tag:blogger.com,1999:blog-6424211413492597169.post6842434651899737210..comments2024-03-02T07:11:56.376+01:00Comments on me nugget: Empirical Orthogonal Function (EOF) Analysis for gappy dataMarc in the boxhttp://www.blogger.com/profile/00459761376667614040noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6424211413492597169.post-51731477672548378172014-12-05T22:21:32.677+01:002014-12-05T22:21:32.677+01:00thanks for this information! thanks for this information! Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6424211413492597169.post-61463229737020373832013-04-23T00:11:33.226+02:002013-04-23T00:11:33.226+02:00can you also post your code for the figures? thank...can you also post your code for the figures? thanks!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6424211413492597169.post-56554583091732727902012-12-15T12:40:57.595+01:002012-12-15T12:40:57.595+01:00Thanks for you comment - what I mean by the "...Thanks for you comment - what I mean by the "leading modes" is just that these are the first ones - i.e. modes 1 & 2. The EOFs are ordered according to the amount of variance that they represent.<br />Marc in the boxhttps://www.blogger.com/profile/00459761376667614040noreply@blogger.comtag:blogger.com,1999:blog-6424211413492597169.post-80329585294200360132012-12-14T22:08:14.499+01:002012-12-14T22:08:14.499+01:00Great post - can you explain what th e "leadi...Great post - can you explain what th e "leading mode" is? I would like to create a similar map and not really sure what you plotted. It "leading mode" the mode of the EOF axis 1, and the mode of EOF axis 2? Thanks, RTAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6424211413492597169.post-67820879220137681462011-11-25T14:04:27.456+01:002011-11-25T14:04:27.456+01:00Hi Brewster,
Thanks for your feedback and complim...Hi Brewster,<br /><br />Thanks for your feedback and compliments regarding the graphics - I sometimes focus too much on making things look good...<br /><br />Regarding explained variance - I might be wrong about this, but I think I have come across references that sometimes present one or the other. I can't really say which is the better one to present, but SCF is basically just squaring the Lambda values before calculating each singular value's contribution to the sum. If I'm wrong about this, let me know.<br /><br />You're right about the calculation of Lambda error. I have just written it in a weird way to accommodate the irlba method - in that case, one is usually calculating a smaller subset of EOFs and thus the length of singular values is smaller than for a full decomposition. Thus the length of Lambda is no longer appropriate to determine the Lambda Errors. By taking the number of columns in your original matrix, you will arrive at the number of singular values in a full matrix decomposition. I haven't got to MCA yet, but in that case the number of singular values would be equal to the number of columns in the smaller of the two field.Marc in the boxhttps://www.blogger.com/profile/00459761376667614040noreply@blogger.comtag:blogger.com,1999:blog-6424211413492597169.post-65432272425796907522011-11-25T09:44:30.128+01:002011-11-25T09:44:30.128+01:00I forgot to mention that the package "clim.pa...I forgot to mention that the package "clim.pact" offers EOF and other related functions for the analysis of climate data. I'm not sure that it handles gappy data though.Marc in the boxhttps://www.blogger.com/profile/00459761376667614040noreply@blogger.comtag:blogger.com,1999:blog-6424211413492597169.post-61582901025524969512011-11-25T00:00:09.388+01:002011-11-25T00:00:09.388+01:00Funny, I was thinking of posting something like th...Funny, I was thinking of posting something like this myself... I really like the graphics you're using for this.<br /><br />I'm a bit curious about about your "mode's explained variance"... I'm working off the top of my head but shouldn't this be the squared singular value of the given mode (i.e. the mode's eigenvalue) divided by the the sum of the squared singular values (i.e. the sum of all modes' eigenvalues) I think you're calling this the SCF? and then you can find the delta lambda for your North test by simply multiplying this by the square root of 2 over the number of singular values...?<br /><br />In code, might it be most efficient to:<br /><br /> explained.var <- L$d^2 / sum(L$d^2)<br /><br /> # Confidence interval for your test based on North et al:<br /> delta.lambda <- (explained.var * sqrt(2/length(L$d)))<br /><br />You may be doing this...? I may not be following your code correctly.<br /><br />At any rate, thank you very much for posting this.Brewsterhttp://ikanb.wordpress.com/noreply@blogger.com