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| {EOF}_1(bold x) {PC}_1(t) + {EOF}_2(bold x) {PC}_2(t) + {EOF}_3(bold x) {PC}_3(t) + ... = mydata(x,t) |
- How many values (numbers) are in your input data array? 144x240 = 34560
- How many values (numbers) are needed to build each term on the left? Each term on the left of the above equation represents a SPACE (EOF) and TIME (PC) component. So, each term requires 144 (space or longitude values) plus 240 (time or monthly values) making a total of 384 values.
- If 5 EOFs capture most of the data's variance, how much smaller (in the above sense) is the EOFxPC representation compared to the full data set? 5 EOFs require 384 values X 5 terms = 1920 values, which in this case is 18 times smaller than the original dataset.
