On Tue, Mar 26, 2013 at 10:25 AM, Rashad M <[email protected]> wrote: > > > > On Tue, Mar 26, 2013 at 2:44 PM, Markus Metz <[email protected]> > wrote: >> >> On Tue, Mar 26, 2013 at 8:39 AM, Rashad M <[email protected]> >> wrote: >> > Hi, >> > >> > i.pca creates 6 raster maps for 6 input raster >> > >> > eg: >> > >> > i.pca >> > >> > in=lsat7_2002_10,lsat7_2002_20,lsat7_2002_30,lsat7_2002_40,lsat7_2002_50,lsat7_2002_70 >> > \ >> > out=lsat7_2002_pca >> > >> > >> > I have a doubt here >> > >> > Does the output lsat7_2002_pca.1 is the corresponding map for >> > lsat7_2002_10? >> >> No. lsat7_2002_pca.1 contains the principal component with the highest >> eigenvalue and is calculated using all input bands. More generally, >> each lsat7_2002_pca.* map is calculated using all input bands. Look at >> the eigenvectors to find out the weights with which each input map >> contributes to a specific PCA. >> > > So by knowing a pixel value of lsat7_2002_pca.1 can I get the value before > transformation in PCA. > consider this: > I have two bands B1,B2 and after i.pca I get PC1,PC2. Then is there a way to > derive value of X,Y where X belongs to B1 and Y belongs to B2 by only > knowing P,Q > where P belongs to PC1 and Q belongs to PC2 > > and also X,Y and P,Q are of same index (row,col)
X = P * eigenvector(P, X) + P * eigenvector(Q, X) + mean(B1) Using the example from the manual: X = 0.2824 * PC1 + 0.2541 * PC2 + 0.3801 * PC3 + 0.1752 * PC4 - 0.6170 * PC5 - 0.5475 * PC6 + mean(band 1) Markus M > > I hope this is clear > >> >> Markus M >> >> > >> > If so the lsat7_2002_10_pca.1 contains PCA transformed pixel for >> > lsat7_2002_10 >> > >> > Is this correct? >> > >> > -- >> > Regards, >> > Rashad >> > >> > _______________________________________________ >> > grass-dev mailing list >> > [email protected] >> > http://lists.osgeo.org/mailman/listinfo/grass-dev > > > > > -- > Regards, > Rashad _______________________________________________ grass-dev mailing list [email protected] http://lists.osgeo.org/mailman/listinfo/grass-dev
