4
$\begingroup$

For an assignment, I have used iPython to create the dendrogram below, using Ward's method and Euclidean distance, from the following data: $$a=(0,0)$$ $$b=(1,2)$$ $$c=(3,4)$$ $$d=(4,1)$$ $$e=(2,2)$$ enter image description here where dist({a},{b,e}) = 2.88, and dist({a,b,e},{c,d})=4.27.

How are these values derived? I've tried using the recursive method given here, but am not getting the same results.

Any help muchos appreciated.

I know that dist(a,b)=$\sqrt{5}$, dist(a,e)=$\sqrt{8}$, so using the formula on the wikipedia page, I have dist({a},{b,e})=$\frac{2}{3}\sqrt{5} + \frac{2}{3}\sqrt{8} - \frac{1}{3} = 3.04$. Where am I going wrong?

Improve this question
$\endgroup$
3
  • $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$
    – gung - Reinstate Monica
    Dec 15, 2016 at 14:26
  • $\begingroup$ done - btw this isn't a homework question, it's a question I've put together myself for a quiz $\endgroup$
    – Harr
    Dec 15, 2016 at 14:35
  • $\begingroup$ The Talk page at the wikipedia article discusses all the necessary details. Please see also in stats.stackexchange.com/a/217742/3277. And remember Ward's method operates on squared distances. $\endgroup$
    – ttnphns
    Dec 15, 2016 at 15:17

2 Answers 2

Reset to default
3
$\begingroup$

Having banged my head on the wall for the last 2 hours on this, I feel your pain.

The result is the square root of the increase in within-cluster sum of squares (vs. cluster means), multiplied by $\sqrt{2}$ for some reason.

$$ \sqrt{ 2 \left( \sum_{i \in (C_1 \cup C_2)} \lVert \vec x_i - \bar x_{C_1 \cup C_2} \rVert^2 - \sum_{i \in (C_1)} \lVert \vec x_i - \bar x_{C_1} \rVert^2 - \sum_{i \in (C_2)} \lVert \vec x_i - \bar x_{C_2} \rVert^2 \right) } $$

For $C_1 = \{ a, b, e \}, C_2 = \{ c,d \}$ the result is $4.27$ as you require.

As for how exactly they are derived, depends on the specifics of the algorithm in use.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Michael, the Talk page of Wikipedia's Ward method article explains that sqrt(2) $\endgroup$
    – ttnphns
    Dec 19, 2018 at 16:05
0
$\begingroup$

The right is: $$dist(\{a\},\{b,e\})=\left(\frac23\right)\cdot 5+\left(\frac23\right)\cdot 8–\frac13\cdot 1=\frac{25}3$$ Then $$dist(\{a\},\{b,e\})=\sqrt{\frac{25}3}=2.88$$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.