remove row with matching string
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I'm trying to eliminate an entire row of data if a string is found in the cell array.
Sample array values:
myarray =
[ 16] '10/05/2011' '11:20' 'waiting' 'nopos' [ 10000]
[ 17] '10/05/2011' '11:25' 'open' 'long' [ 0]
[ 34] '10/05/2011' '12:50' 'open' 'short' [ 0]
so i'd like to have:
my array =
[ 17] '10/05/2011' '11:25' 'open' 'long' [ 0]
[ 34] '10/05/2011' '12:50' 'open' 'short' [ 0]
there can be several occurances in the array, I'm not looking to pop the top row...
you help is greatly appreciated
1 件のコメント
Image Analyst
2012 年 4 月 5 日
So which element in the cell array are you checking for the string? In column 4 or column 5, or something else.
回答 (2 件)
Richard Brown
2012 年 4 月 5 日
If it can occur anywhere, use cellfun to test for equality:
[I, ~] = find(cellfun(@(s) isequal(s, 'waiting'), myarray));
myarray(I, :) = [];
otherwise, you can pull out the column of interest and use ismember
I = ismember(myarray(:, 4), 'waiting')
myarray(I, :) = [];
4 件のコメント
Richard Brown
2012 年 4 月 5 日
And yes, just checked - for a cell array vector of 100000 strings, on my computer strcmp takes 0.01 seconds, whereas ismember takes 0.03 :)
Jan
2012 年 4 月 5 日
0.02 seconds lost. But if you do this 10e6 times... :-)
I've overseen, that your FIND command operates on a matrix. Then the 2nd omitted output does matter, as you have written.
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