How to evaluate the clustering result when cluster numbers are not equal to data set class When apply clustering algorithm with the multi-class data set and class numbers are not equal to the result cluster numbers(For example , When we use K-Means algorithm by setting K = 3 apply with "yeast" data set that has 14 classes).

What is the proper way to evaluate the result in this situation ?
If no pair of your classes are overlapping and they do classify each and every element of the data set, then they constitute a partition of that set. So does your clustering.

In such a case, I would suggest to use a distance metric to compare two partitions, instead of any bona fide index, even if it's popular -unless that index happens to be a metric as well.

One such distances for comparing partitions is the $VI$ index, rediscovered by Maria Meila$\phantom{,}^\dagger$ a few years back, but its origin dating back to the 80's in a physics paper by Zurek$\phantom{,}^{\ddagger}$.

How to calculate distance between two partitions P and Q:


Determine $P\wedge Q, $the "intersection" (I always mix up the words join and meet here) of P and Q. This is simply the partition formed by taking the intersection of each cluster of P against each cluster of Q.
For all three partitions $P$, $Q$, $P\wedge Q$ calculate their entropy $H$ defined as follows: $$H(P)=-\frac{1}{N}\sum_{i=1}^Np_i\log p_i\;,\;p_i\equiv\frac{n_i}{N}$$ where $N$ is the total number of elements you clustered/classified and $n_i$ is the number of elements within cluster/class $i$ of a given partition $P$. 
The distance $d(P,Q)$ between partitions P and Q is then given by $$d(P,Q)=2H(P\wedge Q)-H(P)-H(Q)$$


The reason for using a real metric is simple: a metric satisfy the triangular inequality and our intuitions of "close" and "similar" heavily relies on that. Any ad-hoc measure of dis-similarity (distance) that doesn't satisfy the triangular inequality will eventually give you results like this: A is "$d$ away" from B (d being concrete number), with $d$ smaller than a given bound, say $D$. Then you find A close to C by an amount $d'&lt;D$. If this distance were not a metric, it could then happen that the distance $d''$ between B and C is say $2D$. For a sound metric distance, it should be 
$d''\leq d+d'&lt;2D$. In other words A would be "similar" to B and C, but the last two would be "different"/"not similar" from each other. This is counterintuitive.

The downside is that the VI distance may be computationally more expensive than calculating say a Rand index. The upside is that it's mathematically sound and makes sense.

You can use R to calculate the VI distance. You can also use Partanalyzer a program I wrote some time ago (https://github.com/MASantos/Partanalyzer). It needs a rewrite to make it more efficient, but it's open source, so you can modify it at will.

$\dagger$: Meila, M. Journal of Multivariate Analysis, 98, 873 (2007). 

$\ddagger$: W.H. Zurek, Nature, vol.341, p.119 (1989).