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Filling/Populating a multidimensional array,, with elements of another array, and adding/incrementing values?

I read the documentation on Array more carefully.

An array can also be created by explicitly calling ::new with zero, one (the initial size of the Array) or two arguments (the initial size and a default object).

ary =    #=> []       #=> [nil, nil, nil], true) #=> [true, true, true]

Note that the second argument populates the array with references to the same object. Therefore, it is only recommended in cases when you need to instantiate arrays with natively immutable objects such as Symbols, numbers, true or false.

To create an array with separate objects a block can be passed instead. This method is safe to use with mutable objects such as hashes, strings or other arrays: { } #=> [{},
{}, {}, {}]

This is also a quick way to build up multi-dimensional arrays:

 empty_table = { }
 #=> [[nil, nil, nil], [nil, nil, nil], [nil,
nil, nil]]

Which means that I was modifying the element which was referenced in multiple places. And I had to explicitly create a new Array object for each entry along the new dimensions I was adding, so in a way building the extra dimension incrementally. There seem to be multiple ways of doing this. One is this method of a block sent to Array, and another is to do an iterator on the dimension and make a new Array for each entry.

I also found this post which highlights the problem in very simple terms:>>

The wrong way, and the way I see people trying often is to say 4,, 0) ). In other words, an array of 4 rows, each row being an array of 4 zeroes. And this appears to work at first. However, run the following code:

 a = 4,, 0) )
a[0][0] = 1 pp a 

It looks simple. Make a 4x4 array of zeroes, set the top-left element to 1. But print it and we get…

[[1, 0, 0, 0], [1, 0, 0, 0], [1, 0, 0,
0], [1, 0, 0, 0]] 

It set the entire first column to 1, what gives? When we made the arrays, the inner-most call to gets called first, making a single row. A single reference to this row is then duplicated 4 times to fill the outer-most array. Each row is then referencing the same array. Change one, change them all.

Instead of passing a value to the method, we pass a block. The block is executed every time the method needs a new value. So if you were to say { gets.chomp }, Ruby will stop and ask for input 5 times. So all we need to do is just create a new array inside this block. So we end up with {,0) }. Now let's try that test case again.

 a = {, 0) }
a[0][0] = 1 pp a 

And it does just as you'd expect.

 [[1, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0,
0], [0, 0, 0, 0]]

Categories : Ruby

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