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I would like to construct a measure to calculate the sparseness of a vector of length $k$.

Let $X = [x_i]$ be a vector of length $k$ such that there exist an $x_i \neq 0$ . Assume $x_i \geq 0$ for all $i$.

One such measure I came across is defined as $$\frac{\sqrt{k} - \frac{\|X\|_1}{{\|X\|_2}}} {\sqrt{k} -1}\;,$$ where $\|X\|_1$ is $L_1$ norm and $\|X\|_2$ is $L_2$ norm.

Here, $\operatorname{Sparseness}(X) = 0$ whenever the vector is dense (all components are equal and non-zero) and $\operatorname{Sparseness}(X) = 1$ whenever the vector is sparse (only one component is non zero).

This post only explains the when $0$ and $1$ achieved by the above mentioned measure.

Is there any other function defining the sparseness of the vector.

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Isn't your sparseness function 1 for a sparse vector and 0 for a dense vector? –  Christian Rau Mar 8 '12 at 10:51
    
It doesnt matter.we can achieve that by defining the measure as 1-Sparseness($X$). –  Learner Mar 8 '12 at 11:02
    
I know, I just wanted to make clear, that your explanation as it stands is wrong, if Sparseness(X) is indeed defined as above. –  Christian Rau Mar 8 '12 at 11:20
    
Yeah. I should have used different name. –  Learner Mar 8 '12 at 11:24
    
Sorry if my comments are a bit confusing. It isn't the name sparseness that bothers me, it's the hard fact, that your above function (the one with the $\sqrt{k}$) is 1 for a sparse vector and 0 for a dense vector (no matter how you name it). –  Christian Rau Mar 8 '12 at 11:28
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3 Answers

up vote 4 down vote accepted
+50

You could of course generalize your current measure

\begin{align} S(X) = \frac{\frac{k^{(1/m)}}{k^{(1/n)}} -\frac{\|X\|_m}{\|X\|_n} } {\frac{k^{(1/m)}}{k^{(1/n)}}-1} \end{align}

while preserving your properties you specified.

An interesting special case could be $m = 1, n \to \infty$, in which case the expression simplifies to

\begin{equation} S(X) = \frac{k-\frac{\|X\|_1}{\|X\|_c}}{k-1} \end{equation}

where $c = \infty$, (for some reason, mathjax refused to render when I inserted $\infty$ directly in the fraction)

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Good that you generalized it. I never thought in those lines. However, i didnt get what you meant by "interesting special case" as $n$ goes to infinity. Can you please elaborate more on that. –  Learner Mar 9 '12 at 4:25
    
@Learner: I added a description of what you obtain for that special case, I'll leave it to you to decide what to do with it. –  Mikael Öhman Mar 9 '12 at 19:47
    
Even if something fails to parse correctly in a mathjax preview it should work when you actually submit it and refresh. –  anon Mar 12 '12 at 11:22
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There is a definition of sparsity, which is used (amongst others) in the compressed sensing literature, see e.g. here.

A vector $x\in \mathbb{C}^k$ is called $s$-sparse, if $|| x ||_0 = |\text{supp}(x)| \leq s$, that is, it has at most $s$ non-zero entries. Denote by $\Sigma_s$ the set of all such vectors. Then, the $s$-term approximation error of a vector $x\in \mathbb{C}^k$ is defined as $$ \sigma_s(x)_p = \min_{y\in\Sigma_s} ||x-y||_p. $$

Now this quantity equals $0$, if your vector $x$ is $s$-sparse, and will be greater than $0$ otherwise. Note that you now have two parameters $s$ and $p$ to tune this "measure". Clearly, you get your definition of sparsity if you set $s=1$.

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Disclaimer: This post considers the case in which you do not mind some computational effort to get nice sparseness value. For something new, please skip to part 2.

Part 1

I agree with Mikael, that kind of generalization is nice, what's more, with Mikael's formula it is intuitive from where it came from: the most basic notion for vector sparseness would be $$\frac{\text{number of indices $k$ such that }X_k = 0}{\text{total number of indices}}.$$

However, by this definition $\langle 0, 0, \ldots, 0\rangle$ is sparse, but vector $\langle c, c, \ldots, c \rangle$ is not. Still, it is easy to fix it: $$\frac{\text{number of indices on which }X_k - c = 0}{\text{total number of indices}}\,,$$ where $c$ is some average of $X$, e.g. $c = \|X\|_\infty$. The problem with this measure is that it is not easy to count the number of indices. To alleviate for that, we could approximate the number of indices by $\frac{\|X\|_1}{\|X\|_\infty}$. Naturally we need some normalization, and by that we arrive at Mikael's special case: $$\frac{k-\frac{\|X\|_1}{\|X\|_\infty}}{k-1}.$$

But the average we took as an example $x = \|X\|_\infty$ isn't the only one. Similarly we could approximate the number of indices in different fashions: $\frac{\|X\|_m}{\|X\|_n}$ would do for any $m < n$, and the normalization is just $$\frac{\frac{\|C\|_m}{\|C\|_n}-\frac{\|X\|_1}{\|X\|_\infty}}{R_\max-R_\min}, R_\max = \frac{\|C\|_m}{\|C\|_n}, R_\min = \frac{\|D\|_m}{\|D\|_n},$$ where $C = \langle c, c, \ldots, c \rangle$ and $D = \langle c, 0, 0, \ldots, 0 \rangle$ for any $c \neq 0$.

Part 2

Still, this measure is rather weird, because intuitively it more depends on the sizes of the values, than how many different numbers there are. I am not sure that this is a property we would want to have. There is a different measure that can take that into account, i.e. measure based on entropy. Interpreting $X$ as the samples, one can calculate $$ -\sum_i P(X = i) \log_k P(X = i) .$$

To soften this a bit just pick any distribution you want (best specific to your application), e.g. $F_\mu = N(\mu, \sigma^{2})$, set $$F = \frac{1}{k}\sum_i F_{X_i},$$ and then calculate differential entropy ($f$ is the density function of $F$): $$-\int_\mathbb{R} f(x) \ln f(x) \ dx$$ or even better relative entropy if you have some reference measure (e.g. the very same $F_\mu$ adjusted a bit might do the trick). Of course, all this has to be scaled to $[0,1]$ what makes the formulas even nastier, however, in my opinion, it catches the notion of sparseness pretty good. Finally, you can combine the two approaches in infinitely many ways, to get even more sparseness models!

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