remove elements from cell

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skysky2000
skysky2000 2017-7-12
评论： skysky2000 2017-7-19
Dear all, I know, I've asked alot and I really appreciate your help. I have a cell array:
a = {[1 2 3 55 77 8] [3 77] [ 1 2 5 7 8 9 0 2] [ 3 44 6]}
and I would like to remove the numbers before the index based on vector b=[3 9 6]. The expected answer should be:
cell2 = {[55 77 8] [77] [0 2] []}
Thanks.
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Jan
Jan 2017-7-13
编辑：Jan 2017-7-13
It is very welcome if you ask many questions here. But it would be appreciated, if you explain the problem as detailed as needed to define it uniquely. The explanation "remove the numbers before the index based on vector b=[3 9 6]" is not clear. Then showing any own effort and asking a specific question concerning your code would be useful also, for you and for the readers.

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回答（2 个）

Image Analyst
Image Analyst 2017-7-12
This will do it:
a = {[1 2 3 55 77 8] [3 77] [ 1 2 5 7 8 9 0 2] [ 3 44 6]}
b=[3 9 6]
% cell2 = {[55 77 8] [77] [0 2] []}
for k = 1 : length(a)
thisArray = a{k}; % Extract array from cell.
[inA, inB] = ismember(b, thisArray); % Find where any of b occurs in thisArray.
index = find(inB>0, 1, 'last'); % Find last occurrence.
outputArray = thisArray(inB(index)+1:end); % Extract out to new array.
cell2{k} = outputArray; % Stuff into a new cell array.
end
celldisp(cell2); % Display in command window.
It's easy and intuitive, though I'm sure someone will give you a cryptic one-liner shortly.
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dbmn
dbmn 2017-7-13
skysky could you tell us what didnt work in your case?
image_analyst: one liner coming up
cell2 = cellfun(@(x) x(max([find(ismember(x, b),1, 'last'), 0])+1:end), a, 'uni', 0);
p.s. when you use max([index, 0]) you can get rid of the empty matrices :)
skysky2000
skysky2000 2017-7-19
thanks alot dbmn for your help

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skysky2000
skysky2000 2017-7-13
any help please?
  2 个评论
Image Analyst
Image Analyst 2017-7-13
I gave you help. My code produced exactly the cell array that you said you wanted. What's the problem? Prove to me that it does not give you what you asked for, because I don't know what else to do.
Jan
Jan 2017-7-13
@skysky2000: Please do not bump your thread by posting a pseudo-answer. The question is not really clear, but Image Analysts has posted a method to produce the requested answer. Now it is your turn to explain, which details is not solved yet.

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