A=[{[-0.3368 -0.0329]} ...
{ [-0.5666 -1.2306 -1.9879 -1.6091 -1.0889 -0.6703]} {[-0.5950 -0.7264 -0.2151]} {[ -0.3346 -0.8525 -1.0663 -0.7172 -0.3690 ]} {[-0.3587 -0.2960]} ...
{[-1.0327 -2.4401 -3.1295 -2.8144 -2.3720 -1.5680 -0.6808]} {[-0.3728 -1.0863 -1.0865 -0.6916 -0.4153 -0.0177]} {[-0.3969 -1.2583 -1.9764 -2.7800 -2.4100 -1.6038 -0.4912]}];
whos A
Actually, it's 1x8 but who's counting??? :)
You were on the right track, but matching floating point with "==" is risky. Since you have a cell array with differently sized cells, rearranging into a regular 2D array and doing min by column doesn't work without a lot of additional effort.
So
minA=cellfun(@min,A) % go ahead and find the min of each
[minA,ixA]=min(minA) % now find the min of the min's and return where it is
minC=A{ixA} % then the desired cell is the one at that location.
NOTA BENE: If there might be a need for the other minima, you can "have your cake and eat it, too!"
[~,ixA]=min(minA) % now find the min of the min's and return ONLY where it is
The above would not overwrite the minA array, returning only the index of where the overall minimum is. (This must replace the above line, not follow it.)
