[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

*Subject*: Assignment, Form 5 (what they didn't tell you in the manuals)*From*: David Ritscher <davidNO.ritscherSPAM(at)bigfoot.com>*Date*: Wed, 18 Nov 1998 15:51:13 -0600*Newsgroups*: comp.lang.idl-pvwave*Organization*: University of Alabama at Birmingham*Xref*: news.doit.wisc.edu comp.lang.idl-pvwave:12857

Today's matrix subscripting puzzler: I'm noting that there are some extra rules govorning assignement statements where the right side contains subscripting expressions in more than one of the dimensions of a matrix. Given a 3-D array, array(indgen(5), indgen(5), indgen(5)) yields a 5-element vector array(indgen(5), 0, indgen(5)) yields a 5 x 1 x 5 matrix I'm trying to create an index that allows me to extract vectors of information from a matrix that has been stored. Here is a simple example of the form of what I'm trying to extract. The matrix 'index' will put the matrix into the order I want. threeD = indgen(3,4,5) twoD = indgen(3,5) index = [[indgen(15) mod 3], [indgen(15) / 3]] This 'index' allows me to make a vector out of the ascending elements of the twoD matrix: info, twoD(index(*,0), index(*,1)) <Expression> INT = Array(15) print, twoD(index(*,0), index(*,1)) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 How do I do the same thing, with the threeD matrix?: info, threeD(index(*,0), *, index(*,1)) <Expression> INT = Array(15, 4, 15) info, threeD(index(*,0), 0, index(*,1)) <Expression> INT = Array(15, 1, 15) if I want to get a vector that would extract the first element of the second dimension, as I had hoped the above would do, I can use: info, threeD(index(*,0), intarr(15), index(*,1)) print, threeD(index(*,0), intarr(15), index(*,1)) How do I do this, where I get a 2-D result where the second dimension of the incoming matrix becomes the second (and final) dimension of my outgoing matrix? Here is a solution, done in long-hand: print, [[ threeD(index(*,0), intarr(15), index(*,1))], $ [threeD(index(*,0), intarr(15)+1, index(*,1))], $ [threeD(index(*,0), intarr(15)+2, index(*,1))], $ [threeD(index(*,0), intarr(15)+3, index(*,1))] ] I can also do it with for-loops: out = lonarr(15, 4) for i=0,3 do out(0,i) = threeD(index(*,0), intarr(15)+i, index(*,1)) this does it, too: for i=0,3 do out(0,i) = (reform(threeD(*,i,*)))(index(*,0), index(*,1)) I won't go into the details, but my actual matrices contain information on every heart beat, stored over a 5-day period. I'm trying to string this information together, in a channel-by-channel order. Every now and then, with IDL and PV-Wave, one notes that you just can't quite get there from here... Does anyone see a true matrix solution to this? Thanks, David Ritscher -- Cardiac Rhythm Management Laboratory Department of Medicine University of Alabama at Birmingham B168 Volker Hall - 1670 University Boulevard Birmingham AL 35294-0019 Tel: (205) 975-2122 Fax: (205) 975-4720 Email: davidNO.ritscherSPAM@bigfoot.com

**Follow-Ups**:**Re: Assignment, Form 5 (what they didn't tell you in the manuals)***From:*Stein Vidar Hagfors Haugan

- Prev by Date:
**Re: RSI's priorities** - Next by Date:
**Re: RSI's Priorities (was: GUI Builder...)** - Prev by thread:
**Re: Dumb Dumb Question** - Next by thread:
**Re: Assignment, Form 5 (what they didn't tell you in the manuals)** - Index(es):