how to print result of _.zip.apply()

Question

I am trying underscore.js for the first time and want to transpose my row array into column. As, I need to join it with 2D array

I checked discussion here and found can be done by _.zip.apply(). But, when I am trying it, it is not showing any reuslt

  Final = my 2D array
  dtr = ['s', 's', 'n'];

I need to join dtr with final such that every element in dtr is a column header in final array doing this:

  _.zip.apply(dtr, Final)

but it is not showing any result

expected outcome:

my 2d array:-

and the dtr array need to go at top of the 2d array.. hope it is clear.

Answer 1

In case you want your dtr array to be the first array in the 2d array. One could do this:

var a = ['s','s','n'];
var b = [[50, 50, 50], 
         [50 , 5, 5],
         ['hello', 'beta', 'gama']];

var result = [a].concat(b);

Result:

[['s','s','n'],
 [50, 50, 50], 
 [50 , 5, 5],
 ['hello', 'beta', 'gama']];

Edit

var a = ['s','s','n'];
var b = [[50, 50, 'hello'], 
         [50,  5, 'beta'], 
         [50,  5, 'gama']];

var result = _.map(_.zip(a, b), _.flatten) 
// if native map support _.zip(a, b).map(_.flatten)

Result:

[['s', 50, 50, 'hello'], 
 ['s', 50,  5, 'beta'], 
 ['n', 50,  5, 'gama']];

Answer 2

My answer contains two cases, which are dependent on how array b is organised.

Case one:

var a = ['s','s','n'];
var b = [[50, 50, 'hello'], 
         [50,  5, 'beta'], 
         [50,  5, 'gama']];

_.zip(a, b).map(_.flatten) // -> [Array[4], Array[4], Array[4]]

Case Two

var a = ['s','s','n'];
var c = [[50, 50, 50], 
         [50 , 5, 5],
         ['hello', 'beta', 'gama']];
var b = _.zip.apply(null, c); // content equals to case one variable b

_.zip(a, b).map(_.flatten) // -> [Array[4], Array[4], Array[4]]

Result:

[['s', 50, 50, 'hello'],
 ['s', 50,  5, 'beta'],
 ['n', 50,  5, 'gama']]