For the purposes of this question, assume we have 10-dimensional data - that is, ignore the Overall column.
A)
Explain why we need to scale this data set before performing PCA.
The 10 events are measured in different units and on very different scales, from times like the 100m and 1500m to distances like the high jump and javelin. PCA maximizes explained variance, so without scaling, the events with large values and spreads would dominate the PCs because of their units. Scaling gives every event mean 0 and SD 1, so each one contributes equally.
B)
Use svd() to find the first 2 principal component scores and their loadings. Full credit will only be granted if you use the svd() ingredients u, d, and v. What percent of the overall variability do the first two PCs explain?
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48.415% is the overall variabliity that the first two PCS explain.
C)
Find and print the loadings. Based on the loadings alone, if the first two PCs are plotted in a 2D plane as shown below, which of the four quadrants will the medalists be in? Explain your reasoning.
Quadrant II because PC1s loadings are running events and field events are opposite, and weighting the loadings this way gives −2.10, so better athletes got lower PC1 scores. The same weighting gives +1.05 for PC2, so better performance makes PC2 higher, putting the medalists at negative PC1 and positive PC2.
D)
Add the PCs to the decathlon data set and create a scatterplot of these PCs, with the points labeled by the athletes’ names. Color-code the points on whether or not the athlete was a medalist. Use the ggrepel package for better labeling. Verify that your intuition from C) is correct.
library(ggrepel)# Name column = first text column in the dataname_col <-names(decathlon)[sapply(decathlon, is.character)][1]decathlon <- decathlon %>%mutate(PC1 = first2scores[, 1],PC2 = first2scores[, 2],Medalist =if_else(rank(-Overall, ties.method ="min") <=3,"Medalist", "Non-Medalist"))pc_plot <-ggplot(decathlon, aes(x = PC1, y = PC2)) +geom_hline(yintercept =0, color ="gray60") +geom_vline(xintercept =0, color ="gray60") +geom_point(aes(color = Medalist), size =3) +geom_text_repel(aes(label = .data[[name_col]], color = Medalist), size =3, max.overlaps =20, show.legend =FALSE) +scale_color_manual(values =c("Medalist"="darkgoldenrod2","Non-Medalist"="gray30")) +labs(x ="PC1", y ="PC2", color =NULL,title ="Medalists Vs. Non-Medalists, 2024 Olympic Decathlon") +theme_classic(base_size =14)pc_plot
E)
Canadian Damian Warner won the gold medal in the decathlon in the 2020 Tokyo games. He began the 2024 decathlon but bowed out after three straight missed pole vault attempts.
Would this have won a medal if it had happened in 2024? To answer this, we will compute his PCs with respect to the 2024 athletes and add it to the plot to see where his 2020 gold-medal performance compares to the 2024 athletes. To do this:
Find the mean vector from the 2024 athletes. Call it mean_vec_24.
Find the standard deviation vector from the 2024 athletes. Call it sd_vec_24.
Standardize Warner’s 2020 results with respect to the 2024 athletes: (warner-mean_vec_24)/sd_vec_24
Find Warner’s PC coordinates using the 2024 loadings.
Add his point to the scatterplot.
Do you think his 2020 performance would have won a medal if it had happened in 2024?
warner <-c(10.12, 8.24, 14.8, 2.02, 47.48, 13.46, 48.67, 4.9, 63.44, 271.08)# Mean and SD of the 2024 athletes (saved by scale() in 1B)mean_vec_24 <-attr(decathlon_scaled, "scaled:center")sd_vec_24 <-attr(decathlon_scaled, "scaled:scale")warner_scaled <- (warner - mean_vec_24) / sd_vec_24# Warner's PCs using the 2024 loadingswarner_PCs <- warner_scaled %*% V[, 1:2]warner_PCs
Each data set has an X and a Y column which represent 2-dimensional variables that we need to rotate.
A)
Scale each data set and plot them side-by-side using the patchwork package. Make sure the aspect ratio of each graph is 1 (i.e., make the height and width of each graph equal). At this point, explain why you think I was skeptical. Specifically, do you think the percent variability explained by the first PC of each data set appears to exceed or fall short of the variability I asked it to?
I think you were skeptical because all three scaled data sets look almost the same. Very thin, tight lines. That shape means PC1 explains nearly all of the variability in each data set.
B)
Use SVD to find the first PC for each data set, and find the actual percent of total variability explained by each PC using aggregation methods.
svdA <-svd(claudeA_scaled)svdB <-svd(claudeB_scaled)svdC <-svd(claudeC_scaled)# PC scores = U D (first column = first PC)PCs_A <- svdA$u %*%diag(svdA$d)PCs_B <- svdB$u %*%diag(svdB$d)PCs_C <- svdC$u %*%diag(svdC$d)head(PCs_A[, 1]); head(PCs_B[, 1]); head(PCs_C[, 1])
