Extract substring in right format

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')

Answer 1

[DGM]

Why include the extra _?

str = 'Subject160/VSTMB_160_band_resample/data_22_right_average_210225_1827.mat'

str = 'Subject160/VSTMB_160_band_resample/data_22_right_average_210225_1827.mat'

EventName = regexp(str,'[1-9]+_right','match')

EventName = 1×1 cell array
    {'22_right'}

[DavidL88]

Thanks.