EasyNLPnlp-196

Diarization Segment Count

Problem

Count the number of speaker segments by counting runs of equal labels.

An empty label sequence gives 0, and each change of speaker starts a new segment.

Hint

Compare each label with the previous one and count the changes.

case 1

in: [["a","a","b","b","a"]]

expected: 3

case 2

in: [[]]

expected: 0

case 3

in: [["a"]]

expected: 1

case 4

in: [["a","b","c"]]

expected: 3

Starter codePython

def diarization_segment_count(labels):
    # Your code here
    pass

4 test cases · run your code against them in the editor