Fill cell in second column based on first column cell array

2017 8 17
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更新时间：2017 8 17
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I have a cell array of image names in the first column ('9_X_0_a.bmp', '9_X_0_b.bmp','19_X_0_a.bmp', '19_X_0_b.bmp', etc.). I would like to fill the second column with either '1' or '2' based on if the image names contain 'a' or 'b'.
I currently have the code set to run as an 'if' statement which works but only for one index. In other words, it will place the proper number in the second column for either 'a' or 'b' images, but not for both. Any advice?
for i = length(listOfFiles1);
indexa = find(contains(listOfFiles1, 'a'));
indexb = find(contains(listOfFiles1, 'b.bmp'));
if i == indexa
listOfFiles1{indexa,2} = 1;
elseif i == indexb
listOfFiles1{indexb,2} = 2;
end
end

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 采纳的回答

Stephen23
Stephen23 2017-8-17
编辑：Stephen23 2017-8-17

0 个投票

Method one: regexp and convert using subtraction:
>> C = {'9_X_0_a.bmp';'9_X_0_b.bmp';'19_X_0_a.bmp';'19_X_0_b.bmp'};
>> D = regexp(C,'[ab](?=\.bmp)','match','once');
>> C(:,2) = num2cell([D{:}]-'a'+1)
C =
'9_X_0_a.bmp' 1
'9_X_0_b.bmp' 2
'19_X_0_a.bmp' 1
'19_X_0_b.bmp' 2
Method two: regexprep:
>> C = {'9_X_0_a.bmp';'9_X_0_b.bmp';'19_X_0_a.bmp';'19_X_0_b.bmp'};
>> C(:,2) = regexprep(C,{'.*a\.bmp$','.*b\.bmp'},{'1','2'}) % as string
C =
'9_X_0_a.bmp' '1'
'9_X_0_b.bmp' '2'
'19_X_0_a.bmp' '1'
'19_X_0_b.bmp' '2'
or as double:
D = regexprep(C,{'.*a\.bmp$','.*b\.bmp'},{'1','2'});
C(:,2) = num2cell(str2double(D));

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