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I have a question concerning the colon operator and expansion of vectors in MATLAB. My problem is to understand how the following line of code expands, to be able to use it for other sequences. The line of MATLAB code is: 

a(1:2:5) = 1:-4:-7

Note that a is not defined before the expansion. This returns the vector 

a = 1 0 3 0 -7

I know how the colon operator works with {start}:{step}:{stop}, my problem is to understand how and why the combination of a(1:2:5)and 1:-4:-7 returns a vector of five elements with zeros in position 2 and 5?

Daniel Widdis
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johnhenning
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2 Answers2

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Whenever Matlab detects you're indecing to an element outside the current bounds of the matrix/array, it will automatically pad the missing elements with zeros: 

>> clear b; b(10) = 5 
b =
    0     0     0     0     0     0     0     0     0     5

This feature is both very useful, and very dangerous. It is useful for the fact declarations can be made very easy, such as your own case. You can create a whole array of custom-made classes by issuing something like

myClassArray(500) = myClass(1, 2);

which is infinitely better than something like

% cannot pre-allocate (zeros() or ones() give double/uint8/..., not myClass)
for ii = 1:499
     myClassArray(ii) = myClass; % so, growing array
end
myClassArray(500) = myClass(1,2);

But, growing arrays can be hard to spot: 

a = zeros(10,1);
for ii = 1:10
    a(ii+1) = rand;
end

which can make performance drop tremendously. Also, when you translate code prototyped in Matlab to a statically-typed language like C++, copying this code will result in buffer overflows and thus segfaults.

Now, going back to your case: 

clear a;   a(1:2:5) = 1:-4:-7

The 1:2:5 will expand to the array [1 3 5], and the 1:-4:-7 will give the values [1 -3 -7]. Since the variable a does not exist yet, Matlab will create a new one and fill the elements [1 3 5] with the values [1 -3 -7]. The indices that have been skipped in order to initialize variable a (namely, [2 4]) will then have been initialized automatically to zero.

If you're familiar with Python, it's a bit like the syntax to assign multiple values to multiple variables

x,y = 1,2

But in your Matlab case, these different variables are indices to a non-existent array, which requires "filling the holes with something" to make it a valid, consistent array.

Does this make things clear?

Rody Oldenhuis
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when you define a(1:2:5), it creates a size 5 vector (zero-valued), and selecting odd indexed(3 of them exists) cells. 1:-4:-7 creates three values (not five). Finally your selected three cells are filled with data of 3 values coming from 1:-4:-7

Seçkin Savaşçı
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  • Ok, so I have to be careful to compare `a(1:2:5)` and `1:-4:-7`. The first expression creates five elements, the second creates 3 elements, when combined the three elements are printed on the odd spaces/elements in vector `a`. Thanks for your help, do you have advice on where to find more information about my question? When I search the web I can not find relevant information. Thanks for your support! – johnhenning Aug 24 '12 at 08:17
  • Please accept as an answer if you think it solves your problem. 1:2:5 actually creates 3 values if you write in solely on MATLAB command line, the trick is if you use it for indexing a vector as in the situation of a(1:2:5), it works as creating 5 elements and selecting 3 of them. – Seçkin Savaşçı Aug 24 '12 at 08:20