have the answer. Without the initialization for the arrays of b and c, we can not use push method.

There exists the existential operator ? in Coffeescript but I don't know hot to use it in this problem. Do you have a better method for creating the arrays of b and c without the explicit initialization?

Thank you!

Thanks. You will get the same result for b and c. The answer is wrong. — yun_cn
– yun_cn, Commented Mar 8, 2013 at 8:32
@varnie: No, the arrays need to be separate, and their contents are different. — T.J. Crowder
– T.J. Crowder, Commented Mar 8, 2013 at 8:33
[b,c] =[[],[]] works but is still trapped in the many parens. — yun_cn
– yun_cn, Commented Mar 11, 2013 at 0:36

nl_0 · Accepted Answer · 2013-03-09 01:06:28Z

4

You can use a little help from underscore (or any other lib that provides zip-like functionality):

[b, c] = _.zip ([[i, i * 2], [i, i * 3]] for i in [0..10])...

After executing it we have:

coffee> b 
[ [ 0, 0 ],
  [ 1, 2 ],
  [ 2, 4 ],
  [ 3, 6 ],
  [ 4, 8 ],
  [ 5, 10 ],
  [ 6, 12 ],
  [ 7, 14 ],
  [ 8, 16 ],
  [ 9, 18 ],
  [ 10, 20 ] ]

coffee> c
[ [ 0, 0 ],
  [ 1, 3 ],
  [ 2, 6 ],
  [ 3, 9 ],
  [ 4, 12 ],
  [ 5, 15 ],
  [ 6, 18 ],
  [ 7, 21 ],
  [ 8, 24 ],
  [ 9, 27 ],
  [ 10, 30 ] ]

See the section about splats in CoffeeScript docs for more details and examples.

edited Mar 9, 2013 at 1:06

answered Mar 8, 2013 at 12:53

nl_0

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5 Comments

yun_cn Over a year ago

Very interesting but it is really a challenge. I haven't got the right answer with it. Thank you.

nl_0 Over a year ago

I've updated my answer with the result I get. is it incorrect?

yun_cn Over a year ago

Thanks a lot. I have also got the same results. I forgot the three dots at the end of line. I have not really well understand the usage of the three dots. Thank you very much.

nl_0 Over a year ago

From CoffeeScript docs: "The JavaScript arguments object is a useful way to work with functions that accept variable numbers of arguments. CoffeeScript provides splats ..., both for function definition as well as invocation, making variable numbers of arguments a little bit more palatable". See the docs for details and examples.

yun_cn Over a year ago

Thanks you very much for your kind suggestion. It seems that the splats must be used as an argument of function such as _.zip. I will try to practice more with it.

Sanketh Katta · Accepted Answer · 2013-03-08 10:13:19Z

1

How about this using the existential operator:

for i in [0..10]
    b = [] if not b?.push [i, i*2]
    c = [] if not c?.push [i, i*3]

console.log "b=#{b}"
console.log "c=#{c}"

Or to be a bit more understandable:

for i in [0..10]
    (if b? then b else b = []).push [i, i*2]
    (if c? then c else c = []).push [i, i*3]

console.log "b=#{b}"
console.log "c=#{c}"

EDIT: from comments:

OK but you you have to write so many tedious codes. The same reason is also for ` (b = b or []).push [i, i*2]

It is tedious, so we can wrap it in a function (but beware the variables will be global now):

# for node.js
array = (name) -> global[name] = global[name] or []

# for the browser
array = (name) -> window[name] = window[name] or []

for i in [0..10]
    array('b').push [i, i*2]
    array('c').push [i, i*3]

console.log "b=#{b}"
console.log "c=#{c}"

edited Mar 8, 2013 at 10:13

answered Mar 8, 2013 at 9:26

Sanketh Katta

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4 Comments

yun_cn Over a year ago

OK but you you have to write so many tedious codes. The same reason is also for ` (b = b or []).push [i, i*2] `

yun_cn Over a year ago

Too complex. Still no better way than initializing them separately with b =[] and c = []. I wonder if there are some ways which can combine the 1st and 2nd methods as I mentioned. We perhaps may need to modify the definition for the push method for the array to make it have a initial value.

Sanketh Katta Over a year ago

I agree that the explicit definition is definitely the simplest, and will be the easiest to understand. I wouldn't want to get into extending the .push functionality, seems even messier.

yun_cn Over a year ago

So do I. Thanks a lot for your kindness suggestion.

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