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Re: comparing and filling matrices
Posted:
Jul 26, 2013 2:52 PM


dpb <none@non.net> wrote in message <ksufrj$5ll$1@speranza.aioe.org>... > On 7/26/2013 1:01 PM, Jonathan wrote: > > dpb <none@non.net> wrote in message <ksubob$pnu$1@speranza.aioe.org>... > >> On 7/26/2013 12:13 PM, Jonathan wrote: > >> ... > >> > >> > I am trying to fill a matrix (CompletDat: > >> > with data from a second matrix (RealDat: > >> > (CompletDat) contains discrete dates in the first > >> > column, which are repeated every 15 rows. A unique id is located in the > >> > second column, ... repeated for 15 rows, and then increments to the > >> > next value and so on. ... > >> > The second matrix has the same format except there are dates that are > >> > missing. Hence, the overall goal is to populate the CompletDat matrix > >> > with data from the other matrix, placing them in their correct row > >> based on the same date and unique id. > >> > ... > > >> A short example showing the problem and desired solution would likely > >> help...probably only needs a couple lines of a couple combinations to > >> be sufficient. > >> > ... > > > The dates are identical. Hadn't thought about ismember. I just tried it > > and it's correctly identifying the areas where dates are missing. The > > next step is how to retrieve the matching data from the second array, > > which is the area I have been struggling with. Using interp etc is not > > an option. > > > > Here's a sample of the data... > > > > RealDat = [13*4] > > 1997.66805 2583 21 0 84.85 > > 1998.24915 2583 21 0 80.02 > > 2002.66805 2583 21 0 90.03 > > 2007.741973 2583 21 0 85.25 > > 2008.254626 2583 21 0 79.25 > > 2008.755663 2583 21 0 83.24 > ... > > C =[ > 1997.66805 2583 NaN NaN > 1998.24915 2583 NaN NaN > 2002.66805 2583 NaN NaN > 2007.741973 2583 NaN NaN > 2007.966482 2583 NaN NaN > ... > > Well, first have to augment the C array to full rank of R or would have > to switch to cell array to handle jagged array. > > C=[C C(:,3)]; > > Then, simply use the ismember() logical array... > > >> C(ismember(C(:,1:2),R(:,1:2),'rows'),3:end)=R(:,3:end) > C = > 1.0e+03 * > 1.9977 2.5830 0.0210 0 0.0848 > 1.9982 2.5830 0.0210 0 0.0800 > 2.0027 2.5830 0.0210 0 0.0900 > 2.0077 2.5830 0.0210 0 0.0853 > 2.0080 2.5830 NaN NaN NaN > 2.0083 2.5830 0.0210 0 0.0793 > 2.0088 2.5830 0.0210 0 0.0832 > 2.0092 2.5830 0.0210 0 0.0798 > 2.0097 2.5830 0.0210 0 0.0967 > 2.0100 2.5830 NaN NaN NaN > 2.0101 2.5830 0.0210 0 0.0764 > 2.0108 2.5830 0.0210 0 0.0972 > 2.0112 2.5830 0.0210 0 0.0972 > 2.0128 2.5830 0.0210 0 0.1018 > 2.0130 2.5830 0.0210 0 0.0922 > >> > >  > Thank you, this worked perfectly. Makes a mockery of my loop attempt.
Cheers Jon



