1
$\begingroup$

I am interested in clustering daily gridded data.

Because of the many dimensions (gridpoints), I first perform PCA to reduce the dimensionality and keep the n-first PCs that account for at least 85% of the variation of the actual data. Then I use these n PCs as inputs to k-Means clustering.

My question is if I should use the standardized (mean=0, std=1) PCs as inputs to the k_Means clustering, or adjust the PCs based on the % of total variation that each PC accounts for. The adjustment can be something like PC[i] * Variation[i] /sum(Variation)

The 1st option results in PCs with the same variance, thus the clustering is unbiased. The 2nd option will end up with a bias towards the PCs that explain most of the variation.

Based on my understanding it is better to work with unbiased data. Nevertheless, in the case of PCA, the PCs are by default of varying importance. Would this support the use of the adjusted PCs?

$\endgroup$
1
$\begingroup$

I had a more careful check, and I will try to answer my question based on the additional insights. I would appreciate further answers/comments regarding the accuracy of the provided answer.

The actual results of the PCs are not standardized, and their variance is proportional to the square root of the total variance (variation) explained by each PC. These values are the ones that should be used for the subsequent clustering. In this way, the weighting of the PCs is indeed proportional to the (square root of the) explained variance.

If the standardized PCs (mean=0, std=1) are used, then the values should be adjusted with the square root of the % of variance explained by each PC. In this way, the adjusted PCs will have the same weighting as the actual PCs, and the only difference will be on the range of the values.

I run the analysis for clustering based on 1) Actual PCs, 2) Standardized PCs, 3) Standardized PCs adjusted with the % of the variance, and 4) Standardized PCs adjusted with the square root of the % of the variance. 1) and 4) were practically equal (I assume some small differences are due to the fact that the range of the values affects the instances very close to 0) and the best ones, followed by 3). Option 2) was by far giving the worst performance. The performance was tested by the Sum of Squared Distances between each field and the allocated centroid on the original dataset (daily gridded data).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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