Extract substring in right format
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How can I get the substring 22_right from the string below? The code below give me _22_right. I need it without the first '_'
str = 'Subject160/VSTMB_160_band_resample/data_22_right_average_210225_1827.mat'
EventName = regexp(str,'[1-9_]+_right','match')
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Why include the extra _?
str = 'Subject160/VSTMB_160_band_resample/data_22_right_average_210225_1827.mat'
EventName = regexp(str,'[1-9]+_right','match')
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