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dpb
Posts:
8,424
Registered:
6/7/07


Re: doubts in matlab
Posted:
Feb 18, 2013 12:00 AM


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 fieldthe above compares the subarray to the constant 2, not what I was intending...
[x ix] = min(a(2:end,:),[],2);
doc min




