Annotator Agreement Percent

Problem

Return the percent agreement between two equal-length label lists: 100 * matches / n. Return 0.0 when the lists are empty or their lengths differ.

case 1

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

expected: 66.66666666666667

case 2

in: [[],[]]

expected: 0

case 3

in: [["x"],["x"]]

expected: 100

case 4

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

expected: 0

Starter codePython

def agreement_percent(a, b):
    # Your code here
    pass

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