REPET find_peaks max_num bug

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Function, as written on the page: find_peaks(data, min_thr=0.5, min_dist=None, max_num=1)
Simple example I used:
repet1 = Repet(signal)
data = np.array([0,0,.2,.5,1,1,0,1,0,1,0])
peaks_indices = repet1.find_peaks(data)
output: peaks_indices = [[4]]
peaks_indices = repet1.find_peaks(data, max_num=2)
output: peaks_indices = [[4 7]]
peaks_indices = repet1.find_peaks(data, max_num=3)
output: peaks_indices = [[0 4 7]]
peaks_indices = repet1.find_peaks(data, max_num=4)
output: peaks_indices = [[0 0 4 7]]
peaks_indices = repet1.find_peaks(data, max_num=5)
output: peaks_indices = [[0 0 0 4 7]]
etc.

So when max_num is set to an integer value greater than the number of peaks in the data, it adds the index 0 to the output peaks_indices list. It actually keeps adding index 0 (it does this (max_num - number_of_peaks_found) times). Instead, you should probably list only the peak indices found in data, and if fewer than max_num are found, then that is fine, because it is a max_num, not a min_num or total_num.

So in the example above, peaks_indices = repet1.find_peaks(data, max_num=5) should give an output of peaks_indices = [[4 7]]. Even though max_num=5, the system only found 2 peaks, so it should return just those two rather than adding in an incorrect (and repeated) index of 0.