```Date: Feb 18, 2013 12:00 AM
Author: dpb
Subject: Re: doubts in matlab

On 2/17/2013 9:57 PM, Shobana wrote:> dpb <none@non.net> wrote in message <kfqpf8\$mhc\$1@speranza.aioe.org>...>> On 2/17/2013 5:21 AM, Shobana wrote:>> > Hai...>> > I am tried to find minimum value of particular columns and rows>> > separately but i am getting wrong answers.>> > for example a = [1 2 3 ;>> > 6 5 4;>> > 8 1 7]>> > [x y] = min(a(2:end,2))>> >>> > instead of getting x= 3 y =2 i am getting some other value.how to write>> > correct code for this concept.>> > Thanks in advance>>>> a(2:end,2) ==> [5;1] % 2nd to last rows, 2nd column>>>> so you should find [x,y] == [1,2]>>>> Not sure which you were really after that would return your expected>> results otomh I don't see one.>> sorry its not working.its showing me an errorWell, w/o seeing what you did and the error, can't comment...the above is commentary, not code.  The subset of a given by (2:end,2) is a column vector containing 5,1.  So, the min() of that vector is 1 and the second location.>> If you're trying to write a subset of the array and then work min() on>> the 2nd dimension, the ',2' is misplaced above (and you're missing the>> 2nd dimension for the subscript expression if you simply move it.>>>> Try something like>>>> [x ix] = min(a(2:end,:),2)...Oh, sorry...for the dimension optional argument, need an empty comparison field--the above compares the subarray to the constant 2, not what I was intending...[x ix] = min(a(2:end,:),[],2);doc min--
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