Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


dpb
Posts:
9,822
Registered:
6/7/07


Re: Removing a location from a vector.
Posted:
Mar 24, 2013 9:50 PM


On 3/24/2013 3:50 PM, Skylar wrote: > I am making comparisons between three different vectors, testing to see > if the data at these given locations is the same. > > Here is the program so far: > > Tot_Press = find(suture2(:,3)<60suture2(:,3)>70); %This finds which ...
> > Tot_Temp = find(suture2(:,2)<150suture2(:,2)>170); %This finds which ...
> > Tot_DT = find(suture2(:,4)<2.0suture2(:,4)>2.5); %This finds which of ...
> > for i=0:26 > if (Tot_Press(i) == Tot_Temp(i))  (Tot_Press(i) == Tot_DT(i)) > else > if (Tot_Temp(i) == Tot_DT(i)) > end
Not sure what you want as end result but several problems herefirst, Matlab arrays are _always_1based so the above loop will fail on the first iteration w/ index out of bounds.
Next, the result of each separate find() may well be a different lengthit would seem only by rare happenstance that wouldn't be so. So, you would need to account for that. > > As you see I have made the comparisons, but I am curious as to if there > is a function to delete certain locations from the vectors without > having to create another vector and store all data that is not equal > into it.
How about using the helper function I posted recently...
function flg=iswithin(x,lo,hi) % Logical T if x between lo and hi, inclusive flg = (x>=lo) & (x<=hi);
ixPr = ~iswithin(suture2(:,3),60,70); ixT = ~iswithin(suture2(:,2),150,170); ixDT = ~iswithin(suture2(:,4),2.0,2.5);
Now you have three logical vectors of the same length as the original array but with T (ones) where the condition in each is so. NB the ~ to negate the inclusive within test to get the opposite of your test.
With that you can then find which are true collectively by simply writing the expression
ixAll = all([ixPr ixT ixDT],2);
which will be a logical (column) vector of the same length but T only where each of the three conditions you've tested for is simultaneously T.
Again, not sure where you're headed, but hopefully that gives some ideas. Note that to get the values you then simply can write expressions like
S=suture2(ixAll,:);
to get the full dataset (all columns) for the rows found.




