Having trouble getting this find work properly

I have two cell arrays where the first one is time converted into serial date and the second cell array (bb) which has specific times associated with other data in it. I'm trying to extract the rows in bb in chronological order by using time cell array. BB cell array is a 1x34 cell array where each cell once expanded is a nx7 array.

I am rather new at this, but I came up with this, and I can't get this working correctly as well as do what I want it to do. Not even sure if it is in the right direction.

I'm trying to find the serial date in the test1 file in all columns of bb (1x34) and extract it into another matrix when I call the first row of time (test1). BB's {1x34}(nx7) the column 7 of the nx7 is the serial date for that BB.

for i=1:length(test1)
    for t=1:length(bb)
        for w=1:length(bb{t})
            [a{t} b(w,1:7)]=find(bb{t}(w,1:7)==test1(i)
        end
    end
end

Answer 1

[KSSV]

don't use find, check with ismemebr, ismemebrtol.

[liu James]

Hey KSSV I've tried it with ismember and it's still not correct.

for i=1:length(test1)
      for t=1:length(bb)
          for w=1:length(bb{t})
              [a b]=ismember(bb{t}(w),test1(i));
          end
      end
  end

[KSSV]

clc; clear
load bb.mat ;
load test1.mat ;
for i=1:length(test1)
    for t=1:length(bb)
        for w=1:length(bb{t})
            [aa1, bb1]=find(bb{t}(w,:)==test1(i)) 
            if ~isempty(aa1)
                a{t} = aa1 ; b(w,1:7) = bb1 ;
            end
        end
    end
end

Takes hell lot of time and no values I can find common.

[liu James]

Shouldn't be. Under Test1 row 589 for example provides time series number 735028, while bb{1}{1,7} is also 735028.

[Steven Lord]

If that's the case, test1(589)-bb{1}{1,7} should be exactly equal to 0. From what you've said it will not be exactly 0 but will be a nonzero value with very small magnitude.

The == operator performs exact, down-to-the-last-bit equality testing. This means that two numbers that look the same may not be the same.

format short
x = 1.25
y = x+eps(x)
areTheyEqual = x == y % false
howDifferentAreThey = y-x % small

If you want to confirm that x and y are not the same, display their hex pattern.

format hex
x
y
format % reset the display format to default

[liu James]

If I'm understood what you said, the find will not extract it because of the extra decimals even though they are 0 is that correct? However, if I type this

[a b]=find(test1(589)==bb{1}(1,7))

a =

            1.00

b =

            1.00

it will show that there is a position vs empty matrix.

[a b]=find(test1(590)==bb{1}(1,7))

a =

       []

b =

       []

[KSSV]

You fix a small parameter eps and subtract them.

eps = 10^-6 ;
idx = test1(589)-bb{1}(1,:)<eps

so in idx if any value is 1, then you can say they are equal. Using == for flottant numbers is not advisable. Also, I suggested you to use ismembertol.