问题描述:

I have a vector `v`

of size `1 x 5`

, and a diagonal matrix `D`

of size `5 x 5`

. In my example, I have `v = [0 1 2 3 4].`

**First question:** I want to put the vector v on the diagonal of `D`

, so that `D(1,1) = 0`

, `D(2,2) = 1`

, `D(3,3) = 2`

, `D(4,4) = 3`

and `D(5,5) = 4`

.

I wrote a matlab code for this but I am sure that there is another automatic method much less expensive in computation.

So what I wrote (and to be optimized from you) is the following:

`ii = 1;`

for a = 1 : size(D,1)

for b = 1 : size(D,2)

if(a == b)

D(a,b) = v(1, ii);

ii = ii + 1;

end

end

end

**Second Question:** After finishing the first question, I need now to check if the diagonal values of D are equal to zero. If I can find (in an automatic way) a value on the diagonal of D is equal to zero, so replace it by 0.001.

In fact this can be written in a lot of manners, for example:

`for a = 1 : size(D,1)`

for b = 1 : size(D,2)

if(a == b)

if(D(a,b) == 0)

D(a,b) = 0.001;

end

end

end

end

But I need another automatic method which can be very cheap in terms of computations.

Any help will be very appreciated!

You're making your life harder than necessary: you don't need two loops (one would be enough), since you only need to walk along the diagonal.
You actually don't need any loop, here; only the `diag`

function.

Anyway, it may be easier to replace zero entries in a copy of `v`

, and then use it to build the diagonal matrix:

```
v2 = v; % create a copy of v (if needed)
v2(v2 == 0) = 1e-3; % replace zero entries
```

**Note**: If the entries you want to replace are not exactly zero, but small floating-point values, you'll need something like this instead:

```
v2(abs(v2) < epsilon) = 1e-3; % replace zero entries
```

where `epsilon`

is the threshold below which, as far as you're concerned, the entry is close enough to zero that it can be neglected.

Then, since you write that `D`

already is a diagonal matrix, you can simply overwrite it:

```
D = diag(v2);
```

I don't have matlab on-hand right now, but I'd use a combination of the following functions: - diag - eye - logical

An equivalent question was answered here: How to assign values on the diagonal?

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