vocalpy.metrics.segmentation.ir.concat_starts_and_stops

vocalpy.metrics.segmentation.ir.concat_starts_and_stops(starts: ndarray[Any, dtype[_ScalarType_co]], stops: ndarray[Any, dtype[_ScalarType_co]]) → ndarray[Any, dtype[_ScalarType_co]]

Concatenate arrays of start and stop times into a single array of boundary times.

Some segmenting algorithms return lists of segments denoted by the start and stop times of each segment. (You may also see these times called “onsets” and “offsets”.) Typically, such segmenting algorithms work by setting a threshold on some acoustic feature, e.g. the Root-Mean-Square of the spectral power. This means the segments will be separated by brief “silent gaps” (periods below threshold).

To compute metrics for segmentation like precision and recall, you may want to combine the start and stop times into a single array of boundary times. Such an approach is valid if we think of a “silent gaps” as a segment whose start time is the stop time/offset of the preceding segment.

If you have arrays of start and stop times, you can concatenate into a single array of boundary times with this function. Both starts and stops must be 1-dimensional arrays of non-negative, strictly increasing values, with the same dtype. The two arrays starts and stops must be the same length, and all start times must be less than the corresponding stop times, i.e., np.all(starts < stops) should evaluate to True.

Parameters:

startsnumpy.ndarray

Array of start times of segments.

stopsnumpy.ndarray

Array of stop times of segments.

Returns:

boundariesnumpy.ndarray

The array of boundary times, concatenated and then sorted, so that np.all(boundaries[1:] > boundaries[:-1] evaluates to True.

Examples

>>> starts = np.array([0, 8, 16, 24])
>>> stops = np.array([4, 12, 20, 28])
>>> concat_starts_and_stops(starts, stops)
np.array([0, 4, 8, 12, 16, 20, 24, 28])

>>> starts = np.array([0.000, 8.000, 16.000, 24.000])
>>> stops = np.array([4.000, 12.000, 20.000, 28.000])
>>> concat_starts_and_stops(starts, stops)
np.array([0.000, 4.000, 8.000, 12.000, 16.000, 20.000, 24.000, 28.000])