Fill cell in second column based on first column cell array

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

 採用された回答

Stephen23
Stephen23 2017 年 8 月 17 日
編集済み: Stephen23 2017 年 8 月 17 日

0 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));

1 件のコメント

Lauryn Burleigh
Lauryn Burleigh 2017 年 8 月 17 日
This is definitely easier than what I was trying to do. Thank you so much!!

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