Why does this not vectorize

Simply this sort of loop evaluates my function to the reference correctly:

for t = 2:length(x)
    x(t) = x(t-1)+x(t);
end
    
tf = isequal(x,y);
tf

tf =

  logical

   1
   

While, surprisngly, this does not:

t = 2:length(x);
x(t) = x(t-1)+x(t);
  
tf = isequal(x,y);
tf

tf =

  logical

   0
   

Is there something obvious about Matlab I'm missing here?

Answer 1

[Stephen23]

You are overwriting x values with the loop, so when you then try the vectorized code you have different x values. Clearly different input x values will give different output x values.

It works perfectly for me, when I do NOT overwrite the x vector values:

>> x = randi(9,1,7)
x =
     1     8     9     7     7     7     4
>> for t = 2:length(x), y(t) = x(t-1)+x(t); end % assign to y, not x
>> y
y =
     0     9    17    16    14    14    11
>> t = 2:length(x);
>> z(t) = x(t-1)+x(t);                          % assign to z, not x
>> z
z =
     0     9    17    16    14    14    11

[Oliver Dechant]

Sure this is a fine answer, except that I need to overwrite the values. That behaviour is desired.

[Torsten]

The result is not surprising to me.

While the loop works sequentially and already takes into account previously changed elements in the x-vector, the vectorized code gives the result of the following:

xold = x;
for t=2:length(x)
  x(t) = xold(t-1) + xold(t);
end

[Stephen23]

"... except that I need to overwrite the values. That behaviour is desired."

As far as I can tell, all you need is cumsum:

>> x = [1,8,9,7,7,7,4];
>> for t = 2:length(x), x(t) = x(t-1)+x(t); end, x
x =
     1     9    18    25    32    39    43
>> x = [1,8,9,7,7,7,4];
>> x = cumsum(x)
x =
     1     9    18    25    32    39    43
     

And, in case you were wondering, vectorizing cumsum withthout the cumsum function is certainly possible, but it would not be particularly simple or particularly efficient. cumsum is the way to go (better than using a loop too).

[Torsten]

I think the given recursion was just an example. "cumsum" won't be applicable in a more general case, I guess.

[Jan]

@Torsten: If cumsum does not match the needs, filter is more powerful. E.g. an emultaion of cumsum:

x = randi(10, 1, 10)
a = cumsum(x)
b = filter(1, [1,-1], x)
isequal(a,b)