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 error

Well, 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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